{"id":1381,"date":"2018-08-11T03:04:10","date_gmt":"2018-08-11T03:04:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1381"},"modified":"2025-12-01T20:33:12","modified_gmt":"2025-12-01T20:33:12","slug":"formula-mass-and-mole-concept-from-che100","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/formula-mass-and-mole-concept-from-che100\/","title":{"raw":"7.1 The Mole Concept","rendered":"7.1 The Mole Concept"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define the amount unit mole and the related quantity Avogadro\u2019s number<\/li>\r\n \t<li>Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Mole<\/h2>\r\nThe identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H<sub>2<\/sub>O, and hydrogen peroxide, H<sub>2<\/sub>O<sub>2<\/sub>, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the <em>mole<\/em>, which remains indispensable in modern chemical science.\r\n\r\nThe mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of <em>the number<\/em> of atoms or molecules in a bulk sample of matter. A <strong>mole<\/strong> is defined as <em>the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure <sup>12<\/sup>C weighing exactly 12 g.<\/em> One Latin connotation for the word \u201cmole\u201d is \u201clarge mass\u201d or \u201cbulk,\u201d which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.\r\n\r\nThe number of entities composing a mole has been experimentally determined to be [latex]6.02214179\\times {10}^{23}[\/latex], a fundamental constant named <strong>Avogadro\u2019s number<\/strong> <strong>(<em>N<sub>A<\/sub><\/em>)<\/strong> or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of \u201cper mole,\u201d a conveniently rounded version being [latex]6.022\\times {10}^{23}\\text{\/mol}[\/latex].\r\n\r\nConsistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The<strong> molar mass<\/strong> of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g\/mol) (see Figure 1).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"880\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211119\/CNX_Chem_03_02_moles1.jpg\" alt=\"This figure contains eight different substances displayed on white circles. The amount of each substance is visibly different.\" width=\"880\" height=\"582\" \/> Figure 1. Each sample contains 6.022 \u00d7 10<sup>23<\/sup> atoms\u20141.00 mol of atoms. From left to right (top row): 65.4g zinc, 12.0g carbon, 24.3g magnesium, and 63.5g copper. From left to right (bottom row): 32.1g sulfur, 28.1g silicon, 207g lead, and 118.7g tin. (credit: modification of work by Mark Ott)[\/caption]\r\n\r\nBecause the definitions of both the mole and the atomic mass unit are based on the same reference substance, <sup>12<\/sup>C, <em>the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu<\/em>. Per the amu definition, a single <sup>12<\/sup>C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of <sup>12<\/sup>C contains 1 mole of <sup>12<\/sup>C atoms (its molar mass is 12 g\/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, <sup>12<\/sup>C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211122\/CNX_Chem_03_02_compound1.jpg\" alt=\"This photo shows two vials filled with a colorless liquid. It also shows two bowls: one filled with an off-white powder and one filled with a bright red powder.\" width=\"500\" height=\"333\" \/> Figure 2. Each sample contains 6.02 \u00d7 10<sup>23<\/sup> molecules or formula units\u20141.00 mol of the compound or element. Clock-wise from the upper left: 130.2g of C<sub>8<\/sub>H<sub>17<\/sub>OH (1-octanol, formula mass 130.2 amu), 454.9g of HgI<sub>2<\/sub> (mercury(II) iodide, formula mass 459.9 amu), 32.0g of CH<sub>3<\/sub>OH (methanol, formula mass 32.0 amu) and 256.5g of S<sub>8<\/sub> (sulfur, formula mass 256.6 amu). (credit: Sahar Atwa)[\/caption]\r\n<table id=\"fs-idp17650992\" class=\"medium unnumbered\" summary=\"A table is shown that is made up of four columns and six rows. The header row reads: \u201cElement,\u201d \u201cAverage Atomic Mass (a m u),\u201d \u201cMolar Mass (g \/ m o l),\u201d and \u201cAtoms \/ Mole.\u201d The first column contains the symbols \u201cC,\u201d \u201cH,\u201d \u201cO,\u201d \u201cN a,\u201d and \u201cC l.\u201d The second column contains the values \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d \u201c22.99,\u201d and \u201c33.45.\u201d The third column contains the values \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d \u201c22.99,\u201d and \u201c33.45.\u201d The final column contains the value \u201c6.022 times 10 superscript 23\u201d in each cell.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Average Atomic Mass (amu)<\/th>\r\n<th>Molar Mass (g\/mol)<\/th>\r\n<th>Atoms\/Mole<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>C<\/td>\r\n<td>12.01<\/td>\r\n<td>12.01<\/td>\r\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>1.008<\/td>\r\n<td>1.008<\/td>\r\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>16.00<\/td>\r\n<td>16.00<\/td>\r\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Na<\/td>\r\n<td>22.99<\/td>\r\n<td>22.99<\/td>\r\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Cl<\/td>\r\n<td>33.45<\/td>\r\n<td>33.45<\/td>\r\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"attachment_5625\" align=\"alignright\" width=\"301\"]<img class=\"wp-image-5625\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/218\/2016\/08\/26220520\/CNX_Chem_03_02_water1-e1472249151518.jpg\" alt=\"A close-up photo of a water droplet on a leaf is shown. The water droplet is not perfectly spherical.\" width=\"301\" height=\"213\" \/> Figure 3. A single drop of water.[\/caption]\r\n\r\nWhile atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 3). The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth.\r\n\r\nAlthough this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.\r\n\r\nThe mole is used in chemistry to represent [latex]6.022\\times {10}^{23}[\/latex] of something, but it can be difficult to conceptualize such a large number. Watch this video to learn more.\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n\r\nhttps:\/\/youtu.be\/TEl4jeETVmg\r\n\r\n<\/div>\r\nThe relationships between formula mass, the mole, and Avogadro\u2019s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance\u2019s mass.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: <strong>Deriving Moles from Grams for an Element<\/strong><\/h3>\r\nAccording to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?\r\n\r\n[reveal-answer q=\"670401\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"670401\"]\r\n\r\nThe mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g\/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable \u201cballpark\u201d estimate of the number of moles would be slightly greater than 0.1 mol.\r\n\r\nThe molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g\/mol):\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211127\/CNX_Chem_03_02_potassium_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMass of K atoms ( g )\u201d while the one on the right contains the phrase, \u201cMoles of K atoms ( mol ).\u201d There is a phrase under the arrow that says, \u201cDivide by molar mass (g \/ mol).\u201d\" width=\"650\" height=\"147\" \/>\r\n\r\nThe factor-label method supports this mathematical approach since the unit \u201cg\u201d cancels and the answer has units of \u201cmol:\u201d\r\n<p style=\"text-align: center;\">[latex]4.7\\cancel{\\text{ g K}}\\left(\\frac{\\text{1 mol K}}{\\text{39.10 }\\cancel{\\text{g K}}}\\right)=0.12\\text{ mol K}[\/latex]<\/p>\r\nThe calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nBeryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?\r\n\r\n[reveal-answer q=\"27674\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"27674\"]0.360 mol[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2: <strong>Deriving Grams from Moles for an Element<\/strong><\/h3>\r\nA liter of air contains [latex]9.2\\times {10}^{-4}[\/latex] mol argon. What is the mass of Ar in a liter of air?\r\n\r\n[reveal-answer q=\"363345\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"363345\"]\r\n\r\nThe molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10<sup>\u20133<\/sup>) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211128\/CNX_Chem_03_02_argon_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMoles of A r atoms ( mol )\u201d while the one on the right contains the phrase, \u201cMass of A r atoms ( g ).\u201d There is a phrase under the arrow that says \u201cMultiply by molar mass ( g \/ mol ).\u201d\" width=\"650\" height=\"147\" \/>\r\n\r\nIn this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g\/mol):\r\n<p style=\"text-align: center;\">[latex]9.2\\times {10}^{-4}\\cancel{\\text{mol Ar}}\\left(\\frac{\\text{39.95 }\\text{g Ar}}{\\text{1 }\\cancel{\\text{mol Ar}}}\\right)=\\text{0.037 }\\text{g Ar}[\/latex]<\/p>\r\nThe result is in agreement with our expectations as noted above, around 0.04 g Ar.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat is the mass of 2.561 mol of gold?\r\n\r\n[reveal-answer q=\"363301\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"363301\"]504.4 g[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3: <strong>Deriving Number of Atoms from Mass for an Element<\/strong><\/h3>\r\nCopper is commonly used to fabricate electrical wire (Figure 7). How many copper atoms are in 5.00 g of copper wire?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"299\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211129\/CNX_Chem_03_02_copper1.jpg\" alt=\"A close-up photo of a spool of copper wire is shown.\" width=\"299\" height=\"398\" \/> Figure 7. Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)[\/caption]\r\n\r\n[reveal-answer q=\"383171\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"383171\"]\r\n\r\nThe number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro\u2019s number (<em>N<sub>A<\/sub><\/em>) to convert this molar amount to number of Cu atoms:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211131\/CNX_Chem_03_02_copperMoles_img1.jpg\" alt=\"A diagram of three boxes connected by a right-facing arrow in between each is shown. The box on the left contains the phrase, \u201cMass of C u atoms ( g ),\u201d the middle box reads, \u201cMoles of C u atoms ( mol ),\u201d while the one on the right contains the phrase, \u201cNumber of C u atoms.\u201d There is a phrase under the left arrow that says \u201cDivide by molar mass (g \/ mol),\u201d and under the right arrow it states, \u201cMultiply by Avogadro\u2019s number ( mol superscript negative one ).\u201d\" width=\"881\" height=\"127\" \/>\r\n\r\nConsidering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth <em>N<sub>A<\/sub><\/em>, or approximately 10<sup>22<\/sup> Cu atoms. Carrying out the two-step computation yields:\r\n<p style=\"text-align: center;\">[latex]\\text{5.00 }\\cancel{\\text{g Cu}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol Cu}}}{\\text{63.55 }\\cancel{\\text{g Cu}}}\\right)\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms Cu}}{\\text{1 }\\cancel{\\text{mol Cu}}}\\right)=4.74\\times {10}^{22}\\text{ atoms Cu}[\/latex]<\/p>\r\nThe factor-label method yields the desired cancellation of units, and the computed result is on the order of 10<sup>22<\/sup> as expected.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nA prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?\r\n[reveal-answer q=\"619266\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"619266\"][latex]4.586\\times {10}^{22}\\text{Au atoms}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4: <strong>Deriving Moles from Grams for a Compound<\/strong><\/h3>\r\nOur bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C<sub>2<\/sub>H<sub>5<\/sub>O<sub>2<\/sub>N. How many moles of glycine molecules are contained in 28.35 g of glycine?\r\n\r\n[reveal-answer q=\"129673\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"129673\"]\r\n\r\nWe can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 3:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211132\/CNX_Chem_03_02_glycine_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMass of C subscript 2 H subscript 5 O subscript 2 N ( g )\u201d while the box on the right contains the phrase, \u201cMoles of C subscript 2 H subscript 5 O subscript 2 N ( mol ).\u201d There is a phrase under the arrow that says \u201cDivide by molar mass (g \/ mol).\u201d\" width=\"650\" height=\"147\" \/>\r\n\r\nThe molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C<sub>2<\/sub>H<sub>5<\/sub>O<sub>2<\/sub>N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211134\/CNX_Chem_03_01_glycinemass_img1.jpg\" alt=\"A table is shown that is made up of six columns and six rows. The header row reads: \u201cElement,\u201d \u201cQuantity (mol element \/ mol compound,\u201d a blank space, \u201cMolar mass (g \/ mol element),\u201d a blank space, and \u201cSubtotal (a m u).\u201d The first column contains the symbols \u201cC,\u201d \u201cH,\u201d \u201cO,\u201d \u201cN,\u201d and a merged cell. The merged cell runs the width of the first five columns. The second column contains the numbers \u201c2,\u201d \u201c5,\u201d \u201c2,\u201d and \u201c1\u201d as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d and \u201c14.007\u201d as well as the merged cell. The fifth column contains the symbol \u201c=\u201d in each cell except for the last, merged cell. The sixth column contains the values \u201c24.02,\u201d \u201c5.040,\u201d \u201c32.00,\u201d \u201c14.007,\u201d and \u201c75.07.\u201d There is a thick black line under the number 14.007. The merged cell under the first five columns reads \u201cMolar mass (g \/ mol compound). There is a ball-and-stick drawing to the right of this table. It shows a black sphere that forms a double bond with a slightly smaller red sphere, a single bond with another red sphere, and a single bond with another black sphere. The red sphere that forms a single bond with the black sphere also forms a single bond with a smaller, white sphere. The second black sphere forms a single bond with a smaller, white sphere and a smaller blue sphere. The blue sphere forms a single bond with two smaller, white spheres each.\" width=\"879\" height=\"301\" \/>\r\n\r\nThe provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g\/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound\u2019s mass by its molar mass yields:\r\n<p style=\"text-align: center;\">[latex]\\text{28.35 }\\cancel{\\text{g }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}\\left(\\frac{\\text{1 }\\text{mol }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}{\\text{75.07 }\\cancel{\\text{g }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}}\\right)=0.378\\text{ mol }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}[\/latex]<\/p>\r\nThis result is consistent with our rough estimate.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many moles of sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>, are in a 25-g sample of sucrose?\r\n\r\n[reveal-answer q=\"442280\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"442280\"]0.073 mol[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5: <strong>Deriving Grams from Moles for a Compound<\/strong><\/h3>\r\nVitamin C is a covalent compound with the molecular formula C<sub>6<\/sub>H<sub>8<\/sub>O<sub>6<\/sub>. The recommended daily dietary allowance of vitamin C for children aged 4\u20138 years is [latex]1.42\\times {10}^{-4}\\text{mol.}[\/latex] What is the mass of this allowance in grams?\r\n\r\n[reveal-answer q=\"372136\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"372136\"]\r\n\r\nAs for elements, the mass of a compound can be derived from its molar amount as shown:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211135\/CNX_Chem_03_02_vitC_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMoles of vitamin C ( mol )\u201d while the one the right contains the phrase, \u201cMass of vitamin C ( g )\u201d. There is a phrase under the arrow that says \u201cMultiply by molar mass (g \/ mol).\u201d\" width=\"882\" height=\"133\" \/>\r\n\r\nThe molar mass for this compound is computed to be 176.124 g\/mol. The given number of moles is a very small fraction of a mole (~10<sup>-4<\/sup> or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:\r\n<p style=\"text-align: center;\">[latex]1.42\\times {10}^{-4}\\cancel{\\text{mol }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}\\left(\\frac{\\text{176.124 }\\text{g }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}{\\text{1 }\\cancel{\\text{mol }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}}\\right)=\\text{0.0250 }\\text{g }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}[\/latex]<\/p>\r\nThis is consistent with the anticipated result.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nWhat is the mass of 0.443 mol of hydrazine, N<sub>2<\/sub>H<sub>4<\/sub>?\r\n\r\n[reveal-answer q=\"604947\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"604947\"]14.2 g[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6: <strong>Deriving the Number of Atoms and Molecules from the Mass of a Compound<\/strong><\/h3>\r\nA packet of an artificial sweetener contains 0.0400 g of saccharin (C<sub>7<\/sub>H<sub>5<\/sub>NO<sub>3<\/sub>S). (a) Given that saccharin has a molar mass of 183.18 g\/mol, how many saccharin molecules are in a 0.0400-g sample of saccharin?\r\n\r\n[reveal-answer q=\"490490\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"490490\"]\r\n\r\nThe number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Figure 5, and then multiplying by Avogadro\u2019s number:\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211138\/CNX_Chem_03_02_sacch_img1.jpg\" alt=\"A diagram of three boxes connected by a right-facing arrow in between each is shown. The box on the left contains the phrase, \u201cMass of C subscript seven H subscript five N O subscript three S ( g ),\u201d the middle box reads, \u201cMoles of C subscript seven H subscript five N O subscript three S ( mol ),\u201d while the one on the right contains the phrase, \u201cNumber of C subscript seven H subscript five N O subscript three S molecules.\u201d There is a phrase under the left arrow that says, \u201cDivide by molar mass (g \/ mol),\u201d and under the right arrow it states, \u201cMultiply by Avogadro\u2019s number ( mol superscript negative one).\u201d\" width=\"883\" height=\"131\" \/>\r\n\r\nUsing the provided mass and molar mass for saccharin yields:\r\n<p style=\"text-align: center;\">[latex]\\text{0.0400 }\\cancel{\\text{g }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol }}{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}{\\text{183.18 }\\cancel{\\text{g }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{ molecules}{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S }}{\\text{1 }\\cancel{\\text{mol }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}}\\right)=1.31\\times {10}^{20}\\text{molecules }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many C<sub>4<\/sub>H<sub>10<\/sub> molecules are contained in 9.213 g of this compound?\r\n\r\n[reveal-answer q=\"891154\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"891154\"]\r\n\r\n[latex]9.545\\times {10}^{22}\\text{molecules }{\\text{C}}_{4}{\\text{H}}_{10}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Chemical Formulas and the Mole<\/h2>\r\nSuppose we want to know the number of hydrogen atoms found in a dozen CH<sub>4<\/sub> molecule. For each CH<sub>4<\/sub> molecule there are four hydrogen atoms, we determine this from the chemical formula, there is one carbon atom per one CH<sub>4<\/sub> molecule. As shown below, if we have one dozen molecules of CH<sub>4<\/sub> we can determine that we will have a total of 48 hydrogen\u2019s using dimensional analysis.\r\n<p style=\"text-align: center;\">[latex]12\\text{ molecules CH}_{4}\\left(\\frac{\\text{4 }\\text{atoms H}}{\\text{1 molecule CH}_{4}}\\right)=48\\text{ }\\text{atoms H}[\/latex]<\/p>\r\nSince is unlikely a chemist will be working with a single molecule, and much more likely to be working molecules on the scale of Avogadro's number, it makes more sense for us to relate the number of moles of H to the moles of CH<sub>4<\/sub>.\u00a0\u00a0 If we had 12 mole sample of CH<sub>4<\/sub>, how many moles of H atoms will be present in the sample?\u00a0 We can derive a mole to mole relationship between H and CH<sub>4<\/sub> using the chemical formula.\r\n<p style=\"text-align: center;\">[latex]\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 mol CH}_{4}}\\right)[\/latex]<\/p>\r\nTherefore, if we have a 12 mol sample of CH<sub>4<\/sub>, we can calculate the moles of H present in the sample:\r\n<p style=\"text-align: center;\">[latex]\\text{12 }\\cancel{\\text{mol CH}_{4}}\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 }\\cancel{\\text{mol CH}_{4}}}\\right)=48 \\text{ mol H} [\/latex]<\/p>\r\nAll we need to derive a mol to mol relationship between a compound and an element within that compound is the chemical formula.\r\n<div class=\"textbox examples\">\r\n<h3>Example 7: <strong>Converting between moles of a compound and moles of an element within the compound\r\n<\/strong><\/h3>\r\nAn organic compound, commonly known as strawberry aldehyde, is used in the flavor industry in artificial fruit flavors. Given the formula C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, determine the moles of each element within 2.50 moles of strawberry aldehyde?\r\n\r\n[reveal-answer q=\"49041920\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"49041920\"]\r\n\r\nSince we are being asked to determine the number of moles of each element (C, H, and O) within a mole sample of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, we can use a mole to mole ratio from the chemical formula.\r\n\r\nMoles of carbon: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be twelve moles of C.\r\n\r\n[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{12 }\\text{mol C}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=30 \\text{ mol C} [\/latex]\r\n\r\nMoles of hydrogen: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be fourteen moles of H.\r\n\r\n[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{14 }\\text{mol H}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=35 \\text{ mol H} [\/latex]\r\n\r\nMoles of oxygen: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be three moles of O.\r\n\r\n[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{3 }\\text{mol O}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=7.5 \\text{ mol O} [\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many moles of nitrogen are in 10.0 moles of N<sub>2<\/sub>O<sub>4<\/sub>?\r\n\r\n[reveal-answer q=\"8911541\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"8911541\"]\r\n\r\n[latex]\\text{10.0 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}\\left(\\frac{\\text{2 }\\text{mol N}}{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}\\right)=20.0 \\text{ mol N} [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can use mol to mol ratios in combination with other mole conversions, such as Avogadro's number and molar mass.\r\n<div class=\"textbox examples\">\r\n<h3>Example 8: <strong>Converting between grams of a compound and grams of an element within the compound\r\n<\/strong><\/h3>\r\nIsoamyl acetate, C<sub>7<\/sub>H<sub>14<\/sub>O<sub>2<\/sub>, is an organic ester commonly referred to as banana oil due to it strong banana odor.\u00a0 How many grams of carbon will be in a 0.500 g sample of isoamyl acetate?\r\n\r\n[reveal-answer q=\"190419022\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"190419022\"]\r\n\r\nSince we are given grams of isoamyl acetate, we need the molar mass of isoamyl acetate.\u00a0 Using the periodic table we can calculate the molar mass of isoamyl acetate to be 130.18 g\/mol.\u00a0 Since we are looking for grams of carbon, we also need the molar mass of carbon (12.01 g\/mol).\u00a0 We also need to utilize a mole to mole ratio from a chemical formula since we are looking for \"units\" of carbon within \"units\" of the larger isoamyl acetate sample.\u00a0 Using the provided formula, we can derive the mole to mole:\u00a0 [latex]\\left(\\frac{\\text{7 }\\text{mol C}}{\\text{1 }\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}\\right) [\/latex]\r\n\r\n[latex]\\text{0.500 }\\cancel{\\text{g C}_{7}\\text{H}_{14}\\text{O}_{2}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}}{\\text{130.18 }\\cancel{\\text{g C}_{7}\\text{H}_{14}\\text{O}_{2}}}\\right)\\left(\\frac{\\text{7 }\\cancel{\\text{mol C}}}{\\text{1 }\\cancel{\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}}\\right)\\left(\\frac{\\text{12.01 }\\text{g C}}{\\text{1 }\\cancel{\\text{mol C}}}\\right)=0.323 \\text{ g C} [\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many grams of oxygen are in 5.0 g of N<sub>2<\/sub>O<sub>4<\/sub>?\r\n\r\n[reveal-answer q=\"191154199\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"191154199\"]\r\n\r\n[latex]\\text{5.0 }\\cancel{\\text{g N}_{2}\\text{O}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}{\\text{92.02 }\\cancel{\\text{g N}_{2}\\text{O}_{4}}}\\right)\\left(\\frac{\\text{4 }\\cancel{\\text{mol O}}}{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}\\right)\\left(\\frac{\\text{16.00 }\\cancel{\\text{g O}}}{\\text{1 }\\cancel{\\text{mol O}}}\\right)=3.5 \\text{ g O} [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 9: <strong>Converting between grams of a compound and atoms of an element within the compound\r\n<\/strong><\/h3>\r\nCow's and other livestock are responsible for nearly 40% of global methane (CH<sub>4<\/sub>) emissions.\u00a0 A cow produces on average, 260.0 g of methane per day.\u00a0 How many atoms of hydrogen is present in a 260.0 g sample of methane?\r\n\r\n[reveal-answer q=\"190491023\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"190491023\"]\r\n\r\nSince we have grams of methane given, we will need the molar mass of methane (16.04 g\/mol).\u00a0 We are looking for atoms of H, meaning we will need Avogadro's number, and since we are going from \"units\" of methane to \"units\" of carbon, we will need a mole to mole ratio [latex]\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 }\\text{mol C}\\text{H}_{4}}\\right) [\/latex]\r\n\r\n&nbsp;\r\n\r\n[latex]\\text{260.0 }\\cancel{\\text{g C}\\text{H}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}{\\text{16.04 }\\cancel{\\text{g C}\\text{H}_{4}}}\\right)\\left(\\frac{\\text{4 }\\cancel{\\text{mol H}}}{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{atoms H}}{\\text{1 }\\cancel{\\text{mol H}}}\\right)=3.905\\times {10}^{25} \\text{ atoms H} [\/latex]\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/strong><\/h4>\r\nHow many atoms of carbon is present in a 260.0 g sample of methane?\r\n\r\n[reveal-answer q=\"132911541\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"132911541\"]\r\n\r\n[latex]\\text{260.0 }\\cancel{\\text{g C}\\text{H}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}{\\text{16.04 }\\cancel{\\text{g C}\\text{H}_{4}}}\\right)\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}}}{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{atoms C}}{\\text{1 }\\cancel{\\text{mol C}}}\\right)=9.761\\times {10}^{24} \\text{ atoms C} [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nA convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be [latex]6.022\\times {10}^{23}[\/latex], a quantity called Avogadro\u2019s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g\/mol) for any substance are numerically equivalent (for example, one H<sub>2<\/sub>O molecule weighs approximately18 amu and 1 mole of H<sub>2<\/sub>O molecules weighs approximately 18 g).\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.<\/li>\r\n \t<li>Compare 1 mole of H<sub>2<\/sub>, 1 mole of O<sub>2<\/sub>, and 1 mole of F<sub>2<\/sub>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Which has the largest number of molecules? Explain why.<\/li>\r\n \t<li>Which has the greatest mass? Explain why.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the number of indicated particles in the following;\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>number of atoms of F in 1.5 moles of F<\/li>\r\n \t<li>number of atoms of Al in 5.2 mol Al<\/li>\r\n \t<li>number of molecules of CO<sub>2<\/sub> in 0.67 mol of CO<sub>2<\/sub><\/li>\r\n \t<li>number of molecules of C<sub>2<\/sub>H<sub>5<\/sub>OH in 0.0250 mol of C<sub>2<\/sub>H<sub>5<\/sub>OH<\/li>\r\n \t<li>number of formula units of NaCl in 0.050 mol of NaCl<\/li>\r\n \t<li>number of formula units of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in 3.40 mol of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the molar mass of each of the following compounds:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>HF<\/li>\r\n \t<li>NH<sub>3 <\/sub><\/li>\r\n \t<li>HNO<sub>3 <\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4 <\/sub><\/li>\r\n \t<li>B(OH)<sub>3<\/sub><\/li>\r\n \t<li>S<sub>8 <\/sub><\/li>\r\n \t<li>C<sub>5<\/sub>H<sub>12 <\/sub><\/li>\r\n \t<li>Sc<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3 <\/sub><\/li>\r\n \t<li>CH<sub>3<\/sub>COCH<sub>3<\/sub> (acetone)<\/li>\r\n \t<li>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> (glucose)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the number of moles in the following;\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>number of moles of Cu in 5.50 g Cu<\/li>\r\n \t<li>number of moles of S\u00a0in 30.2 g S<\/li>\r\n \t<li>number of moles of CCl<sub>4<\/sub> in 0.250 g CCl<sub>4<\/sub><\/li>\r\n \t<li>number of moles of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> in 100.0 g C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub><\/li>\r\n \t<li>number of moles of Na<sub>2<\/sub>S in 12.0 g Na<sub>2<\/sub>S<\/li>\r\n \t<li>number of moles of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in 20.0 g\u00a0Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C<sub>2<\/sub>H<sub>5<\/sub>OH), 0.60 mol of formic acid (HCO<sub>2<\/sub>H), or 1.0 mol of water (H<sub>2<\/sub>O)? Explain why.<\/li>\r\n \t<li>Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C<sub>2<\/sub>H<sub>5<\/sub>OH), 1 mol of formic acid (HCO<sub>2<\/sub>H), or 1 mol of water (H<sub>2<\/sub>O)? Explain why.<\/li>\r\n \t<li>Determine the number of moles of compound and the number of moles of each type of atom in each of the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>25.0 g of propylene, C<sub>3<\/sub>H<sub>6 <\/sub><\/li>\r\n \t<li>[latex]3.06\\times {10}^{-3}\\text{g}[\/latex] of the amino acid glycine, C<sub>2<\/sub>H<sub>5<\/sub>NO<sub>2 <\/sub><\/li>\r\n \t<li>25 lb of the herbicide Treflan, C<sub>13<\/sub>H<sub>16<\/sub>N<sub>2<\/sub>O<sub>4<\/sub>F (1 lb = 454 g)<\/li>\r\n \t<li>0.125 kg of the insecticide Paris Green, Cu<sub>4<\/sub>(AsO<sub>3<\/sub>)<sub>2<\/sub>(CH<sub>3<\/sub>CO<sub>2<\/sub>)<sub>2 <\/sub><\/li>\r\n \t<li>325 mg of aspirin, C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)(CO<sub>2<\/sub>CH<sub>3<\/sub>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>2.12 g of potassium bromide, KBr<\/li>\r\n \t<li>0.1488 g of phosphoric acid, H<sub>3<\/sub>PO<sub>4 <\/sub><\/li>\r\n \t<li>23 kg of calcium carbonate, CaCO<sub>3 <\/sub><\/li>\r\n \t<li>78.452 g of aluminum sulfate, Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3 <\/sub><\/li>\r\n \t<li>0.1250 mg of caffeine, C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The approximate minimum daily dietary requirement of the amino acid leucine, C<sub>6<\/sub>H<sub>13<\/sub>NO<sub>2<\/sub>, is 1.1 g. What is this requirement in moles?<\/li>\r\n \t<li>Determine the mass in grams of each of the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>0.600 mol of oxygen atoms<\/li>\r\n \t<li>0.600 mol of oxygen molecules, O<sub>2 <\/sub><\/li>\r\n \t<li>0.600 mol of ozone molecules, O<sub>3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A 55-kg woman has [latex]7.5\\times {10}^{-3}\\text{mol}[\/latex] of hemoglobin (molar mass = 64,456 g\/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?<\/li>\r\n \t<li>Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO<sub>4<\/sub>, a semiprecious stone.<\/li>\r\n \t<li>Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH<sub>4<\/sub>, 0.6 mol of C<sub>6<\/sub>H<sub>6<\/sub>, or 0.4 mol of C<sub>3<\/sub>H<sub>8<\/sub>.<\/li>\r\n \t<li>Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO<sub>4<\/sub>, 266 g of Al<sub>2<\/sub>Cl<sub>6<\/sub>, or 225 g of Al<sub>2<\/sub>S<sub>3<\/sub>.<\/li>\r\n \t<li>Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?<\/li>\r\n \t<li>The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone<\/li>\r\n \t<li>One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?<\/li>\r\n \t<li>A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?<\/li>\r\n \t<li>A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na<sub>2<\/sub>PO<sub>3<\/sub>F) in 100 mL\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What mass of fluorine atoms in mg was present?<\/li>\r\n \t<li>How many fluorine atoms were present?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following represents the least number of molecules?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>20.0 g of H<sub>2<\/sub>O (18.02 g\/mol)<\/li>\r\n \t<li>77.0 g of CH<sub>4<\/sub> (16.06 g\/mol)<\/li>\r\n \t<li>68.0 g of CaH<sub>2<\/sub> (42.09 g\/mol)<\/li>\r\n \t<li>100.0 g of N<sub>2<\/sub>O (44.02 g\/mol)<\/li>\r\n \t<li>84.0 g of HF (20.01 g\/mol)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"649796\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"649796\"]\r\n\r\n1. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.\r\n\r\n2. Compare 1 mole of H<sub>2<\/sub>, 1 mole of O<sub>2<\/sub>, and 1 mole of F<sub>2<\/sub>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>They all have the same number of molecules since each is 1 mol.\u00a0 A mol of anything is equal to Avogadro's number.<\/li>\r\n \t<li>Fluorine, F<sub>2<\/sub>, has the greatest mass since it has the greatest molar mass.<\/li>\r\n<\/ol>\r\n3.The number of \"particles\" of each substance is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>atoms of F:\r\n[latex]\\text{1.5 }\\cancel{\\text{mol F}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms F}}{\\text{1 }\\cancel{\\text{mol F}}}\\right)=9.0\\times {10}^{23}\\text{ atoms F}[\/latex]<\/li>\r\n \t<li>atoms of Al:\r\n[latex]\\text{5.2 }\\cancel{\\text{mol Al}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms Al}}{\\text{1 }\\cancel{\\text{mol Al}}}\\right)=3.1\\times {10}^{24}\\text{ atoms Al}[\/latex]<\/li>\r\n \t<li>molecules of CO<sub>2<\/sub>:\r\n[latex]\\text{0.67 }\\cancel{\\text{mol CO}_{2}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{ molecules CO}_{2}}{\\text{1 }\\cancel{\\text{mol CO}_{2}}}\\right)=4.0\\times {10}^{23}\\text{ molecules CO}_{2}[\/latex]<\/li>\r\n \t<li>molecules of C<sub>2<\/sub>H<sub>5<\/sub>OH:\r\n[latex]\\text{0.0250 }\\cancel{\\text{mol C}_{2}\\text{H}_{5}\\text{OH}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{molecules C}_{2}\\text{H}_{5}\\text{OH}}{\\text{1 }\\cancel{\\text{mol C}_{2}\\text{H}_{5}\\text{OH}}}\\right)=1.51\\times {10}^{22}\\text{ molecules C}_{2}\\text{H}_{5}\\text{OH}[\/latex]<\/li>\r\n \t<li>formula units of NaCl:\r\n[latex]\\text{0.050 }\\cancel{\\text{mol NaCl}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{ formula units NaCl}}{\\text{1 }\\cancel{\\text{mol NaCl}}}\\right)=3.0\\times {10}^{22}\\text{ formula units NaCl}[\/latex]<\/li>\r\n \t<li>formula units of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:\r\n[latex]\\text{3.40 }\\cancel{\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{formula units Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}{\\text{1 }\\cancel{\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}}\\right)=2.05\\times {10}^{24}\\text{ formula units Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}[\/latex]<\/li>\r\n<\/ol>\r\n4. The molar mass of each substance is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>HF molar mass = 20.01 g\/mol<\/li>\r\n \t<li>NH<sub>3<\/sub>\u00a0molar mass = 17.03 g\/mol<\/li>\r\n \t<li>HNO<sub>3<\/sub>\u00a0molar mass = 63.02 g\/mol<\/li>\r\n \t<li>\u00a0Ag<sub>2<\/sub>SO<sub>4<\/sub>\u00a0molar mass = 311.87 g\/mol<\/li>\r\n \t<li>B(OH)<sub>3<\/sub>\u00a0molar mass = 61.83 g\/mol<\/li>\r\n \t<li>S<sub>8<\/sub>\u00a0molar mass = 256.56 g\/mol<\/li>\r\n \t<li>C<sub>5<\/sub>H<sub>12<\/sub>\u00a0molar mass = 72.15 g\/mol<\/li>\r\n \t<li>Sc<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>\u00a0\u00a0molar mass = 378.13 g\/mol<\/li>\r\n \t<li>CH<sub>3<\/sub>COCH<sub>3<\/sub> (acetone)\u00a0\u00a0molar mass = 58.08 g\/mol<\/li>\r\n \t<li>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> (glucose)\u00a0\u00a0molar mass = 180.16 g\/mol<\/li>\r\n<\/ol>\r\n5. The moles of each substance is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>moles of Cu:\r\n[latex]\\text{5.50 }\\cancel{\\text{g Cu}}\\left(\\frac{\\text{1 mol Cu}}{\\text{63.55 }\\cancel{\\text{g Cu}}}\\right)=0.0865\\text{ mol Cu}[\/latex]<\/li>\r\n \t<li>moles of S:\r\n[latex]\\text{30.2 }\\cancel{\\text{g S}}\\left(\\frac{\\text{1 mol S}}{\\text{32.07 }\\cancel{\\text{g S}}}\\right)=0.942\\text{ mol S}[\/latex]<\/li>\r\n \t<li>moles of CCl<sub>4<\/sub>:\r\n[latex]\\text{0.250 }\\cancel{\\text{g CCl}_{4}}\\left(\\frac{\\text{1 mol CCl}_{4}}{\\text{153.81 }\\cancel{\\text{g CCl}_{4}}}\\right)=0.00163\\text{ mol CCl}_{4}[\/latex]<\/li>\r\n \t<li>moles of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>:\r\n[latex]\\text{100.0 }\\cancel{\\text{g C}_{12}\\text{H}_{22}\\text{O}_{11}}\\left(\\frac{\\text{1 mol C}_{12}\\text{H}_{22}\\text{O}_{11}}{\\text{342.30 }\\cancel{\\text{g C}_{12}\\text{H}_{22}\\text{O}_{11}}}\\right)=0.2921\\text{ mol C}_{12}\\text{H}_{22}\\text{O}_{11}[\/latex]<\/li>\r\n \t<li>moles of Na<sub>2<\/sub>S:\r\n[latex]\\text{12.0 }\\cancel{\\text{g Na}_{2}\\text{S}}\\left(\\frac{\\text{1 mol Na}_{2}\\text{S}}{\\text{78.05 }\\cancel{\\text{g Na}_{2}\\text{S}}}\\right)=0.154\\text{ mol Na}_{2}\\text{S}[\/latex]<\/li>\r\n \t<li>moles of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:\r\n[latex]\\text{20.0 }\\cancel{\\text{g Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}\\left(\\frac{\\text{1}\\text{ }\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}{\\text{310.18 }\\cancel{\\text{g Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}}\\right)=0.0645 \\text{ mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}[\/latex]<\/li>\r\n<\/ol>\r\n6. Since the molar mass of oxygen is the same for each (16.00 g per mole), the determining factor has to be the moles of each sample and the number of moles of oxygen contained in each.\u00a0 For ethanol, it contains 1 mol of O per 1 mol of ethanol. For formic acid, it contains 2 mol of O per 1 mol of formic acid.\u00a0 For water, it contains 1 mol of O per 1 mol of water.\u00a0 Thus, there are more moles of O per compound for the formic acid compared to the other samples.\u00a0 \u00a0Even though there are more moles of ethanol (0.75 mol) and water (1.0 mol) than formic acid (0.60 mol), the fact that there are 2 mol of O in formic acid means there will be more grams of O present (the larger moles of ethanol and water are not enough to overcome the mol to mol ratio).\u00a0 Confirm this result by performing the calculations (you should get 12 grams of O in ethanol, 19 grams of O in formic acid, and 16.00 grams of O in water).\r\n\r\n7.\r\n\r\n8.\r\n\r\n13. The mass of each compound is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]0.600\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=9.60\\text{g}[\/latex]<\/li>\r\n \t<li>[latex]0.600\\cancel{\\text{mol}}\\times 2\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=19.2\\text{g}[\/latex]<\/li>\r\n \t<li>[latex]0.600\\cancel{\\text{mol}}\\times 3\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=28.8\\text{g}[\/latex]<\/li>\r\n<\/ol>\r\n15. Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times 1023\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 91.224\\text{g\/}\\cancel{\\text{mol}}=30.87\\text{g;}[\/latex] Silicon: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times {10}^{23}\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 28.0855\\text{g\/}\\cancel{\\text{mol}}=9.504\\text{g;}[\/latex] Oxygen: [latex]4\\times 0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=8.151\\times {10}^{23}\\text{atoms;}4\\times 0.3384\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=21.66\\text{g}[\/latex]\r\n\r\n17. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.\r\n<ul>\r\n \t<li>Molar mass AlPO<sub>4<\/sub>: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g\/mol<\/li>\r\n \t<li>Molar mass Al<sub>2<\/sub>Cl<sub>6<\/sub>: 2(26.981539) + 6(35.4527) = 266.6793 g\/mol<\/li>\r\n \t<li>Molar mass Al<sub>2<\/sub>S<sub>3<\/sub>: 2(26.981539) + 3(32.066) = 150.161 g\/mol<\/li>\r\n<\/ul>\r\nAlPO<sub>4<\/sub>: [latex]\\frac{122\\cancel{\\text{g}}}{121.9529\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.000\\text{mol.}[\/latex]\r\n\r\n[latex]\\text{mol Al}=1\\times 1.000\\text{mol}=1.000\\text{mol}[\/latex]\r\n\r\nAl<sub>2<\/sub>Cl<sub>6<\/sub>: [latex]\\frac{266\\text{g}}{266.6793\\text{g}{\\text{mol}}^{-1}}=0.997\\text{mol}[\/latex]\r\n\r\n[latex]\\text{mol Al}=2\\times 0.997\\text{mol}=1.994\\text{mol}[\/latex]\r\n\r\nAl<sub>2<\/sub>S<sub>3<\/sub>: [latex]\\frac{225\\cancel{\\text{g}}}{150.161\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.50\\text{mol}[\/latex]\r\n\r\n[latex]\\text{mol Al}=2\\times 1.50\\text{mol}=3.00\\text{mol}[\/latex]\r\n\r\n19. Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro\u2019s number by the number of moles:\r\n<p style=\"text-align: center;\">[latex]\\frac{3104\\cancel{\\text{carats}}\\times \\frac{200\\cancel{\\text{mg}}}{1\\cancel{\\text{carat}}}\\times \\frac{1\\cancel{\\text{g}}}{1000\\cancel{\\text{mg}}}}{12.011\\cancel{\\text{g}}\\cancel{{\\text{mol}}^{-1}}\\left(6.022\\times {10}^{23}\\cancel{{\\text{mol}}^{-1}}\\right)}=3.113\\times {10}^{25}\\text{C atoms}[\/latex]<\/p>\r\n21. Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g\/mol; Then [latex]0.0278\\text{mol}\\times 342.300\\text{g\/mol}=9.52\\text{g sugar.}[\/latex] This 9.52 g of sugar represents [latex]\\frac{11.0}{60.0}[\/latex] of one serving or\r\n<p style=\"text-align: center;\">[latex]\\frac{60.0\\text{g serving}}{11.0\\cancel{\\text{g sugar}}}\\times 9.52\\cancel{\\text{g sugar}}=51.9\\text{g cereal.}[\/latex]<\/p>\r\nThis amount is [latex]\\frac{51.9\\text{g cereal}}{60.0\\text{g serving}}=0.865[\/latex] servings, or about 1 serving.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>Avogadro\u2019s number (<em>N<sub>A<\/sub><\/em>): <\/strong>experimentally determined value of the number of entities comprising 1 mole of substance, equal to [latex]6.022\\times {10}^{23}{\\text{mol}}^{-1}[\/latex]\r\n\r\n<strong>molar mass: <\/strong>mass in grams of 1 mole of a substance\r\n\r\n<strong>mole: <\/strong>amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of <sup>12<\/sup>C","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define the amount unit mole and the related quantity Avogadro\u2019s number<\/li>\n<li>Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another<\/li>\n<\/ul>\n<\/div>\n<h2>The Mole<\/h2>\n<p>The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H<sub>2<\/sub>O, and hydrogen peroxide, H<sub>2<\/sub>O<sub>2<\/sub>, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the <em>mole<\/em>, which remains indispensable in modern chemical science.<\/p>\n<p>The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of <em>the number<\/em> of atoms or molecules in a bulk sample of matter. A <strong>mole<\/strong> is defined as <em>the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure <sup>12<\/sup>C weighing exactly 12 g.<\/em> One Latin connotation for the word \u201cmole\u201d is \u201clarge mass\u201d or \u201cbulk,\u201d which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.<\/p>\n<p>The number of entities composing a mole has been experimentally determined to be [latex]6.02214179\\times {10}^{23}[\/latex], a fundamental constant named <strong>Avogadro\u2019s number<\/strong> <strong>(<em>N<sub>A<\/sub><\/em>)<\/strong> or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of \u201cper mole,\u201d a conveniently rounded version being [latex]6.022\\times {10}^{23}\\text{\/mol}[\/latex].<\/p>\n<p>Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The<strong> molar mass<\/strong> of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g\/mol) (see Figure 1).<\/p>\n<div style=\"width: 890px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211119\/CNX_Chem_03_02_moles1.jpg\" alt=\"This figure contains eight different substances displayed on white circles. The amount of each substance is visibly different.\" width=\"880\" height=\"582\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Each sample contains 6.022 \u00d7 10<sup>23<\/sup> atoms\u20141.00 mol of atoms. From left to right (top row): 65.4g zinc, 12.0g carbon, 24.3g magnesium, and 63.5g copper. From left to right (bottom row): 32.1g sulfur, 28.1g silicon, 207g lead, and 118.7g tin. (credit: modification of work by Mark Ott)<\/p>\n<\/div>\n<p>Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, <sup>12<\/sup>C, <em>the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu<\/em>. Per the amu definition, a single <sup>12<\/sup>C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of <sup>12<\/sup>C contains 1 mole of <sup>12<\/sup>C atoms (its molar mass is 12 g\/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, <sup>12<\/sup>C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2).<\/p>\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211122\/CNX_Chem_03_02_compound1.jpg\" alt=\"This photo shows two vials filled with a colorless liquid. It also shows two bowls: one filled with an off-white powder and one filled with a bright red powder.\" width=\"500\" height=\"333\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Each sample contains 6.02 \u00d7 10<sup>23<\/sup> molecules or formula units\u20141.00 mol of the compound or element. Clock-wise from the upper left: 130.2g of C<sub>8<\/sub>H<sub>17<\/sub>OH (1-octanol, formula mass 130.2 amu), 454.9g of HgI<sub>2<\/sub> (mercury(II) iodide, formula mass 459.9 amu), 32.0g of CH<sub>3<\/sub>OH (methanol, formula mass 32.0 amu) and 256.5g of S<sub>8<\/sub> (sulfur, formula mass 256.6 amu). (credit: Sahar Atwa)<\/p>\n<\/div>\n<table id=\"fs-idp17650992\" class=\"medium unnumbered\" summary=\"A table is shown that is made up of four columns and six rows. The header row reads: \u201cElement,\u201d \u201cAverage Atomic Mass (a m u),\u201d \u201cMolar Mass (g \/ m o l),\u201d and \u201cAtoms \/ Mole.\u201d The first column contains the symbols \u201cC,\u201d \u201cH,\u201d \u201cO,\u201d \u201cN a,\u201d and \u201cC l.\u201d The second column contains the values \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d \u201c22.99,\u201d and \u201c33.45.\u201d The third column contains the values \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d \u201c22.99,\u201d and \u201c33.45.\u201d The final column contains the value \u201c6.022 times 10 superscript 23\u201d in each cell.\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Average Atomic Mass (amu)<\/th>\n<th>Molar Mass (g\/mol)<\/th>\n<th>Atoms\/Mole<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>C<\/td>\n<td>12.01<\/td>\n<td>12.01<\/td>\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>1.008<\/td>\n<td>1.008<\/td>\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>16.00<\/td>\n<td>16.00<\/td>\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Na<\/td>\n<td>22.99<\/td>\n<td>22.99<\/td>\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Cl<\/td>\n<td>33.45<\/td>\n<td>33.45<\/td>\n<td>[latex]6.022\\times {10}^{23}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"attachment_5625\" style=\"width: 311px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5625\" class=\"wp-image-5625\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/218\/2016\/08\/26220520\/CNX_Chem_03_02_water1-e1472249151518.jpg\" alt=\"A close-up photo of a water droplet on a leaf is shown. The water droplet is not perfectly spherical.\" width=\"301\" height=\"213\" \/><\/p>\n<p id=\"caption-attachment-5625\" class=\"wp-caption-text\">Figure 3. A single drop of water.<\/p>\n<\/div>\n<p>While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 3). The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth.<\/p>\n<p>Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.<\/p>\n<p>The mole is used in chemistry to represent [latex]6.022\\times {10}^{23}[\/latex] of something, but it can be difficult to conceptualize such a large number. Watch this video to learn more.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"How big is a mole? (Not the animal, the other one.) - Daniel Dulek\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TEl4jeETVmg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>The relationships between formula mass, the mole, and Avogadro\u2019s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance\u2019s mass.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1: <strong>Deriving Moles from Grams for an Element<\/strong><\/h3>\n<p>According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q670401\">Show Answer<\/span><\/p>\n<div id=\"q670401\" class=\"hidden-answer\" style=\"display: none\">\n<p>The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g\/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable \u201cballpark\u201d estimate of the number of moles would be slightly greater than 0.1 mol.<\/p>\n<p>The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g\/mol):<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211127\/CNX_Chem_03_02_potassium_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMass of K atoms ( g )\u201d while the one on the right contains the phrase, \u201cMoles of K atoms ( mol ).\u201d There is a phrase under the arrow that says, \u201cDivide by molar mass (g \/ mol).\u201d\" width=\"650\" height=\"147\" \/><\/p>\n<p>The factor-label method supports this mathematical approach since the unit \u201cg\u201d cancels and the answer has units of \u201cmol:\u201d<\/p>\n<p style=\"text-align: center;\">[latex]4.7\\cancel{\\text{ g K}}\\left(\\frac{\\text{1 mol K}}{\\text{39.10 }\\cancel{\\text{g K}}}\\right)=0.12\\text{ mol K}[\/latex]<\/p>\n<p>The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q27674\">Show Answer<\/span><\/p>\n<div id=\"q27674\" class=\"hidden-answer\" style=\"display: none\">0.360 mol<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2: <strong>Deriving Grams from Moles for an Element<\/strong><\/h3>\n<p>A liter of air contains [latex]9.2\\times {10}^{-4}[\/latex] mol argon. What is the mass of Ar in a liter of air?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q363345\">Show Answer<\/span><\/p>\n<div id=\"q363345\" class=\"hidden-answer\" style=\"display: none\">\n<p>The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10<sup>\u20133<\/sup>) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211128\/CNX_Chem_03_02_argon_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMoles of A r atoms ( mol )\u201d while the one on the right contains the phrase, \u201cMass of A r atoms ( g ).\u201d There is a phrase under the arrow that says \u201cMultiply by molar mass ( g \/ mol ).\u201d\" width=\"650\" height=\"147\" \/><\/p>\n<p>In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g\/mol):<\/p>\n<p style=\"text-align: center;\">[latex]9.2\\times {10}^{-4}\\cancel{\\text{mol Ar}}\\left(\\frac{\\text{39.95 }\\text{g Ar}}{\\text{1 }\\cancel{\\text{mol Ar}}}\\right)=\\text{0.037 }\\text{g Ar}[\/latex]<\/p>\n<p>The result is in agreement with our expectations as noted above, around 0.04 g Ar.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What is the mass of 2.561 mol of gold?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q363301\">Show Answer<\/span><\/p>\n<div id=\"q363301\" class=\"hidden-answer\" style=\"display: none\">504.4 g<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3: <strong>Deriving Number of Atoms from Mass for an Element<\/strong><\/h3>\n<p>Copper is commonly used to fabricate electrical wire (Figure 7). How many copper atoms are in 5.00 g of copper wire?<\/p>\n<div style=\"width: 309px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211129\/CNX_Chem_03_02_copper1.jpg\" alt=\"A close-up photo of a spool of copper wire is shown.\" width=\"299\" height=\"398\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q383171\">Show Answer<\/span><\/p>\n<div id=\"q383171\" class=\"hidden-answer\" style=\"display: none\">\n<p>The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro\u2019s number (<em>N<sub>A<\/sub><\/em>) to convert this molar amount to number of Cu atoms:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211131\/CNX_Chem_03_02_copperMoles_img1.jpg\" alt=\"A diagram of three boxes connected by a right-facing arrow in between each is shown. The box on the left contains the phrase, \u201cMass of C u atoms ( g ),\u201d the middle box reads, \u201cMoles of C u atoms ( mol ),\u201d while the one on the right contains the phrase, \u201cNumber of C u atoms.\u201d There is a phrase under the left arrow that says \u201cDivide by molar mass (g \/ mol),\u201d and under the right arrow it states, \u201cMultiply by Avogadro\u2019s number ( mol superscript negative one ).\u201d\" width=\"881\" height=\"127\" \/><\/p>\n<p>Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth <em>N<sub>A<\/sub><\/em>, or approximately 10<sup>22<\/sup> Cu atoms. Carrying out the two-step computation yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{5.00 }\\cancel{\\text{g Cu}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol Cu}}}{\\text{63.55 }\\cancel{\\text{g Cu}}}\\right)\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms Cu}}{\\text{1 }\\cancel{\\text{mol Cu}}}\\right)=4.74\\times {10}^{22}\\text{ atoms Cu}[\/latex]<\/p>\n<p>The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10<sup>22<\/sup> as expected.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q619266\">Show Answer<\/span><\/p>\n<div id=\"q619266\" class=\"hidden-answer\" style=\"display: none\">[latex]4.586\\times {10}^{22}\\text{Au atoms}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4: <strong>Deriving Moles from Grams for a Compound<\/strong><\/h3>\n<p>Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C<sub>2<\/sub>H<sub>5<\/sub>O<sub>2<\/sub>N. How many moles of glycine molecules are contained in 28.35 g of glycine?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129673\">Show Answer<\/span><\/p>\n<div id=\"q129673\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 3:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211132\/CNX_Chem_03_02_glycine_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMass of C subscript 2 H subscript 5 O subscript 2 N ( g )\u201d while the box on the right contains the phrase, \u201cMoles of C subscript 2 H subscript 5 O subscript 2 N ( mol ).\u201d There is a phrase under the arrow that says \u201cDivide by molar mass (g \/ mol).\u201d\" width=\"650\" height=\"147\" \/><\/p>\n<p>The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C<sub>2<\/sub>H<sub>5<\/sub>O<sub>2<\/sub>N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211134\/CNX_Chem_03_01_glycinemass_img1.jpg\" alt=\"A table is shown that is made up of six columns and six rows. The header row reads: \u201cElement,\u201d \u201cQuantity (mol element \/ mol compound,\u201d a blank space, \u201cMolar mass (g \/ mol element),\u201d a blank space, and \u201cSubtotal (a m u).\u201d The first column contains the symbols \u201cC,\u201d \u201cH,\u201d \u201cO,\u201d \u201cN,\u201d and a merged cell. The merged cell runs the width of the first five columns. The second column contains the numbers \u201c2,\u201d \u201c5,\u201d \u201c2,\u201d and \u201c1\u201d as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers \u201c12.01,\u201d \u201c1.008,\u201d \u201c16.00,\u201d and \u201c14.007\u201d as well as the merged cell. The fifth column contains the symbol \u201c=\u201d in each cell except for the last, merged cell. The sixth column contains the values \u201c24.02,\u201d \u201c5.040,\u201d \u201c32.00,\u201d \u201c14.007,\u201d and \u201c75.07.\u201d There is a thick black line under the number 14.007. The merged cell under the first five columns reads \u201cMolar mass (g \/ mol compound). There is a ball-and-stick drawing to the right of this table. It shows a black sphere that forms a double bond with a slightly smaller red sphere, a single bond with another red sphere, and a single bond with another black sphere. The red sphere that forms a single bond with the black sphere also forms a single bond with a smaller, white sphere. The second black sphere forms a single bond with a smaller, white sphere and a smaller blue sphere. The blue sphere forms a single bond with two smaller, white spheres each.\" width=\"879\" height=\"301\" \/><\/p>\n<p>The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g\/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound\u2019s mass by its molar mass yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{28.35 }\\cancel{\\text{g }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}\\left(\\frac{\\text{1 }\\text{mol }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}{\\text{75.07 }\\cancel{\\text{g }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}}}\\right)=0.378\\text{ mol }\\text{C}_{2}\\text{H}_{5}\\text{O}_{2}\\text{N}[\/latex]<\/p>\n<p>This result is consistent with our rough estimate.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many moles of sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>, are in a 25-g sample of sucrose?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q442280\">Show Answer<\/span><\/p>\n<div id=\"q442280\" class=\"hidden-answer\" style=\"display: none\">0.073 mol<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5: <strong>Deriving Grams from Moles for a Compound<\/strong><\/h3>\n<p>Vitamin C is a covalent compound with the molecular formula C<sub>6<\/sub>H<sub>8<\/sub>O<sub>6<\/sub>. The recommended daily dietary allowance of vitamin C for children aged 4\u20138 years is [latex]1.42\\times {10}^{-4}\\text{mol.}[\/latex] What is the mass of this allowance in grams?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q372136\">Show Answer<\/span><\/p>\n<div id=\"q372136\" class=\"hidden-answer\" style=\"display: none\">\n<p>As for elements, the mass of a compound can be derived from its molar amount as shown:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211135\/CNX_Chem_03_02_vitC_img1.jpg\" alt=\"A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, \u201cMoles of vitamin C ( mol )\u201d while the one the right contains the phrase, \u201cMass of vitamin C ( g )\u201d. There is a phrase under the arrow that says \u201cMultiply by molar mass (g \/ mol).\u201d\" width=\"882\" height=\"133\" \/><\/p>\n<p>The molar mass for this compound is computed to be 176.124 g\/mol. The given number of moles is a very small fraction of a mole (~10<sup>-4<\/sup> or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:<\/p>\n<p style=\"text-align: center;\">[latex]1.42\\times {10}^{-4}\\cancel{\\text{mol }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}\\left(\\frac{\\text{176.124 }\\text{g }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}{\\text{1 }\\cancel{\\text{mol }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}}}\\right)=\\text{0.0250 }\\text{g }\\text{C}_{6}\\text{H}_{8}\\text{O}_{6}[\/latex]<\/p>\n<p>This is consistent with the anticipated result.<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>What is the mass of 0.443 mol of hydrazine, N<sub>2<\/sub>H<sub>4<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604947\">Show Answer<\/span><\/p>\n<div id=\"q604947\" class=\"hidden-answer\" style=\"display: none\">14.2 g<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6: <strong>Deriving the Number of Atoms and Molecules from the Mass of a Compound<\/strong><\/h3>\n<p>A packet of an artificial sweetener contains 0.0400 g of saccharin (C<sub>7<\/sub>H<sub>5<\/sub>NO<sub>3<\/sub>S). (a) Given that saccharin has a molar mass of 183.18 g\/mol, how many saccharin molecules are in a 0.0400-g sample of saccharin?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q490490\">Show Answer<\/span><\/p>\n<div id=\"q490490\" class=\"hidden-answer\" style=\"display: none\">\n<p>The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Figure 5, and then multiplying by Avogadro\u2019s number:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211138\/CNX_Chem_03_02_sacch_img1.jpg\" alt=\"A diagram of three boxes connected by a right-facing arrow in between each is shown. The box on the left contains the phrase, \u201cMass of C subscript seven H subscript five N O subscript three S ( g ),\u201d the middle box reads, \u201cMoles of C subscript seven H subscript five N O subscript three S ( mol ),\u201d while the one on the right contains the phrase, \u201cNumber of C subscript seven H subscript five N O subscript three S molecules.\u201d There is a phrase under the left arrow that says, \u201cDivide by molar mass (g \/ mol),\u201d and under the right arrow it states, \u201cMultiply by Avogadro\u2019s number ( mol superscript negative one).\u201d\" width=\"883\" height=\"131\" \/><\/p>\n<p>Using the provided mass and molar mass for saccharin yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{0.0400 }\\cancel{\\text{g }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol }}{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}{\\text{183.18 }\\cancel{\\text{g }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{ molecules}{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S }}{\\text{1 }\\cancel{\\text{mol }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}}}\\right)=1.31\\times {10}^{20}\\text{molecules }{\\text{C}}_{7}{\\text{H}}_{5}{\\text{NO}}_{3}\\text{S}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many C<sub>4<\/sub>H<sub>10<\/sub> molecules are contained in 9.213 g of this compound?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q891154\">Show Answer<\/span><\/p>\n<div id=\"q891154\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]9.545\\times {10}^{22}\\text{molecules }{\\text{C}}_{4}{\\text{H}}_{10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Chemical Formulas and the Mole<\/h2>\n<p>Suppose we want to know the number of hydrogen atoms found in a dozen CH<sub>4<\/sub> molecule. For each CH<sub>4<\/sub> molecule there are four hydrogen atoms, we determine this from the chemical formula, there is one carbon atom per one CH<sub>4<\/sub> molecule. As shown below, if we have one dozen molecules of CH<sub>4<\/sub> we can determine that we will have a total of 48 hydrogen\u2019s using dimensional analysis.<\/p>\n<p style=\"text-align: center;\">[latex]12\\text{ molecules CH}_{4}\\left(\\frac{\\text{4 }\\text{atoms H}}{\\text{1 molecule CH}_{4}}\\right)=48\\text{ }\\text{atoms H}[\/latex]<\/p>\n<p>Since is unlikely a chemist will be working with a single molecule, and much more likely to be working molecules on the scale of Avogadro&#8217;s number, it makes more sense for us to relate the number of moles of H to the moles of CH<sub>4<\/sub>.\u00a0\u00a0 If we had 12 mole sample of CH<sub>4<\/sub>, how many moles of H atoms will be present in the sample?\u00a0 We can derive a mole to mole relationship between H and CH<sub>4<\/sub> using the chemical formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 mol CH}_{4}}\\right)[\/latex]<\/p>\n<p>Therefore, if we have a 12 mol sample of CH<sub>4<\/sub>, we can calculate the moles of H present in the sample:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{12 }\\cancel{\\text{mol CH}_{4}}\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 }\\cancel{\\text{mol CH}_{4}}}\\right)=48 \\text{ mol H}[\/latex]<\/p>\n<p>All we need to derive a mol to mol relationship between a compound and an element within that compound is the chemical formula.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7: <strong>Converting between moles of a compound and moles of an element within the compound<br \/>\n<\/strong><\/h3>\n<p>An organic compound, commonly known as strawberry aldehyde, is used in the flavor industry in artificial fruit flavors. Given the formula C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, determine the moles of each element within 2.50 moles of strawberry aldehyde?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q49041920\">Show Answer<\/span><\/p>\n<div id=\"q49041920\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since we are being asked to determine the number of moles of each element (C, H, and O) within a mole sample of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, we can use a mole to mole ratio from the chemical formula.<\/p>\n<p>Moles of carbon: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be twelve moles of C.<\/p>\n<p>[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{12 }\\text{mol C}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=30 \\text{ mol C}[\/latex]<\/p>\n<p>Moles of hydrogen: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be fourteen moles of H.<\/p>\n<p>[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{14 }\\text{mol H}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=35 \\text{ mol H}[\/latex]<\/p>\n<p>Moles of oxygen: For one mole of C<sub>12<\/sub>H<sub>14<\/sub>O<sub>3<\/sub>, there will be three moles of O.<\/p>\n<p>[latex]\\text{2.50 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}\\left(\\frac{\\text{3 }\\text{mol O}}{\\text{1 }\\cancel{\\text{mol C}_{12}\\text{H}_{14}\\text{O}_{3}}}\\right)=7.5 \\text{ mol O}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many moles of nitrogen are in 10.0 moles of N<sub>2<\/sub>O<sub>4<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8911541\">Show Answer<\/span><\/p>\n<div id=\"q8911541\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{10.0 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}\\left(\\frac{\\text{2 }\\text{mol N}}{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}\\right)=20.0 \\text{ mol N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can use mol to mol ratios in combination with other mole conversions, such as Avogadro&#8217;s number and molar mass.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 8: <strong>Converting between grams of a compound and grams of an element within the compound<br \/>\n<\/strong><\/h3>\n<p>Isoamyl acetate, C<sub>7<\/sub>H<sub>14<\/sub>O<sub>2<\/sub>, is an organic ester commonly referred to as banana oil due to it strong banana odor.\u00a0 How many grams of carbon will be in a 0.500 g sample of isoamyl acetate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190419022\">Show Answer<\/span><\/p>\n<div id=\"q190419022\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since we are given grams of isoamyl acetate, we need the molar mass of isoamyl acetate.\u00a0 Using the periodic table we can calculate the molar mass of isoamyl acetate to be 130.18 g\/mol.\u00a0 Since we are looking for grams of carbon, we also need the molar mass of carbon (12.01 g\/mol).\u00a0 We also need to utilize a mole to mole ratio from a chemical formula since we are looking for &#8220;units&#8221; of carbon within &#8220;units&#8221; of the larger isoamyl acetate sample.\u00a0 Using the provided formula, we can derive the mole to mole:\u00a0 [latex]\\left(\\frac{\\text{7 }\\text{mol C}}{\\text{1 }\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}\\right)[\/latex]<\/p>\n<p>[latex]\\text{0.500 }\\cancel{\\text{g C}_{7}\\text{H}_{14}\\text{O}_{2}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}}{\\text{130.18 }\\cancel{\\text{g C}_{7}\\text{H}_{14}\\text{O}_{2}}}\\right)\\left(\\frac{\\text{7 }\\cancel{\\text{mol C}}}{\\text{1 }\\cancel{\\text{mol C}_{7}\\text{H}_{14}\\text{O}_{2}}}\\right)\\left(\\frac{\\text{12.01 }\\text{g C}}{\\text{1 }\\cancel{\\text{mol C}}}\\right)=0.323 \\text{ g C}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many grams of oxygen are in 5.0 g of N<sub>2<\/sub>O<sub>4<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q191154199\">Show Answer<\/span><\/p>\n<div id=\"q191154199\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{5.0 }\\cancel{\\text{g N}_{2}\\text{O}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}{\\text{92.02 }\\cancel{\\text{g N}_{2}\\text{O}_{4}}}\\right)\\left(\\frac{\\text{4 }\\cancel{\\text{mol O}}}{\\text{1 }\\cancel{\\text{mol N}_{2}\\text{O}_{4}}}\\right)\\left(\\frac{\\text{16.00 }\\cancel{\\text{g O}}}{\\text{1 }\\cancel{\\text{mol O}}}\\right)=3.5 \\text{ g O}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 9: <strong>Converting between grams of a compound and atoms of an element within the compound<br \/>\n<\/strong><\/h3>\n<p>Cow&#8217;s and other livestock are responsible for nearly 40% of global methane (CH<sub>4<\/sub>) emissions.\u00a0 A cow produces on average, 260.0 g of methane per day.\u00a0 How many atoms of hydrogen is present in a 260.0 g sample of methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190491023\">Show Answer<\/span><\/p>\n<div id=\"q190491023\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since we have grams of methane given, we will need the molar mass of methane (16.04 g\/mol).\u00a0 We are looking for atoms of H, meaning we will need Avogadro&#8217;s number, and since we are going from &#8220;units&#8221; of methane to &#8220;units&#8221; of carbon, we will need a mole to mole ratio [latex]\\left(\\frac{\\text{4 }\\text{mol H}}{\\text{1 }\\text{mol C}\\text{H}_{4}}\\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\text{260.0 }\\cancel{\\text{g C}\\text{H}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}{\\text{16.04 }\\cancel{\\text{g C}\\text{H}_{4}}}\\right)\\left(\\frac{\\text{4 }\\cancel{\\text{mol H}}}{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{atoms H}}{\\text{1 }\\cancel{\\text{mol H}}}\\right)=3.905\\times {10}^{25} \\text{ atoms H}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/strong><\/h4>\n<p>How many atoms of carbon is present in a 260.0 g sample of methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q132911541\">Show Answer<\/span><\/p>\n<div id=\"q132911541\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{260.0 }\\cancel{\\text{g C}\\text{H}_{4}}\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}{\\text{16.04 }\\cancel{\\text{g C}\\text{H}_{4}}}\\right)\\left(\\frac{\\text{1 }\\cancel{\\text{mol C}}}{\\text{1 }\\cancel{\\text{mol C}\\text{H}_{4}}}\\right)\\left(\\frac{6.022\\times {10}^{23}\\text{atoms C}}{\\text{1 }\\cancel{\\text{mol C}}}\\right)=9.761\\times {10}^{24} \\text{ atoms C}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be [latex]6.022\\times {10}^{23}[\/latex], a quantity called Avogadro\u2019s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g\/mol) for any substance are numerically equivalent (for example, one H<sub>2<\/sub>O molecule weighs approximately18 amu and 1 mole of H<sub>2<\/sub>O molecules weighs approximately 18 g).<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.<\/li>\n<li>Compare 1 mole of H<sub>2<\/sub>, 1 mole of O<sub>2<\/sub>, and 1 mole of F<sub>2<\/sub>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Which has the largest number of molecules? Explain why.<\/li>\n<li>Which has the greatest mass? Explain why.<\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of indicated particles in the following;\n<ol style=\"list-style-type: lower-alpha;\">\n<li>number of atoms of F in 1.5 moles of F<\/li>\n<li>number of atoms of Al in 5.2 mol Al<\/li>\n<li>number of molecules of CO<sub>2<\/sub> in 0.67 mol of CO<sub>2<\/sub><\/li>\n<li>number of molecules of C<sub>2<\/sub>H<sub>5<\/sub>OH in 0.0250 mol of C<sub>2<\/sub>H<sub>5<\/sub>OH<\/li>\n<li>number of formula units of NaCl in 0.050 mol of NaCl<\/li>\n<li>number of formula units of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in 3.40 mol of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molar mass of each of the following compounds:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>HF<\/li>\n<li>NH<sub>3 <\/sub><\/li>\n<li>HNO<sub>3 <\/sub><\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4 <\/sub><\/li>\n<li>B(OH)<sub>3<\/sub><\/li>\n<li>S<sub>8 <\/sub><\/li>\n<li>C<sub>5<\/sub>H<sub>12 <\/sub><\/li>\n<li>Sc<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3 <\/sub><\/li>\n<li>CH<sub>3<\/sub>COCH<sub>3<\/sub> (acetone)<\/li>\n<li>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> (glucose)<\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles in the following;\n<ol style=\"list-style-type: lower-alpha;\">\n<li>number of moles of Cu in 5.50 g Cu<\/li>\n<li>number of moles of S\u00a0in 30.2 g S<\/li>\n<li>number of moles of CCl<sub>4<\/sub> in 0.250 g CCl<sub>4<\/sub><\/li>\n<li>number of moles of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> in 100.0 g C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub><\/li>\n<li>number of moles of Na<sub>2<\/sub>S in 12.0 g Na<sub>2<\/sub>S<\/li>\n<li>number of moles of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in 20.0 g\u00a0Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C<sub>2<\/sub>H<sub>5<\/sub>OH), 0.60 mol of formic acid (HCO<sub>2<\/sub>H), or 1.0 mol of water (H<sub>2<\/sub>O)? Explain why.<\/li>\n<li>Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C<sub>2<\/sub>H<sub>5<\/sub>OH), 1 mol of formic acid (HCO<sub>2<\/sub>H), or 1 mol of water (H<sub>2<\/sub>O)? Explain why.<\/li>\n<li>Determine the number of moles of compound and the number of moles of each type of atom in each of the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>25.0 g of propylene, C<sub>3<\/sub>H<sub>6 <\/sub><\/li>\n<li>[latex]3.06\\times {10}^{-3}\\text{g}[\/latex] of the amino acid glycine, C<sub>2<\/sub>H<sub>5<\/sub>NO<sub>2 <\/sub><\/li>\n<li>25 lb of the herbicide Treflan, C<sub>13<\/sub>H<sub>16<\/sub>N<sub>2<\/sub>O<sub>4<\/sub>F (1 lb = 454 g)<\/li>\n<li>0.125 kg of the insecticide Paris Green, Cu<sub>4<\/sub>(AsO<sub>3<\/sub>)<sub>2<\/sub>(CH<sub>3<\/sub>CO<sub>2<\/sub>)<sub>2 <\/sub><\/li>\n<li>325 mg of aspirin, C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)(CO<sub>2<\/sub>CH<sub>3<\/sub>)<\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>2.12 g of potassium bromide, KBr<\/li>\n<li>0.1488 g of phosphoric acid, H<sub>3<\/sub>PO<sub>4 <\/sub><\/li>\n<li>23 kg of calcium carbonate, CaCO<sub>3 <\/sub><\/li>\n<li>78.452 g of aluminum sulfate, Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3 <\/sub><\/li>\n<li>0.1250 mg of caffeine, C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>The approximate minimum daily dietary requirement of the amino acid leucine, C<sub>6<\/sub>H<sub>13<\/sub>NO<sub>2<\/sub>, is 1.1 g. What is this requirement in moles?<\/li>\n<li>Determine the mass in grams of each of the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>0.600 mol of oxygen atoms<\/li>\n<li>0.600 mol of oxygen molecules, O<sub>2 <\/sub><\/li>\n<li>0.600 mol of ozone molecules, O<sub>3<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>A 55-kg woman has [latex]7.5\\times {10}^{-3}\\text{mol}[\/latex] of hemoglobin (molar mass = 64,456 g\/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?<\/li>\n<li>Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO<sub>4<\/sub>, a semiprecious stone.<\/li>\n<li>Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH<sub>4<\/sub>, 0.6 mol of C<sub>6<\/sub>H<sub>6<\/sub>, or 0.4 mol of C<sub>3<\/sub>H<sub>8<\/sub>.<\/li>\n<li>Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO<sub>4<\/sub>, 266 g of Al<sub>2<\/sub>Cl<sub>6<\/sub>, or 225 g of Al<sub>2<\/sub>S<sub>3<\/sub>.<\/li>\n<li>Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?<\/li>\n<li>The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone<\/li>\n<li>One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?<\/li>\n<li>A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?<\/li>\n<li>A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na<sub>2<\/sub>PO<sub>3<\/sub>F) in 100 mL\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What mass of fluorine atoms in mg was present?<\/li>\n<li>How many fluorine atoms were present?<\/li>\n<\/ol>\n<\/li>\n<li>Which of the following represents the least number of molecules?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>20.0 g of H<sub>2<\/sub>O (18.02 g\/mol)<\/li>\n<li>77.0 g of CH<sub>4<\/sub> (16.06 g\/mol)<\/li>\n<li>68.0 g of CaH<sub>2<\/sub> (42.09 g\/mol)<\/li>\n<li>100.0 g of N<sub>2<\/sub>O (44.02 g\/mol)<\/li>\n<li>84.0 g of HF (20.01 g\/mol)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q649796\">Show Selected Answers<\/span><\/p>\n<div id=\"q649796\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.<\/p>\n<p>2. Compare 1 mole of H<sub>2<\/sub>, 1 mole of O<sub>2<\/sub>, and 1 mole of F<sub>2<\/sub>.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>They all have the same number of molecules since each is 1 mol.\u00a0 A mol of anything is equal to Avogadro&#8217;s number.<\/li>\n<li>Fluorine, F<sub>2<\/sub>, has the greatest mass since it has the greatest molar mass.<\/li>\n<\/ol>\n<p>3.The number of &#8220;particles&#8221; of each substance is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>atoms of F:<br \/>\n[latex]\\text{1.5 }\\cancel{\\text{mol F}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms F}}{\\text{1 }\\cancel{\\text{mol F}}}\\right)=9.0\\times {10}^{23}\\text{ atoms F}[\/latex]<\/li>\n<li>atoms of Al:<br \/>\n[latex]\\text{5.2 }\\cancel{\\text{mol Al}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{atoms Al}}{\\text{1 }\\cancel{\\text{mol Al}}}\\right)=3.1\\times {10}^{24}\\text{ atoms Al}[\/latex]<\/li>\n<li>molecules of CO<sub>2<\/sub>:<br \/>\n[latex]\\text{0.67 }\\cancel{\\text{mol CO}_{2}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{ molecules CO}_{2}}{\\text{1 }\\cancel{\\text{mol CO}_{2}}}\\right)=4.0\\times {10}^{23}\\text{ molecules CO}_{2}[\/latex]<\/li>\n<li>molecules of C<sub>2<\/sub>H<sub>5<\/sub>OH:<br \/>\n[latex]\\text{0.0250 }\\cancel{\\text{mol C}_{2}\\text{H}_{5}\\text{OH}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{molecules C}_{2}\\text{H}_{5}\\text{OH}}{\\text{1 }\\cancel{\\text{mol C}_{2}\\text{H}_{5}\\text{OH}}}\\right)=1.51\\times {10}^{22}\\text{ molecules C}_{2}\\text{H}_{5}\\text{OH}[\/latex]<\/li>\n<li>formula units of NaCl:<br \/>\n[latex]\\text{0.050 }\\cancel{\\text{mol NaCl}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{ formula units NaCl}}{\\text{1 }\\cancel{\\text{mol NaCl}}}\\right)=3.0\\times {10}^{22}\\text{ formula units NaCl}[\/latex]<\/li>\n<li>formula units of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:<br \/>\n[latex]\\text{3.40 }\\cancel{\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}\\left(\\frac{\\text{6.022}\\times{10}^{23}\\text{ }\\text{formula units Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}{\\text{1 }\\cancel{\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}}\\right)=2.05\\times {10}^{24}\\text{ formula units Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}[\/latex]<\/li>\n<\/ol>\n<p>4. The molar mass of each substance is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>HF molar mass = 20.01 g\/mol<\/li>\n<li>NH<sub>3<\/sub>\u00a0molar mass = 17.03 g\/mol<\/li>\n<li>HNO<sub>3<\/sub>\u00a0molar mass = 63.02 g\/mol<\/li>\n<li>\u00a0Ag<sub>2<\/sub>SO<sub>4<\/sub>\u00a0molar mass = 311.87 g\/mol<\/li>\n<li>B(OH)<sub>3<\/sub>\u00a0molar mass = 61.83 g\/mol<\/li>\n<li>S<sub>8<\/sub>\u00a0molar mass = 256.56 g\/mol<\/li>\n<li>C<sub>5<\/sub>H<sub>12<\/sub>\u00a0molar mass = 72.15 g\/mol<\/li>\n<li>Sc<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>\u00a0\u00a0molar mass = 378.13 g\/mol<\/li>\n<li>CH<sub>3<\/sub>COCH<sub>3<\/sub> (acetone)\u00a0\u00a0molar mass = 58.08 g\/mol<\/li>\n<li>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> (glucose)\u00a0\u00a0molar mass = 180.16 g\/mol<\/li>\n<\/ol>\n<p>5. The moles of each substance is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>moles of Cu:<br \/>\n[latex]\\text{5.50 }\\cancel{\\text{g Cu}}\\left(\\frac{\\text{1 mol Cu}}{\\text{63.55 }\\cancel{\\text{g Cu}}}\\right)=0.0865\\text{ mol Cu}[\/latex]<\/li>\n<li>moles of S:<br \/>\n[latex]\\text{30.2 }\\cancel{\\text{g S}}\\left(\\frac{\\text{1 mol S}}{\\text{32.07 }\\cancel{\\text{g S}}}\\right)=0.942\\text{ mol S}[\/latex]<\/li>\n<li>moles of CCl<sub>4<\/sub>:<br \/>\n[latex]\\text{0.250 }\\cancel{\\text{g CCl}_{4}}\\left(\\frac{\\text{1 mol CCl}_{4}}{\\text{153.81 }\\cancel{\\text{g CCl}_{4}}}\\right)=0.00163\\text{ mol CCl}_{4}[\/latex]<\/li>\n<li>moles of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>:<br \/>\n[latex]\\text{100.0 }\\cancel{\\text{g C}_{12}\\text{H}_{22}\\text{O}_{11}}\\left(\\frac{\\text{1 mol C}_{12}\\text{H}_{22}\\text{O}_{11}}{\\text{342.30 }\\cancel{\\text{g C}_{12}\\text{H}_{22}\\text{O}_{11}}}\\right)=0.2921\\text{ mol C}_{12}\\text{H}_{22}\\text{O}_{11}[\/latex]<\/li>\n<li>moles of Na<sub>2<\/sub>S:<br \/>\n[latex]\\text{12.0 }\\cancel{\\text{g Na}_{2}\\text{S}}\\left(\\frac{\\text{1 mol Na}_{2}\\text{S}}{\\text{78.05 }\\cancel{\\text{g Na}_{2}\\text{S}}}\\right)=0.154\\text{ mol Na}_{2}\\text{S}[\/latex]<\/li>\n<li>moles of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:<br \/>\n[latex]\\text{20.0 }\\cancel{\\text{g Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}\\left(\\frac{\\text{1}\\text{ }\\text{mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}{\\text{310.18 }\\cancel{\\text{g Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}}}\\right)=0.0645 \\text{ mol Ca}_{3}\\left(\\text{PO}_{4}\\right )\\text{}_{2}[\/latex]<\/li>\n<\/ol>\n<p>6. Since the molar mass of oxygen is the same for each (16.00 g per mole), the determining factor has to be the moles of each sample and the number of moles of oxygen contained in each.\u00a0 For ethanol, it contains 1 mol of O per 1 mol of ethanol. For formic acid, it contains 2 mol of O per 1 mol of formic acid.\u00a0 For water, it contains 1 mol of O per 1 mol of water.\u00a0 Thus, there are more moles of O per compound for the formic acid compared to the other samples.\u00a0 \u00a0Even though there are more moles of ethanol (0.75 mol) and water (1.0 mol) than formic acid (0.60 mol), the fact that there are 2 mol of O in formic acid means there will be more grams of O present (the larger moles of ethanol and water are not enough to overcome the mol to mol ratio).\u00a0 Confirm this result by performing the calculations (you should get 12 grams of O in ethanol, 19 grams of O in formic acid, and 16.00 grams of O in water).<\/p>\n<p>7.<\/p>\n<p>8.<\/p>\n<p>13. The mass of each compound is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]0.600\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=9.60\\text{g}[\/latex]<\/li>\n<li>[latex]0.600\\cancel{\\text{mol}}\\times 2\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=19.2\\text{g}[\/latex]<\/li>\n<li>[latex]0.600\\cancel{\\text{mol}}\\times 3\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=28.8\\text{g}[\/latex]<\/li>\n<\/ol>\n<p>15. Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times 1023\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 91.224\\text{g\/}\\cancel{\\text{mol}}=30.87\\text{g;}[\/latex] Silicon: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times {10}^{23}\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 28.0855\\text{g\/}\\cancel{\\text{mol}}=9.504\\text{g;}[\/latex] Oxygen: [latex]4\\times 0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=8.151\\times {10}^{23}\\text{atoms;}4\\times 0.3384\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=21.66\\text{g}[\/latex]<\/p>\n<p>17. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.<\/p>\n<ul>\n<li>Molar mass AlPO<sub>4<\/sub>: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g\/mol<\/li>\n<li>Molar mass Al<sub>2<\/sub>Cl<sub>6<\/sub>: 2(26.981539) + 6(35.4527) = 266.6793 g\/mol<\/li>\n<li>Molar mass Al<sub>2<\/sub>S<sub>3<\/sub>: 2(26.981539) + 3(32.066) = 150.161 g\/mol<\/li>\n<\/ul>\n<p>AlPO<sub>4<\/sub>: [latex]\\frac{122\\cancel{\\text{g}}}{121.9529\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.000\\text{mol.}[\/latex]<\/p>\n<p>[latex]\\text{mol Al}=1\\times 1.000\\text{mol}=1.000\\text{mol}[\/latex]<\/p>\n<p>Al<sub>2<\/sub>Cl<sub>6<\/sub>: [latex]\\frac{266\\text{g}}{266.6793\\text{g}{\\text{mol}}^{-1}}=0.997\\text{mol}[\/latex]<\/p>\n<p>[latex]\\text{mol Al}=2\\times 0.997\\text{mol}=1.994\\text{mol}[\/latex]<\/p>\n<p>Al<sub>2<\/sub>S<sub>3<\/sub>: [latex]\\frac{225\\cancel{\\text{g}}}{150.161\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.50\\text{mol}[\/latex]<\/p>\n<p>[latex]\\text{mol Al}=2\\times 1.50\\text{mol}=3.00\\text{mol}[\/latex]<\/p>\n<p>19. Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro\u2019s number by the number of moles:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{3104\\cancel{\\text{carats}}\\times \\frac{200\\cancel{\\text{mg}}}{1\\cancel{\\text{carat}}}\\times \\frac{1\\cancel{\\text{g}}}{1000\\cancel{\\text{mg}}}}{12.011\\cancel{\\text{g}}\\cancel{{\\text{mol}}^{-1}}\\left(6.022\\times {10}^{23}\\cancel{{\\text{mol}}^{-1}}\\right)}=3.113\\times {10}^{25}\\text{C atoms}[\/latex]<\/p>\n<p>21. Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g\/mol; Then [latex]0.0278\\text{mol}\\times 342.300\\text{g\/mol}=9.52\\text{g sugar.}[\/latex] This 9.52 g of sugar represents [latex]\\frac{11.0}{60.0}[\/latex] of one serving or<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{60.0\\text{g serving}}{11.0\\cancel{\\text{g sugar}}}\\times 9.52\\cancel{\\text{g sugar}}=51.9\\text{g cereal.}[\/latex]<\/p>\n<p>This amount is [latex]\\frac{51.9\\text{g cereal}}{60.0\\text{g serving}}=0.865[\/latex] servings, or about 1 serving.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>Avogadro\u2019s number (<em>N<sub>A<\/sub><\/em>): <\/strong>experimentally determined value of the number of entities comprising 1 mole of substance, equal to [latex]6.022\\times {10}^{23}{\\text{mol}}^{-1}[\/latex]<\/p>\n<p><strong>molar mass: <\/strong>mass in grams of 1 mole of a substance<\/p>\n<p><strong>mole: <\/strong>amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of <sup>12<\/sup>C<\/p>\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-1381\">\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. <strong>Provided by<\/strong>: OpenStaxCollege. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><li>Water drop on a leaf. <strong>Authored by<\/strong>: tanakawho. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Water_drop_on_a_leaf.jpg\">https:\/\/commons.wikimedia.org\/wiki\/File:Water_drop_on_a_leaf.jpg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>How big is a mole? (Not the animal, the other one.) - Daniel Dulek. <strong>Authored by<\/strong>: TED-Ed. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/TEl4jeETVmg\">https:\/\/youtu.be\/TEl4jeETVmg<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":6181,"menu_order":1,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"How big is a mole? (Not the animal, the other one.) - Daniel Dulek\",\"author\":\"TED-Ed\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/TEl4jeETVmg\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStaxCollege\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"},{\"type\":\"cc\",\"description\":\"Water drop on a leaf\",\"author\":\"tanakawho\",\"organization\":\"\",\"url\":\"https:\/\/commons.wikimedia.org\/wiki\/File:Water_drop_on_a_leaf.jpg\",\"project\":\"\",\"license\":\"cc-by\",\"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-1381","chapter","type-chapter","status-publish","hentry"],"part":114,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1381","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":36,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1381\/revisions"}],"predecessor-version":[{"id":1977,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1381\/revisions\/1977"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/114"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1381\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1381"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1381"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1381"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1381"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}