{"id":657,"date":"2014-08-12T03:11:28","date_gmt":"2014-08-12T03:11:28","guid":{"rendered":"https:\/\/courses.candelalearning.com\/cheminter\/?post_type=chapter&#038;p=657"},"modified":"2017-08-29T20:35:15","modified_gmt":"2017-08-29T20:35:15","slug":"the-mole","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/the-mole\/","title":{"raw":"The Mole","rendered":"The Mole"},"content":{"raw":"<h1 id=\"x-ck12-VGhlIE1vbGU.-chapter\">The Mole<\/h1>\r\n<div class=\"x-ck12-data\"><\/div>\r\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBOdW1iZXI.\">Avogadro's Number<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YTYxYzI1MTQ4MGY4MDQ2NTk0MjU4NWE0NmI2NjEzYWE.-xoq\">\r\n \t<li>Define mole.<\/li>\r\n \t<li>Define Avogadro\u2019s number.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MzkyYjkxNDgyYzA4ZGYyZDcyNjdlZjViNDRlYmQ5ZDE.-tkc\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211607\/20140811155325897191.jpeg\" alt=\"It is often convenient to have a unit for large amounts, such as a mole\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZGVkOGIzY2YxNTJjNGY4NjMzZGRhYTk3YTJiYTI3YzE.-bci\"><strong> Is there an easier way to load this truck? <\/strong><\/p>\r\n<p id=\"x-ck12-N2I5MTBiNzMzN2ZiNjMyYzA2ZmJhMmVkZmMzODJjODY.-8sa\">When the weather is nice, many people begin to work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add to other materials. You could order up twenty million grains of sand and have people really stare at you. You could order by the pound, but that takes a lot of time weighing out. The best bet is to order by the yard, meaning a cubic yard. The loader can easily scoop up what you need and put it directly in your truck.<\/p>\r\n\r\n<h3>Avogadro\u2019s Number<\/h3>\r\n<p id=\"x-ck12-MTkwNDE0OGNhYTljYjgwYjNkZWRlNDAwNmM4MGU1ODE.-emq\">It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small \u2013 far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 10 <sup> 22 <\/sup> molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6\u00a0\u00d7\u00a010 <sup> 10 <\/sup> molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.<\/p>\r\n<p id=\"x-ck12-YWU3M2U3ZWI5OTczMDA5NjY4ZDFkNTVlZDMyMjA5NmY.-4tg\">Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 - 1856), an Italian scientist, provided just such a number. He is responsible for the counting unit of measure called the mole. A <strong> mole <\/strong> (mol) is the amount of a substance that contains 6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> representative particles of that substance. The mole is the SI unit for amount of a substance.\u00a0 Just like the dozen and the gross, it is a name that stands for a number. There are therefore\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> water molecules in a mole of water molecules. There also would be\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> bananas in a mole of bananas, if such a huge number of bananas ever existed.<\/p>\r\n\r\n<div id=\"x-ck12-NTlhNGU0NWE2MzU3ODc3NWVkY2NhMWIzZDM2NDAzNDA.-sf5\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-85y\"><img id=\"x-ck12-OTgwNDUtMTM2MjEzODkzOS00My04MC0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211608\/20140811155326053064.jpeg\" alt=\"Portrait of Amedeo Avogadro\" longdesc=\"Italian%20scientist%20Amedeo%20Avogadro%2C%20whose%20work%20led%20to%20the%20concept%20of%20the%20mole%20as%20a%20counting%20unit%20in%20chemistry.\" \/><\/p>\r\n<strong> Figure 10.1 <\/strong>\r\n<p id=\"x-ck12-ZjY3MWE1ZWM1M2IwMzQ1YTRjYzdkN2IyOWY1Yzc5NmU.-ly2\">Italian scientist Amedeo Avogadro, whose work led to the concept of the mole as a counting unit in chemistry.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YjZiYjc2MjdlOTU5ZTU0NTQ4NzdmY2MxZjc3M2ZkMzY.-heu\">The number\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> is called <strong> Avogadro\u2019s number <\/strong> , the number of representative particles in a mole. It is an experimentally determined number. A <strong> representative particle <\/strong> is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are H <sub> 2 <\/sub> , N <sub> 2 <\/sub> , O <sub> 2 <\/sub> , F <sub> 2 <\/sub> , Cl <sub> 2 <\/sub> , Br <sub> 2 <\/sub> , and I <sub> 2 <\/sub> . The representative particle for these elements is the molecule. Likewise, all molecular compounds such as H <sub> 2 <\/sub> O and CO <sub> 2 <\/sub> exist as molecules and so the molecule is their representative particle.\u00a0 For ionic compounds such as NaCl and Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub> , the representative particle is the formula unit. A mole of any substance contains Avogadro\u2019s number (6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> ) of representative particles.<\/p>\r\n\r\n<div id=\"x-ck12-NWNhNmVjZjZhMTVlODJhNmM4NmIxZDRlODQ4N2U1NDE.-kp5\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-2wh\"><img id=\"x-ck12-OTgwNDUtMTM2MjEzODk3My0yOS0yOS0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211609\/20140811155326222771.jpeg\" alt=\"The animal mole is very different from the unit mole\" longdesc=\"The%20animal%20mole%20is%20very%20different%20than%20the%20counting%20unit%20of%20the%20mole.%20Chemists%20nonetheless%20have%20adopted%20the%20mole%20as%20their%20unofficial%20mascot.%20National%20Mole%20Day%20is%20a%20celebration%20of%20chemistry%20that%20occurs%20on%20October%2023rd%20%2810\/23%29%20of%20each%20year.\" \/><\/p>\r\n<strong> Figure 10.2 <\/strong>\r\n<p id=\"x-ck12-ZjllNjY2MmQ1NTFlMDFkNmQ2YjEzMTQwZmU0MzY2NDg.-2go\">The animal mole is very different than the counting unit of the mole. Chemists nonetheless have adopted the mole as their unofficial mascot. National Mole Day is a celebration of chemistry that occurs on October 23rd (10\/23) of each year.<\/p>\r\n\r\n<\/div>\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-YzU4ZjM4NWVjOWU1NjFhYjMwOGU0ZWUxNjA3YTk1ZjE.-kgh\">\r\n \t<li>A mole of any substance contains Avogadro\u2019s number\u00a0(6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> ) of representative particles.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-nwf\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-uya\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-OTliODQ0ODNlOWVkMTU0YmVhODAwNzFlNmVhYmRhODg.-boc\"><a href=\"http:\/\/www.scientificamerican.com\/article.cfm?id=how-was-avogadros-number\"> http:\/\/www.scientificamerican.com\/article.cfm?id=how-was-avogadros-number <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-NjU2NmM2OTg3ZGY5YWVkZDBhMmZiYjI4MjI2MmE0ODM.-oqa\">\r\n \t<li>What was Avogadro\u2019s hypothesis?<\/li>\r\n \t<li>Who first calculated this number?<\/li>\r\n \t<li>Who coined the term \u201cAvogadro\u2019s number\u201d?<\/li>\r\n \t<li>What contribution did Robert Millikan make to the determination for the value for the number?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-acu\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-YjkwNjRlMTY2OWNkYjk4ZDJlMTJmMGM5MDE3NDA1OWQ.-vf1\">\r\n \t<li>What is the SI unit for amount of a substance?<\/li>\r\n \t<li>What is the representative particle for an element?<\/li>\r\n \t<li>The formula unit is the representative particle for what?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-YTljNGM0N2U5MmE3ZWI0ZjgxMGE2YTFmNGJmMTkyNTI.-unp\">\r\n \t<li><strong> Avogadro\u2019s number: <\/strong> The number of representative particles in a mole, 6.02 \u00d7 10 <sup> 23 <\/sup> .<\/li>\r\n \t<li><strong> mole <\/strong> (mol): The amount of a substance that contains 6.02 \u00d7 10 <sup> 23 <\/sup> representative particles of that substance.<\/li>\r\n \t<li><strong> representative particle <\/strong> : The smallest unit in which a substance naturally exists.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgQmV0d2VlbiBNb2xlcyBhbmQgQXRvbXM.\">Conversions Between Moles and Atoms<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MDE2NzM2NjdlODM3Mjk0MDQ0NTFhOWMwNDg0ODgzOTU.-zug\">\r\n \t<li>Perform calculations involving conversions between number of moles and number of atoms or molecules.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YWQ5OWZhOTRkZGVmMzg4YjA2YjljMDkyZTdjODJhZDI.-urd\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211610\/20140811155326442657.jpeg\" alt=\"Using moles allows us to avoid superscripts and subscripts\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NGFhOGQyOTg3NWMxNGY3YTJiYWQ4OTgxZjlhMGVjMGQ.-7oa\"><strong> Big numbers or little numbers? <\/strong><\/p>\r\n<p id=\"x-ck12-ZmQ4OWI4MGFhYzkyMGUxMWY1Y2Q3OTM1ZmFkOTIwMjk.-pc4\">Do you hate to type subscripts and superscripts? Even with a good word-processing program, having to click on an icon to get a superscript and then remembering to click off after you type the number can be a real hassle.\u00a0 If we did not know about moles and just knew about numbers of atoms or molecules (those big numbers that require lots of superscripts), life would be much more complicated and we would make many more typing errors.<\/p>\r\n\r\n<h3>Conversions Between Moles and Atoms<\/h3>\r\n<h4>Conversions Between Moles and Number of Particles<\/h4>\r\n<p id=\"x-ck12-YTEwZjU1MWUzYTY3ZGVhMTRkMGRlMDhiYTk4ZDE4MjA.-cva\">Using our unit conversion techniques, we can use the mole label to convert back and forth between the number of particles and moles.<\/p>\r\n\r\n<h4>Sample Problem 1: Converting Number of Particles to Moles<\/h4>\r\n<p id=\"x-ck12-ZTdjNDdkMDYyMTI2Yjc2YmE1OGY4YjQ4NjJkYjFkYzg.-nji\">The element carbon exists in two primary forms: graphite and diamond.\u00a0 How many moles of carbon atoms is 4.72\u00a0\u00d7 10 <sup> 24 <\/sup> atoms of carbon?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-i1n\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-oid\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Y2I2ZmE3OWU2ODYxZGJjN2I4NjI0NTIyYWI2NWZkZmI.-d5o\">\r\n \t<li>number of C atoms = 4.72\u00a0\u00d7 10 <sup> 24 <\/sup><\/li>\r\n \t<li>1 mole = 6.02\u00a0\u00d7 10 <sup> 23 <\/sup> atoms<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hjj\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Nzg3MWZlYmM3YTZjYzYxYmM4OTBlNjE3MzA1NTM4M2M.-iqb\">\r\n \t<li>4.72\u00a0\u00d7 10 <sup> 24 <\/sup> = ? mol C<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YmU2MmVlMTZjOGY4ODU5NTRmYTYzNWYxMWIzZjgwZDc.-a5e\">One conversion factor will allow us to convert from the number of C atoms to moles of C atoms.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-lvd\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-rya\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211612\/3e4ed0b6835f34a4c97240ed40be954b.png\" alt=\"4.72 times 10^{24} text{atoms C} times frac{1 text{mol C}}{6.02 times 10^{23} text{atoms C}}=7.84 text{mol C}\" width=\"449\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-syd\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-YThhOTJlMDZlMmY5NzlhZTViMTI4OWQ3YTRkM2U0ZDU.-27m\">The given number of carbon atoms was greater than Avogadro\u2019s number, so the number of moles of C atoms is greater than 1 mole.\u00a0 Since Avogadro\u2019s number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.<\/p>\r\n<p id=\"x-ck12-OTkxMTA0MTAwYzAzYmNkMmYzMjVhMGIxZjJlYmM5NGU.-osb\">Suppose that you wanted to know how many hydrogen atoms were in a mole of water molecules.\u00a0 First, you would need to know the chemical formula for water, which is H <sub> 2 <\/sub> O.\u00a0 There are two atoms of hydrogen in each molecule of water.\u00a0 How many atoms of hydrogen would there be in two water molecules?\u00a0 There would be 2\u00a0\u00d7 2 = 4 hydrogen atoms. How about in a dozen?\u00a0 In that case a dozen is 12 so 12\u00a0\u00d7 2 = 24 hydrogen atoms in a dozen water molecules.\u00a0 To get the answers, (4 and 24) you had to multiply the given number of molecules by two atoms of hydrogen per molecule.\u00a0 So to find the number of hydrogen atoms in a mole of water molecules, the problem could be solved using conversion factors.<\/p>\r\n<p id=\"x-ck12-q8j\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211613\/7c2e42b7b8dfdd5355ba9d2fc2377155.png\" alt=\"1 text{mol} text{H}_2text{O} times frac{6.02 times 10^{23} text{molecules} text{H}_2text{O}}{1 text{mol} text{H}_2text{O}} times frac{2 text{atoms H}}{1 text{molecule} text{H}_2text{O}}=1.20 times 10^{24} text{atoms H}\" width=\"647\" height=\"43\" \/><\/p>\r\n<p id=\"x-ck12-ZGNjY2IxM2UwNjYxNTJhNDY3M2M2ZjZjMGY0YzNhNTM.-al6\">The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.<\/p>\r\n\r\n<div id=\"x-ck12-NTkzY2Q1YzEyNjNiMWNjYTExMDVlZDE3MDJmMmM2YmE.-jfh\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-zgo\"><img id=\"x-ck12-OTgwNDUtMTM2MjEzOTMyOS02LTE3LTU.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211615\/20140811155326541708.png\" alt=\"Two water molecules\" longdesc=\"Two%20water%20molecules%20contain%204%20hydrogen%20atoms%20and%202%20oxygen%20atoms.%20A%20mole%20of%20water%20molecules%20contains%202%20moles%20of%20hydrogen%20atoms%20and%201%20mole%20of%20oxygen%20atoms.\" \/><\/p>\r\n<strong> Figure 10.3 <\/strong>\r\n<p id=\"x-ck12-Y2QxZGNkMDY2ZTQyN2RiZjE5YTcyNGY1MTA1YTM2N2Y.-yqf\">Two water molecules contain 4 hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.<\/p>\r\n\r\n<\/div>\r\n<h4>Sample Problem 2: Atoms, Molecules, and Moles<\/h4>\r\n<p id=\"x-ck12-YWU2Y2Y3ZWRmNzgyOTAxZDk5NDgyZDBlZTRjYTk4ZTk.-k1s\">Sulfuric acid has the chemical formula H <sub> 2 <\/sub> SO <sub> 4 <\/sub> .\u00a0 A certain quantity of sulfuric acid contains 4.89\u00a0\u00d7\u00a010 <sup> 25 <\/sup> atoms of oxygen.\u00a0 How many moles of sulfuric acid is the sample?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-oqv\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-vle\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Zjk2MjRmMDMyYzNiMjMzMzc4Y2U0MjJjYzdjZDI3NDA.-rfs\">\r\n \t<li>4.89\u00a0\u00d7\u00a010 <sup> 25 <\/sup> = O atoms<\/li>\r\n \t<li>1 mole = 6.02 \u00d7 10 <sup> 23 <\/sup> molecules H <sub> 2 <\/sub> SO <sub> 4 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-g7w\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-M2QzYzRmYjdjOTlkOWQ4NWIxYTE3NjAxMDFmYzJiNjc.-z4e\">\r\n \t<li>mol of H <sub> 2 <\/sub> SO <sub> 4 <\/sub> molecules<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZmE2ZDZiMDllYjJjMDI2MzdhOWIxYmMyZjVlOTAyYjY.-nmx\">Two conversion factors will be used.\u00a0 First, convert atoms of oxygen to molecules of sulfuric acid.\u00a0 Then, convert molecules of sulfuric acid to moles of sulfuric acid.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-jng\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-32m\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211615\/d90613a834f279fea6ad319f41ad697d.png\" alt=\"4.89 times 10^{25} text{atoms O} times frac{1 text{molecule} text{H}_2text{SO}_4}{4 text{atoms O}} times frac{1 text{mol} text{H}_2text{SO}_4}{6.02 times 10^{23} text{molecules} text{H}_2text{SO}4}=20.3 text{mol} text{H}_2text{SO}_4\" width=\"730\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ilv\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NWUxZTczY2NiZDI1YWU0MWJlMTMwOTkxYmJmNjBlYmI.-pbd\">The original number of oxygen atoms was about 80 times larger than Avogadro\u2019s number.\u00a0 Since each sulfuric acid molecule contains 4 oxygen atoms, there are about 20 moles of sulfuric acid molecules.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-NmVlZWZkMWI1NjI3ZjUxNWZhNThlMzRiODE1OWYwZTQ.-vzu\">\r\n \t<li>Methods are described for conversions between moles, atoms, and molecules.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-OGZlOWQ0OWRiYTlmOGMyYWJlOTc0MmRiZTM2NmM1Zjc.-x0f\">Read the relevant portions of the following article and do problems 3, 5, 9, 13, and 18.\u00a0 Do not worry about the calculations involving conversions dealing with molar mass (that will come next).<\/p>\r\n<p id=\"x-ck12-ZGI2OWE5MjAzYWQzZjVhMGNkMDMwYjllMjA0YjUyNWQ.-tbg\"><a href=\"http:\/\/faculty.rcc.edu\/freitas\/1AWorksheets\/14GramsToMolesToMolecules.pdf\" target=\"_blank\" rel=\"noopener\">Grams To Moles To Molecules<\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-rz3\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-NmE0MDMzYWMxYWQyNWZiMWIyMzlmMWQ2MDcxMGYyOWY.-erx\">\r\n \t<li>What important number do we need to know to do these conversions?<\/li>\r\n \t<li>I want to convert atoms to moles. My friend tells me to multiply the number of atoms by\u00a06.02\u00a0\u00d7 10 <sup> 23 <\/sup> atoms\/mole.\u00a0 Is this correct?<\/li>\r\n \t<li>Why should I know the formula for a molecule in order to calculate the number of moles of one of the atoms?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-TW9sYXIgTWFzcw..\">Molar Mass<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ODEyMDUzZWM2ZmZjM2M5NTIwMGRlY2QwNWJmOTY4ZmQ.-bp2\">\r\n \t<li>Define molar mass.<\/li>\r\n \t<li>Perform calculations involving molar mass.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-OGYzODA3OGJjMmQzMTQxYjg3NTE3YjMzYWQxYTYyOTA.-oba\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211617\/20140811155326668657.jpeg\" alt=\"Pile of potassium dichromate\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MGVlY2U3OTcwOTk2ZjUzODkwODU4M2YxYTdhMTAzNTU.-mvm\"><strong> When creating a solution, how do I know how much of each substance to put in? <\/strong><\/p>\r\n<p id=\"x-ck12-YTAzMzBkYTE3MjM0ZjcwMzYyY2NlNTViNGFjMjRhODE.-3ec\">I want to make a solution that contains 1.8 moles of potassium dichromate.\u00a0 I don\u2019t have a balance calibrated in molecules, but I do have one calibrated in grams.\u00a0 If I know the relationship between moles and the number of grams in a mole, I can use my balance to measure out the needed amount of material.<\/p>\r\n\r\n<h3>Molar Mass<\/h3>\r\n<p id=\"x-ck12-NTVmYjBlYmUwMjdhMzY3MGU2YWExNmFkY2Q3MjQ1Njg.-ft5\"><strong> Molar mass <\/strong> is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is 6.94 g, the molar mass of zinc is 65.38 g, and the molar mass of gold is 196.97 g.\u00a0 Each of these quantities contains 6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> atoms of that particular element.\u00a0 The units for molar mass are grams per mole or g\/mol.<\/p>\r\n\r\n<h4>Molar Masses of Compounds<\/h4>\r\n<p id=\"x-ck12-OGI0NTYxYzMwYmQ2ZTY3MTYwNDU2MGVjYjc4MWEwNjU.-0wz\">A molecular formula of the compound carbon dioxide is CO <sub> 2 <\/sub> .\u00a0 One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen.\u00a0 We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen.<\/p>\r\n<p id=\"x-ck12-wz7\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211618\/5d7614bb108ac0aff5ef7ec1971d7660.png\" alt=\"12.01 text{amu} + 2(16.00 text{amu})=44.01 text{amu}\" width=\"308\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-YzRhZmQ5OTA5NDgxOWFmMDcyNDAzM2QwYzNiNmNhYTk.-gws\">The <strong> molecular mass <\/strong> of a compound is the mass of one molecule of that compound.\u00a0 The molecular mass of carbon dioxide is 44.01 amu.<\/p>\r\n<p id=\"x-ck12-ZTA3ODBlNDIxZTAwMTIzZWQ0ZThhM2JhMTJhNWJlYWI.-c3k\">The molar mass of any compound is the mass in grams of one mole of that compound.\u00a0 One mole of carbon dioxide molecules has a mass of 44.01 g, while one mole of sodium sulfide formula units has a mass of 78.04 g.\u00a0 The molar masses are 44.01 g\/mol and 78.04 g\/mol respectively.\u00a0 In both cases, that is the mass of\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> representative particles.\u00a0 The representative particle of CO <sub> 2 <\/sub> is the molecule, while for Na <sub> 2 <\/sub> S, it is the formula unit.<\/p>\r\n\r\n<h4>Sample Problem:\u00a0 Molar Mass of\u00a0 a Compound<\/h4>\r\n<p id=\"x-ck12-ZGYwZTBmZWRhODdlYTQwZmY1MGE1YWFkNDUyODk2MDA.-1si\">Calcium nitrate, Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub> , is used as a component in fertilizer Determine the molar mass of calcium nitrate.<\/p>\r\n<p id=\"x-ck12-ZmZmN2ZhMGNjMmI5MGRiYmU2NTc1YmFmN2M0MjFiMDY.-n5r\"><em> Step 1:\u00a0 List the known and unknown quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-uib\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OGQ5NzMwYWU1YWM1ZDgxNjUxNDg5ZTZlMzkyYTEwMDE.-wkr\">\r\n \t<li>formula = Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub><\/li>\r\n \t<li>molar mass Ca = 40.08 g\/mol<\/li>\r\n \t<li>molar mass N = 14.01 g\/mol<\/li>\r\n \t<li>molar mass O = 16.00 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-8p9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MjdjYmEzZjYzOTkzMDQwMjczYjhhOGI4MWE3MTQyODg.-1pt\">\r\n \t<li>molar mass Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MTA5MzYxNzEwZDI3YmQ3NzYzODY4MTc0ZDNjMzYzMzM.-xcm\">First we need to analyze the formula.\u00a0 Since the Ca lacks a subscript, there is one Ca atom per formula unit.\u00a0 The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of one nitrogen atom and three oxygen atoms.\u00a0 Therefore, there are a total of 1 \u00d7 2 = 2 nitrogen atoms and 3\u00a0\u00d7\u00a02\u00a0=\u00a06 oxygen atoms per formula unit.\u00a0 Thus, 1 mol of calcium nitrate contains 1 mol of Ca atoms, 2 mol of N atoms, and 6 mol of O atoms.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-l5a\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-NDI0MjIzMDFiYjlkYWU5ZDJjNzc1NWMzYzlmNmZhYmQ.-swz\">Use the molar masses of each atom together with the number of atoms in the formula and add together.<\/p>\r\n<p id=\"x-ck12-ete\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211620\/4522d977ad371139a59cf20a71f1ef84.png\" alt=\"&amp; 1 text{mol Ca} times frac{40.08 text{g Ca}}{1 text{mol Ca}}=40.08 text{g Ca} \\&amp; 2 text{mol N} times frac{14.01 text{g N}}{1 text{mol N}}=28.02 text{g N} \\&amp; 6 text{mol O} times frac{16.00 text{g O}}{1 text{mol O}}=96.00 text{g O} \\&amp; text{molar mass of } text{Ca(NO}_3)_2=40.08 text{g} + 28.02 text{g} + 96.00 text{g} = 164.10 text{g} \/ text{mol}\" width=\"554\" height=\"150\" \/><\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ODAzYzUxZGQ3Y2MwNmZiMWVhZDBlNzJiM2E0MDk1NDA.-ig1\">\r\n \t<li>Calculations are described for the determination of molar mass of an atom or a compound.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-MDZiNTc4ODQ1OTFlZWE1ZGY5ZDhkN2FiNjQ3MjA3Yjc.-npp\">Read the material at the link below and work the problems at the end:<\/p>\r\n<p id=\"x-ck12-ZjA0NGJlM2RiOGVhNTYzYzNhZjIwZGFhOTRkNmQ2NDY.-4ae\"><a href=\"http:\/\/misterguch.brinkster.net\/molarmass.html\"> http:\/\/misterguch.brinkster.net\/molarmass.html <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-xwq\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-ZmRkNGUzMWQ3NTE0NGM0NWU3YzMxY2EzZTYyMmMyZWQ.-vtl\">\r\n \t<li>What is the molar mass of Pb?<\/li>\r\n \t<li>Why do we need to include the units in our answer?<\/li>\r\n \t<li>I want to calculate the molar mass of CaCl <sub> 2 <\/sub> .\u00a0 How many moles of Cl are in one mole of the compound?<\/li>\r\n \t<li>How many moles of H are in the compound (NH <sub> 4 <\/sub> ) <sub> 3 <\/sub> PO <sub> 4 <\/sub> ?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-Mjc0YjY2ODFjYTBkZTY0YjAzNzU3YWY4YTYyZGVhNjg.-ffr\">\r\n \t<li><strong> molar mass <\/strong> : The mass of one mole of representative particles of a substance.<\/li>\r\n \t<li><strong> molecular mass <\/strong> : The mass of one molecule of that compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNb2xlcyBhbmQgTWFzcw..\">Conversions between Moles and Mass<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NTIzNDkxNmM3YTljY2MwZmZjMzBlNDk1ZDc3ZDFlZDU.-sfp\">\r\n \t<li>Perform calculations dealing with conversions between moles and mass.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YjcxNWU1MGJkYWQwMjY3OTdlNzY0NGJiZTVmMTU0MjQ.-z9j\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211621\/20140811155326808566.jpeg\" alt=\"Chemical plants need to know how to convert between moles and mass\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-YzA4ZjQ3ZDI0ZjQ3NjRjMWQ1ODU3ZDYxMDZkN2RhZDg.-jui\"><strong> How can we get more product? <\/strong><\/p>\r\n<p id=\"x-ck12-NzhhMDE1YjNmOGI4ODE2Y2IzZjQyNzdiZDExYmQ0NmQ.-alu\">Chemical manufacturing plants are always seeking to improve their processes.\u00a0 One of the ways this improvement comes about is through measuring the amount of material produced in a reaction.\u00a0 By knowing how much is made, the scientists and engineers can try different ways of getting more product at less cost.<\/p>\r\n\r\n<h3>Conversions Between Moles and Mass<\/h3>\r\n<p id=\"x-ck12-YmI2NmU2YTMyYjY0MTI2MWIzZTFjNGUwOGRhMTJjMWE.-c7q\">The molar mass of any substance is the mass in grams of one mole of representative particles of that substance.\u00a0 The representative particles can be atoms, molecules, or formula units of ionic compounds.\u00a0 This relationship is frequently used in the laboratory.\u00a0 Suppose that for a certain experiment you need 3.00 moles of calcium chloride (CaCl <sub> 2 <\/sub> ).\u00a0 Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed.\u00a0 The molar mass of CaCl <sub> 2 <\/sub> is 110.98 g\/mol.\u00a0 The conversion factor that can be used is then based on the equality that 1 mol = 110.98 g CaCl <sub> 2 <\/sub> .\u00a0 Dimensional analysis will allow you to calculate the mass of CaCl <sub> 2 <\/sub> that you should measure.<\/p>\r\n<p id=\"x-ck12-dpp\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211622\/0f8dd645623eeef9bf7da078647392c8.png\" alt=\"3.00 text{mol} text{CaCl}_2 times frac{110.98 text{g} text{CaCl}_2}{1 text{mol} text{CaCl}_2}=333 text{g} text{CaCl}_2\" width=\"381\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-NjM3OTcwMmFmMTEyNTMwNzNlZDVhZDE1NTA1NjkxNzg.-4iy\">When you measure the mass of 333 g of CaCl <sub> 2 <\/sub> , you are measuring 3.00 moles of CaCl <sub> 2 <\/sub> .<\/p>\r\n\r\n<div id=\"x-ck12-ZjRlM2I3MzYwN2FmYTIxNGQ0ZWRjMTM0YTZlYjk1M2Q.-csu\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-v8t\"><img id=\"x-ck12-OTgwNDUtMTM2MjEzOTczNy05Mi0xOS04\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211624\/20140811155326951109.jpeg\" alt=\"Calcium chloride is used as a drying agent and as a road deicer\" longdesc=\"Calcium%20chloride%20is%20used%20as%20a%20drying%20agent%20and%20as%20a%20road%20deicer.\" \/><\/p>\r\n<strong> Figure 10.4 <\/strong>\r\n<p id=\"x-ck12-ZmM2NzM1ZWM5MDIwYWRjNmQyZjdlMzUxZWU4N2UxZWM.-2j8\">Calcium chloride is used as a drying agent and as a road deicer.<\/p>\r\n\r\n<\/div>\r\n<h4>Sample Problem: Converting Moles to Mass<\/h4>\r\n<p id=\"x-ck12-YmU1NzVhZmVlMTMyOTVkMTMyOTIwMmQzOWQ2MjA2NjU.-4za\">Chromium metal is used for decorative electroplating of car bumpers and other surfaces.\u00a0 Find the mass of 0.560 moles of chromium.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qrj\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-dgp\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZWEzODk1MmJlMDlmMjMyZDQ0YTNmZDU0Yzc2YzYzOTk.-uyd\">\r\n \t<li>molar mass of Cr = 52.00 g mol<\/li>\r\n \t<li>0.560 mol Cr<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-9yb\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZDAwMTEzNjA3ZTQwYjhmYmE0NjkyZjU1OTljM2EzZTk.-yeo\">\r\n \t<li>0.560 mol Cr = ? g<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-OWY1NDdiNTAxMmEzN2VjOTNiNDk4M2MyMDUzYTlmNmU.-hpe\">One conversion factor will allow us to convert from the moles of Cr to mass.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-tro\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-v4c\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211624\/cc47cd6884a6fe79190aa322be8c07f4.png\" alt=\"0.560 text{mol Cr} times frac{52.00 text{g} text{Cr}}{1 text{mol Cr}}=29.1 text{g Cr}\" width=\"306\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-NzhjOWE5NTVlYTMxYTViNTExNTViMDY3ODFmOTgyZDM.-1rw\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NzhjOWE5NTVlYTMxYTViNTExNTViMDY3ODFmOTgyZDM.-zsp\">Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass.\u00a0 The answer has three significant figures because of the 0.560 mol.<\/p>\r\n<p id=\"x-ck12-YTY3ZWJmZWYxNjQ0NTk0MjlhY2I0NTE1NzkxNzJkNmU.-jcs\">A similar conversion factor utilizing molar mass can be used to convert from the mass of an substance to moles.\u00a0 In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed.\u00a0 It will often then be necessary to determine the number of moles of the product that was formed.\u00a0 The next problem illustrates this situation.<\/p>\r\n\r\n<h4>Sample Problem: Converting Mass to Moles<\/h4>\r\n<p id=\"x-ck12-NGFjYTU5NGUxYTA0ZjRlYjI2NWRkMmUzNTg3NWExNTE.-ttg\">A certain reaction produces 2.81 g of copper(II) hydroxide, Cu(OH) <sub> 2 <\/sub> .\u00a0 Determine the number of moles produced in the reaction.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qy2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-z4f\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MDA3YjU0OGQ4MDE2NTQxZGJhNTU5NDRjYTFlYzcyNDE.-ns0\">\r\n \t<li>mass = 2.81 g<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-3e2\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OTRmMTE1MjJmM2U0Yjg5YmFjZjA5ZmI1YjFjNGM5ZDc.-dd6\">\r\n \t<li>mol Cu(OH) <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NzcyYWY5ZGFiNzg0NzUzMjZkNjU0NzZhYzM1ZDY1YWQ.-0t4\">One conversion factor will allow us to convert from mass to moles.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-9el\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-OWEzMzgwMzE3MjRiZDJmMmFmZTFjNGFiOGFlYjk5MzA.-4y5\">First, it is necessary to calculate the molar mass of Cu(OH) <sub> 2 <\/sub> from the molar masses of Cu, O, and H.\u00a0 The molar mass is 97.57 g\/mol.<\/p>\r\n<p id=\"x-ck12-ujr\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211625\/573a40438243112b5bef60459f4ca487.png\" alt=\"2.81 text{g} text{Cu(OH)}_2 times frac{1 text{mol} text{Cu(OH)}_2}{97.57 text{g} text{Cu(OH)}_2}=0.0288 text{mol} text{Cu(OH)}_2\" width=\"467\" height=\"44\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-njz\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-MjVjMzE1MTBmZTU2N2IxODA2MjU0OTUwM2QxZmUxZmQ.-ixo\">The relatively small mass of product formed results in a small number of moles.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-Y2I0OGFhMzI3NDdjMGE5NzZlN2Y2N2ZhMGFjMWFiNTc.-z8w\">\r\n \t<li>Calculations involving conversions between moles of a material and the mass of that material are described.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-N2IzZjFlODZlYzZmOTQ3MDA5YmRlYzNjNWFmMTNiZmU.-drt\">Read the material in the link below and work the problems at the end.<\/p>\r\n<p id=\"x-ck12-Mzk5YThjNTA2YjA3ZWE4OGU1OWRkYzRiNzRjMjE5NjM.-tsw\"><a href=\"http:\/\/www.occc.edu\/kmbailey\/chem1115tutorials\/Stoichiometry_Molar_Mass_Calculations.htm\"> http:\/\/www.occc.edu\/kmbailey\/chem1115tutorials\/Stoichiometry_Molar_Mass_Calculations.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-euk\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-M2I1ZTVlN2M5ZGUwMmU3MTE0NDNmMzRlZDk5ODJjZGY.-isk\">\r\n \t<li>Why would you want to calculate the mass of a material?<\/li>\r\n \t<li>Why would you want to determine how many moles of material you produced in a reaction?<\/li>\r\n \t<li>You have 19.7 grams of a material and wonder how many moles were formed. Your friend tells you to multiply the mass by grams\/mole.\u00a0 Is your friend correct?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNYXNzIGFuZCBOdW1iZXIgb2YgUGFydGljbGVz\">Conversions between Mass and Number of Particles<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MDU0YjM4MGFlOWZiMWQwZmQzYzY5NjhkYjY1MzdhMjY.-adn\">\r\n \t<li>Perform calculations involving conversions of mass and number of particles.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-M2Q4NjhiMDlmYWM4ZjIwYWY5MzQ0MDdkMTgxYzFmMjk.-tkp\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211627\/20140811155327444990.jpeg\" alt=\"Equal volumes of gas under the same conditions contain the same number of particles\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MTgyMDVjZmIwM2UwM2JiYWFlZjY5NTAwNGUzYTAzOTE.-dmb\"><strong> How much gas is there? <\/strong><\/p>\r\n<p id=\"x-ck12-YWU2ZDQyZjlmNDc4NWQ3NTBmZDVlZDZkY2ZmODkxNzE.-lfp\">Avogadro was interested in studying gases.\u00a0 He theorized that equal volumes of gases under the same conditions contained the same number of particles.\u00a0 Other researchers studied how many gas particles were in a specific volume of gas. Eventually, scientists were able to develop the relationship between number of particles and mass using the idea of moles.<\/p>\r\n\r\n<h3>Conversions Between Mass and Number of Particles<\/h3>\r\n<p id=\"x-ck12-NmU5OTkwMDEwZTQ3MDA1MjE3YTg4MGVmZjE1MTVhMGI.-ty5\">In \"Conversions between Moles and Mass\", you learned how to convert back and forth between moles and the number of representative particles.\u00a0 Now you have seen how to convert back and forth between moles and mass of a substance in grams.\u00a0 We can combine the two types of problems into one.\u00a0 Mass and number of particles are both related to grams.\u00a0 In order to convert from mass to number of particles or vice-versa, it will first require a conversion to moles.<\/p>\r\n\r\n<div id=\"x-ck12-NjJiYmI0YWI4ZGI1ZmQyY2UwOGQwN2UzYWE1MTU4MTE.-gtb\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-c6u\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MDA5Mi0yNS01OC0xMA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211628\/20140811155327594002.png\" alt=\"Flowchart of conversion between number of particles, moles, and mass\" longdesc=\"Conversion%20from%20number%20of%20particles%20to%20mass%20or%20from%20mass%20to%20number%20of%20particles%20requires%20two%20steps\" \/><\/p>\r\n<strong> Figure 10.5 <\/strong>\r\n<p id=\"x-ck12-MjMzZWEyMzNiMmFmMGJmMDgzNDc2N2FjY2VkYjFhZTg.-4y1\">Conversion from number of particles to mass or from mass to number of particles requires two steps<\/p>\r\n\r\n<\/div>\r\n<h4>Sample Problem: Converting Mass to Particles<\/h4>\r\n<p id=\"x-ck12-ODc1Y2MyMGIwZTNmZTBmZTc2YzU0OWE2YjRiMDcxMDA.-m5d\">How many molecules is 20.0 g of chlorine gas, Cl <sub> 2 <\/sub> ?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ink\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-m5c\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ODc1NmY3MWZkYjRlYzZjMjUzZjY3M2IwY2JhNWVmYWY.-6v2\">\r\n \t<li>molar mass Cl <sub> 2 <\/sub> = 70.90 g\/mol<\/li>\r\n \t<li>20.0 g Cl <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-oqz\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MDFjNjgxOTBhOWU3MDk2MjkxN2U4Y2QxYmRkNDdhMzE.-jdy\">\r\n \t<li>number of molecules of Cl <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YzM3NDRlNTk4MjM1YWI2ZmE3MDQ1ZDMxNzk0M2M1NGU.-b5y\">Use two conversion factors.\u00a0 The first converts grams of Cl <sub> 2 <\/sub> to moles.\u00a0 The second converts moles of Cl <sub> 2 <\/sub> to the number of molecules.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-wg1\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-dcj\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211630\/7485c4d2caf2013df3630b023c68c14f.png\" alt=\"20.0 text{g} text{Cl}_2 times frac{1 text{mol} text{Cl}_2}{70.90 text{g} text{Cl}_2} times frac{6.02 times 10^{23} text{molecules Cl}_2}{1 text{mol} text{Cl}_2}=1.70 times 10^{23} text{molecules Cl}_2\" width=\"634\" height=\"44\" \/><\/p>\r\n<p id=\"x-ck12-NTZkMTU0ZDdjNjE3NjU0MzY3MGIxYjdjYWE5NzlhOGQ.-gap\">The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of Cl <sub> 2 <\/sub> .<\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-idt\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ODM0ZGY1YmMxMTk2N2EwNzAxYzMzYTkxZGVkNDVmZWM.-dxg\">Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro\u2019s number.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-MTQzZDcxNjA5ZTdjOWVmNWMyNTM5ZmJlYjE3Njc4MmI.-mxr\">\r\n \t<li>Calculations are illustrated for conversions between mass and number of particles.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NTNlMzE4NDEwZDM5MGQyYjNkOWQ2ZDY0YTgyMzVhNWE.-lqg\">Read the material at the link below and then do practice problems on page 9 and the problem on page 17 (don\u2019t peak at the answers until you have tried the problems).<\/p>\r\n<p id=\"x-ck12-Yjg1YzhhNzcwNGFlNzkxN2VkYzAzMGI3NDhmMzJlMDg.-ehm\"><a href=\"http:\/\/schools.fwps.org\/decatur-old\/staff\/adewaraja\/chemistry\/curriculum%20units\/chapter%2010\/Lesson5\/particle_mole_mass_calcu.pdf\"> http:\/\/schools.fwps.org\/decatur-old\/staff\/adewaraja\/chemistry\/curriculum%20units\/chapter%2010\/Lesson5\/particle_mole_mass_calcu.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-jkn\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-YzY3MDIwMTI2N2I1NjQwZWEyZDllNjM1YWQ1OTM3MmU.-qv6\">\r\n \t<li>Why can\u2019t we convert directly from number of particles to grams?<\/li>\r\n \t<li>How many atoms of chlorine are present in the problem above?<\/li>\r\n \t<li>The periodic table says the atomic weight of chlorine is 35.5. Why can\u2019t I use that value in my calculations?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBIeXBvdGhlc2lzIGFuZCBNb2xhciBWb2x1bWU.\">Avogadro's Hypothesis and Molar Volume<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ZjVhZWY3OGQ1OTc5ZDYxZjMzYjRlOThlNzAzYjZhNzE.-keq\">\r\n \t<li>State Avogadro\u2019s hypothesis.<\/li>\r\n \t<li>Define standard temperature and pressure.<\/li>\r\n \t<li>Define molar volume.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-Nzc4YzcxYzI1ZTY1ZDdmNzQwZjExNzBmOGU5YmI3OTE.-t0m\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211631\/20140811155327709392.jpeg\" alt=\"Gas laws are used to determine the amount of air in a scuba tank\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MDE4ODNhNmVlOWQ5Njc0NmM0ODhlN2ZkYmUwZWZkYWQ.-dig\"><strong> How do scuba divers know if they will run out of gas? <\/strong><\/p>\r\n<p id=\"x-ck12-YTg2YWQ1NDk3ZjcxYzBmN2M2ZmFhODI5NDhlMjg4OTQ.-bpe\">Knowing how much gas is available for a dive is crucial to the survival of the diver.\u00a0 The tank on the diver\u2019s back is equipped with gauges to tell how much gas is present and what the pressure is.\u00a0 A basic knowledge of gas behavior allows the diver to assess how long to stay under water without developing problems.<\/p>\r\n\r\n<h3>Avogadro\u2019s Hypothesis and Molar Volume<\/h3>\r\n<p id=\"x-ck12-N2EyYmRmYTE0ZWEzNzI4NDE3MjVlNjczODU4ZGM4NzQ.-vf0\">Volume is a third way to measure the amount of matter, after item count and mass.\u00a0 With liquids and solids, volume varies greatly depending on the density of the substance.\u00a0 This is because solid and liquid particles are packed close together with very little space in between the particles.\u00a0 However, gases are largely composed of empty space between the actual gas particles (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDQ1NS05Mi00NC0xMg..\"> below <\/a> ).<\/p>\r\n\r\n<div id=\"x-ck12-MmJjMzM1MDQ4NDQ0ZjE1M2MwOTY5NTdhZmI5NTU4MDA.-tnk\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-10u\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MDQ1NS05Mi00NC0xMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211633\/20140811155327899786.png\" alt=\"Gas particles are small compared to the empty space between them\" longdesc=\"Gas%20particles%20are%20very%20small%20compared%20to%20the%20large%20amounts%20of%20empty%20space%20between%20them.\" \/><\/p>\r\n<strong> Figure 10.6 <\/strong>\r\n<p id=\"x-ck12-ZDk5NjZhNmIzMDU3MTFkYmJiYzY1NmM4MjIyYzQzYWM.-w7b\">Gas particles are very small compared to the large amounts of empty space between them.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YWI1ZjEyYWFlMTE2ZWE2OTIxOTExN2YwY2MwNWQ3YmY.-xms\">In 1811, Amedeo Avogadro explained that the volumes of all gases can be easily determined. <strong> Avogadro\u2019s hypothesis <\/strong> states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.\u00a0 Since the total volume that a gas occupies is made up primarily of the empty space between the particles, the actual size of the particles themselves is nearly negligible.\u00a0 A given volume of a gas with small light particles such as hydrogen (H <sub> 2 <\/sub> ) contains the same number of particles as the same volume of a heavy gas with large particles such as sulfur hexafluoride, SF <sub> 6 <\/sub> .<\/p>\r\n<p id=\"x-ck12-NTEyMjljMWY2ZDQ0YjRlZTNkNzljYzkwMjUwYzFmMGQ.-vkt\">Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another.\u00a0 This decreases the amount of empty space and reduces the volume of the gas.\u00a0 Gas volume is also affected by temperature.\u00a0 When a gas is heated, its molecules move faster and the gas expands.\u00a0 Because of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at one standard temperature and pressure.\u00a0 <strong> Standard temperature and pressure (STP) <\/strong> is defined as\u00a00\u00b0C\u00a0(273.15 K) and 1 atm pressure. The <strong> molar volume <\/strong> of a gas is the volume of one mole of a gas at STP.\u00a0 At STP, one mole (6.02\u00a0\u00d7 10 <sup> 23 <\/sup> representative particles) of any gas occupies a volume of 22.4 L ( <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDQ5Ny0zNS0yLTEz\"> below <\/a> ).<\/p>\r\n\r\n<div id=\"x-ck12-ZjY5NTM1NmFlYmQ4NzM2NGI0NTA1MzdiNTM4MmI3M2I.-dhc\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-syr\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MDQ5Ny0zNS0yLTEz\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211633\/20140811155328052846.png\" alt=\"Avogadro's hypothesis states that one mole of gas at STP occupies 22.4 liters\" longdesc=\"A%20mole%20of%20any%20gas%20occupies%2022.4%20L%20at%20standard%20temperature%20and%20pressure%20(0%C2%B0C%20and%201%20atm).\" \/><\/p>\r\n<strong> Figure 10.7 <\/strong>\r\n<p id=\"x-ck12-MWQ0YTUyMTFjN2FlNDE5YWQxMDcxNGY1ODBkMzg2NTE.-guq\">A mole of any gas occupies 22.4 L at standard temperature and pressure (0\u00b0C and 1 atm).<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NGVkM2M2Mjc3NjcwNmMwOWEyODU5ZGRmZGE3YWFhMDc.-9wd\">The <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDUzMS0xNy05OS0xNA..\"> below <\/a> illustrates how molar volume can be seen when comparing different gases. Samples of helium (He), nitrogen (N <sub> 2 <\/sub> ), and methane (CH <sub> 4 <\/sub> ) are at STP.\u00a0 Each contains 1 mole or 6.02\u00a0\u00d7 10 <sup> 23 <\/sup> particles.\u00a0 However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g\/mol for He, 28.0 g\/mol for N <sub> 2 <\/sub> , and 16.0 g\/mol for CH <sub> 4 <\/sub> .<\/p>\r\n\r\n<div id=\"x-ck12-YTBkYTA4ZDg2Y2Q2M2QyNDEzMTAxYTA5MjE1Mzk4Njk.-3tf\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-5kv\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MDUzMS0xNy05OS0xNA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211634\/20140811155328170493.png\" alt=\"Avogadro's hypothesis means that \u00a0one mole of neon, nitrogen, or methane occupies 22.4 liters at STP\" longdesc=\"Avogadro%E2%80%99s%20hypothesis%20states%20that%20equal%20volumes%20of%20any%20gas%20at%20the%20same%20temperature%20and%20pressure%20contain%20the%20same%20number%20of%20particles.%20At%20standard%20temperature%20and%20pressure%2C%201%20mole%20of%20any%20gas%20occupies%2022.4%20L.\" \/><\/p>\r\n<strong> Figure 10.8 <\/strong>\r\n<p id=\"x-ck12-ZGY2MDRkMmRmNjY1MTdjNjk5ZWQyZDhlZjdjMzk2MmY.-pyq\">Avogadro\u2019s hypothesis states that equal volumes of any gas at the same temperature and pressure contain the same number of particles. At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.<\/p>\r\n\r\n<\/div>\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-YWJhNjNlY2Q2MzI3NDBkYzVkNTIwNzkyZDQ5Y2RlNDE.-jvn\">\r\n \t<li>Equal volumes of gases at the same conditions contain the same number of particles.<\/li>\r\n \t<li>Standard temperature and pressure are defined.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-vba\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-xkw\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-NTY1YTUxZjJiM2I3MmZkOTI2MTIxNWYxMGQyZTJiODM.-hwa\"><a href=\"http:\/\/chemed.chem.purdue.edu\/demos\/main_pages\/4.6.html\"> http:\/\/chemed.chem.purdue.edu\/demos\/main_pages\/4.6.html <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-MWZkMzk3YjBiNWMzNjQ2ODk0OWU2ODZkZDExOGE0ZGY.-zrg\">\r\n \t<li>What was the volume of each gas that was weighed?<\/li>\r\n \t<li>What did the experiment find?<\/li>\r\n \t<li>What was the relationship between gas weight and molecular weight?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-qdh\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MjQ4ZDdhZGY1YjY2YmMyN2UwY2M4ZDBjMzU1NzQ4YmQ.-y1x\">\r\n \t<li>What do we know about the space actually taken up by a gas?<\/li>\r\n \t<li>Why do we need to do all our comparisons at the same temperature and pressure?<\/li>\r\n \t<li>How can we use this information?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-NzJhZjI3YzRmMGZjNmU4OTFhMzk3ZGMwNDA4YWI5ZmI.-bja\">\r\n \t<li><strong> Avogadro\u2019s hypothesis: <\/strong> Equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.<\/li>\r\n \t<li><strong> molar volume: <\/strong> The volume of one mole of a gas at STP.<\/li>\r\n \t<li><strong> standard temperature and pressure (STP): <\/strong> \u00a00\u00b0C (273.15 K) and 1 atm pressure.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNb2xlcyBhbmQgR2FzIFZvbHVtZQ..\">Conversions between Moles and Gas Volume<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MThlOGI0YTY5MjA4NWIzYzg1N2Y2MzU3NTMzMzVhOGU.-ns3\">\r\n \t<li>Make conversions between the volume of a gas and the number of moles of that gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-Y2ZmMThkZDkyYzBmZmNkMzFmZTBjNDVmM2IzNGQ2NDY.-xto\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211635\/20140811155328513045.jpeg\" alt=\"Gas volumes can be used to determine the moles of gas in these tanks\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZTgyYmZkMjZmMDQ0NmM5ZjA0NjYwZmNlNDdhYTZjNmY.-xgk\"><strong> How can you tell how much gas is in these containers? <\/strong><\/p>\r\n<p id=\"x-ck12-MjA5ODgwZGUzODExOGFiZjNmN2FkOWM0MWM0ZGNiZmQ.-e3b\">Small gas tanks are often used to supply gases for chemistry reactions.\u00a0 A gas gauge will give some information about how much is in the tank, but quantitative estimates are needed so the reaction will be able to proceed to completion.\u00a0 Knowing how to calculate needed parameters for gases is very helpful to avoid running out too early.<\/p>\r\n\r\n<h3>Conversions Between Moles and Gas Volume<\/h3>\r\n<p id=\"x-ck12-NGIyNDVlMmZjMGQxOGIwM2VlY2ExMTU4MjJkYzhmMzY.-65e\">Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles.\u00a0 The equality of 1 mole = 22.4 L is the basis for the conversion factor.<\/p>\r\n\r\n<h4>Sample Problem One: Converting Gas Volume to Moles<\/h4>\r\n<p id=\"x-ck12-NjM1YjNmNDJkZjA0YzlkODExMjUyOGU5NDQ3NWI3YmM.-vhc\">Many metals react with acids to produce hydrogen gas.\u00a0 A certain reaction produces 86.5 L of hydrogen gas at STP.\u00a0 How many moles of hydrogen were produced?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-6jk\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-xiy\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NTk3ODc2MTlmMzBhMDllN2M1N2MxMDkyNjdjZDkxODQ.-edq\">\r\n \t<li>86.5 L H <sub> 2 <\/sub><\/li>\r\n \t<li>1 mol = 22.4 L <sub>\r\n<\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-h7n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MzYyMjhkNTRkMmIwZDg2YTcwNzExNjNhN2M1MDc5Njk.-hrl\">\r\n \t<li>moles of H <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NTJhZDAzODU4N2M0MjRkZDY0YTA5NTIwYzE0N2E2OWQ.-jvh\">Apply a conversion factor to convert from liters to moles.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-u6t\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-xv2\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211636\/6a8f0b1206fc84696b7358e4eaf8cf7b.png\" alt=\"86.5 text{L H}_2 times frac{1 text{mol H}_2}{22.4 text{L H}_2}=3.86 text{mol H}_2\" width=\"293\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-by6\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NWJjYjg2OGFjMzEwMjYyMTAxNDc4Y2JjMzA1MTNlZTg.-owf\">The volume of gas produced is nearly four times larger than the molar volume.\u00a0 The fact that the gas is hydrogen plays no role in the calculation.<\/p>\r\n\r\n<h4>Sample Problem Two: Converting Moles to Gas Volume<\/h4>\r\n<p id=\"x-ck12-MGE0M2U1MmE1MTI5YTIzMTA5YTJlNmE2ZWU4NjZiZDQ.-jwf\">What volume does 4.96 moles of O <sub> 2 <\/sub> occupy at STP?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-mev\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-f78\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OTA1ODg4ZGNkYmI1OGM1ZDE0YWZmZmY1YjMwYjY2Y2M.-apu\">\r\n \t<li>4.96 moles O <sub> 2 <\/sub><\/li>\r\n \t<li>1 mol = 22.4 L <sub>\r\n<\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-dah\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OGU3OWM3YTgwYjI0NTc3MTM0ZmVkYjU0NTFkOTEzNjQ.-b3x\">\r\n \t<li>volume of O <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-pf1\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-x4x\" class=\"x-ck12-indent\"><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211637\/e94b42a3d26348b83abd955d3d7b651d.png\" alt=\"4.96 text{moles} times {22.4 text{liters\/mole}}=111.1 text{liters}\" width=\"348\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-lzv\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-MWYyNGNhYzA0YThhMjRkMWMzZjFjMzc3ZDdjYmM4Zjk.-8ay\">The volume seems correct given the number of moles.<\/p>\r\n\r\n<h4 id=\"x-ck12-ZWQ4ZjY3NzA4NTEyMTYwMWY1ZjNmZGMzY2IzMjM5MGI.-pht_6-2yv\">Sample Problem Three: Converting Volume to Mass<\/h4>\r\n<p id=\"x-ck12-YzVhZDY2ZDg0YmViZWYyNzU0NTU4ODhiOTMyZGFkYWE.-bl8\">If we know the volume of a gas sample at STP, we can determine how much mass is present.\u00a0 Assume we have 867 liters of N <sub> 2 <\/sub> at STP.\u00a0 What is the mass of the nitrogen gas?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ap2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-dr5\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MjgwYzRkYjc3ZWY5ZjVkNmQ5Y2U5NzIwZjgxNjFlODY.-25f\">\r\n \t<li>867 L N <sub> 2 <\/sub><\/li>\r\n \t<li>1 mol = 22.4 L<\/li>\r\n \t<li>molar mass of N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-yc5\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OTVjOGU5OWFiM2YzMTQzZWUzY2UxNDQzNWJlNTY4NDk.-8pc\">\r\n \t<li>mass of N <sub> 2 <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ltw\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-NDhjNzI0ZjgwMWYxOTM4YjI2NzcxODQ2ODQ0N2IyM2E.-fea\">We start by determining the number of moles of gas present. We know that 22.4 liters of a gas at STP equals one mole, so:<\/p>\r\n<p id=\"x-ck12-wkq\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211639\/64de3fd328a183a96d0c9ea65751ae7d.png\" alt=\"867 text{litres} times frac{1 text{mole}}{22.4 text{liters}}=3.87 text{moles}\" width=\"280\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-ODBjZWUxZWJkMGZmODE4MDI2MDVjMzFhNjU2NjFiYzA.-kmp\">We also know the molecular weight of N <sub> 2 <\/sub> (28.0 grams\/mole), so we can then calculate the weight of nitrogen gas in 867 liters:<\/p>\r\n<p id=\"x-ck12-2sg\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211640\/f5e3f79dc20e669e05dc062476193cbb.png\" alt=\"38.7 text{moles} times frac{28 text{grams}}{text{mole}}=1083.6 text{grams N}_2\" width=\"330\" height=\"38\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-8im\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ODVlMTM2MjhlNjA2MmYzMmJlZGJjYjJjNDRmODA2OGM.-eqq\">In a multi-step problem, be sure that the units check.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-MWE1YTNkMGVhYzgxNWE4YTU3NTRlNjk1NGE3YjgzOWY.-clt\">\r\n \t<li>Conversions between moles and volume of a gas are shown.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-ZTg5NGIyM2VjZDY5N2ExYWY2ZjZmNzgyMTY0ZjU4MDI.-slu\">Work the practice problems at the link below. Focus on conversions between volume and moles, but try some of the others:<\/p>\r\n<p id=\"x-ck12-NjYyYWM0NDE0MmE1MDM1Yzc1ZmJhMDdkY2QxN2UyNzQ.-lbp\"><a href=\"http:\/\/www.sciencegeek.net\/Chemistry\/taters\/Unit4GramMoleVolume.htm\"> http:\/\/www.sciencegeek.net\/Chemistry\/taters\/Unit4GramMoleVolume.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-mvq\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MDMyYTJmYmZlYjc5ZGE1YmM1ZTkyYzc3MjExMzExODQ.-iog\">\r\n \t<li>Why do the gases need to be at STP?<\/li>\r\n \t<li>When does the identity of the gas become important?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-R2FzIERlbnNpdHk.\">Gas Density<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YjEwMDNkNjFiZGUzMTI4MGFkMmM3YTI4NWQwMzdmYjY.-ko5\">\r\n \t<li>Make calculations dealing with molar mass and density of a gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MTYyZTUyMDViNjg3NTgzY2UwNWYxNDBkYTQzMmZiODI.-b0p\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211641\/20140811155328601836.jpeg\" alt=\"Carbon dioxide sinks in air\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NmQxYmFhZmM2MzlmZjAwNjMxOGY0NzhjMGRkZjI2Mjk.-m4s\"><strong> Why does carbon dioxide sink in air? <\/strong><\/p>\r\n<p id=\"x-ck12-M2VjMTFjYzFiYTRiZWQ5NzQwZWIzZmZlNjJiZDQxMGM.-3my\">When we run a reaction to produce a gas, we expect it to rise into the air.\u00a0 Many students have done experiments where gases such as hydrogen are formed.\u00a0 The gas can be trapped in a test tube held upside-down over the reaction.\u00a0 Carbon dioxide, on the other hand, sinks when it is released.\u00a0 Carbon dioxide has a density greater that air, so it will not rise like these other gases would.<\/p>\r\n\r\n<h3>Gas Density<\/h3>\r\n<p id=\"x-ck12-ZTBlOTNhNTQyY2ExMGUwZDM4ODU4MTExODcyZjExNzE.-bi5\">As you know, density is defined as the mass per unit volume of a substance.\u00a0 Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass.\u00a0 A gas with a small molar mass will have a lower density than a gas with a large molar mass.\u00a0 Gas densities are typically reported in g\/L.\u00a0 Gas density can be calculated from molar mass and molar volume.<\/p>\r\n\r\n<div id=\"x-ck12-NWMwNGY3ZGRlM2U5OGEyYzIzMGYwZTQyMzM4NGJlMWM.-mra\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-gzp\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MDg4NS02Mi04Mi0xNw..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211643\/20140811155328777663.jpeg\" alt=\"Balloons float because they contain helium, which is lighter than air\" longdesc=\"Balloons%20filled%20with%20helium%20gas%20float%20in%20air%20because%20the%20density%20of%20helium%20is%20less%20than%20the%20density%20of%20air.\" \/><\/p>\r\n<strong> Figure 10.9 <\/strong>\r\n<p id=\"x-ck12-MGVlMTBhODU2NWU4MjJlNTEyODg1ODMxNjIzZWJmY2U.-jml\">Balloons filled with helium gas float in air because the density of helium is less than the density of air.<\/p>\r\n\r\n<\/div>\r\n<h4>Sample Problem One: Gas Density<\/h4>\r\n<p id=\"x-ck12-YWIxZDk5YmMyZGRjZjNmZjAzZTRhMzI3MTNiNzE3MDk.-vlz\">What is the density of nitrogen gas at STP?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ekt\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-0ib\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Y2EyYWJkM2Q3MmEyM2E0MGIyMjlkMzI4MDcyZjM2ZDc.-0gu\">\r\n \t<li>N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\r\n \t<li>1 mol = 22.4 L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-tqy\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NmYxZmZhM2E1Yjk0ZGE3NGEyNTZjZDdkYmZjNWJkNjE.-ang\">\r\n \t<li>density = ? g\/L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-OGNiODMxZWU3NjE4MzY3ZjMzODU0ZTgzOTFiYzE3ODg.-ee9\">Molar mass divided by molar volume yields the gas density at STP.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ljo\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-ae0\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211644\/126428d64cdb8fd62bc7401cf20e7a56.png\" alt=\"frac{28.02 text{g}}{1 text{mol}} times frac{1 text{mol}}{22.4 text{L}}=1.25 text{g} \/ text{L}\" width=\"224\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-Y2ZmMzU1YWFiZWJiMzlmMjcyODg0MmFkY2Y2MWIxMjE.-huj\">When set up with a conversion factor, the mol unit cancels, leaving g\/L as the unit in the result.<\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-hqd\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ZTNkYjczMzM1NjcwZWI1YmUyOTkwMTBkMDU3M2VjNDI.-uj6\">The molar mass of nitrogen is slightly larger than molar volume, so the density is slightly greater than 1 g\/L.<\/p>\r\n<p id=\"x-ck12-YTAzMTZiYjQ4MjlhOGEwZGVkYWMxODk4ZjVkYjJmNDI.-r9s\">Alternatively, the molar mass of a gas can be determined if the density of the gas at STP is known.<\/p>\r\n\r\n<h4>Sample Problem Two: Molar Mass from Gas Density<\/h4>\r\n<p id=\"x-ck12-NDZiYjBhNzdmNWIyMzY0N2JjNzU5NTNmMzQ3ZjM3YmU.-lmx\">What is the molar mass of a gas whose density is 0.761 g\/L at STP?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-1z8\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-q4z\"><span class=\"x-ck12-underline\"> Known\r\n<\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Y2EyYWJkM2Q3MmEyM2E0MGIyMjlkMzI4MDcyZjM2ZDc.-2cm\">\r\n \t<li>N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\r\n \t<li>1 mol = 22.4 L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hov\"><span class=\"x-ck12-underline\"> Unknown <\/span> <em>\r\n<\/em><\/p>\r\n\r\n<ul id=\"x-ck12-OTM0YzliMjIwMjJmMzMwYjNlYWFlZWFkZDkzN2VjNDE.-mzo\">\r\n \t<li>molar mass = ? g\/L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-M2RhMzQzZmY1NTQxNzM0YjBjMjRlZTdmMDI5MDBjMjM.-vui\">Molar mass is equal to density multiplied by molar volume.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-08f\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-e6g\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211645\/a6dd8262edcc127b90582136e771b11f.png\" alt=\"frac{0.761 text{g}}{1 text{L}} times frac{22.4 text{L}}{1 text{mol}}=17.0 text{g} \/ text{mol}\" width=\"241\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-m7i\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-MTQ2MmRkZTU4NDY5MTg4NzhkMDVjZWYyYmI3MjMzNWQ.-1lq\">Because the density of the gas is less than 1 g\/L, the molar mass is less than 22.4.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-OGZiMGUxZDRkOGYyODRkMzM1MDZhNGQ1N2JjZDY0NzU.-xm5\">\r\n \t<li>Calculations are described showing conversions between molar mass and density for gases.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-5x2\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-ias\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-MWU0MTNjYmI5MDQ4NWVmZWUxNWFiZjVkZGJhMTI5ZDM.-4lu\"><a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/gas_density_s.html\"> http:\/\/employees.oneonta.edu\/viningwj\/sims\/gas_density_s.html <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-MmJlZmE2ZmM3NjlmYTdlYjkzNDM2MGU5NzIwZGQ0NjQ.-zpj\">\r\n \t<li>Which of the gases has the highest density?<\/li>\r\n \t<li>Which gas has the lowest density?<\/li>\r\n \t<li>Would you expect nitrogen to have a higher or lower density that oxygen? Why?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-uca\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-OTQzMmFhZTNlMGI4MTk1YzQ4ZjZkNzI0YmJiY2QyZDU.-wn5\">\r\n \t<li>How is density calculated?<\/li>\r\n \t<li>How is molar mass calculated?<\/li>\r\n \t<li>What would be the volume of 3.5 moles of a gas?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-TW9sZSBSb2FkIE1hcA..\">Mole Road Map<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YmNiYzQ3ZTliYjRjMWIwNDM3YTA2ODhlYTNmYTc3NzM.-eus\">\r\n \t<li>Perform calculations involving interconversions of mass, moles, and volume of a gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MGQ1NDE4ZGZiMDkwOGUwODc0ZmFjYTRjMzllYzRiOTc.-nid\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211646\/20140811155328853790.png\" alt=\"Chemistry road maps are similar to actual maps\" width=\"250\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MjViOGQ5NjcyYzM0NzRmZDNiNjcxODEyNTExYjI5YWY.-mey\"><strong> How do I get from here to there? <\/strong><\/p>\r\n<p id=\"x-ck12-ZTRkMDQ3ZDk1NDkxNzk1ZjE0Nzc3YjZkYTdjN2RhN2I.-mqh\">If I want to visit the town of Manteo, North Carolina, out on the coast, I will need a map of how to get there.\u00a0 I may have a printed map or I may download directions from the internet, but I need something to get me going in the right direction.\u00a0 Chemistry road maps serve the same purpose.\u00a0 How do I handle a certain type of calculation? There is a process and a set of directions to help.<\/p>\r\n\r\n<h3>Mole Road Map<\/h3>\r\n<p id=\"x-ck12-MDlkNTY1NmU2OTA1OTNhZGVhMTNiODYxNjE4Y2U2OTI.-kln\">Previously, we saw how the conversions between mass and number of particles required two steps, with moles as the intermediate.\u00a0 This concept can now be extended to also include gas volume at STP.\u00a0 The resulting diagram is referred to as a mole road map (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MTA3MS00MS03MS0xOQ..\"> below <\/a> ).<\/p>\r\n\r\n<div id=\"x-ck12-Mjk3N2YwYmJjNmM0MWIzNTgxZjZiYTQ5N2NlNjIxOGY.-j41\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-x3k\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MTA3MS00MS03MS0xOQ..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211648\/20140811155329093869.png\" alt=\"How to convert between moles, mass, number of particles, and volume of a gas\" longdesc=\"The%20mole%20road%20map%20shows%20the%20conversion%20factors%20needed%20to%20interconvert%20between%20mass%2C%20number%20of%20particles%2C%20and%20volume%20of%20a%20gas.\" \/><\/p>\r\n<strong> Figure 10.10 <\/strong>\r\n<p id=\"x-ck12-NGU3MzNmOGRiZTM5MzMxNDc1NzBmMGE5MTc3MjVlMDc.-cmc\">The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and volume of a gas.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-ZTg5OGE0ODIxZDk5ZTg2NGZmZjU4MzJkZDFiNGRlZjI.-fiy\">The mole is at the center of any calculation involving amount of a substance.\u00a0 The sample problem below is one of many different problems that can be solved using the mole road map.<\/p>\r\n\r\n<h4>Sample Problem One:\u00a0 Mole Road Map<\/h4>\r\n<p id=\"x-ck12-YWQ4NjAwY2EwZjcxOWI0NTdhNGI0NzYwMWYwNGZkZDY.-izn\">What is the volume of 79.3 g of neon gas at STP?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-isb\"><em> Step 1:\u00a0 List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-8na\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NWQyMjJkMWQ4MTQ4YmJiOTM2YTU5MDg2ZGVlOWMwMGI.-pbn\">\r\n \t<li>Ne = 20.18 g\/mol<\/li>\r\n \t<li>1 mol = 22.4 L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-vp9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ODVjMGMyNTI0MDRmOGI2NDhjOWVkOTY3Mjc4ZDhmNWI.-wh0\">\r\n \t<li>volume = ? L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MGU5NWI1NzE5MTllODY2NmM2MmZmMThiNTRhOTIwMzM.-qu4\">The conversion factors will be grams\u00a0\u2192 moles\u00a0\u2192 gas volume.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ehp\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-uj9\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211650\/a2f5d08ba0aa75f64bba645de62a7cc0.png\" alt=\"79.3 text{g Ne} times frac{1 text{mol Ne}}{20.18 text{g Ne}} times frac{22.4 text{L Ne}}{1 text{mol Ne}}=88.0 text{L Ne}\" width=\"386\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-bcc\"><em> Step 3:\u00a0 Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NzUxYjhhYTQ0ZWNhNTQ1NjQwMWI0MzQxOGNkMjJiNzA.-voc\">The given mass of neon is equal to about 4 moles, resulting in a volume that is about 4 times larger than molar volume.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-MjQ1MTJiOWEzYmUwNTMxYTBjMDZkMWIyMTUzNjczNWU.-ti4\">\r\n \t<li>An overall process is given for calculations involving moles, grams, and gas volume.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-YmU2ZWFjN2FiNWJkZjhmYzMyYzhjODk3YjI1MjM0OWE.-rlt\">Use the link below to carry out some practice calculations.\u00a0 Do problems 1, 2, and 5 (you can try the others if you are feeling especially brave):<\/p>\r\n<p id=\"x-ck12-MmQzYjgwNmVlYmM3YjRkZDIyMDAxNjBlNjk0NjE4Y2Q.-x1d\"><a href=\"http:\/\/www.docbrown.info\/page04\/4_73calcs\/MVGmcTEST.htm\"> http:\/\/www.docbrown.info\/page04\/4_73calcs\/MVGmcTEST.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-qm5\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-Njk4MDY3MmU1MDAxZWUzMjQyZDhmMWUxOGU4NmVjMjM.-kw7\">\r\n \t<li>In the problem above, what is the formula weight of neon?<\/li>\r\n \t<li>What value is at the center of all the calculations?<\/li>\r\n \t<li>If we had 79.3 grams of Xe, would we expect a volume that is greater than or less than that obtained with neon?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-UGVyY2VudCBDb21wb3NpdGlvbg..\">Percent Composition<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ODNhNzFiMGFhNjhmMGZiZjhjYmExOWRhMTFiZmI5MDE.-urd\">\r\n \t<li>Define percent composition.<\/li>\r\n \t<li>Perform percent composition calculations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MjI1MGUxYzE0OGE4ODAyODVhODExNGE4NjAwZWY0ZDc.-291\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211652\/20140811155329231868.jpeg\" alt=\"Nutritional information labels can let you know the percent composition\" width=\"500\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZmFjOWEzYThmMWEyODQyOWEwODY0YWNlNDg3MmM1ODY.-75u\"><strong> Is there anything healthy in this jar? <\/strong><\/p>\r\n<p id=\"x-ck12-ZWQyMGM2YzQ5ZmRmNTVlNGVjNGE5MmNiZWM1MjA1ZWU.-cxi\">Packaged foods that you eat typically have nutritional information provided on the label.\u00a0 The label on a jar of peanut butter (shown above) reveals that one serving size is considered to be 32 g.\u00a0 The label also gives the masses of various types of compounds that are present in each serving.\u00a0 One serving contains 7 g of protein, 15 g of fat, and 3 g of sugar.\u00a0 By calculating the fraction of protein, fat, or sugar in one serving of size of peanut butter and converting to percent values, we can determine the composition of the peanut butter on a percent by mass basis.<\/p>\r\n\r\n<h3>Percent Composition<\/h3>\r\n<p id=\"x-ck12-OTViMjJlMGExNTg3NDUzYTNlNWQ1YTk1NmFlYzE0OWU.-kwb\">Chemists often need to know what elements are present in a compound and in what percentage.\u00a0 The <strong> percent composition <\/strong> is the percent by mass of each element in a compound.\u00a0 It is calculated in a similar way that we just indicated for the peanut butter.<\/p>\r\n<p id=\"x-ck12-xxu\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211653\/b92d82fcecec402f2b5cdc6e1c2bb85d.png\" alt=\"% text{by mass}=frac{text{mass of element}}{text{mass of compound}} times 100 %\" width=\"320\" height=\"42\" \/><\/p>\r\n\r\n<h4>Percent Composition from Mass Data<\/h4>\r\n<p id=\"x-ck12-M2JmODNmY2MyZWUzNTE3M2M3MTUxMTYzOWM4MDYyYWQ.-coz\">The sample problem below shows the calculation of the percent composition of a compound based on mass data.<\/p>\r\n<p id=\"x-ck12-NjYxZmM0OTlkYTgyYjA5YzBkZGI0N2QyMTY2OWQ5NTE.-4ib\"><strong> Sample Problem One: Percent Composition from Mass <\/strong><\/p>\r\n<p id=\"x-ck12-YzMxNzBhODRmOTQyMjAzMzlkODBjZTVmZGI3M2NlNTA.-zl0\">A certain newly synthesized compound is known to contain the elements zinc and oxygen.\u00a0 When a 20.00 g sample of the sample is decomposed, 16.07 g of zinc remains.\u00a0 Determine the percent composition of the compound.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wb2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-x6p\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZTMyNDJhZmZhN2Q4YTBkZDBhY2NkMzI4NWJiYjUxNWY.-fag\">\r\n \t<li>mass of compound = 20.00 g<\/li>\r\n \t<li>mass of Zn = 16.07 g<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-a7n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YjUxMWRkOGZhYmVhNjczMmIxZTI2MmExZjQ5OTdlNjA.-gju\">\r\n \t<li>percent Zn = ? %<\/li>\r\n \t<li>percent O = ? %<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZTZmOTQ4YmE0Y2FjN2RmN2NkMzA0OTkzYTNhNjU4NzU.-iod\">Subtract to find the mass of oxygen in the compound.\u00a0 Divide each element\u2019s mass by the mass of the compound to find the percent by mass.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-a1m\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-u7m\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211654\/647f0368deb4e44c4f3d666318504d8d.png\" alt=\"text{Mass of oxygen}&amp;=20.00 text{g} - 16.07 text{g}=3.93 text{g O} \\% text{Zn}&amp;=frac{16.07 text{g Zn}}{20.00 text{g}} times 100 %=80.35 % text{Zn} \\% text{O}&amp;=frac{3.93 text{g O}}{20.00 text{g}} times 100 %=19.65 % text{O} \\\" width=\"401\" height=\"113\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-yha\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NDE0OGI5MjYzNjA5ZjE3MTQ1MWJiMTYyNDExMWU4M2Y.-k90\">The calculations make sense because the sum of the two percentages adds up to 100%.\u00a0 By mass, the compound is mostly zinc.<\/p>\r\n\r\n<h4 id=\"x-ck12-cxz_9-j53\">Percent Composition from a Chemical Formula<\/h4>\r\n<p id=\"x-ck12-ZjAwODE1ZTdiYmJkZTA5NzcyODQ2YWZhMzU0MDZiZDM.-ums\">The percent composition of a compound can also be determined from the formula of the compound.\u00a0 The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound.\u00a0 That is divided by the molar mass of the compound and multiplied by 100%.<\/p>\r\n<p id=\"x-ck12-gdp\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211656\/3b76b799ec06e735db961ababcef9120.png\" alt=\"% text{by mass }=frac{text{mass of element in} 1 text{mol}}{text{molar mass of compound}}times 100 %\" width=\"376\" height=\"42\" \/><\/p>\r\n<p id=\"x-ck12-YTQwNDE3NzBhYzRmNGUxNTZhOTYzMjhmZDA4M2UwODg.-pa4\">The percent composition of a given compound is always the same as long as the compound is pure.<\/p>\r\n<p id=\"x-ck12-MjQ5ZTI3Mjc2ZmUyOGFjMGUwNGJiNDEzNjg5MWNlODQ.-6zm\"><strong> Sample Problem Two: Percent Composition from Chemical Formula <\/strong><\/p>\r\n<p id=\"x-ck12-MTI4OWNiNDZmMTAwZTBiNDUyODlkNTZmNjUyMzY4OTA.-x58\">Dichlorineheptoxide (Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> ) is a highly reactive compound used in some organic synthesis reactions.\u00a0 Calculate the percent composition of dichlorineheptoxide.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wwu\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-xrx\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Y2JjNTVhZTI3OTQxNTBlOWE1YzIxZjE1YTZhMmYyNmM.-mor\">\r\n \t<li>mass of Cl in 1 mol Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 70.90 g<\/li>\r\n \t<li>mass of O in 1 mol Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 112.00 g<\/li>\r\n \t<li>molar mass of Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 182.90 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-so9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NWRlOGU1MzFjZjc3YTRmMTgwNjlmMGUzMzIzOGE3NmU.-cvp\">\r\n \t<li>percent Cl = ? %<\/li>\r\n \t<li>percent O = ? %<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NmEwZWUxMGU3NTZmYmNkZjliZTM2MTBiYTJhNmM3MzY.-wbo\">Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-xlh\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-cv0\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211657\/2cde2d521ee2c9531e0bb2c7e4e9fccf.png\" alt=\"% text{Cl}&amp;=frac{70.90 text{g Cl}}{182.90 text{g}} times 100 %=38.76 % text{Cl} \\% text{O}&amp;=frac{112.00 text{g O}}{182.90 text{g}} times 100 %=61.24 % text{O} \\\" width=\"314\" height=\"88\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-q10\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ZDQ5OTlmMzQ0YTExODZhYTY0OTExNDgzODUyZTczNTg.-15g\">The percentages add up to 100%.<\/p>\r\n<p id=\"x-ck12-ZmMyYjRhOTBmN2MzZTcxNzY5YjlhOTg0YjM0OTBmODQ.-gjt\">Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound.\u00a0 In the previous sample problem, it was found that the percent composition of dichlorineheptoxide is 38.76% Cl and 61.24% O.\u00a0 Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 g sample of dichlorineheptoxide.\u00a0 You can set up a conversion factor based on the percent by mass of each element.<\/p>\r\n<p id=\"x-ck12-jzl\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211659\/fc60c15a707115b10a71875f71e4c9fa.png\" alt=\"12.50 text{g Cl}_2 text{O}_7 times frac{38.76 text{g Cl}}{100 text{g Cl}_2text{O}_7}&amp;=4.845 text{g Cl} \\12.50 text{g Cl}_2 text{O}_7 times frac{61.24 text{g O}}{100 text{g Cl}_2text{O}_7}&amp;=7.655 text{g O} \\\" width=\"330\" height=\"90\" \/><\/p>\r\n<p id=\"x-ck12-NTA5NmJlZjdjOTUyM2EzZTE5ZmZlY2M3OGM5ZWMxYjA.-mqt\">The sum of the two masses is 12.50 g, the mass of the sample size.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZThmYjBmOWNhZDNkMWNlZGU4ZjU5NzhiMWI1NzI1NzA.-ui4\">\r\n \t<li>Processes are described for calculating the percent composition of a material based on mass or on chemical composition.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-ODU0YzM5ZWMxYzZmMzg1YzQ1MTBjMGQ2ODc4OWM3Yzc.-6sm\">Use the link below to review material and do calculations. Read both parts of the lesson and do as many calculations as you have time for.<\/p>\r\n<p id=\"x-ck12-YjZmZWNkOTg5NmM2MDE0MGNhNzkwN2U0OWIzMjRkNzI.-3si\"><a href=\"http:\/\/www.chemteam.info\/Mole\/Percent-Composition-Part1.html\"> http:\/\/www.chemteam.info\/Mole\/Percent-Composition-Part1.html <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-en4\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MjQ3OWQ1YTg4ZDk5ZjlmZDk1YTJjM2I4Zjk2OTUyMzk.-wgm\">\r\n \t<li>What is the formula for calculating percent composition?<\/li>\r\n \t<li>What information do you need to calculate percent composition by mass?<\/li>\r\n \t<li>What do subscripts in a chemical formula tell you?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MjY4YzM5MGJjMWZiMWQ5Nzg1YTk0MmEzMzcyNjUyNmM.-rxd\">\r\n \t<li><strong> percent composition: <\/strong> The percent by mass of each element in a compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-UGVyY2VudCBvZiBXYXRlciBpbiBhIEh5ZHJhdGU.\">Percent of Water in a Hydrate<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ZjU4MmEwOTI1ZWQwZTYyNzA3MDNkZTAyNDgwZjM3Y2Y.-xjt\">\r\n \t<li>Define hydrate.<\/li>\r\n \t<li>Calculate the percent water in hydrate when give relevant data.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-OGZkYzA3Y2Y0YmM4OGIzZGRjMzVmZDQ1MDg0ZjgwYTk.-yq5\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211700\/20140811155329359963.jpeg\" alt=\"Copper sulfate changes color when hydrated\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-M2Y5YzNlZWM2NzY4ZmEzYmEyNWEyNGZkZjNhOWJjMTg.-htu\"><strong> Why does the color change? <\/strong><\/p>\r\n<p id=\"x-ck12-YTk2MzRhMTQ3NDZmZGZiY2U2OTgxYTYxYWY4MWNhZjY.-zds\">If you look at a typical bottle of copper sulfate, it will be a bluish-green.\u00a0 If someone tells you that copper sulfate is white, you won\u2019t believe them.\u00a0 You are both right; it just depends on the copper sulfate.\u00a0 Your blue-green copper sulfate has several water molecules attached to it while your friend\u2019s copper sulfate is anhydrous (no water attached).\u00a0 Why the difference? The water molecules interact with some of the d electrons in the copper ion and produce the color.\u00a0 When the water is removed, the electron configuration changes and the color disappears.<\/p>\r\n\r\n<h3>Percent of Water in a Hydrate<\/h3>\r\n<p id=\"x-ck12-YTI1YzI0NDM3Mzg5Y2UxY2I1OWI5M2ExOWM1NzIzNDk.-kpn\">Many ionic compounds naturally contain water as part of the crystal lattice structure.\u00a0 A <strong> hydrate <\/strong> is a compound that has one or more water molecules bound to each formula unit.\u00a0 Ionic compounds that contain a transition metal are often highly colored.\u00a0 Interestingly, it is common for the hydrated form of a compound to be of a different color than the <strong> anhydrous <\/strong> form, which has no water in its structure.\u00a0 A hydrate can usually be converted to the anhydrous compound by heating. \u00a0For example,\u00a0the anhydrous compound cobalt(II) chloride is blue, while the hydrate is a distinctive magenta color.<\/p>\r\n\r\n<div id=\"x-ck12-ZTk2YzYyMzdlZjYxZTBiNzBmNDQ1YTk4MDRkYWUwY2M.-6wt\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-bdr\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MTUwNy0zNC04Mi0yMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211702\/20140811155329517119.png\" alt=\"Anhydrous cobalt chloride is blue, while hydrated cobalt chloride is red\" longdesc=\"On%20the%20left%20is%20anhydrous%20cobalt(II)%20chloride%2C%20CoCl%3Csub%3E2%3C%2Fsub%3E.%20On%20the%20right%20is%20the%20hydrated%20form%20of%20the%20compound%20called%20cobalt(II)%20chloride%20hexahydrate%2C%20CoCl%3Csub%3E2%3C%2Fsub%3E%E2%80%A26H%3Csub%3E2%3C%2Fsub%3EO.\" \/><\/p>\r\n<strong> Figure 10.11 <\/strong>\r\n<p id=\"x-ck12-NzVlYmFhNGYxYTUzN2NhNGFjYmM5ZDMwNWE3YWFiZGE.-43v\">On the left is anhydrous cobalt(II) chloride, CoCl <sub> 2 <\/sub> . On the right is the hydrated form of the compound called cobalt(II) chloride hexahydrate, CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NTBiZjIxYTQxMmVjYmZmYmU0NzYyZjA5ZDBjZjllZGQ.-sut\">The hydrated form of cobalt(II) chloride contains six water molecules in each formula unit.\u00a0 The name of the compound is cobalt(II) chloride hexahydrate and its formula is CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.\u00a0 The formula for water is set apart at the end of the formula with a dot, followed by a coefficient that represents the number of water molecules per formula unit.<\/p>\r\n<p id=\"x-ck12-YTYyNDI4OWNkNzZlZjI3M2FlMzhkM2FhZDc3NzU2NTY.-4uv\">It is useful to know the percent of water contained within a hydrate.\u00a0 The sample problem below demonstrates the procedure.<\/p>\r\n\r\n<h4>Sample Problem One: Percent of Water in a Hydrate<\/h4>\r\n<p id=\"x-ck12-YzQzZDQ3MDI2NGY3YjFjYmZlYzIxNmZhYjU4NmY1YzA.-13j\">Find the percent water in cobalt(II) chloride hexahydrate, CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-6s4\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-ZDRmODhiMDJiYzgyYzhmNWUzNDhjZmViZWJhNDA4YmE.-jbc\">The mass of water in the hydrate is the coefficient (6) multiplied by the molar mass of H <sub> 2 <\/sub> O.\u00a0 The molar mass of the hydrate is the molar mass of the CoCl <sub> 2 <\/sub> plus the mass of water.<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-rw0\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-N2EzMzg1ZDlkOTY0NDI0Yjg1NjQyZTI2ODYxNGM5NjQ.-q0s\">\r\n \t<li>mass of H <sub> 2 <\/sub> O in 1 mol hydrate = 108.12 g<\/li>\r\n \t<li>molar mass of hydrate = 237.95 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-efs\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZTc3NjJjMDFlYTJhMzc5NmUwYjA2ZjFiYTMzODVlM2M.-cnv\">\r\n \t<li>percent H <sub> 2 <\/sub> O = ? %<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZDliZGQ0M2M5ODY1MGY5YTdkYjllYTY1NDZmYzFhZWI.-hxv\">Calculate the percent by mass of water by dividing the mass of H <sub> 2 <\/sub> O in 1 mole of the hydrate by the molar mass of the hydrate and multiplying by 100%.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-i34\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-lkr\" class=\"x-ck12-indent\"><img id=\"x-ck12-MTQwMDUyNjEzMTk1MA..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211703\/8eb891808f0b1fa0db4291b307a7b03a.png\" alt=\"% text{H}_2text{O}=frac{108.12 text{g H}_2 text{O}}{237.95 text{g}} times 100 %=45.44 % text{H}_2text{O}\" width=\"372\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-pf8\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NTY3MzJjNGE2Mzg4NjYyYjY3ZWJhN2JiNDFlYjI3ZGU.-ea1\">Nearly half of the mass of the hydrate is composed of water molecules within the crystal.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZDRjOGQ3MDg5MGZjNjg5MjQxMDUzNzRhMGI1Y2JlMjI.-j25\">\r\n \t<li>The process of calculating the percent water in a hydrate is described.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-N2RmYjM3YzZjYjgwMTM0ZWVkODI5ZWZjMzIzNDNiNDU.-qco\">Use the following link to practice calculating percent water in a hydrate:<\/p>\r\n<p id=\"x-ck12-MjAxM2U1YmZmOThiYTYxNTMzNThjNzVlZmIzN2U0Yzg.-gd4\"><a href=\"https:\/\/web.archive.org\/web\/20150314182712\/http:\/\/www.sd84.k12.id.us\/shs\/departments\/science\/martz\/2007_ssem2\/Chemistry\/hydrate.htm\" target=\"_blank\" rel=\"noopener\">Hydrate Problems<\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ok5\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-YWM5MzhmYmE1YTg5NjBkZTJjMmQ0ODA1YjIwZDI1YjM.-ubb\">\r\n \t<li>What is a hydrate?<\/li>\r\n \t<li>How can you convert a hydrate to an anhydrous compound?<\/li>\r\n \t<li>What does hexahydrate mean?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-Njg3NTJjMzFjN2ViNzEzNDg4NWUyMjZhOWY3MTBmYjE.-dgv\">\r\n \t<li><strong> Anhydrous: <\/strong> Without water.<\/li>\r\n \t<li><strong> Hydrate: <\/strong> A compound that has one or more water molecules bound to each formula unit<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-RGV0ZXJtaW5pbmcgRW1waXJpY2FsIEZvcm11bGFz\">Determining Empirical Formulas<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NjgwNWFkZmJjN2RmZmYxZWQ5OGRlZDc0OTUwOWNlOTc.-mj0\">\r\n \t<li>Define empirical formula.<\/li>\r\n \t<li>Calculate the empirical formula for a compound when given the elemental analysis of the compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-NGYzYTZkMGE0NDUyODdiMmQ4NTUwYmZhNTU0ZDJlYzY.-0zt\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211705\/20140811155329603191.jpeg\" alt=\"Scientist trying to determine the formula of a compound in the early days of chemistry\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-YmVhNTkxMDgwODZkMjMxNzE5Zjc1NTFlZDJhOTc5OGE.-0as\"><strong> What is occuring in this picture? <\/strong><\/p>\r\n<p id=\"x-ck12-MjcwMTM1NzM4ZTI0M2M4ZDFmODJlMDAyZDgzNTlhMTY.-npe\">In the early days of chemistry, there were few tools for the detailed study of compounds.\u00a0 Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials.\u00a0 The \u201cnew\u201d field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely.\u00a0 The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios.\u00a0 We did not know exactly how many of these atoms were actually in a specific molecule.<\/p>\r\n\r\n<h3>Determining Empirical Formulas<\/h3>\r\n<p id=\"x-ck12-OTU4NTdiYmJmNmU1YmNiYzdmMGE3ZGYxZTc1MmM4MTk.-09t\">An <strong> empirical formula <\/strong> is one that shows the lowest whole-number ratio of the elements in a compound.\u00a0 Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical.\u00a0 However, we can also consider the empirical formula of a molecular compound.\u00a0 Ethene is a small hydrocarbon compound with the formula C <sub> 2 <\/sub> H <sub> 4 <\/sub> (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MTc3NS01OS03Ni0yNA..\"> below <\/a> ).\u00a0 While C <sub> 2 <\/sub> H <sub> 4 <\/sub> is its molecular formula and represents its true molecular structure, it has an empirical formula of CH <sub> 2 <\/sub> .\u00a0 The simplest ratio of carbon to hydrogen in ethene is 1:2.\u00a0 There are two ways to view that ratio.\u00a0 Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen.\u00a0 Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen.\u00a0 So the subscripts in a formula represent the mole ratio of the elements in that formula.<\/p>\r\n\r\n<div id=\"x-ck12-ZDc4ZGU1MjM1ZDI0NzRkMDI2NzNlMzEwOTE1MDhjMDA.-xni\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-bfo\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MTc3NS01OS03Ni0yNA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211707\/20140811155329734238.png\" alt=\"Ball and stick model of ethene\" longdesc=\"Ball-and-stick%20model%20of%20ethene%2C%20C%3Csub%3E2%3C\/sub%3EH%3Csub%3E4%3C\/sub%3E.\" \/><\/p>\r\n<strong> Figure 10.12 <\/strong>\r\n<p id=\"x-ck12-Zjg2YmVhMDc5OTlmZGFkMTI4N2NmZjc5N2RlYmI1ZjA.-vkq\">Ball-and-stick model of ethene, C <sub> 2 <\/sub> H <sub> 4 <\/sub> .<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-MGIwNTU0MWQ1Y2ZmYjNiNDgyYTQxODQxYTc3NTlmZTY.-lpo\">In a procedure called <strong> elemental analysis <\/strong> , an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it.\u00a0 These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula.\u00a0 The steps to be taken are outlined below.<\/p>\r\n\r\n<ol id=\"x-ck12-NzY4OGI5OTNiNDBiZDIzYzU0MTU5MzIxZDQyMGUyYjY.-kuz\">\r\n \t<li>Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.<\/li>\r\n \t<li>Use each element\u2019s molar mass to convert the grams of each element to moles.<\/li>\r\n \t<li>In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.<\/li>\r\n \t<li>If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.<\/li>\r\n \t<li>In some cases, one or more of the moles calculated in step 3 will not be whole numbers.\u00a0 Multiply each of the moles by the smallest whole number that will convert each into a whole number.\u00a0 Write the empirical formula.<\/li>\r\n<\/ol>\r\n<h4>Sample Problem One: Determining the Empirical Formula of a Compound<\/h4>\r\n<p id=\"x-ck12-OTY0ZjU3YmNjNTNhMGRhYzYzNmQwYjFmYTcwNjAzM2U.-qxb\">A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen.\u00a0 Find the empirical formula of the compound.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qt9\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-mry\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZjA3NmYwZmQ4YmM4YzIzNmI5YWQ5NmE2MGVmNzhhOTQ.-4lo\">\r\n \t<li>% of Fe = 69.94%<\/li>\r\n \t<li>% of O = 30.06%<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-md4\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MmU1NTM2NzNhMmYyMzU2MWRkNTkyMTY1ZGYzOGUyNjg.-srb\">\r\n \t<li>Empirical formula = Fe <sub> ? <\/sub> O <sub> ? <\/sub><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-Y2UwNGJmZDM4ZWVhMTQyYTYxMDViYmQxN2UzN2I4NmU.-23z\">Steps to follow are outlined in the text.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-d07\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-YTc1MGZkMjRmYTgwYTExMTM5NmI3YTdiYzRmZGJmZjI.-ckw\">1. Assume a 100 g sample.<\/p>\r\n<p id=\"x-ck12-8fp\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211708\/a12004a26997f9bc06927cf8a99d6235.png\" alt=\"&amp; 69.94 text{g Fe} \\&amp; 30.06 text{g O}\" width=\"80\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-OGJkYjc0ZTNhN2E4NGJkN2NmNmY1MTU4NGM0YmQyODY.-lgc\">2. Convert to moles.<\/p>\r\n<p id=\"x-ck12-npt\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211708\/fa055626cb3aa7cfc7f1f24cebfbb9da.png\" alt=\"69.94 text{g Fe} times frac{1 text{mol Fe}}{55.85 text{g Fe}}&amp;=1.252 text{mol Fe} \\30.06 text{g O} times frac{1 text{mol O}}{16.00 text{g O}}&amp;=1.879 text{mol O}\" width=\"309\" height=\"88\" \/><\/p>\r\n<p id=\"x-ck12-NWYzYzU4MjE5NDZhN2U2ZDkzMzc5MzEwODVkZWEyYTI.-awb\">3. Divide both moles by the smallest of the results.<\/p>\r\n<p id=\"x-ck12-jtc\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211709\/a3f8d99609e28a14fff832e9c8ba2fe3.png\" alt=\"frac{1.252 text{mol Fe}}{1.252}=1 text{mol Fe} qquad qquad frac{1.879 text{mol O}}{1.252}=1.501 text{mol O}\" width=\"483\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MTk5YjJjM2NkNzUyZjNmYTcwMjFhZTQyNDNkMDJhZjg.-ld1\">4\/5. Since the moles of O, is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.<\/p>\r\n<p id=\"x-ck12-bw1\" class=\"x-ck12-indent\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211711\/c2909a1f6b81830213b3cfe0e81413a9.png\" alt=\"1 text{mol Fe} times 2=2 text{mol Fe} qquad qquad 1.501 text{mol O} times 2=3 text{mol O}\" width=\"475\" height=\"14\" \/><\/p>\r\n<p id=\"x-ck12-N2Q2Y2ZmNWJmYTkwYzU0MzQ0NTZjZjNiMzIzYzQ2NzA.-rh1\">The empirical formula of the compound is Fe <sub> 2 <\/sub> O <sub> 3 <\/sub> .<\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-rym\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NjI0NzhiYmVmZTJmM2E1M2ZlMzY5ODBkM2Y2NjVkMDY.-3ml\">The subscripts are whole numbers and represent the mole ratio of the elements in the compound.\u00a0 The compound is the ionic compound iron(III) oxide.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-YzA2OTg4Y2NjZDQ4MzQzYTBlYmI0YmVjNWEyMGYwYmE.-omu\">\r\n \t<li>A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NjIwM2QyZjBiODY5NjU2MTdiYjU0NDM5ZjE1M2E4MTI.-kje\">Use the link below to read about calculating empirical formulas and practice working some problems:<\/p>\r\n<p id=\"x-ck12-OGM1YzEwYjM1ZmQzNGQyOGM2ZmI4MjJjNDU2Mzc3ZTY.-yqd\"><a href=\"http:\/\/www.chemteam.info\/Mole\/EmpiricalFormula.html\"> http:\/\/www.chemteam.info\/Mole\/EmpiricalFormula.html <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-25c\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-OTA1MjIyYjBjZjVhM2E3NGZlNWI1N2UyY2YzYjMxZGQ.-21w\">\r\n \t<li>What is an empirical formula?<\/li>\r\n \t<li>What does an empirical formula tell you?<\/li>\r\n \t<li>What does it not tell you?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-NDM4YWMxMzk3N2VlZjllOWE2ZjliYjQ5ZWU5OTEyNDg.-ktb\">\r\n \t<li><strong> elemental analysis: <\/strong> Determines the percentages of each element in a compound.<\/li>\r\n \t<li><strong> empirical formula: <\/strong> Shows the lowest whole-number ratio of the elements in a compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-RGV0ZXJtaW5pbmcgTW9sZWN1bGFyIEZvcm11bGFz\">Determining Molecular Formulas<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ODQ5ZjAyNGE0MWVjNjllNzhhMWI3MDAzODkzNDc1MTA.-7v2\">\r\n \t<li>Define molecular formula.<\/li>\r\n \t<li>Determine the molecular formula when give the empirical formula and the molar mass of the compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-N2NlOTRkYmFlMDFjNTI5MTIxMDFiNmM3OTA3MmM4NzQ.-swg\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211712\/20140811155329857990.jpg\" alt=\"Fischer projection of glucose\" width=\"100\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZmRhMGQ4Mzg4YzRkZTdjM2QwOWEzMjM2ZjYwNGIwYzE.-wqv\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211713\/20140811155329998495.png\" alt=\"Structure of sucrose\" width=\"300\" \/><\/span><\/p>\r\n<p id=\"x-ck12-OGJlMjRhMDUxYWY0YzAwNDY1YmM0YTRhNzhkY2IxMjg.-kpz\"><strong> How can you determine the differences between these two molecules? <\/strong><\/p>\r\n<p id=\"x-ck12-OWIyNThlYmIyM2EyMTM1YTljOWQ0OTgzNjAyZDdjZjY.-4md\">Above we see two carbohydrates: glucose and sucrose.\u00a0 Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar.\u00a0 Some people could distinguish them on the basis of taste, but it\u2019s not a good idea to go around tasting chemicals. The best way is to determine the molecular weights \u2013 this approach allows you to easily tell which compound is which.<\/p>\r\n\r\n<h3>Molecular Formulas<\/h3>\r\n<p id=\"x-ck12-ZGM1MDRmYWRhMWM5NjZlZWJmZmU1ODBjNGMzN2FlN2I.-pk4\"><strong> Molecular formulas <\/strong> give the kind and number of atoms of each element present in a molecular compound.\u00a0 In many cases, the molecular formula is the same as the empirical formula.\u00a0 The molecular formula of methane is CH <sub> 4 <\/sub> and because it contains only one carbon atom, that is also its empirical formula.\u00a0 Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula.\u00a0 Acetic acid\u00a0 is an organic acid that is the main component of vinegar.\u00a0 Its molecular formula is C <sub> 2 <\/sub> H <sub> 4 <\/sub> O <sub> 2 <\/sub> .\u00a0 Glucose is\u00a0 a simple sugar that cells use as a primary source of energy.\u00a0 Its molecular formula is C <sub> 6 <\/sub> H <sub> 12 <\/sub> O <sub> 6 <\/sub> .\u00a0 The structures of both molecules are shown in the figure below.\u00a0 They are very different compounds, yet both have the same empirical formula of CH <sub> 2 <\/sub> O.<\/p>\r\n\r\n<div id=\"x-ck12-M2VkYmVmYzM2MGRhM2I1NzZmMGEyN2Y0YzQ4MzhhNzQ.-sae\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-m8v\"><img id=\"x-ck12-OTgwNDUtMTM2MjE0MjAzOS05NS05NC0yNw..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211714\/20140811155330084912.png\" alt=\"Acetic acid and glucose both the same empirical formula\" longdesc=\"Acetic%20acid%20%28left%29%20has%20a%20molecular%20formula%20of%20C%3Csub%3E2%3C\/sub%3EH%3Csub%3E4%3C\/sub%3EO%3Csub%3E2%3C\/sub%3E%2C%20while%20glucose%20%28right%29%20has%20a%20molecular%20formula%20of%20C%3Csub%3E6%3C\/sub%3EH%3Csub%3E12%3C\/sub%3EO%3Csub%3E6%3C\/sub%3E.%20Both%20have%20the%20empirical%20formula%20CH%3Csub%3E2%3C\/sub%3EO.\" \/><\/p>\r\n<strong> Figure 10.13 <\/strong>\r\n<p id=\"x-ck12-ZTI2OTAwNGQyMDBjYzhhNzM3NzAwMDk4OGU0Y2FlOGM.-xtu\">Acetic acid (left) has a molecular formula of C <sub> 2 <\/sub> H <sub> 4 <\/sub> O <sub> 2 <\/sub> , while glucose (right) has a molecular formula of C <sub> 6 <\/sub> H <sub> 12 <\/sub> O <sub> 6 <\/sub> . Both have the empirical formula CH <sub> 2 <\/sub> O.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NDFiODQyMzRiNTY3NWIwNGVlN2E1OTAxYjczZmRmMTM.-gty\">Empirical formulas can be determined from the percent composition of a compound.\u00a0 In order to determine its molecular formula, it is necessary to know the molar mass of the compound.\u00a0 Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds.\u00a0 In order to go from the empirical formula to the molecular formula, follow these steps:<\/p>\r\n\r\n<ol id=\"x-ck12-MWNlODE1ZTM5NTJhNDJkYjBiODg3MmIwODhjZTM0YmE.-x9t\">\r\n \t<li>Calculate the <strong> empirical formula mass (EFM) <\/strong> , which is simply the molar mass represented by the empirical formula.<\/li>\r\n \t<li>Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.<\/li>\r\n \t<li>Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.<\/li>\r\n<\/ol>\r\n<h4>Sample Problem One: Determining the Molecular Formula of a Compound<\/h4>\r\n<p id=\"x-ck12-NDE5MWIyMzIzYzc0NTI5NDBhZGRkYmE0MGRmZmJmMTg.-1lt\">The empirical formula of a compound of boron and hydrogen is BH <sub> 3 <\/sub> .\u00a0 Its molar mass is 27.7 g\/mol.\u00a0 Determine the molecular formula of the compound.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-noc\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-4yc\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-Yzc1Yjg5YjIwMGVjMDA5MTg3NzA1YWVmMjMwMTYyZDM.-hnl\">\r\n \t<li>empirical formula = BH <sub> 3 <\/sub><\/li>\r\n \t<li>molar mass = 27.7 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-4yo\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZGM3MzlhMjY5YjliMjA5YWIyYWJlNWM5NDUwMDg2YzY.-f3h\">\r\n \t<li>molecular formula = ?<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-Y2UwNGJmZDM4ZWVhMTQyYTYxMDViYmQxN2UzN2I4NmU.-ted\">Steps to follow are outlined in the text.<\/p>\r\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-7l1\"><em> Step 2: Calculate. <\/em><\/p>\r\n<p id=\"x-ck12-OGNlZTZhMDVkNGMxZTA2ZTQ2MDk4NWYyMWM2ZTYyZjc.-mcs\">1. The empirical formula mass (EFM) = 13.84 g\/mol<\/p>\r\n<p id=\"x-ck12-YzgxZTcyOGQ5ZDRjMmY2MzZmMDY3Zjg5Y2MxNDg2MmM.-mkd\">2. <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211715\/f1f79692abe6d70305d06ef262d09b50.png\" alt=\"frac{text{molar mass}}{text{EFM}}=frac{27.7}{13.84}=2\" width=\"156\" height=\"25\" \/><\/p>\r\n<p id=\"x-ck12-ZWNjYmM4N2U0YjVjZTJmZTI4MzA4ZmQ5ZjJhN2JhZjM.-vim\">3. <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211716\/d925b5855abf5be5b1bcd2a7c049b06f.png\" alt=\"text{BH}_3 times 2 =text{B}_2text{H}_6\" width=\"128\" height=\"16\" \/><\/p>\r\n<p id=\"x-ck12-NDMwODQwZDVkOTM2ODc1YTUzZGJkOGQzZTQ1YmRmN2Q.-at7\">The molecular formula of the compound is B <sub> 2 <\/sub> H <sub> 6 <\/sub> .<\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-euc\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-Y2JhOWY3YWE1MmM3OTVhZDRiYjIyODA4NzUzODJiMjQ.-n64\">The molar mass of the molecular formula matches the molar mass of the compound.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ODc4MzI5NmI2MmIzMWNiY2E4NzgzYTU0ZDFmZGE2MjM.-kat\">\r\n \t<li>A procedure is described that allows the calculation of the exact molecular formula for a compound.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-N2M0NWJjMDA5MWJiNzNhMDMyNmE0NDQ2ZDQ1NWU0NGY.-rlr\">Use the link below to access practice problems.\u00a0 Try as many as you have time for:<\/p>\r\n<p id=\"x-ck12-NDYyOWY0MTZiZTU1N2E1MTIzOGI0MmY2ZWUwNTIwNjc.-r43\"><a href=\"http:\/\/chemistry.about.com\/od\/chemistry-test-questions\/tp\/Molecular-Formula-Practice-Test-Questions.htm\"> http:\/\/chemistry.about.com\/od\/chemistry-test-questions\/tp\/Molecular-Formula-Practice-Test-Questions.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-mrr\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-N2QwNGRlNTM3N2U5OGU2ZDlhZDVhMTdkODY3ZDdjY2Y.-ld1\">\r\n \t<li>What is the difference between an empirical formula and a molecular formula?<\/li>\r\n \t<li>In addition to the elemental analysis, what do you need to know to calculate the molecular formula?<\/li>\r\n \t<li>What does the empirical formula mass tell you?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ODc4OTcwMTA1NzVjMDQ4NWViNWZjZjI5ODU0MjNkNTI.-i9v\">\r\n \t<li><strong> empirical formula mass (EFM): <\/strong> The molar mass represented by the empirical formula.<\/li>\r\n \t<li><strong> molecular formula: <\/strong> Gives the kind and number of atoms of each element present in a molecular compound.<\/li>\r\n<\/ul>\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>User:OSX\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:1983-1988_Toyota_Hilux_4-door_utility_01.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:1983-1988_Toyota_Hilux_4-door_utility_01.jpg <\/a> .<\/li>\r\n \t<li>C. Sentier. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Amedeo_Avogadro2.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Amedeo_Avogadro2.jpg <\/a> .<\/li>\r\n \t<li>Left: Michael David Hill, 2005 (Mikiwikipikidikipedia); Right: chrisbb@prodigy.net. Left: http:\/\/commons.wikimedia.org\/wiki\/File:Close-up_of_mole.jpg; Right: http:\/\/www.flickr.com\/photos\/chrisbrenschmidt\/436990097\/ .<\/li>\r\n \t<li>Courtesy of Diane A. Reid\/National Cancer Institute. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Man_responds_to_telephone_call.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Man_responds_to_telephone_call.jpg <\/a> .<\/li>\r\n \t<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Water-3D-balls-A.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Water-3D-balls-A.png <\/a> .<\/li>\r\n \t<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Potassium-dichromate-sample.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Potassium-dichromate-sample.jpg <\/a> .<\/li>\r\n \t<li>User:Estormiz\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Fermion_Plant_Oulu_2007_01_20.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Fermion_Plant_Oulu_2007_01_20.JPG <\/a> .<\/li>\r\n \t<li>Martin Walker (Wikimedia: Walkerma). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Calcium_chloride.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Calcium_chloride.jpg <\/a> .<\/li>\r\n \t<li>Laura Guerin. CK-12 Foundation .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>User:Mark.murphy\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Diving_-_scubadiver.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Diving_-_scubadiver.JPG <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. CK-12 Foundation .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>Image copyright Marynchenko Oleksandr, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\r\n \t<li>Image copyright md8speed, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\r\n \t<li>Photographer: Warren Denning, courtesy of the Pioneer Balloon Company. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Congrats_bqt.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Congrats_bqt.jpg <\/a> .<\/li>\r\n \t<li>Doug Smith (Wikimedia: Xsmith). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:CurrituckSoundMap.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:CurrituckSoundMap.png <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Joy Sheng. .<\/li>\r\n \t<li>User:Chemicalinterest\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper%28II%29_sulfate.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Copper%28II%29_sulfate.JPG <\/a> .<\/li>\r\n \t<li>(A) Martin Walker (Wikimedia: Walkerma); (B) Ben Mills (Wikimedia: Benjah-bmm27). (A) http:\/\/commons.wikimedia.org\/wiki\/File:Cobalt%28II%29_chloride.jpg; (B) http:\/\/commons.wikimedia.org\/wiki\/File:Cobalt%28II%29-chloride-hexahydrate-sample.jpg .<\/li>\r\n \t<li>Harriet Moore. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:M_Faraday_Lab_H_Moore.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:M_Faraday_Lab_H_Moore.jpg <\/a> .<\/li>\r\n \t<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Ethylene-CRC-MW-3D-balls.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Ethylene-CRC-MW-3D-balls.png <\/a> .<\/li>\r\n \t<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png <\/a> .<\/li>\r\n \t<li>User:glycoform\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Sucrose_3Dprojection.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Sucrose_3Dprojection.png <\/a> .<\/li>\r\n \t<li>(left) Ben Mills (Wikimedia: Benjah-bmm27); (right) Ben Mills (Wikimedia: Benjah-bmm27), User:Yikrazuul\/Wikimedia Commons. Acetic acid: http:\/\/commons.wikimedia.org\/wiki\/File:Acetic-acid-2D-flat.png; Glucose: http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png .<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<h1 id=\"x-ck12-VGhlIE1vbGU.-chapter\">The Mole<\/h1>\n<div class=\"x-ck12-data\"><\/div>\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBOdW1iZXI.\">Avogadro&#8217;s Number<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YTYxYzI1MTQ4MGY4MDQ2NTk0MjU4NWE0NmI2NjEzYWE.-xoq\">\n<li>Define mole.<\/li>\n<li>Define Avogadro\u2019s number.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MzkyYjkxNDgyYzA4ZGYyZDcyNjdlZjViNDRlYmQ5ZDE.-tkc\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211607\/20140811155325897191.jpeg\" alt=\"It is often convenient to have a unit for large amounts, such as a mole\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-ZGVkOGIzY2YxNTJjNGY4NjMzZGRhYTk3YTJiYTI3YzE.-bci\"><strong> Is there an easier way to load this truck? <\/strong><\/p>\n<p id=\"x-ck12-N2I5MTBiNzMzN2ZiNjMyYzA2ZmJhMmVkZmMzODJjODY.-8sa\">When the weather is nice, many people begin to work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add to other materials. You could order up twenty million grains of sand and have people really stare at you. You could order by the pound, but that takes a lot of time weighing out. The best bet is to order by the yard, meaning a cubic yard. The loader can easily scoop up what you need and put it directly in your truck.<\/p>\n<h3>Avogadro\u2019s Number<\/h3>\n<p id=\"x-ck12-MTkwNDE0OGNhYTljYjgwYjNkZWRlNDAwNmM4MGU1ODE.-emq\">It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small \u2013 far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 10 <sup> 22 <\/sup> molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6\u00a0\u00d7\u00a010 <sup> 10 <\/sup> molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.<\/p>\n<p id=\"x-ck12-YWU3M2U3ZWI5OTczMDA5NjY4ZDFkNTVlZDMyMjA5NmY.-4tg\">Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 &#8211; 1856), an Italian scientist, provided just such a number. He is responsible for the counting unit of measure called the mole. A <strong> mole <\/strong> (mol) is the amount of a substance that contains 6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> representative particles of that substance. The mole is the SI unit for amount of a substance.\u00a0 Just like the dozen and the gross, it is a name that stands for a number. There are therefore\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> water molecules in a mole of water molecules. There also would be\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> bananas in a mole of bananas, if such a huge number of bananas ever existed.<\/p>\n<div id=\"x-ck12-NTlhNGU0NWE2MzU3ODc3NWVkY2NhMWIzZDM2NDAzNDA.-sf5\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-85y\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjEzODkzOS00My04MC0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211608\/20140811155326053064.jpeg\" alt=\"Portrait of Amedeo Avogadro\" longdesc=\"Italian%20scientist%20Amedeo%20Avogadro%2C%20whose%20work%20led%20to%20the%20concept%20of%20the%20mole%20as%20a%20counting%20unit%20in%20chemistry.\" \/><\/p>\n<p><strong> Figure 10.1 <\/strong><\/p>\n<p id=\"x-ck12-ZjY3MWE1ZWM1M2IwMzQ1YTRjYzdkN2IyOWY1Yzc5NmU.-ly2\">Italian scientist Amedeo Avogadro, whose work led to the concept of the mole as a counting unit in chemistry.<\/p>\n<\/div>\n<p id=\"x-ck12-YjZiYjc2MjdlOTU5ZTU0NTQ4NzdmY2MxZjc3M2ZkMzY.-heu\">The number\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> is called <strong> Avogadro\u2019s number <\/strong> , the number of representative particles in a mole. It is an experimentally determined number. A <strong> representative particle <\/strong> is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are H <sub> 2 <\/sub> , N <sub> 2 <\/sub> , O <sub> 2 <\/sub> , F <sub> 2 <\/sub> , Cl <sub> 2 <\/sub> , Br <sub> 2 <\/sub> , and I <sub> 2 <\/sub> . The representative particle for these elements is the molecule. Likewise, all molecular compounds such as H <sub> 2 <\/sub> O and CO <sub> 2 <\/sub> exist as molecules and so the molecule is their representative particle.\u00a0 For ionic compounds such as NaCl and Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub> , the representative particle is the formula unit. A mole of any substance contains Avogadro\u2019s number (6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> ) of representative particles.<\/p>\n<div id=\"x-ck12-NWNhNmVjZjZhMTVlODJhNmM4NmIxZDRlODQ4N2U1NDE.-kp5\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-2wh\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjEzODk3My0yOS0yOS0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211609\/20140811155326222771.jpeg\" alt=\"The animal mole is very different from the unit mole\" longdesc=\"The%20animal%20mole%20is%20very%20different%20than%20the%20counting%20unit%20of%20the%20mole.%20Chemists%20nonetheless%20have%20adopted%20the%20mole%20as%20their%20unofficial%20mascot.%20National%20Mole%20Day%20is%20a%20celebration%20of%20chemistry%20that%20occurs%20on%20October%2023rd%20%2810\/23%29%20of%20each%20year.\" \/><\/p>\n<p><strong> Figure 10.2 <\/strong><\/p>\n<p id=\"x-ck12-ZjllNjY2MmQ1NTFlMDFkNmQ2YjEzMTQwZmU0MzY2NDg.-2go\">The animal mole is very different than the counting unit of the mole. Chemists nonetheless have adopted the mole as their unofficial mascot. National Mole Day is a celebration of chemistry that occurs on October 23rd (10\/23) of each year.<\/p>\n<\/div>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-YzU4ZjM4NWVjOWU1NjFhYjMwOGU0ZWUxNjA3YTk1ZjE.-kgh\">\n<li>A mole of any substance contains Avogadro\u2019s number\u00a0(6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> ) of representative particles.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-nwf\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-uya\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-OTliODQ0ODNlOWVkMTU0YmVhODAwNzFlNmVhYmRhODg.-boc\"><a href=\"http:\/\/www.scientificamerican.com\/article.cfm?id=how-was-avogadros-number\"> http:\/\/www.scientificamerican.com\/article.cfm?id=how-was-avogadros-number <\/a><\/p>\n<ol id=\"x-ck12-NjU2NmM2OTg3ZGY5YWVkZDBhMmZiYjI4MjI2MmE0ODM.-oqa\">\n<li>What was Avogadro\u2019s hypothesis?<\/li>\n<li>Who first calculated this number?<\/li>\n<li>Who coined the term \u201cAvogadro\u2019s number\u201d?<\/li>\n<li>What contribution did Robert Millikan make to the determination for the value for the number?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-acu\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-YjkwNjRlMTY2OWNkYjk4ZDJlMTJmMGM5MDE3NDA1OWQ.-vf1\">\n<li>What is the SI unit for amount of a substance?<\/li>\n<li>What is the representative particle for an element?<\/li>\n<li>The formula unit is the representative particle for what?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-YTljNGM0N2U5MmE3ZWI0ZjgxMGE2YTFmNGJmMTkyNTI.-unp\">\n<li><strong> Avogadro\u2019s number: <\/strong> The number of representative particles in a mole, 6.02 \u00d7 10 <sup> 23 <\/sup> .<\/li>\n<li><strong> mole <\/strong> (mol): The amount of a substance that contains 6.02 \u00d7 10 <sup> 23 <\/sup> representative particles of that substance.<\/li>\n<li><strong> representative particle <\/strong> : The smallest unit in which a substance naturally exists.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgQmV0d2VlbiBNb2xlcyBhbmQgQXRvbXM.\">Conversions Between Moles and Atoms<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MDE2NzM2NjdlODM3Mjk0MDQ0NTFhOWMwNDg0ODgzOTU.-zug\">\n<li>Perform calculations involving conversions between number of moles and number of atoms or molecules.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YWQ5OWZhOTRkZGVmMzg4YjA2YjljMDkyZTdjODJhZDI.-urd\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211610\/20140811155326442657.jpeg\" alt=\"Using moles allows us to avoid superscripts and subscripts\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-NGFhOGQyOTg3NWMxNGY3YTJiYWQ4OTgxZjlhMGVjMGQ.-7oa\"><strong> Big numbers or little numbers? <\/strong><\/p>\n<p id=\"x-ck12-ZmQ4OWI4MGFhYzkyMGUxMWY1Y2Q3OTM1ZmFkOTIwMjk.-pc4\">Do you hate to type subscripts and superscripts? Even with a good word-processing program, having to click on an icon to get a superscript and then remembering to click off after you type the number can be a real hassle.\u00a0 If we did not know about moles and just knew about numbers of atoms or molecules (those big numbers that require lots of superscripts), life would be much more complicated and we would make many more typing errors.<\/p>\n<h3>Conversions Between Moles and Atoms<\/h3>\n<h4>Conversions Between Moles and Number of Particles<\/h4>\n<p id=\"x-ck12-YTEwZjU1MWUzYTY3ZGVhMTRkMGRlMDhiYTk4ZDE4MjA.-cva\">Using our unit conversion techniques, we can use the mole label to convert back and forth between the number of particles and moles.<\/p>\n<h4>Sample Problem 1: Converting Number of Particles to Moles<\/h4>\n<p id=\"x-ck12-ZTdjNDdkMDYyMTI2Yjc2YmE1OGY4YjQ4NjJkYjFkYzg.-nji\">The element carbon exists in two primary forms: graphite and diamond.\u00a0 How many moles of carbon atoms is 4.72\u00a0\u00d7 10 <sup> 24 <\/sup> atoms of carbon?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-i1n\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-oid\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-Y2I2ZmE3OWU2ODYxZGJjN2I4NjI0NTIyYWI2NWZkZmI.-d5o\">\n<li>number of C atoms = 4.72\u00a0\u00d7 10 <sup> 24 <\/sup><\/li>\n<li>1 mole = 6.02\u00a0\u00d7 10 <sup> 23 <\/sup> atoms<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hjj\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-Nzg3MWZlYmM3YTZjYzYxYmM4OTBlNjE3MzA1NTM4M2M.-iqb\">\n<li>4.72\u00a0\u00d7 10 <sup> 24 <\/sup> = ? mol C<\/li>\n<\/ul>\n<p id=\"x-ck12-YmU2MmVlMTZjOGY4ODU5NTRmYTYzNWYxMWIzZjgwZDc.-a5e\">One conversion factor will allow us to convert from the number of C atoms to moles of C atoms.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-lvd\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-rya\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211612\/3e4ed0b6835f34a4c97240ed40be954b.png\" alt=\"4.72 times 10^{24} text{atoms C} times frac{1 text{mol C}}{6.02 times 10^{23} text{atoms C}}=7.84 text{mol C}\" width=\"449\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-syd\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-YThhOTJlMDZlMmY5NzlhZTViMTI4OWQ3YTRkM2U0ZDU.-27m\">The given number of carbon atoms was greater than Avogadro\u2019s number, so the number of moles of C atoms is greater than 1 mole.\u00a0 Since Avogadro\u2019s number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.<\/p>\n<p id=\"x-ck12-OTkxMTA0MTAwYzAzYmNkMmYzMjVhMGIxZjJlYmM5NGU.-osb\">Suppose that you wanted to know how many hydrogen atoms were in a mole of water molecules.\u00a0 First, you would need to know the chemical formula for water, which is H <sub> 2 <\/sub> O.\u00a0 There are two atoms of hydrogen in each molecule of water.\u00a0 How many atoms of hydrogen would there be in two water molecules?\u00a0 There would be 2\u00a0\u00d7 2 = 4 hydrogen atoms. How about in a dozen?\u00a0 In that case a dozen is 12 so 12\u00a0\u00d7 2 = 24 hydrogen atoms in a dozen water molecules.\u00a0 To get the answers, (4 and 24) you had to multiply the given number of molecules by two atoms of hydrogen per molecule.\u00a0 So to find the number of hydrogen atoms in a mole of water molecules, the problem could be solved using conversion factors.<\/p>\n<p id=\"x-ck12-q8j\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211613\/7c2e42b7b8dfdd5355ba9d2fc2377155.png\" alt=\"1 text{mol} text{H}_2text{O} times frac{6.02 times 10^{23} text{molecules} text{H}_2text{O}}{1 text{mol} text{H}_2text{O}} times frac{2 text{atoms H}}{1 text{molecule} text{H}_2text{O}}=1.20 times 10^{24} text{atoms H}\" width=\"647\" height=\"43\" \/><\/p>\n<p id=\"x-ck12-ZGNjY2IxM2UwNjYxNTJhNDY3M2M2ZjZjMGY0YzNhNTM.-al6\">The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.<\/p>\n<div id=\"x-ck12-NTkzY2Q1YzEyNjNiMWNjYTExMDVlZDE3MDJmMmM2YmE.-jfh\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-zgo\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjEzOTMyOS02LTE3LTU.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211615\/20140811155326541708.png\" alt=\"Two water molecules\" longdesc=\"Two%20water%20molecules%20contain%204%20hydrogen%20atoms%20and%202%20oxygen%20atoms.%20A%20mole%20of%20water%20molecules%20contains%202%20moles%20of%20hydrogen%20atoms%20and%201%20mole%20of%20oxygen%20atoms.\" \/><\/p>\n<p><strong> Figure 10.3 <\/strong><\/p>\n<p id=\"x-ck12-Y2QxZGNkMDY2ZTQyN2RiZjE5YTcyNGY1MTA1YTM2N2Y.-yqf\">Two water molecules contain 4 hydrogen atoms and 2 oxygen atoms. A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.<\/p>\n<\/div>\n<h4>Sample Problem 2: Atoms, Molecules, and Moles<\/h4>\n<p id=\"x-ck12-YWU2Y2Y3ZWRmNzgyOTAxZDk5NDgyZDBlZTRjYTk4ZTk.-k1s\">Sulfuric acid has the chemical formula H <sub> 2 <\/sub> SO <sub> 4 <\/sub> .\u00a0 A certain quantity of sulfuric acid contains 4.89\u00a0\u00d7\u00a010 <sup> 25 <\/sup> atoms of oxygen.\u00a0 How many moles of sulfuric acid is the sample?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-oqv\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-vle\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-Zjk2MjRmMDMyYzNiMjMzMzc4Y2U0MjJjYzdjZDI3NDA.-rfs\">\n<li>4.89\u00a0\u00d7\u00a010 <sup> 25 <\/sup> = O atoms<\/li>\n<li>1 mole = 6.02 \u00d7 10 <sup> 23 <\/sup> molecules H <sub> 2 <\/sub> SO <sub> 4 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-g7w\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-M2QzYzRmYjdjOTlkOWQ4NWIxYTE3NjAxMDFmYzJiNjc.-z4e\">\n<li>mol of H <sub> 2 <\/sub> SO <sub> 4 <\/sub> molecules<\/li>\n<\/ul>\n<p id=\"x-ck12-ZmE2ZDZiMDllYjJjMDI2MzdhOWIxYmMyZjVlOTAyYjY.-nmx\">Two conversion factors will be used.\u00a0 First, convert atoms of oxygen to molecules of sulfuric acid.\u00a0 Then, convert molecules of sulfuric acid to moles of sulfuric acid.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-jng\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-32m\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211615\/d90613a834f279fea6ad319f41ad697d.png\" alt=\"4.89 times 10^{25} text{atoms O} times frac{1 text{molecule} text{H}_2text{SO}_4}{4 text{atoms O}} times frac{1 text{mol} text{H}_2text{SO}_4}{6.02 times 10^{23} text{molecules} text{H}_2text{SO}4}=20.3 text{mol} text{H}_2text{SO}_4\" width=\"730\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ilv\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NWUxZTczY2NiZDI1YWU0MWJlMTMwOTkxYmJmNjBlYmI.-pbd\">The original number of oxygen atoms was about 80 times larger than Avogadro\u2019s number.\u00a0 Since each sulfuric acid molecule contains 4 oxygen atoms, there are about 20 moles of sulfuric acid molecules.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-NmVlZWZkMWI1NjI3ZjUxNWZhNThlMzRiODE1OWYwZTQ.-vzu\">\n<li>Methods are described for conversions between moles, atoms, and molecules.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-OGZlOWQ0OWRiYTlmOGMyYWJlOTc0MmRiZTM2NmM1Zjc.-x0f\">Read the relevant portions of the following article and do problems 3, 5, 9, 13, and 18.\u00a0 Do not worry about the calculations involving conversions dealing with molar mass (that will come next).<\/p>\n<p id=\"x-ck12-ZGI2OWE5MjAzYWQzZjVhMGNkMDMwYjllMjA0YjUyNWQ.-tbg\"><a href=\"http:\/\/faculty.rcc.edu\/freitas\/1AWorksheets\/14GramsToMolesToMolecules.pdf\" target=\"_blank\" rel=\"noopener\">Grams To Moles To Molecules<\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-rz3\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-NmE0MDMzYWMxYWQyNWZiMWIyMzlmMWQ2MDcxMGYyOWY.-erx\">\n<li>What important number do we need to know to do these conversions?<\/li>\n<li>I want to convert atoms to moles. My friend tells me to multiply the number of atoms by\u00a06.02\u00a0\u00d7 10 <sup> 23 <\/sup> atoms\/mole.\u00a0 Is this correct?<\/li>\n<li>Why should I know the formula for a molecule in order to calculate the number of moles of one of the atoms?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-TW9sYXIgTWFzcw..\">Molar Mass<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ODEyMDUzZWM2ZmZjM2M5NTIwMGRlY2QwNWJmOTY4ZmQ.-bp2\">\n<li>Define molar mass.<\/li>\n<li>Perform calculations involving molar mass.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-OGYzODA3OGJjMmQzMTQxYjg3NTE3YjMzYWQxYTYyOTA.-oba\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211617\/20140811155326668657.jpeg\" alt=\"Pile of potassium dichromate\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-MGVlY2U3OTcwOTk2ZjUzODkwODU4M2YxYTdhMTAzNTU.-mvm\"><strong> When creating a solution, how do I know how much of each substance to put in? <\/strong><\/p>\n<p id=\"x-ck12-YTAzMzBkYTE3MjM0ZjcwMzYyY2NlNTViNGFjMjRhODE.-3ec\">I want to make a solution that contains 1.8 moles of potassium dichromate.\u00a0 I don\u2019t have a balance calibrated in molecules, but I do have one calibrated in grams.\u00a0 If I know the relationship between moles and the number of grams in a mole, I can use my balance to measure out the needed amount of material.<\/p>\n<h3>Molar Mass<\/h3>\n<p id=\"x-ck12-NTVmYjBlYmUwMjdhMzY3MGU2YWExNmFkY2Q3MjQ1Njg.-ft5\"><strong> Molar mass <\/strong> is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is 6.94 g, the molar mass of zinc is 65.38 g, and the molar mass of gold is 196.97 g.\u00a0 Each of these quantities contains 6.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> atoms of that particular element.\u00a0 The units for molar mass are grams per mole or g\/mol.<\/p>\n<h4>Molar Masses of Compounds<\/h4>\n<p id=\"x-ck12-OGI0NTYxYzMwYmQ2ZTY3MTYwNDU2MGVjYjc4MWEwNjU.-0wz\">A molecular formula of the compound carbon dioxide is CO <sub> 2 <\/sub> .\u00a0 One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen.\u00a0 We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen.<\/p>\n<p id=\"x-ck12-wz7\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211618\/5d7614bb108ac0aff5ef7ec1971d7660.png\" alt=\"12.01 text{amu} + 2(16.00 text{amu})=44.01 text{amu}\" width=\"308\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-YzRhZmQ5OTA5NDgxOWFmMDcyNDAzM2QwYzNiNmNhYTk.-gws\">The <strong> molecular mass <\/strong> of a compound is the mass of one molecule of that compound.\u00a0 The molecular mass of carbon dioxide is 44.01 amu.<\/p>\n<p id=\"x-ck12-ZTA3ODBlNDIxZTAwMTIzZWQ0ZThhM2JhMTJhNWJlYWI.-c3k\">The molar mass of any compound is the mass in grams of one mole of that compound.\u00a0 One mole of carbon dioxide molecules has a mass of 44.01 g, while one mole of sodium sulfide formula units has a mass of 78.04 g.\u00a0 The molar masses are 44.01 g\/mol and 78.04 g\/mol respectively.\u00a0 In both cases, that is the mass of\u00a06.02\u00a0\u00d7\u00a010 <sup> 23 <\/sup> representative particles.\u00a0 The representative particle of CO <sub> 2 <\/sub> is the molecule, while for Na <sub> 2 <\/sub> S, it is the formula unit.<\/p>\n<h4>Sample Problem:\u00a0 Molar Mass of\u00a0 a Compound<\/h4>\n<p id=\"x-ck12-ZGYwZTBmZWRhODdlYTQwZmY1MGE1YWFkNDUyODk2MDA.-1si\">Calcium nitrate, Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub> , is used as a component in fertilizer Determine the molar mass of calcium nitrate.<\/p>\n<p id=\"x-ck12-ZmZmN2ZhMGNjMmI5MGRiYmU2NTc1YmFmN2M0MjFiMDY.-n5r\"><em> Step 1:\u00a0 List the known and unknown quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-uib\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-OGQ5NzMwYWU1YWM1ZDgxNjUxNDg5ZTZlMzkyYTEwMDE.-wkr\">\n<li>formula = Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub><\/li>\n<li>molar mass Ca = 40.08 g\/mol<\/li>\n<li>molar mass N = 14.01 g\/mol<\/li>\n<li>molar mass O = 16.00 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-8p9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-MjdjYmEzZjYzOTkzMDQwMjczYjhhOGI4MWE3MTQyODg.-1pt\">\n<li>molar mass Ca(NO <sub> 3 <\/sub> ) <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-MTA5MzYxNzEwZDI3YmQ3NzYzODY4MTc0ZDNjMzYzMzM.-xcm\">First we need to analyze the formula.\u00a0 Since the Ca lacks a subscript, there is one Ca atom per formula unit.\u00a0 The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of one nitrogen atom and three oxygen atoms.\u00a0 Therefore, there are a total of 1 \u00d7 2 = 2 nitrogen atoms and 3\u00a0\u00d7\u00a02\u00a0=\u00a06 oxygen atoms per formula unit.\u00a0 Thus, 1 mol of calcium nitrate contains 1 mol of Ca atoms, 2 mol of N atoms, and 6 mol of O atoms.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-l5a\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\n<p id=\"x-ck12-NDI0MjIzMDFiYjlkYWU5ZDJjNzc1NWMzYzlmNmZhYmQ.-swz\">Use the molar masses of each atom together with the number of atoms in the formula and add together.<\/p>\n<p id=\"x-ck12-ete\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211620\/4522d977ad371139a59cf20a71f1ef84.png\" alt=\"&amp; 1 text{mol Ca} times frac{40.08 text{g Ca}}{1 text{mol Ca}}=40.08 text{g Ca} \\&amp; 2 text{mol N} times frac{14.01 text{g N}}{1 text{mol N}}=28.02 text{g N} \\&amp; 6 text{mol O} times frac{16.00 text{g O}}{1 text{mol O}}=96.00 text{g O} \\&amp; text{molar mass of } text{Ca(NO}_3)_2=40.08 text{g} + 28.02 text{g} + 96.00 text{g} = 164.10 text{g} \/ text{mol}\" width=\"554\" height=\"150\" \/><\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ODAzYzUxZGQ3Y2MwNmZiMWVhZDBlNzJiM2E0MDk1NDA.-ig1\">\n<li>Calculations are described for the determination of molar mass of an atom or a compound.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-MDZiNTc4ODQ1OTFlZWE1ZGY5ZDhkN2FiNjQ3MjA3Yjc.-npp\">Read the material at the link below and work the problems at the end:<\/p>\n<p id=\"x-ck12-ZjA0NGJlM2RiOGVhNTYzYzNhZjIwZGFhOTRkNmQ2NDY.-4ae\"><a href=\"http:\/\/misterguch.brinkster.net\/molarmass.html\"> http:\/\/misterguch.brinkster.net\/molarmass.html <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-xwq\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-ZmRkNGUzMWQ3NTE0NGM0NWU3YzMxY2EzZTYyMmMyZWQ.-vtl\">\n<li>What is the molar mass of Pb?<\/li>\n<li>Why do we need to include the units in our answer?<\/li>\n<li>I want to calculate the molar mass of CaCl <sub> 2 <\/sub> .\u00a0 How many moles of Cl are in one mole of the compound?<\/li>\n<li>How many moles of H are in the compound (NH <sub> 4 <\/sub> ) <sub> 3 <\/sub> PO <sub> 4 <\/sub> ?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-Mjc0YjY2ODFjYTBkZTY0YjAzNzU3YWY4YTYyZGVhNjg.-ffr\">\n<li><strong> molar mass <\/strong> : The mass of one mole of representative particles of a substance.<\/li>\n<li><strong> molecular mass <\/strong> : The mass of one molecule of that compound.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNb2xlcyBhbmQgTWFzcw..\">Conversions between Moles and Mass<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NTIzNDkxNmM3YTljY2MwZmZjMzBlNDk1ZDc3ZDFlZDU.-sfp\">\n<li>Perform calculations dealing with conversions between moles and mass.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YjcxNWU1MGJkYWQwMjY3OTdlNzY0NGJiZTVmMTU0MjQ.-z9j\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211621\/20140811155326808566.jpeg\" alt=\"Chemical plants need to know how to convert between moles and mass\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-YzA4ZjQ3ZDI0ZjQ3NjRjMWQ1ODU3ZDYxMDZkN2RhZDg.-jui\"><strong> How can we get more product? <\/strong><\/p>\n<p id=\"x-ck12-NzhhMDE1YjNmOGI4ODE2Y2IzZjQyNzdiZDExYmQ0NmQ.-alu\">Chemical manufacturing plants are always seeking to improve their processes.\u00a0 One of the ways this improvement comes about is through measuring the amount of material produced in a reaction.\u00a0 By knowing how much is made, the scientists and engineers can try different ways of getting more product at less cost.<\/p>\n<h3>Conversions Between Moles and Mass<\/h3>\n<p id=\"x-ck12-YmI2NmU2YTMyYjY0MTI2MWIzZTFjNGUwOGRhMTJjMWE.-c7q\">The molar mass of any substance is the mass in grams of one mole of representative particles of that substance.\u00a0 The representative particles can be atoms, molecules, or formula units of ionic compounds.\u00a0 This relationship is frequently used in the laboratory.\u00a0 Suppose that for a certain experiment you need 3.00 moles of calcium chloride (CaCl <sub> 2 <\/sub> ).\u00a0 Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed.\u00a0 The molar mass of CaCl <sub> 2 <\/sub> is 110.98 g\/mol.\u00a0 The conversion factor that can be used is then based on the equality that 1 mol = 110.98 g CaCl <sub> 2 <\/sub> .\u00a0 Dimensional analysis will allow you to calculate the mass of CaCl <sub> 2 <\/sub> that you should measure.<\/p>\n<p id=\"x-ck12-dpp\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211622\/0f8dd645623eeef9bf7da078647392c8.png\" alt=\"3.00 text{mol} text{CaCl}_2 times frac{110.98 text{g} text{CaCl}_2}{1 text{mol} text{CaCl}_2}=333 text{g} text{CaCl}_2\" width=\"381\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-NjM3OTcwMmFmMTEyNTMwNzNlZDVhZDE1NTA1NjkxNzg.-4iy\">When you measure the mass of 333 g of CaCl <sub> 2 <\/sub> , you are measuring 3.00 moles of CaCl <sub> 2 <\/sub> .<\/p>\n<div id=\"x-ck12-ZjRlM2I3MzYwN2FmYTIxNGQ0ZWRjMTM0YTZlYjk1M2Q.-csu\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-v8t\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjEzOTczNy05Mi0xOS04\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211624\/20140811155326951109.jpeg\" alt=\"Calcium chloride is used as a drying agent and as a road deicer\" longdesc=\"Calcium%20chloride%20is%20used%20as%20a%20drying%20agent%20and%20as%20a%20road%20deicer.\" \/><\/p>\n<p><strong> Figure 10.4 <\/strong><\/p>\n<p id=\"x-ck12-ZmM2NzM1ZWM5MDIwYWRjNmQyZjdlMzUxZWU4N2UxZWM.-2j8\">Calcium chloride is used as a drying agent and as a road deicer.<\/p>\n<\/div>\n<h4>Sample Problem: Converting Moles to Mass<\/h4>\n<p id=\"x-ck12-YmU1NzVhZmVlMTMyOTVkMTMyOTIwMmQzOWQ2MjA2NjU.-4za\">Chromium metal is used for decorative electroplating of car bumpers and other surfaces.\u00a0 Find the mass of 0.560 moles of chromium.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qrj\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-dgp\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ZWEzODk1MmJlMDlmMjMyZDQ0YTNmZDU0Yzc2YzYzOTk.-uyd\">\n<li>molar mass of Cr = 52.00 g mol<\/li>\n<li>0.560 mol Cr<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-9yb\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ZDAwMTEzNjA3ZTQwYjhmYmE0NjkyZjU1OTljM2EzZTk.-yeo\">\n<li>0.560 mol Cr = ? g<\/li>\n<\/ul>\n<p id=\"x-ck12-OWY1NDdiNTAxMmEzN2VjOTNiNDk4M2MyMDUzYTlmNmU.-hpe\">One conversion factor will allow us to convert from the moles of Cr to mass.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-tro\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-v4c\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211624\/cc47cd6884a6fe79190aa322be8c07f4.png\" alt=\"0.560 text{mol Cr} times frac{52.00 text{g} text{Cr}}{1 text{mol Cr}}=29.1 text{g Cr}\" width=\"306\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-NzhjOWE5NTVlYTMxYTViNTExNTViMDY3ODFmOTgyZDM.-1rw\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NzhjOWE5NTVlYTMxYTViNTExNTViMDY3ODFmOTgyZDM.-zsp\">Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass.\u00a0 The answer has three significant figures because of the 0.560 mol.<\/p>\n<p id=\"x-ck12-YTY3ZWJmZWYxNjQ0NTk0MjlhY2I0NTE1NzkxNzJkNmU.-jcs\">A similar conversion factor utilizing molar mass can be used to convert from the mass of an substance to moles.\u00a0 In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed.\u00a0 It will often then be necessary to determine the number of moles of the product that was formed.\u00a0 The next problem illustrates this situation.<\/p>\n<h4>Sample Problem: Converting Mass to Moles<\/h4>\n<p id=\"x-ck12-NGFjYTU5NGUxYTA0ZjRlYjI2NWRkMmUzNTg3NWExNTE.-ttg\">A certain reaction produces 2.81 g of copper(II) hydroxide, Cu(OH) <sub> 2 <\/sub> .\u00a0 Determine the number of moles produced in the reaction.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qy2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-z4f\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-MDA3YjU0OGQ4MDE2NTQxZGJhNTU5NDRjYTFlYzcyNDE.-ns0\">\n<li>mass = 2.81 g<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-3e2\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-OTRmMTE1MjJmM2U0Yjg5YmFjZjA5ZmI1YjFjNGM5ZDc.-dd6\">\n<li>mol Cu(OH) <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-NzcyYWY5ZGFiNzg0NzUzMjZkNjU0NzZhYzM1ZDY1YWQ.-0t4\">One conversion factor will allow us to convert from mass to moles.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-9el\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-OWEzMzgwMzE3MjRiZDJmMmFmZTFjNGFiOGFlYjk5MzA.-4y5\">First, it is necessary to calculate the molar mass of Cu(OH) <sub> 2 <\/sub> from the molar masses of Cu, O, and H.\u00a0 The molar mass is 97.57 g\/mol.<\/p>\n<p id=\"x-ck12-ujr\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211625\/573a40438243112b5bef60459f4ca487.png\" alt=\"2.81 text{g} text{Cu(OH)}_2 times frac{1 text{mol} text{Cu(OH)}_2}{97.57 text{g} text{Cu(OH)}_2}=0.0288 text{mol} text{Cu(OH)}_2\" width=\"467\" height=\"44\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-njz\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-MjVjMzE1MTBmZTU2N2IxODA2MjU0OTUwM2QxZmUxZmQ.-ixo\">The relatively small mass of product formed results in a small number of moles.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-Y2I0OGFhMzI3NDdjMGE5NzZlN2Y2N2ZhMGFjMWFiNTc.-z8w\">\n<li>Calculations involving conversions between moles of a material and the mass of that material are described.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-N2IzZjFlODZlYzZmOTQ3MDA5YmRlYzNjNWFmMTNiZmU.-drt\">Read the material in the link below and work the problems at the end.<\/p>\n<p id=\"x-ck12-Mzk5YThjNTA2YjA3ZWE4OGU1OWRkYzRiNzRjMjE5NjM.-tsw\"><a href=\"http:\/\/www.occc.edu\/kmbailey\/chem1115tutorials\/Stoichiometry_Molar_Mass_Calculations.htm\"> http:\/\/www.occc.edu\/kmbailey\/chem1115tutorials\/Stoichiometry_Molar_Mass_Calculations.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-euk\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-M2I1ZTVlN2M5ZGUwMmU3MTE0NDNmMzRlZDk5ODJjZGY.-isk\">\n<li>Why would you want to calculate the mass of a material?<\/li>\n<li>Why would you want to determine how many moles of material you produced in a reaction?<\/li>\n<li>You have 19.7 grams of a material and wonder how many moles were formed. Your friend tells you to multiply the mass by grams\/mole.\u00a0 Is your friend correct?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNYXNzIGFuZCBOdW1iZXIgb2YgUGFydGljbGVz\">Conversions between Mass and Number of Particles<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MDU0YjM4MGFlOWZiMWQwZmQzYzY5NjhkYjY1MzdhMjY.-adn\">\n<li>Perform calculations involving conversions of mass and number of particles.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-M2Q4NjhiMDlmYWM4ZjIwYWY5MzQ0MDdkMTgxYzFmMjk.-tkp\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211627\/20140811155327444990.jpeg\" alt=\"Equal volumes of gas under the same conditions contain the same number of particles\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-MTgyMDVjZmIwM2UwM2JiYWFlZjY5NTAwNGUzYTAzOTE.-dmb\"><strong> How much gas is there? <\/strong><\/p>\n<p id=\"x-ck12-YWU2ZDQyZjlmNDc4NWQ3NTBmZDVlZDZkY2ZmODkxNzE.-lfp\">Avogadro was interested in studying gases.\u00a0 He theorized that equal volumes of gases under the same conditions contained the same number of particles.\u00a0 Other researchers studied how many gas particles were in a specific volume of gas. Eventually, scientists were able to develop the relationship between number of particles and mass using the idea of moles.<\/p>\n<h3>Conversions Between Mass and Number of Particles<\/h3>\n<p id=\"x-ck12-NmU5OTkwMDEwZTQ3MDA1MjE3YTg4MGVmZjE1MTVhMGI.-ty5\">In &#8220;Conversions between Moles and Mass&#8221;, you learned how to convert back and forth between moles and the number of representative particles.\u00a0 Now you have seen how to convert back and forth between moles and mass of a substance in grams.\u00a0 We can combine the two types of problems into one.\u00a0 Mass and number of particles are both related to grams.\u00a0 In order to convert from mass to number of particles or vice-versa, it will first require a conversion to moles.<\/p>\n<div id=\"x-ck12-NjJiYmI0YWI4ZGI1ZmQyY2UwOGQwN2UzYWE1MTU4MTE.-gtb\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-c6u\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MDA5Mi0yNS01OC0xMA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211628\/20140811155327594002.png\" alt=\"Flowchart of conversion between number of particles, moles, and mass\" longdesc=\"Conversion%20from%20number%20of%20particles%20to%20mass%20or%20from%20mass%20to%20number%20of%20particles%20requires%20two%20steps\" \/><\/p>\n<p><strong> Figure 10.5 <\/strong><\/p>\n<p id=\"x-ck12-MjMzZWEyMzNiMmFmMGJmMDgzNDc2N2FjY2VkYjFhZTg.-4y1\">Conversion from number of particles to mass or from mass to number of particles requires two steps<\/p>\n<\/div>\n<h4>Sample Problem: Converting Mass to Particles<\/h4>\n<p id=\"x-ck12-ODc1Y2MyMGIwZTNmZTBmZTc2YzU0OWE2YjRiMDcxMDA.-m5d\">How many molecules is 20.0 g of chlorine gas, Cl <sub> 2 <\/sub> ?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ink\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-m5c\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ODc1NmY3MWZkYjRlYzZjMjUzZjY3M2IwY2JhNWVmYWY.-6v2\">\n<li>molar mass Cl <sub> 2 <\/sub> = 70.90 g\/mol<\/li>\n<li>20.0 g Cl <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-oqz\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-MDFjNjgxOTBhOWU3MDk2MjkxN2U4Y2QxYmRkNDdhMzE.-jdy\">\n<li>number of molecules of Cl <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-YzM3NDRlNTk4MjM1YWI2ZmE3MDQ1ZDMxNzk0M2M1NGU.-b5y\">Use two conversion factors.\u00a0 The first converts grams of Cl <sub> 2 <\/sub> to moles.\u00a0 The second converts moles of Cl <sub> 2 <\/sub> to the number of molecules.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-wg1\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-dcj\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211630\/7485c4d2caf2013df3630b023c68c14f.png\" alt=\"20.0 text{g} text{Cl}_2 times frac{1 text{mol} text{Cl}_2}{70.90 text{g} text{Cl}_2} times frac{6.02 times 10^{23} text{molecules Cl}_2}{1 text{mol} text{Cl}_2}=1.70 times 10^{23} text{molecules Cl}_2\" width=\"634\" height=\"44\" \/><\/p>\n<p id=\"x-ck12-NTZkMTU0ZDdjNjE3NjU0MzY3MGIxYjdjYWE5NzlhOGQ.-gap\">The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of Cl <sub> 2 <\/sub> .<\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-idt\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ODM0ZGY1YmMxMTk2N2EwNzAxYzMzYTkxZGVkNDVmZWM.-dxg\">Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro\u2019s number.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-MTQzZDcxNjA5ZTdjOWVmNWMyNTM5ZmJlYjE3Njc4MmI.-mxr\">\n<li>Calculations are illustrated for conversions between mass and number of particles.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NTNlMzE4NDEwZDM5MGQyYjNkOWQ2ZDY0YTgyMzVhNWE.-lqg\">Read the material at the link below and then do practice problems on page 9 and the problem on page 17 (don\u2019t peak at the answers until you have tried the problems).<\/p>\n<p id=\"x-ck12-Yjg1YzhhNzcwNGFlNzkxN2VkYzAzMGI3NDhmMzJlMDg.-ehm\"><a href=\"http:\/\/schools.fwps.org\/decatur-old\/staff\/adewaraja\/chemistry\/curriculum%20units\/chapter%2010\/Lesson5\/particle_mole_mass_calcu.pdf\"> http:\/\/schools.fwps.org\/decatur-old\/staff\/adewaraja\/chemistry\/curriculum%20units\/chapter%2010\/Lesson5\/particle_mole_mass_calcu.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-jkn\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-YzY3MDIwMTI2N2I1NjQwZWEyZDllNjM1YWQ1OTM3MmU.-qv6\">\n<li>Why can\u2019t we convert directly from number of particles to grams?<\/li>\n<li>How many atoms of chlorine are present in the problem above?<\/li>\n<li>The periodic table says the atomic weight of chlorine is 35.5. Why can\u2019t I use that value in my calculations?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBIeXBvdGhlc2lzIGFuZCBNb2xhciBWb2x1bWU.\">Avogadro&#8217;s Hypothesis and Molar Volume<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ZjVhZWY3OGQ1OTc5ZDYxZjMzYjRlOThlNzAzYjZhNzE.-keq\">\n<li>State Avogadro\u2019s hypothesis.<\/li>\n<li>Define standard temperature and pressure.<\/li>\n<li>Define molar volume.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-Nzc4YzcxYzI1ZTY1ZDdmNzQwZjExNzBmOGU5YmI3OTE.-t0m\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211631\/20140811155327709392.jpeg\" alt=\"Gas laws are used to determine the amount of air in a scuba tank\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-MDE4ODNhNmVlOWQ5Njc0NmM0ODhlN2ZkYmUwZWZkYWQ.-dig\"><strong> How do scuba divers know if they will run out of gas? <\/strong><\/p>\n<p id=\"x-ck12-YTg2YWQ1NDk3ZjcxYzBmN2M2ZmFhODI5NDhlMjg4OTQ.-bpe\">Knowing how much gas is available for a dive is crucial to the survival of the diver.\u00a0 The tank on the diver\u2019s back is equipped with gauges to tell how much gas is present and what the pressure is.\u00a0 A basic knowledge of gas behavior allows the diver to assess how long to stay under water without developing problems.<\/p>\n<h3>Avogadro\u2019s Hypothesis and Molar Volume<\/h3>\n<p id=\"x-ck12-N2EyYmRmYTE0ZWEzNzI4NDE3MjVlNjczODU4ZGM4NzQ.-vf0\">Volume is a third way to measure the amount of matter, after item count and mass.\u00a0 With liquids and solids, volume varies greatly depending on the density of the substance.\u00a0 This is because solid and liquid particles are packed close together with very little space in between the particles.\u00a0 However, gases are largely composed of empty space between the actual gas particles (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDQ1NS05Mi00NC0xMg..\"> below <\/a> ).<\/p>\n<div id=\"x-ck12-MmJjMzM1MDQ4NDQ0ZjE1M2MwOTY5NTdhZmI5NTU4MDA.-tnk\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-10u\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MDQ1NS05Mi00NC0xMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211633\/20140811155327899786.png\" alt=\"Gas particles are small compared to the empty space between them\" longdesc=\"Gas%20particles%20are%20very%20small%20compared%20to%20the%20large%20amounts%20of%20empty%20space%20between%20them.\" \/><\/p>\n<p><strong> Figure 10.6 <\/strong><\/p>\n<p id=\"x-ck12-ZDk5NjZhNmIzMDU3MTFkYmJiYzY1NmM4MjIyYzQzYWM.-w7b\">Gas particles are very small compared to the large amounts of empty space between them.<\/p>\n<\/div>\n<p id=\"x-ck12-YWI1ZjEyYWFlMTE2ZWE2OTIxOTExN2YwY2MwNWQ3YmY.-xms\">In 1811, Amedeo Avogadro explained that the volumes of all gases can be easily determined. <strong> Avogadro\u2019s hypothesis <\/strong> states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.\u00a0 Since the total volume that a gas occupies is made up primarily of the empty space between the particles, the actual size of the particles themselves is nearly negligible.\u00a0 A given volume of a gas with small light particles such as hydrogen (H <sub> 2 <\/sub> ) contains the same number of particles as the same volume of a heavy gas with large particles such as sulfur hexafluoride, SF <sub> 6 <\/sub> .<\/p>\n<p id=\"x-ck12-NTEyMjljMWY2ZDQ0YjRlZTNkNzljYzkwMjUwYzFmMGQ.-vkt\">Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another.\u00a0 This decreases the amount of empty space and reduces the volume of the gas.\u00a0 Gas volume is also affected by temperature.\u00a0 When a gas is heated, its molecules move faster and the gas expands.\u00a0 Because of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at one standard temperature and pressure.\u00a0 <strong> Standard temperature and pressure (STP) <\/strong> is defined as\u00a00\u00b0C\u00a0(273.15 K) and 1 atm pressure. The <strong> molar volume <\/strong> of a gas is the volume of one mole of a gas at STP.\u00a0 At STP, one mole (6.02\u00a0\u00d7 10 <sup> 23 <\/sup> representative particles) of any gas occupies a volume of 22.4 L ( <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDQ5Ny0zNS0yLTEz\"> below <\/a> ).<\/p>\n<div id=\"x-ck12-ZjY5NTM1NmFlYmQ4NzM2NGI0NTA1MzdiNTM4MmI3M2I.-dhc\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-syr\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MDQ5Ny0zNS0yLTEz\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211633\/20140811155328052846.png\" alt=\"Avogadro's hypothesis states that one mole of gas at STP occupies 22.4 liters\" longdesc=\"A%20mole%20of%20any%20gas%20occupies%2022.4%20L%20at%20standard%20temperature%20and%20pressure%20(0%C2%B0C%20and%201%20atm).\" \/><\/p>\n<p><strong> Figure 10.7 <\/strong><\/p>\n<p id=\"x-ck12-MWQ0YTUyMTFjN2FlNDE5YWQxMDcxNGY1ODBkMzg2NTE.-guq\">A mole of any gas occupies 22.4 L at standard temperature and pressure (0\u00b0C and 1 atm).<\/p>\n<\/div>\n<p id=\"x-ck12-NGVkM2M2Mjc3NjcwNmMwOWEyODU5ZGRmZGE3YWFhMDc.-9wd\">The <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MDUzMS0xNy05OS0xNA..\"> below <\/a> illustrates how molar volume can be seen when comparing different gases. Samples of helium (He), nitrogen (N <sub> 2 <\/sub> ), and methane (CH <sub> 4 <\/sub> ) are at STP.\u00a0 Each contains 1 mole or 6.02\u00a0\u00d7 10 <sup> 23 <\/sup> particles.\u00a0 However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g\/mol for He, 28.0 g\/mol for N <sub> 2 <\/sub> , and 16.0 g\/mol for CH <sub> 4 <\/sub> .<\/p>\n<div id=\"x-ck12-YTBkYTA4ZDg2Y2Q2M2QyNDEzMTAxYTA5MjE1Mzk4Njk.-3tf\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-5kv\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MDUzMS0xNy05OS0xNA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211634\/20140811155328170493.png\" alt=\"Avogadro's hypothesis means that \u00a0one mole of neon, nitrogen, or methane occupies 22.4 liters at STP\" longdesc=\"Avogadro%E2%80%99s%20hypothesis%20states%20that%20equal%20volumes%20of%20any%20gas%20at%20the%20same%20temperature%20and%20pressure%20contain%20the%20same%20number%20of%20particles.%20At%20standard%20temperature%20and%20pressure%2C%201%20mole%20of%20any%20gas%20occupies%2022.4%20L.\" \/><\/p>\n<p><strong> Figure 10.8 <\/strong><\/p>\n<p id=\"x-ck12-ZGY2MDRkMmRmNjY1MTdjNjk5ZWQyZDhlZjdjMzk2MmY.-pyq\">Avogadro\u2019s hypothesis states that equal volumes of any gas at the same temperature and pressure contain the same number of particles. At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.<\/p>\n<\/div>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-YWJhNjNlY2Q2MzI3NDBkYzVkNTIwNzkyZDQ5Y2RlNDE.-jvn\">\n<li>Equal volumes of gases at the same conditions contain the same number of particles.<\/li>\n<li>Standard temperature and pressure are defined.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-vba\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-xkw\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-NTY1YTUxZjJiM2I3MmZkOTI2MTIxNWYxMGQyZTJiODM.-hwa\"><a href=\"http:\/\/chemed.chem.purdue.edu\/demos\/main_pages\/4.6.html\"> http:\/\/chemed.chem.purdue.edu\/demos\/main_pages\/4.6.html <\/a><\/p>\n<ol id=\"x-ck12-MWZkMzk3YjBiNWMzNjQ2ODk0OWU2ODZkZDExOGE0ZGY.-zrg\">\n<li>What was the volume of each gas that was weighed?<\/li>\n<li>What did the experiment find?<\/li>\n<li>What was the relationship between gas weight and molecular weight?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-qdh\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MjQ4ZDdhZGY1YjY2YmMyN2UwY2M4ZDBjMzU1NzQ4YmQ.-y1x\">\n<li>What do we know about the space actually taken up by a gas?<\/li>\n<li>Why do we need to do all our comparisons at the same temperature and pressure?<\/li>\n<li>How can we use this information?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-NzJhZjI3YzRmMGZjNmU4OTFhMzk3ZGMwNDA4YWI5ZmI.-bja\">\n<li><strong> Avogadro\u2019s hypothesis: <\/strong> Equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.<\/li>\n<li><strong> molar volume: <\/strong> The volume of one mole of a gas at STP.<\/li>\n<li><strong> standard temperature and pressure (STP): <\/strong> \u00a00\u00b0C (273.15 K) and 1 atm pressure.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q29udmVyc2lvbnMgYmV0d2VlbiBNb2xlcyBhbmQgR2FzIFZvbHVtZQ..\">Conversions between Moles and Gas Volume<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MThlOGI0YTY5MjA4NWIzYzg1N2Y2MzU3NTMzMzVhOGU.-ns3\">\n<li>Make conversions between the volume of a gas and the number of moles of that gas.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-Y2ZmMThkZDkyYzBmZmNkMzFmZTBjNDVmM2IzNGQ2NDY.-xto\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211635\/20140811155328513045.jpeg\" alt=\"Gas volumes can be used to determine the moles of gas in these tanks\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-ZTgyYmZkMjZmMDQ0NmM5ZjA0NjYwZmNlNDdhYTZjNmY.-xgk\"><strong> How can you tell how much gas is in these containers? <\/strong><\/p>\n<p id=\"x-ck12-MjA5ODgwZGUzODExOGFiZjNmN2FkOWM0MWM0ZGNiZmQ.-e3b\">Small gas tanks are often used to supply gases for chemistry reactions.\u00a0 A gas gauge will give some information about how much is in the tank, but quantitative estimates are needed so the reaction will be able to proceed to completion.\u00a0 Knowing how to calculate needed parameters for gases is very helpful to avoid running out too early.<\/p>\n<h3>Conversions Between Moles and Gas Volume<\/h3>\n<p id=\"x-ck12-NGIyNDVlMmZjMGQxOGIwM2VlY2ExMTU4MjJkYzhmMzY.-65e\">Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles.\u00a0 The equality of 1 mole = 22.4 L is the basis for the conversion factor.<\/p>\n<h4>Sample Problem One: Converting Gas Volume to Moles<\/h4>\n<p id=\"x-ck12-NjM1YjNmNDJkZjA0YzlkODExMjUyOGU5NDQ3NWI3YmM.-vhc\">Many metals react with acids to produce hydrogen gas.\u00a0 A certain reaction produces 86.5 L of hydrogen gas at STP.\u00a0 How many moles of hydrogen were produced?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-6jk\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-xiy\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NTk3ODc2MTlmMzBhMDllN2M1N2MxMDkyNjdjZDkxODQ.-edq\">\n<li>86.5 L H <sub> 2 <\/sub><\/li>\n<li>1 mol = 22.4 L <sub><br \/>\n<\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-h7n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-MzYyMjhkNTRkMmIwZDg2YTcwNzExNjNhN2M1MDc5Njk.-hrl\">\n<li>moles of H <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-NTJhZDAzODU4N2M0MjRkZDY0YTA5NTIwYzE0N2E2OWQ.-jvh\">Apply a conversion factor to convert from liters to moles.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-u6t\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-xv2\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211636\/6a8f0b1206fc84696b7358e4eaf8cf7b.png\" alt=\"86.5 text{L H}_2 times frac{1 text{mol H}_2}{22.4 text{L H}_2}=3.86 text{mol H}_2\" width=\"293\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-by6\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NWJjYjg2OGFjMzEwMjYyMTAxNDc4Y2JjMzA1MTNlZTg.-owf\">The volume of gas produced is nearly four times larger than the molar volume.\u00a0 The fact that the gas is hydrogen plays no role in the calculation.<\/p>\n<h4>Sample Problem Two: Converting Moles to Gas Volume<\/h4>\n<p id=\"x-ck12-MGE0M2U1MmE1MTI5YTIzMTA5YTJlNmE2ZWU4NjZiZDQ.-jwf\">What volume does 4.96 moles of O <sub> 2 <\/sub> occupy at STP?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-mev\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-f78\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-OTA1ODg4ZGNkYmI1OGM1ZDE0YWZmZmY1YjMwYjY2Y2M.-apu\">\n<li>4.96 moles O <sub> 2 <\/sub><\/li>\n<li>1 mol = 22.4 L <sub><br \/>\n<\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-dah\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-OGU3OWM3YTgwYjI0NTc3MTM0ZmVkYjU0NTFkOTEzNjQ.-b3x\">\n<li>volume of O <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-pf1\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\n<p id=\"x-ck12-x4x\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211637\/e94b42a3d26348b83abd955d3d7b651d.png\" alt=\"4.96 text{moles} times {22.4 text{liters\/mole}}=111.1 text{liters}\" width=\"348\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-lzv\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-MWYyNGNhYzA0YThhMjRkMWMzZjFjMzc3ZDdjYmM4Zjk.-8ay\">The volume seems correct given the number of moles.<\/p>\n<h4 id=\"x-ck12-ZWQ4ZjY3NzA4NTEyMTYwMWY1ZjNmZGMzY2IzMjM5MGI.-pht_6-2yv\">Sample Problem Three: Converting Volume to Mass<\/h4>\n<p id=\"x-ck12-YzVhZDY2ZDg0YmViZWYyNzU0NTU4ODhiOTMyZGFkYWE.-bl8\">If we know the volume of a gas sample at STP, we can determine how much mass is present.\u00a0 Assume we have 867 liters of N <sub> 2 <\/sub> at STP.\u00a0 What is the mass of the nitrogen gas?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ap2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-dr5\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-MjgwYzRkYjc3ZWY5ZjVkNmQ5Y2U5NzIwZjgxNjFlODY.-25f\">\n<li>867 L N <sub> 2 <\/sub><\/li>\n<li>1 mol = 22.4 L<\/li>\n<li>molar mass of N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-yc5\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-OTVjOGU5OWFiM2YzMTQzZWUzY2UxNDQzNWJlNTY4NDk.-8pc\">\n<li>mass of N <sub> 2 <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ltw\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-NDhjNzI0ZjgwMWYxOTM4YjI2NzcxODQ2ODQ0N2IyM2E.-fea\">We start by determining the number of moles of gas present. We know that 22.4 liters of a gas at STP equals one mole, so:<\/p>\n<p id=\"x-ck12-wkq\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211639\/64de3fd328a183a96d0c9ea65751ae7d.png\" alt=\"867 text{litres} times frac{1 text{mole}}{22.4 text{liters}}=3.87 text{moles}\" width=\"280\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-ODBjZWUxZWJkMGZmODE4MDI2MDVjMzFhNjU2NjFiYzA.-kmp\">We also know the molecular weight of N <sub> 2 <\/sub> (28.0 grams\/mole), so we can then calculate the weight of nitrogen gas in 867 liters:<\/p>\n<p id=\"x-ck12-2sg\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211640\/f5e3f79dc20e669e05dc062476193cbb.png\" alt=\"38.7 text{moles} times frac{28 text{grams}}{text{mole}}=1083.6 text{grams N}_2\" width=\"330\" height=\"38\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-8im\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ODVlMTM2MjhlNjA2MmYzMmJlZGJjYjJjNDRmODA2OGM.-eqq\">In a multi-step problem, be sure that the units check.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-MWE1YTNkMGVhYzgxNWE4YTU3NTRlNjk1NGE3YjgzOWY.-clt\">\n<li>Conversions between moles and volume of a gas are shown.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-ZTg5NGIyM2VjZDY5N2ExYWY2ZjZmNzgyMTY0ZjU4MDI.-slu\">Work the practice problems at the link below. Focus on conversions between volume and moles, but try some of the others:<\/p>\n<p id=\"x-ck12-NjYyYWM0NDE0MmE1MDM1Yzc1ZmJhMDdkY2QxN2UyNzQ.-lbp\"><a href=\"http:\/\/www.sciencegeek.net\/Chemistry\/taters\/Unit4GramMoleVolume.htm\"> http:\/\/www.sciencegeek.net\/Chemistry\/taters\/Unit4GramMoleVolume.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-mvq\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MDMyYTJmYmZlYjc5ZGE1YmM1ZTkyYzc3MjExMzExODQ.-iog\">\n<li>Why do the gases need to be at STP?<\/li>\n<li>When does the identity of the gas become important?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-R2FzIERlbnNpdHk.\">Gas Density<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YjEwMDNkNjFiZGUzMTI4MGFkMmM3YTI4NWQwMzdmYjY.-ko5\">\n<li>Make calculations dealing with molar mass and density of a gas.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MTYyZTUyMDViNjg3NTgzY2UwNWYxNDBkYTQzMmZiODI.-b0p\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211641\/20140811155328601836.jpeg\" alt=\"Carbon dioxide sinks in air\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-NmQxYmFhZmM2MzlmZjAwNjMxOGY0NzhjMGRkZjI2Mjk.-m4s\"><strong> Why does carbon dioxide sink in air? <\/strong><\/p>\n<p id=\"x-ck12-M2VjMTFjYzFiYTRiZWQ5NzQwZWIzZmZlNjJiZDQxMGM.-3my\">When we run a reaction to produce a gas, we expect it to rise into the air.\u00a0 Many students have done experiments where gases such as hydrogen are formed.\u00a0 The gas can be trapped in a test tube held upside-down over the reaction.\u00a0 Carbon dioxide, on the other hand, sinks when it is released.\u00a0 Carbon dioxide has a density greater that air, so it will not rise like these other gases would.<\/p>\n<h3>Gas Density<\/h3>\n<p id=\"x-ck12-ZTBlOTNhNTQyY2ExMGUwZDM4ODU4MTExODcyZjExNzE.-bi5\">As you know, density is defined as the mass per unit volume of a substance.\u00a0 Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass.\u00a0 A gas with a small molar mass will have a lower density than a gas with a large molar mass.\u00a0 Gas densities are typically reported in g\/L.\u00a0 Gas density can be calculated from molar mass and molar volume.<\/p>\n<div id=\"x-ck12-NWMwNGY3ZGRlM2U5OGEyYzIzMGYwZTQyMzM4NGJlMWM.-mra\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-gzp\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MDg4NS02Mi04Mi0xNw..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211643\/20140811155328777663.jpeg\" alt=\"Balloons float because they contain helium, which is lighter than air\" longdesc=\"Balloons%20filled%20with%20helium%20gas%20float%20in%20air%20because%20the%20density%20of%20helium%20is%20less%20than%20the%20density%20of%20air.\" \/><\/p>\n<p><strong> Figure 10.9 <\/strong><\/p>\n<p id=\"x-ck12-MGVlMTBhODU2NWU4MjJlNTEyODg1ODMxNjIzZWJmY2U.-jml\">Balloons filled with helium gas float in air because the density of helium is less than the density of air.<\/p>\n<\/div>\n<h4>Sample Problem One: Gas Density<\/h4>\n<p id=\"x-ck12-YWIxZDk5YmMyZGRjZjNmZjAzZTRhMzI3MTNiNzE3MDk.-vlz\">What is the density of nitrogen gas at STP?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-ekt\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-0ib\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-Y2EyYWJkM2Q3MmEyM2E0MGIyMjlkMzI4MDcyZjM2ZDc.-0gu\">\n<li>N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\n<li>1 mol = 22.4 L<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-tqy\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-NmYxZmZhM2E1Yjk0ZGE3NGEyNTZjZDdkYmZjNWJkNjE.-ang\">\n<li>density = ? g\/L<\/li>\n<\/ul>\n<p id=\"x-ck12-OGNiODMxZWU3NjE4MzY3ZjMzODU0ZTgzOTFiYzE3ODg.-ee9\">Molar mass divided by molar volume yields the gas density at STP.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ljo\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-ae0\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211644\/126428d64cdb8fd62bc7401cf20e7a56.png\" alt=\"frac{28.02 text{g}}{1 text{mol}} times frac{1 text{mol}}{22.4 text{L}}=1.25 text{g} \/ text{L}\" width=\"224\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-Y2ZmMzU1YWFiZWJiMzlmMjcyODg0MmFkY2Y2MWIxMjE.-huj\">When set up with a conversion factor, the mol unit cancels, leaving g\/L as the unit in the result.<\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-hqd\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ZTNkYjczMzM1NjcwZWI1YmUyOTkwMTBkMDU3M2VjNDI.-uj6\">The molar mass of nitrogen is slightly larger than molar volume, so the density is slightly greater than 1 g\/L.<\/p>\n<p id=\"x-ck12-YTAzMTZiYjQ4MjlhOGEwZGVkYWMxODk4ZjVkYjJmNDI.-r9s\">Alternatively, the molar mass of a gas can be determined if the density of the gas at STP is known.<\/p>\n<h4>Sample Problem Two: Molar Mass from Gas Density<\/h4>\n<p id=\"x-ck12-NDZiYjBhNzdmNWIyMzY0N2JjNzU5NTNmMzQ3ZjM3YmU.-lmx\">What is the molar mass of a gas whose density is 0.761 g\/L at STP?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-1z8\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-q4z\"><span class=\"x-ck12-underline\"> Known<br \/>\n<\/span><\/p>\n<ul id=\"x-ck12-Y2EyYWJkM2Q3MmEyM2E0MGIyMjlkMzI4MDcyZjM2ZDc.-2cm\">\n<li>N <sub> 2 <\/sub> = 28.02 g\/mol<\/li>\n<li>1 mol = 22.4 L<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hov\"><span class=\"x-ck12-underline\"> Unknown <\/span> <em><br \/>\n<\/em><\/p>\n<ul id=\"x-ck12-OTM0YzliMjIwMjJmMzMwYjNlYWFlZWFkZDkzN2VjNDE.-mzo\">\n<li>molar mass = ? g\/L<\/li>\n<\/ul>\n<p id=\"x-ck12-M2RhMzQzZmY1NTQxNzM0YjBjMjRlZTdmMDI5MDBjMjM.-vui\">Molar mass is equal to density multiplied by molar volume.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-08f\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-e6g\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211645\/a6dd8262edcc127b90582136e771b11f.png\" alt=\"frac{0.761 text{g}}{1 text{L}} times frac{22.4 text{L}}{1 text{mol}}=17.0 text{g} \/ text{mol}\" width=\"241\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-m7i\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-MTQ2MmRkZTU4NDY5MTg4NzhkMDVjZWYyYmI3MjMzNWQ.-1lq\">Because the density of the gas is less than 1 g\/L, the molar mass is less than 22.4.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-OGZiMGUxZDRkOGYyODRkMzM1MDZhNGQ1N2JjZDY0NzU.-xm5\">\n<li>Calculations are described showing conversions between molar mass and density for gases.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-5x2\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-ias\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-MWU0MTNjYmI5MDQ4NWVmZWUxNWFiZjVkZGJhMTI5ZDM.-4lu\"><a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/gas_density_s.html\"> http:\/\/employees.oneonta.edu\/viningwj\/sims\/gas_density_s.html <\/a><\/p>\n<ol id=\"x-ck12-MmJlZmE2ZmM3NjlmYTdlYjkzNDM2MGU5NzIwZGQ0NjQ.-zpj\">\n<li>Which of the gases has the highest density?<\/li>\n<li>Which gas has the lowest density?<\/li>\n<li>Would you expect nitrogen to have a higher or lower density that oxygen? Why?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-uca\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-OTQzMmFhZTNlMGI4MTk1YzQ4ZjZkNzI0YmJiY2QyZDU.-wn5\">\n<li>How is density calculated?<\/li>\n<li>How is molar mass calculated?<\/li>\n<li>What would be the volume of 3.5 moles of a gas?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-TW9sZSBSb2FkIE1hcA..\">Mole Road Map<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YmNiYzQ3ZTliYjRjMWIwNDM3YTA2ODhlYTNmYTc3NzM.-eus\">\n<li>Perform calculations involving interconversions of mass, moles, and volume of a gas.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MGQ1NDE4ZGZiMDkwOGUwODc0ZmFjYTRjMzllYzRiOTc.-nid\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211646\/20140811155328853790.png\" alt=\"Chemistry road maps are similar to actual maps\" width=\"250\" \/><\/span><\/p>\n<p id=\"x-ck12-MjViOGQ5NjcyYzM0NzRmZDNiNjcxODEyNTExYjI5YWY.-mey\"><strong> How do I get from here to there? <\/strong><\/p>\n<p id=\"x-ck12-ZTRkMDQ3ZDk1NDkxNzk1ZjE0Nzc3YjZkYTdjN2RhN2I.-mqh\">If I want to visit the town of Manteo, North Carolina, out on the coast, I will need a map of how to get there.\u00a0 I may have a printed map or I may download directions from the internet, but I need something to get me going in the right direction.\u00a0 Chemistry road maps serve the same purpose.\u00a0 How do I handle a certain type of calculation? There is a process and a set of directions to help.<\/p>\n<h3>Mole Road Map<\/h3>\n<p id=\"x-ck12-MDlkNTY1NmU2OTA1OTNhZGVhMTNiODYxNjE4Y2U2OTI.-kln\">Previously, we saw how the conversions between mass and number of particles required two steps, with moles as the intermediate.\u00a0 This concept can now be extended to also include gas volume at STP.\u00a0 The resulting diagram is referred to as a mole road map (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MTA3MS00MS03MS0xOQ..\"> below <\/a> ).<\/p>\n<div id=\"x-ck12-Mjk3N2YwYmJjNmM0MWIzNTgxZjZiYTQ5N2NlNjIxOGY.-j41\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-x3k\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MTA3MS00MS03MS0xOQ..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211648\/20140811155329093869.png\" alt=\"How to convert between moles, mass, number of particles, and volume of a gas\" longdesc=\"The%20mole%20road%20map%20shows%20the%20conversion%20factors%20needed%20to%20interconvert%20between%20mass%2C%20number%20of%20particles%2C%20and%20volume%20of%20a%20gas.\" \/><\/p>\n<p><strong> Figure 10.10 <\/strong><\/p>\n<p id=\"x-ck12-NGU3MzNmOGRiZTM5MzMxNDc1NzBmMGE5MTc3MjVlMDc.-cmc\">The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and volume of a gas.<\/p>\n<\/div>\n<p id=\"x-ck12-ZTg5OGE0ODIxZDk5ZTg2NGZmZjU4MzJkZDFiNGRlZjI.-fiy\">The mole is at the center of any calculation involving amount of a substance.\u00a0 The sample problem below is one of many different problems that can be solved using the mole road map.<\/p>\n<h4>Sample Problem One:\u00a0 Mole Road Map<\/h4>\n<p id=\"x-ck12-YWQ4NjAwY2EwZjcxOWI0NTdhNGI0NzYwMWYwNGZkZDY.-izn\">What is the volume of 79.3 g of neon gas at STP?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-isb\"><em> Step 1:\u00a0 List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-8na\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NWQyMjJkMWQ4MTQ4YmJiOTM2YTU5MDg2ZGVlOWMwMGI.-pbn\">\n<li>Ne = 20.18 g\/mol<\/li>\n<li>1 mol = 22.4 L<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-vp9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ODVjMGMyNTI0MDRmOGI2NDhjOWVkOTY3Mjc4ZDhmNWI.-wh0\">\n<li>volume = ? L<\/li>\n<\/ul>\n<p id=\"x-ck12-MGU5NWI1NzE5MTllODY2NmM2MmZmMThiNTRhOTIwMzM.-qu4\">The conversion factors will be grams\u00a0\u2192 moles\u00a0\u2192 gas volume.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-ehp\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\n<p id=\"x-ck12-uj9\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211650\/a2f5d08ba0aa75f64bba645de62a7cc0.png\" alt=\"79.3 text{g Ne} times frac{1 text{mol Ne}}{20.18 text{g Ne}} times frac{22.4 text{L Ne}}{1 text{mol Ne}}=88.0 text{L Ne}\" width=\"386\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-bcc\"><em> Step 3:\u00a0 Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NzUxYjhhYTQ0ZWNhNTQ1NjQwMWI0MzQxOGNkMjJiNzA.-voc\">The given mass of neon is equal to about 4 moles, resulting in a volume that is about 4 times larger than molar volume.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-MjQ1MTJiOWEzYmUwNTMxYTBjMDZkMWIyMTUzNjczNWU.-ti4\">\n<li>An overall process is given for calculations involving moles, grams, and gas volume.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-YmU2ZWFjN2FiNWJkZjhmYzMyYzhjODk3YjI1MjM0OWE.-rlt\">Use the link below to carry out some practice calculations.\u00a0 Do problems 1, 2, and 5 (you can try the others if you are feeling especially brave):<\/p>\n<p id=\"x-ck12-MmQzYjgwNmVlYmM3YjRkZDIyMDAxNjBlNjk0NjE4Y2Q.-x1d\"><a href=\"http:\/\/www.docbrown.info\/page04\/4_73calcs\/MVGmcTEST.htm\"> http:\/\/www.docbrown.info\/page04\/4_73calcs\/MVGmcTEST.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-qm5\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-Njk4MDY3MmU1MDAxZWUzMjQyZDhmMWUxOGU4NmVjMjM.-kw7\">\n<li>In the problem above, what is the formula weight of neon?<\/li>\n<li>What value is at the center of all the calculations?<\/li>\n<li>If we had 79.3 grams of Xe, would we expect a volume that is greater than or less than that obtained with neon?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-UGVyY2VudCBDb21wb3NpdGlvbg..\">Percent Composition<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ODNhNzFiMGFhNjhmMGZiZjhjYmExOWRhMTFiZmI5MDE.-urd\">\n<li>Define percent composition.<\/li>\n<li>Perform percent composition calculations.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MjI1MGUxYzE0OGE4ODAyODVhODExNGE4NjAwZWY0ZDc.-291\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211652\/20140811155329231868.jpeg\" alt=\"Nutritional information labels can let you know the percent composition\" width=\"500\" \/><\/span><\/p>\n<p id=\"x-ck12-ZmFjOWEzYThmMWEyODQyOWEwODY0YWNlNDg3MmM1ODY.-75u\"><strong> Is there anything healthy in this jar? <\/strong><\/p>\n<p id=\"x-ck12-ZWQyMGM2YzQ5ZmRmNTVlNGVjNGE5MmNiZWM1MjA1ZWU.-cxi\">Packaged foods that you eat typically have nutritional information provided on the label.\u00a0 The label on a jar of peanut butter (shown above) reveals that one serving size is considered to be 32 g.\u00a0 The label also gives the masses of various types of compounds that are present in each serving.\u00a0 One serving contains 7 g of protein, 15 g of fat, and 3 g of sugar.\u00a0 By calculating the fraction of protein, fat, or sugar in one serving of size of peanut butter and converting to percent values, we can determine the composition of the peanut butter on a percent by mass basis.<\/p>\n<h3>Percent Composition<\/h3>\n<p id=\"x-ck12-OTViMjJlMGExNTg3NDUzYTNlNWQ1YTk1NmFlYzE0OWU.-kwb\">Chemists often need to know what elements are present in a compound and in what percentage.\u00a0 The <strong> percent composition <\/strong> is the percent by mass of each element in a compound.\u00a0 It is calculated in a similar way that we just indicated for the peanut butter.<\/p>\n<p id=\"x-ck12-xxu\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211653\/b92d82fcecec402f2b5cdc6e1c2bb85d.png\" alt=\"% text{by mass}=frac{text{mass of element}}{text{mass of compound}} times 100 %\" width=\"320\" height=\"42\" \/><\/p>\n<h4>Percent Composition from Mass Data<\/h4>\n<p id=\"x-ck12-M2JmODNmY2MyZWUzNTE3M2M3MTUxMTYzOWM4MDYyYWQ.-coz\">The sample problem below shows the calculation of the percent composition of a compound based on mass data.<\/p>\n<p id=\"x-ck12-NjYxZmM0OTlkYTgyYjA5YzBkZGI0N2QyMTY2OWQ5NTE.-4ib\"><strong> Sample Problem One: Percent Composition from Mass <\/strong><\/p>\n<p id=\"x-ck12-YzMxNzBhODRmOTQyMjAzMzlkODBjZTVmZGI3M2NlNTA.-zl0\">A certain newly synthesized compound is known to contain the elements zinc and oxygen.\u00a0 When a 20.00 g sample of the sample is decomposed, 16.07 g of zinc remains.\u00a0 Determine the percent composition of the compound.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wb2\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-x6p\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ZTMyNDJhZmZhN2Q4YTBkZDBhY2NkMzI4NWJiYjUxNWY.-fag\">\n<li>mass of compound = 20.00 g<\/li>\n<li>mass of Zn = 16.07 g<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-a7n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-YjUxMWRkOGZhYmVhNjczMmIxZTI2MmExZjQ5OTdlNjA.-gju\">\n<li>percent Zn = ? %<\/li>\n<li>percent O = ? %<\/li>\n<\/ul>\n<p id=\"x-ck12-ZTZmOTQ4YmE0Y2FjN2RmN2NkMzA0OTkzYTNhNjU4NzU.-iod\">Subtract to find the mass of oxygen in the compound.\u00a0 Divide each element\u2019s mass by the mass of the compound to find the percent by mass.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-a1m\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-u7m\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211654\/647f0368deb4e44c4f3d666318504d8d.png\" alt=\"text{Mass of oxygen}&amp;=20.00 text{g} - 16.07 text{g}=3.93 text{g O} \\% text{Zn}&amp;=frac{16.07 text{g Zn}}{20.00 text{g}} times 100 %=80.35 % text{Zn} \\% text{O}&amp;=frac{3.93 text{g O}}{20.00 text{g}} times 100 %=19.65 % text{O} \\\" width=\"401\" height=\"113\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-yha\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NDE0OGI5MjYzNjA5ZjE3MTQ1MWJiMTYyNDExMWU4M2Y.-k90\">The calculations make sense because the sum of the two percentages adds up to 100%.\u00a0 By mass, the compound is mostly zinc.<\/p>\n<h4 id=\"x-ck12-cxz_9-j53\">Percent Composition from a Chemical Formula<\/h4>\n<p id=\"x-ck12-ZjAwODE1ZTdiYmJkZTA5NzcyODQ2YWZhMzU0MDZiZDM.-ums\">The percent composition of a compound can also be determined from the formula of the compound.\u00a0 The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound.\u00a0 That is divided by the molar mass of the compound and multiplied by 100%.<\/p>\n<p id=\"x-ck12-gdp\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211656\/3b76b799ec06e735db961ababcef9120.png\" alt=\"% text{by mass }=frac{text{mass of element in} 1 text{mol}}{text{molar mass of compound}}times 100 %\" width=\"376\" height=\"42\" \/><\/p>\n<p id=\"x-ck12-YTQwNDE3NzBhYzRmNGUxNTZhOTYzMjhmZDA4M2UwODg.-pa4\">The percent composition of a given compound is always the same as long as the compound is pure.<\/p>\n<p id=\"x-ck12-MjQ5ZTI3Mjc2ZmUyOGFjMGUwNGJiNDEzNjg5MWNlODQ.-6zm\"><strong> Sample Problem Two: Percent Composition from Chemical Formula <\/strong><\/p>\n<p id=\"x-ck12-MTI4OWNiNDZmMTAwZTBiNDUyODlkNTZmNjUyMzY4OTA.-x58\">Dichlorineheptoxide (Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> ) is a highly reactive compound used in some organic synthesis reactions.\u00a0 Calculate the percent composition of dichlorineheptoxide.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wwu\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-xrx\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-Y2JjNTVhZTI3OTQxNTBlOWE1YzIxZjE1YTZhMmYyNmM.-mor\">\n<li>mass of Cl in 1 mol Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 70.90 g<\/li>\n<li>mass of O in 1 mol Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 112.00 g<\/li>\n<li>molar mass of Cl <sub> 2 <\/sub> O <sub> 7 <\/sub> = 182.90 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-so9\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-NWRlOGU1MzFjZjc3YTRmMTgwNjlmMGUzMzIzOGE3NmU.-cvp\">\n<li>percent Cl = ? %<\/li>\n<li>percent O = ? %<\/li>\n<\/ul>\n<p id=\"x-ck12-NmEwZWUxMGU3NTZmYmNkZjliZTM2MTBiYTJhNmM3MzY.-wbo\">Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-xlh\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-cv0\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211657\/2cde2d521ee2c9531e0bb2c7e4e9fccf.png\" alt=\"% text{Cl}&amp;=frac{70.90 text{g Cl}}{182.90 text{g}} times 100 %=38.76 % text{Cl} \\% text{O}&amp;=frac{112.00 text{g O}}{182.90 text{g}} times 100 %=61.24 % text{O} \\\" width=\"314\" height=\"88\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-q10\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ZDQ5OTlmMzQ0YTExODZhYTY0OTExNDgzODUyZTczNTg.-15g\">The percentages add up to 100%.<\/p>\n<p id=\"x-ck12-ZmMyYjRhOTBmN2MzZTcxNzY5YjlhOTg0YjM0OTBmODQ.-gjt\">Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound.\u00a0 In the previous sample problem, it was found that the percent composition of dichlorineheptoxide is 38.76% Cl and 61.24% O.\u00a0 Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 g sample of dichlorineheptoxide.\u00a0 You can set up a conversion factor based on the percent by mass of each element.<\/p>\n<p id=\"x-ck12-jzl\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211659\/fc60c15a707115b10a71875f71e4c9fa.png\" alt=\"12.50 text{g Cl}_2 text{O}_7 times frac{38.76 text{g Cl}}{100 text{g Cl}_2text{O}_7}&amp;=4.845 text{g Cl} \\12.50 text{g Cl}_2 text{O}_7 times frac{61.24 text{g O}}{100 text{g Cl}_2text{O}_7}&amp;=7.655 text{g O} \\\" width=\"330\" height=\"90\" \/><\/p>\n<p id=\"x-ck12-NTA5NmJlZjdjOTUyM2EzZTE5ZmZlY2M3OGM5ZWMxYjA.-mqt\">The sum of the two masses is 12.50 g, the mass of the sample size.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZThmYjBmOWNhZDNkMWNlZGU4ZjU5NzhiMWI1NzI1NzA.-ui4\">\n<li>Processes are described for calculating the percent composition of a material based on mass or on chemical composition.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-ODU0YzM5ZWMxYzZmMzg1YzQ1MTBjMGQ2ODc4OWM3Yzc.-6sm\">Use the link below to review material and do calculations. Read both parts of the lesson and do as many calculations as you have time for.<\/p>\n<p id=\"x-ck12-YjZmZWNkOTg5NmM2MDE0MGNhNzkwN2U0OWIzMjRkNzI.-3si\"><a href=\"http:\/\/www.chemteam.info\/Mole\/Percent-Composition-Part1.html\"> http:\/\/www.chemteam.info\/Mole\/Percent-Composition-Part1.html <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-en4\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MjQ3OWQ1YTg4ZDk5ZjlmZDk1YTJjM2I4Zjk2OTUyMzk.-wgm\">\n<li>What is the formula for calculating percent composition?<\/li>\n<li>What information do you need to calculate percent composition by mass?<\/li>\n<li>What do subscripts in a chemical formula tell you?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MjY4YzM5MGJjMWZiMWQ5Nzg1YTk0MmEzMzcyNjUyNmM.-rxd\">\n<li><strong> percent composition: <\/strong> The percent by mass of each element in a compound.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-UGVyY2VudCBvZiBXYXRlciBpbiBhIEh5ZHJhdGU.\">Percent of Water in a Hydrate<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ZjU4MmEwOTI1ZWQwZTYyNzA3MDNkZTAyNDgwZjM3Y2Y.-xjt\">\n<li>Define hydrate.<\/li>\n<li>Calculate the percent water in hydrate when give relevant data.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-OGZkYzA3Y2Y0YmM4OGIzZGRjMzVmZDQ1MDg0ZjgwYTk.-yq5\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211700\/20140811155329359963.jpeg\" alt=\"Copper sulfate changes color when hydrated\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-M2Y5YzNlZWM2NzY4ZmEzYmEyNWEyNGZkZjNhOWJjMTg.-htu\"><strong> Why does the color change? <\/strong><\/p>\n<p id=\"x-ck12-YTk2MzRhMTQ3NDZmZGZiY2U2OTgxYTYxYWY4MWNhZjY.-zds\">If you look at a typical bottle of copper sulfate, it will be a bluish-green.\u00a0 If someone tells you that copper sulfate is white, you won\u2019t believe them.\u00a0 You are both right; it just depends on the copper sulfate.\u00a0 Your blue-green copper sulfate has several water molecules attached to it while your friend\u2019s copper sulfate is anhydrous (no water attached).\u00a0 Why the difference? The water molecules interact with some of the d electrons in the copper ion and produce the color.\u00a0 When the water is removed, the electron configuration changes and the color disappears.<\/p>\n<h3>Percent of Water in a Hydrate<\/h3>\n<p id=\"x-ck12-YTI1YzI0NDM3Mzg5Y2UxY2I1OWI5M2ExOWM1NzIzNDk.-kpn\">Many ionic compounds naturally contain water as part of the crystal lattice structure.\u00a0 A <strong> hydrate <\/strong> is a compound that has one or more water molecules bound to each formula unit.\u00a0 Ionic compounds that contain a transition metal are often highly colored.\u00a0 Interestingly, it is common for the hydrated form of a compound to be of a different color than the <strong> anhydrous <\/strong> form, which has no water in its structure.\u00a0 A hydrate can usually be converted to the anhydrous compound by heating. \u00a0For example,\u00a0the anhydrous compound cobalt(II) chloride is blue, while the hydrate is a distinctive magenta color.<\/p>\n<div id=\"x-ck12-ZTk2YzYyMzdlZjYxZTBiNzBmNDQ1YTk4MDRkYWUwY2M.-6wt\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-bdr\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MTUwNy0zNC04Mi0yMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211702\/20140811155329517119.png\" alt=\"Anhydrous cobalt chloride is blue, while hydrated cobalt chloride is red\" longdesc=\"On%20the%20left%20is%20anhydrous%20cobalt(II)%20chloride%2C%20CoCl%3Csub%3E2%3C%2Fsub%3E.%20On%20the%20right%20is%20the%20hydrated%20form%20of%20the%20compound%20called%20cobalt(II)%20chloride%20hexahydrate%2C%20CoCl%3Csub%3E2%3C%2Fsub%3E%E2%80%A26H%3Csub%3E2%3C%2Fsub%3EO.\" \/><\/p>\n<p><strong> Figure 10.11 <\/strong><\/p>\n<p id=\"x-ck12-NzVlYmFhNGYxYTUzN2NhNGFjYmM5ZDMwNWE3YWFiZGE.-43v\">On the left is anhydrous cobalt(II) chloride, CoCl <sub> 2 <\/sub> . On the right is the hydrated form of the compound called cobalt(II) chloride hexahydrate, CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.<\/p>\n<\/div>\n<p id=\"x-ck12-NTBiZjIxYTQxMmVjYmZmYmU0NzYyZjA5ZDBjZjllZGQ.-sut\">The hydrated form of cobalt(II) chloride contains six water molecules in each formula unit.\u00a0 The name of the compound is cobalt(II) chloride hexahydrate and its formula is CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.\u00a0 The formula for water is set apart at the end of the formula with a dot, followed by a coefficient that represents the number of water molecules per formula unit.<\/p>\n<p id=\"x-ck12-YTYyNDI4OWNkNzZlZjI3M2FlMzhkM2FhZDc3NzU2NTY.-4uv\">It is useful to know the percent of water contained within a hydrate.\u00a0 The sample problem below demonstrates the procedure.<\/p>\n<h4>Sample Problem One: Percent of Water in a Hydrate<\/h4>\n<p id=\"x-ck12-YzQzZDQ3MDI2NGY3YjFjYmZlYzIxNmZhYjU4NmY1YzA.-13j\">Find the percent water in cobalt(II) chloride hexahydrate, CoCl <sub> 2 <\/sub> \u20226H <sub> 2 <\/sub> O.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-6s4\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-ZDRmODhiMDJiYzgyYzhmNWUzNDhjZmViZWJhNDA4YmE.-jbc\">The mass of water in the hydrate is the coefficient (6) multiplied by the molar mass of H <sub> 2 <\/sub> O.\u00a0 The molar mass of the hydrate is the molar mass of the CoCl <sub> 2 <\/sub> plus the mass of water.<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-rw0\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-N2EzMzg1ZDlkOTY0NDI0Yjg1NjQyZTI2ODYxNGM5NjQ.-q0s\">\n<li>mass of H <sub> 2 <\/sub> O in 1 mol hydrate = 108.12 g<\/li>\n<li>molar mass of hydrate = 237.95 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-efs\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ZTc3NjJjMDFlYTJhMzc5NmUwYjA2ZjFiYTMzODVlM2M.-cnv\">\n<li>percent H <sub> 2 <\/sub> O = ? %<\/li>\n<\/ul>\n<p id=\"x-ck12-ZDliZGQ0M2M5ODY1MGY5YTdkYjllYTY1NDZmYzFhZWI.-hxv\">Calculate the percent by mass of water by dividing the mass of H <sub> 2 <\/sub> O in 1 mole of the hydrate by the molar mass of the hydrate and multiplying by 100%.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-i34\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-lkr\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDUyNjEzMTk1MA..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211703\/8eb891808f0b1fa0db4291b307a7b03a.png\" alt=\"% text{H}_2text{O}=frac{108.12 text{g H}_2 text{O}}{237.95 text{g}} times 100 %=45.44 % text{H}_2text{O}\" width=\"372\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-pf8\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NTY3MzJjNGE2Mzg4NjYyYjY3ZWJhN2JiNDFlYjI3ZGU.-ea1\">Nearly half of the mass of the hydrate is composed of water molecules within the crystal.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZDRjOGQ3MDg5MGZjNjg5MjQxMDUzNzRhMGI1Y2JlMjI.-j25\">\n<li>The process of calculating the percent water in a hydrate is described.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-N2RmYjM3YzZjYjgwMTM0ZWVkODI5ZWZjMzIzNDNiNDU.-qco\">Use the following link to practice calculating percent water in a hydrate:<\/p>\n<p id=\"x-ck12-MjAxM2U1YmZmOThiYTYxNTMzNThjNzVlZmIzN2U0Yzg.-gd4\"><a href=\"https:\/\/web.archive.org\/web\/20150314182712\/http:\/\/www.sd84.k12.id.us\/shs\/departments\/science\/martz\/2007_ssem2\/Chemistry\/hydrate.htm\" target=\"_blank\" rel=\"noopener\">Hydrate Problems<\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ok5\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-YWM5MzhmYmE1YTg5NjBkZTJjMmQ0ODA1YjIwZDI1YjM.-ubb\">\n<li>What is a hydrate?<\/li>\n<li>How can you convert a hydrate to an anhydrous compound?<\/li>\n<li>What does hexahydrate mean?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-Njg3NTJjMzFjN2ViNzEzNDg4NWUyMjZhOWY3MTBmYjE.-dgv\">\n<li><strong> Anhydrous: <\/strong> Without water.<\/li>\n<li><strong> Hydrate: <\/strong> A compound that has one or more water molecules bound to each formula unit<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-RGV0ZXJtaW5pbmcgRW1waXJpY2FsIEZvcm11bGFz\">Determining Empirical Formulas<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NjgwNWFkZmJjN2RmZmYxZWQ5OGRlZDc0OTUwOWNlOTc.-mj0\">\n<li>Define empirical formula.<\/li>\n<li>Calculate the empirical formula for a compound when given the elemental analysis of the compound.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-NGYzYTZkMGE0NDUyODdiMmQ4NTUwYmZhNTU0ZDJlYzY.-0zt\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211705\/20140811155329603191.jpeg\" alt=\"Scientist trying to determine the formula of a compound in the early days of chemistry\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-YmVhNTkxMDgwODZkMjMxNzE5Zjc1NTFlZDJhOTc5OGE.-0as\"><strong> What is occuring in this picture? <\/strong><\/p>\n<p id=\"x-ck12-MjcwMTM1NzM4ZTI0M2M4ZDFmODJlMDAyZDgzNTlhMTY.-npe\">In the early days of chemistry, there were few tools for the detailed study of compounds.\u00a0 Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials.\u00a0 The \u201cnew\u201d field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely.\u00a0 The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios.\u00a0 We did not know exactly how many of these atoms were actually in a specific molecule.<\/p>\n<h3>Determining Empirical Formulas<\/h3>\n<p id=\"x-ck12-OTU4NTdiYmJmNmU1YmNiYzdmMGE3ZGYxZTc1MmM4MTk.-09t\">An <strong> empirical formula <\/strong> is one that shows the lowest whole-number ratio of the elements in a compound.\u00a0 Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical.\u00a0 However, we can also consider the empirical formula of a molecular compound.\u00a0 Ethene is a small hydrocarbon compound with the formula C <sub> 2 <\/sub> H <sub> 4 <\/sub> (see <strong> Figure <\/strong> <a href=\"#x-ck12-OTgwNDUtMTM2MjE0MTc3NS01OS03Ni0yNA..\"> below <\/a> ).\u00a0 While C <sub> 2 <\/sub> H <sub> 4 <\/sub> is its molecular formula and represents its true molecular structure, it has an empirical formula of CH <sub> 2 <\/sub> .\u00a0 The simplest ratio of carbon to hydrogen in ethene is 1:2.\u00a0 There are two ways to view that ratio.\u00a0 Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen.\u00a0 Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen.\u00a0 So the subscripts in a formula represent the mole ratio of the elements in that formula.<\/p>\n<div id=\"x-ck12-ZDc4ZGU1MjM1ZDI0NzRkMDI2NzNlMzEwOTE1MDhjMDA.-xni\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-bfo\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MTc3NS01OS03Ni0yNA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211707\/20140811155329734238.png\" alt=\"Ball and stick model of ethene\" longdesc=\"Ball-and-stick%20model%20of%20ethene%2C%20C%3Csub%3E2%3C\/sub%3EH%3Csub%3E4%3C\/sub%3E.\" \/><\/p>\n<p><strong> Figure 10.12 <\/strong><\/p>\n<p id=\"x-ck12-Zjg2YmVhMDc5OTlmZGFkMTI4N2NmZjc5N2RlYmI1ZjA.-vkq\">Ball-and-stick model of ethene, C <sub> 2 <\/sub> H <sub> 4 <\/sub> .<\/p>\n<\/div>\n<p id=\"x-ck12-MGIwNTU0MWQ1Y2ZmYjNiNDgyYTQxODQxYTc3NTlmZTY.-lpo\">In a procedure called <strong> elemental analysis <\/strong> , an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it.\u00a0 These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula.\u00a0 The steps to be taken are outlined below.<\/p>\n<ol id=\"x-ck12-NzY4OGI5OTNiNDBiZDIzYzU0MTU5MzIxZDQyMGUyYjY.-kuz\">\n<li>Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.<\/li>\n<li>Use each element\u2019s molar mass to convert the grams of each element to moles.<\/li>\n<li>In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.<\/li>\n<li>If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.<\/li>\n<li>In some cases, one or more of the moles calculated in step 3 will not be whole numbers.\u00a0 Multiply each of the moles by the smallest whole number that will convert each into a whole number.\u00a0 Write the empirical formula.<\/li>\n<\/ol>\n<h4>Sample Problem One: Determining the Empirical Formula of a Compound<\/h4>\n<p id=\"x-ck12-OTY0ZjU3YmNjNTNhMGRhYzYzNmQwYjFmYTcwNjAzM2U.-qxb\">A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen.\u00a0 Find the empirical formula of the compound.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-qt9\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-mry\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ZjA3NmYwZmQ4YmM4YzIzNmI5YWQ5NmE2MGVmNzhhOTQ.-4lo\">\n<li>% of Fe = 69.94%<\/li>\n<li>% of O = 30.06%<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-md4\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-MmU1NTM2NzNhMmYyMzU2MWRkNTkyMTY1ZGYzOGUyNjg.-srb\">\n<li>Empirical formula = Fe <sub> ? <\/sub> O <sub> ? <\/sub><\/li>\n<\/ul>\n<p id=\"x-ck12-Y2UwNGJmZDM4ZWVhMTQyYTYxMDViYmQxN2UzN2I4NmU.-23z\">Steps to follow are outlined in the text.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-d07\"><em> Step 2:\u00a0 Calculate. <\/em><\/p>\n<p id=\"x-ck12-YTc1MGZkMjRmYTgwYTExMTM5NmI3YTdiYzRmZGJmZjI.-ckw\">1. Assume a 100 g sample.<\/p>\n<p id=\"x-ck12-8fp\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211708\/a12004a26997f9bc06927cf8a99d6235.png\" alt=\"&amp; 69.94 text{g Fe} \\&amp; 30.06 text{g O}\" width=\"80\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-OGJkYjc0ZTNhN2E4NGJkN2NmNmY1MTU4NGM0YmQyODY.-lgc\">2. Convert to moles.<\/p>\n<p id=\"x-ck12-npt\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211708\/fa055626cb3aa7cfc7f1f24cebfbb9da.png\" alt=\"69.94 text{g Fe} times frac{1 text{mol Fe}}{55.85 text{g Fe}}&amp;=1.252 text{mol Fe} \\30.06 text{g O} times frac{1 text{mol O}}{16.00 text{g O}}&amp;=1.879 text{mol O}\" width=\"309\" height=\"88\" \/><\/p>\n<p id=\"x-ck12-NWYzYzU4MjE5NDZhN2U2ZDkzMzc5MzEwODVkZWEyYTI.-awb\">3. Divide both moles by the smallest of the results.<\/p>\n<p id=\"x-ck12-jtc\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211709\/a3f8d99609e28a14fff832e9c8ba2fe3.png\" alt=\"frac{1.252 text{mol Fe}}{1.252}=1 text{mol Fe} qquad qquad frac{1.879 text{mol O}}{1.252}=1.501 text{mol O}\" width=\"483\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MTk5YjJjM2NkNzUyZjNmYTcwMjFhZTQyNDNkMDJhZjg.-ld1\">4\/5. Since the moles of O, is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.<\/p>\n<p id=\"x-ck12-bw1\" class=\"x-ck12-indent\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211711\/c2909a1f6b81830213b3cfe0e81413a9.png\" alt=\"1 text{mol Fe} times 2=2 text{mol Fe} qquad qquad 1.501 text{mol O} times 2=3 text{mol O}\" width=\"475\" height=\"14\" \/><\/p>\n<p id=\"x-ck12-N2Q2Y2ZmNWJmYTkwYzU0MzQ0NTZjZjNiMzIzYzQ2NzA.-rh1\">The empirical formula of the compound is Fe <sub> 2 <\/sub> O <sub> 3 <\/sub> .<\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-rym\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NjI0NzhiYmVmZTJmM2E1M2ZlMzY5ODBkM2Y2NjVkMDY.-3ml\">The subscripts are whole numbers and represent the mole ratio of the elements in the compound.\u00a0 The compound is the ionic compound iron(III) oxide.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-YzA2OTg4Y2NjZDQ4MzQzYTBlYmI0YmVjNWEyMGYwYmE.-omu\">\n<li>A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NjIwM2QyZjBiODY5NjU2MTdiYjU0NDM5ZjE1M2E4MTI.-kje\">Use the link below to read about calculating empirical formulas and practice working some problems:<\/p>\n<p id=\"x-ck12-OGM1YzEwYjM1ZmQzNGQyOGM2ZmI4MjJjNDU2Mzc3ZTY.-yqd\"><a href=\"http:\/\/www.chemteam.info\/Mole\/EmpiricalFormula.html\"> http:\/\/www.chemteam.info\/Mole\/EmpiricalFormula.html <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-25c\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-OTA1MjIyYjBjZjVhM2E3NGZlNWI1N2UyY2YzYjMxZGQ.-21w\">\n<li>What is an empirical formula?<\/li>\n<li>What does an empirical formula tell you?<\/li>\n<li>What does it not tell you?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-NDM4YWMxMzk3N2VlZjllOWE2ZjliYjQ5ZWU5OTEyNDg.-ktb\">\n<li><strong> elemental analysis: <\/strong> Determines the percentages of each element in a compound.<\/li>\n<li><strong> empirical formula: <\/strong> Shows the lowest whole-number ratio of the elements in a compound.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-RGV0ZXJtaW5pbmcgTW9sZWN1bGFyIEZvcm11bGFz\">Determining Molecular Formulas<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ODQ5ZjAyNGE0MWVjNjllNzhhMWI3MDAzODkzNDc1MTA.-7v2\">\n<li>Define molecular formula.<\/li>\n<li>Determine the molecular formula when give the empirical formula and the molar mass of the compound.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-N2NlOTRkYmFlMDFjNTI5MTIxMDFiNmM3OTA3MmM4NzQ.-swg\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211712\/20140811155329857990.jpg\" alt=\"Fischer projection of glucose\" width=\"100\" \/><\/span><\/p>\n<p id=\"x-ck12-ZmRhMGQ4Mzg4YzRkZTdjM2QwOWEzMjM2ZjYwNGIwYzE.-wqv\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211713\/20140811155329998495.png\" alt=\"Structure of sucrose\" width=\"300\" \/><\/span><\/p>\n<p id=\"x-ck12-OGJlMjRhMDUxYWY0YzAwNDY1YmM0YTRhNzhkY2IxMjg.-kpz\"><strong> How can you determine the differences between these two molecules? <\/strong><\/p>\n<p id=\"x-ck12-OWIyNThlYmIyM2EyMTM1YTljOWQ0OTgzNjAyZDdjZjY.-4md\">Above we see two carbohydrates: glucose and sucrose.\u00a0 Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar.\u00a0 Some people could distinguish them on the basis of taste, but it\u2019s not a good idea to go around tasting chemicals. The best way is to determine the molecular weights \u2013 this approach allows you to easily tell which compound is which.<\/p>\n<h3>Molecular Formulas<\/h3>\n<p id=\"x-ck12-ZGM1MDRmYWRhMWM5NjZlZWJmZmU1ODBjNGMzN2FlN2I.-pk4\"><strong> Molecular formulas <\/strong> give the kind and number of atoms of each element present in a molecular compound.\u00a0 In many cases, the molecular formula is the same as the empirical formula.\u00a0 The molecular formula of methane is CH <sub> 4 <\/sub> and because it contains only one carbon atom, that is also its empirical formula.\u00a0 Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula.\u00a0 Acetic acid\u00a0 is an organic acid that is the main component of vinegar.\u00a0 Its molecular formula is C <sub> 2 <\/sub> H <sub> 4 <\/sub> O <sub> 2 <\/sub> .\u00a0 Glucose is\u00a0 a simple sugar that cells use as a primary source of energy.\u00a0 Its molecular formula is C <sub> 6 <\/sub> H <sub> 12 <\/sub> O <sub> 6 <\/sub> .\u00a0 The structures of both molecules are shown in the figure below.\u00a0 They are very different compounds, yet both have the same empirical formula of CH <sub> 2 <\/sub> O.<\/p>\n<div id=\"x-ck12-M2VkYmVmYzM2MGRhM2I1NzZmMGEyN2Y0YzQ4MzhhNzQ.-sae\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-m8v\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MjE0MjAzOS05NS05NC0yNw..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211714\/20140811155330084912.png\" alt=\"Acetic acid and glucose both the same empirical formula\" longdesc=\"Acetic%20acid%20%28left%29%20has%20a%20molecular%20formula%20of%20C%3Csub%3E2%3C\/sub%3EH%3Csub%3E4%3C\/sub%3EO%3Csub%3E2%3C\/sub%3E%2C%20while%20glucose%20%28right%29%20has%20a%20molecular%20formula%20of%20C%3Csub%3E6%3C\/sub%3EH%3Csub%3E12%3C\/sub%3EO%3Csub%3E6%3C\/sub%3E.%20Both%20have%20the%20empirical%20formula%20CH%3Csub%3E2%3C\/sub%3EO.\" \/><\/p>\n<p><strong> Figure 10.13 <\/strong><\/p>\n<p id=\"x-ck12-ZTI2OTAwNGQyMDBjYzhhNzM3NzAwMDk4OGU0Y2FlOGM.-xtu\">Acetic acid (left) has a molecular formula of C <sub> 2 <\/sub> H <sub> 4 <\/sub> O <sub> 2 <\/sub> , while glucose (right) has a molecular formula of C <sub> 6 <\/sub> H <sub> 12 <\/sub> O <sub> 6 <\/sub> . Both have the empirical formula CH <sub> 2 <\/sub> O.<\/p>\n<\/div>\n<p id=\"x-ck12-NDFiODQyMzRiNTY3NWIwNGVlN2E1OTAxYjczZmRmMTM.-gty\">Empirical formulas can be determined from the percent composition of a compound.\u00a0 In order to determine its molecular formula, it is necessary to know the molar mass of the compound.\u00a0 Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds.\u00a0 In order to go from the empirical formula to the molecular formula, follow these steps:<\/p>\n<ol id=\"x-ck12-MWNlODE1ZTM5NTJhNDJkYjBiODg3MmIwODhjZTM0YmE.-x9t\">\n<li>Calculate the <strong> empirical formula mass (EFM) <\/strong> , which is simply the molar mass represented by the empirical formula.<\/li>\n<li>Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.<\/li>\n<li>Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.<\/li>\n<\/ol>\n<h4>Sample Problem One: Determining the Molecular Formula of a Compound<\/h4>\n<p id=\"x-ck12-NDE5MWIyMzIzYzc0NTI5NDBhZGRkYmE0MGRmZmJmMTg.-1lt\">The empirical formula of a compound of boron and hydrogen is BH <sub> 3 <\/sub> .\u00a0 Its molar mass is 27.7 g\/mol.\u00a0 Determine the molecular formula of the compound.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-noc\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-4yc\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-Yzc1Yjg5YjIwMGVjMDA5MTg3NzA1YWVmMjMwMTYyZDM.-hnl\">\n<li>empirical formula = BH <sub> 3 <\/sub><\/li>\n<li>molar mass = 27.7 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-4yo\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ZGM3MzlhMjY5YjliMjA5YWIyYWJlNWM5NDUwMDg2YzY.-f3h\">\n<li>molecular formula = ?<\/li>\n<\/ul>\n<p id=\"x-ck12-Y2UwNGJmZDM4ZWVhMTQyYTYxMDViYmQxN2UzN2I4NmU.-ted\">Steps to follow are outlined in the text.<\/p>\n<p id=\"x-ck12-YWQ0ODVjMzM3ODc2OTlmZGQ5MDJmNDg4ZDM1YzU3Mzc.-7l1\"><em> Step 2: Calculate. <\/em><\/p>\n<p id=\"x-ck12-OGNlZTZhMDVkNGMxZTA2ZTQ2MDk4NWYyMWM2ZTYyZjc.-mcs\">1. The empirical formula mass (EFM) = 13.84 g\/mol<\/p>\n<p id=\"x-ck12-YzgxZTcyOGQ5ZDRjMmY2MzZmMDY3Zjg5Y2MxNDg2MmM.-mkd\">2. <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211715\/f1f79692abe6d70305d06ef262d09b50.png\" alt=\"frac{text{molar mass}}{text{EFM}}=frac{27.7}{13.84}=2\" width=\"156\" height=\"25\" \/><\/p>\n<p id=\"x-ck12-ZWNjYmM4N2U0YjVjZTJmZTI4MzA4ZmQ5ZjJhN2JhZjM.-vim\">3. <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211716\/d925b5855abf5be5b1bcd2a7c049b06f.png\" alt=\"text{BH}_3 times 2 =text{B}_2text{H}_6\" width=\"128\" height=\"16\" \/><\/p>\n<p id=\"x-ck12-NDMwODQwZDVkOTM2ODc1YTUzZGJkOGQzZTQ1YmRmN2Q.-at7\">The molecular formula of the compound is B <sub> 2 <\/sub> H <sub> 6 <\/sub> .<\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-euc\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-Y2JhOWY3YWE1MmM3OTVhZDRiYjIyODA4NzUzODJiMjQ.-n64\">The molar mass of the molecular formula matches the molar mass of the compound.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ODc4MzI5NmI2MmIzMWNiY2E4NzgzYTU0ZDFmZGE2MjM.-kat\">\n<li>A procedure is described that allows the calculation of the exact molecular formula for a compound.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-N2M0NWJjMDA5MWJiNzNhMDMyNmE0NDQ2ZDQ1NWU0NGY.-rlr\">Use the link below to access practice problems.\u00a0 Try as many as you have time for:<\/p>\n<p id=\"x-ck12-NDYyOWY0MTZiZTU1N2E1MTIzOGI0MmY2ZWUwNTIwNjc.-r43\"><a href=\"http:\/\/chemistry.about.com\/od\/chemistry-test-questions\/tp\/Molecular-Formula-Practice-Test-Questions.htm\"> http:\/\/chemistry.about.com\/od\/chemistry-test-questions\/tp\/Molecular-Formula-Practice-Test-Questions.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-mrr\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-N2QwNGRlNTM3N2U5OGU2ZDlhZDVhMTdkODY3ZDdjY2Y.-ld1\">\n<li>What is the difference between an empirical formula and a molecular formula?<\/li>\n<li>In addition to the elemental analysis, what do you need to know to calculate the molecular formula?<\/li>\n<li>What does the empirical formula mass tell you?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ODc4OTcwMTA1NzVjMDQ4NWViNWZjZjI5ODU0MjNkNTI.-i9v\">\n<li><strong> empirical formula mass (EFM): <\/strong> The molar mass represented by the empirical formula.<\/li>\n<li><strong> molecular formula: <\/strong> Gives the kind and number of atoms of each element present in a molecular compound.<\/li>\n<\/ul>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>User:OSX\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:1983-1988_Toyota_Hilux_4-door_utility_01.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:1983-1988_Toyota_Hilux_4-door_utility_01.jpg <\/a> .<\/li>\n<li>C. Sentier. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Amedeo_Avogadro2.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Amedeo_Avogadro2.jpg <\/a> .<\/li>\n<li>Left: Michael David Hill, 2005 (Mikiwikipikidikipedia); Right: chrisbb@prodigy.net. Left: http:\/\/commons.wikimedia.org\/wiki\/File:Close-up_of_mole.jpg; Right: http:\/\/www.flickr.com\/photos\/chrisbrenschmidt\/436990097\/ .<\/li>\n<li>Courtesy of Diane A. Reid\/National Cancer Institute. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Man_responds_to_telephone_call.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Man_responds_to_telephone_call.jpg <\/a> .<\/li>\n<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Water-3D-balls-A.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Water-3D-balls-A.png <\/a> .<\/li>\n<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Potassium-dichromate-sample.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Potassium-dichromate-sample.jpg <\/a> .<\/li>\n<li>User:Estormiz\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Fermion_Plant_Oulu_2007_01_20.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Fermion_Plant_Oulu_2007_01_20.JPG <\/a> .<\/li>\n<li>Martin Walker (Wikimedia: Walkerma). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Calcium_chloride.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Calcium_chloride.jpg <\/a> .<\/li>\n<li>Laura Guerin. CK-12 Foundation .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>User:Mark.murphy\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Diving_-_scubadiver.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Diving_-_scubadiver.JPG <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. CK-12 Foundation .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>Image copyright Marynchenko Oleksandr, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\n<li>Image copyright md8speed, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\n<li>Photographer: Warren Denning, courtesy of the Pioneer Balloon Company. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Congrats_bqt.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Congrats_bqt.jpg <\/a> .<\/li>\n<li>Doug Smith (Wikimedia: Xsmith). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:CurrituckSoundMap.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:CurrituckSoundMap.png <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Joy Sheng. .<\/li>\n<li>User:Chemicalinterest\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Copper%28II%29_sulfate.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Copper%28II%29_sulfate.JPG <\/a> .<\/li>\n<li>(A) Martin Walker (Wikimedia: Walkerma); (B) Ben Mills (Wikimedia: Benjah-bmm27). (A) http:\/\/commons.wikimedia.org\/wiki\/File:Cobalt%28II%29_chloride.jpg; (B) http:\/\/commons.wikimedia.org\/wiki\/File:Cobalt%28II%29-chloride-hexahydrate-sample.jpg .<\/li>\n<li>Harriet Moore. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:M_Faraday_Lab_H_Moore.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:M_Faraday_Lab_H_Moore.jpg <\/a> .<\/li>\n<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Ethylene-CRC-MW-3D-balls.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Ethylene-CRC-MW-3D-balls.png <\/a> .<\/li>\n<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png <\/a> .<\/li>\n<li>User:glycoform\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Sucrose_3Dprojection.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Sucrose_3Dprojection.png <\/a> .<\/li>\n<li>(left) Ben Mills (Wikimedia: Benjah-bmm27); (right) Ben Mills (Wikimedia: Benjah-bmm27), User:Yikrazuul\/Wikimedia Commons. Acetic acid: http:\/\/commons.wikimedia.org\/wiki\/File:Acetic-acid-2D-flat.png; Glucose: http:\/\/commons.wikimedia.org\/wiki\/File:D-glucose-chain-2D-Fischer.png .<\/li>\n<\/ol>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-657\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":1507,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-657","chapter","type-chapter","status-publish","hentry"],"part":2331,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/657","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/1507"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/657\/revisions"}],"predecessor-version":[{"id":3666,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/657\/revisions\/3666"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2331"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/657\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=657"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=657"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=657"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}