{"id":615,"date":"2019-05-16T17:51:46","date_gmt":"2019-05-16T17:51:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio\/chapter\/determining-empirical-and-molecular-formulas\/"},"modified":"2021-02-15T22:50:10","modified_gmt":"2021-02-15T22:50:10","slug":"determining-empirical-and-molecular-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/chapter\/determining-empirical-and-molecular-formulas\/","title":{"raw":"Determining Empirical and Molecular Formulas","rendered":"Determining Empirical and Molecular Formulas"},"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>Compute the percent composition of a compound<\/li>\r\n \t<li>Determine the empirical formula of a compound<\/li>\r\n \t<li>Determine the molecular formula of a compound<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.\r\n<h2>Percent Composition<\/h2>\r\nThe elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound\u2019s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound\u2019s <strong>percent composition<\/strong>, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:\r\n<p style=\"text-align: center\">[latex]\\%\\text{H}=\\frac{\\text{mass H}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\%\\text{C}=\\frac{\\text{mass C}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/p>\r\nIf analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:\r\n<p style=\"text-align: center\">[latex]\\%\\text{H}=\\frac{2.5\\text{g H}}{10.0\\text{g compound}}\\times 100\\%=25\\%[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\%\\text{C}=\\frac{7.5\\text{g C}}{10.0\\text{g compound}}\\times 100\\%=75\\%[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0<strong>Calculation of Percent Composition<\/strong><\/h3>\r\nAnalysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?\r\n[reveal-answer q=\"82650\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"82650\"]\r\n\r\nTo calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\\\ \\%\\text{C}=\\frac{7.34\\text{g C}}{12.04\\text{g compound}}\\times 100\\%=61.0\\%\\\\ \\%\\text{H}=\\frac{1.85\\text{g H}}{12.04\\text{g compound}}\\times 100\\%=15.4\\%\\\\ \\%\\text{N}=\\frac{2.85\\text{g N}}{12.04\\text{g compound}}\\times 100\\%=23.7\\%\\end{array}[\/latex]<\/p>\r\nThe analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Determining Percent Composition from Formula Mass<\/h3>\r\nPercent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH<sub>3<\/sub>), ammonium nitrate (NH<sub>4<\/sub>NO<sub>3<\/sub>), and urea (CH<sub>4<\/sub>N<sub>2<\/sub>O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH<sub>3<\/sub> contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3\u00a0\u00d7\u00a01.008\u00a0amu)\u00a0=\u00a03.024\u00a0amu The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\\\ \\%\\text{N}=\\frac{14.01\\text{amu N}}{17.03\\text{amu}{\\text{NH}}_{3}}\\times 100\\%=82.27\\%\\\\ \\%\\text{H}=\\frac{3.024\\text{amu N}}{17.03\\text{amu}{\\text{NH}}_{3}}\\times 100\\%=17.76\\%\\end{array}[\/latex]<\/p>\r\nThis same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated in the example problem below. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0<strong>Determining Percent Composition from a Molecular Formula<\/strong><\/h3>\r\nAspirin is a compound with the molecular formula C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub>. What is its percent composition?\r\n\r\n[reveal-answer q=\"97551\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"97551\"]\r\n\r\nTo calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub>. It is convenient to consider 1 mol of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> and use its molar mass (180.159 g\/mole, determined from the chemical formula) to calculate the percentages of each of its elements:\r\n<p style=\"text-align: left\">[latex]\\%\\text{C}=\\frac{9\\text{ mol C}\\times \\text{ molar mass C}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{9\\times 12.01\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{108.09\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]<\/p>\r\n<p style=\"text-align: left\">%C =60.00% C<\/p>\r\n[latex]\\%\\text{H}=\\frac{8\\text{ mol H }\\times \\text{ molar mass H}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{8\\times 1.008\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{8.064\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]\r\n\r\n%H = 4.476% H\r\n\r\n[latex]\\%\\text{O}=\\frac{4\\text{mol O}\\times \\text{molar mass O}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{4\\times 16.00\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{64.00\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]\r\n\r\n%O = 35.52% O\r\n\r\nNote that these percentages sum to equal 100.00% when appropriately rounded.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Determination of Empirical Formulas<\/h2>\r\nAs previously mentioned, the most common approach to determining a compound\u2019s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative <em>numbers<\/em>, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}1.17\\text{g C}\\times \\frac{1\\text{mol C}}{12.01\\text{g C}}=0.142\\text{mol C}\\\\ 0.287\\text{g H}\\times \\frac{1\\text{mol H}}{1.008\\text{g H}}=0.284\\text{mol H}\\end{array}[\/latex]<\/p>\r\nThus, we can accurately represent this compound with the formula C<sub>0.142<\/sub>H<sub>0.248<\/sub>. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:\r\n<p style=\"text-align: center\">[latex]\\displaystyle{\\text{C}}_{\\frac{0.142}{0.142}}{\\text{H}}_{\\frac{0.248}{0.142}}\\text{ or }{\\text{CH}}_{2}[\/latex]<\/p>\r\n(Recall that subscripts of \u201c1\u201d are not written but rather assumed if no other number is present.)\r\n\r\nThe empirical formula for this compound is thus CH<sub>2<\/sub>. This may or not be the compound\u2019s <em>molecular formula<\/em> as well; however, we would need additional information to make that determination (as discussed later in this section).\r\n\r\nConsider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:\r\n<p style=\"text-align: center\">[latex]{\\text{C1}}_{0.150}{\\text{O}}_{0.525}={\\text{Cl}}_{\\frac{0.150}{0.150}}{\\text{O}}_{\\frac{0.525}{0.150}}={\\text{ClO}}_{3.5}[\/latex]<\/p>\r\nIn this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl<sub>2<\/sub>O<sub>7<\/sub> as the final empirical formula.\r\n\r\nIn summary, empirical formulas are derived from experimentally measured element masses by:\r\n<ol>\r\n \t<li>Deriving the number of moles of each element from its mass<\/li>\r\n \t<li>Dividing each element\u2019s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula<\/li>\r\n \t<li>Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained<\/li>\r\n<\/ol>\r\nFigure\u00a01\u00a0outlines this procedure in flow chart fashion for a substance containing elements A and X.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"878\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211155\/CNX_Chem_03_03_empform1.jpg\" alt=\"A flow chart is shown that is composed of six boxes, two of which are connected together by a right facing arrow and located above two more that are also connected by a right-facing arrow. These two rows of boxes are connected vertically by a line that leads to a right-facing arrow and the last two boxes, connected by a final right facing arrow. The first two upper boxes have the phrases, \u201cMass of A atoms\u201d and \u201cMoles of A atoms\u201d respectively, while the arrow that connects them has the phrase, \u201cDivide by molar mass,\u201d written below it. The second two bottom boxes have the phrases, \u201cMass of X atoms\u201d and \u201cMoles of X atoms\u201d respectively, while the arrow that connects them has the phrase, \u201cDivide by molar mass\u201d written below it. The arrow that connects the upper and lower boxes to the last two boxes has the phrase \u201cDivide by lowest number of moles\u201d written below it. The last two boxes have the phrases, \u201cA to X mole ratio\u201d and \u201cEmpirical formula\u201d respectively, while the arrow that connects them has the phrase, \u201cConvert ratio to lowest whole numbers\u201d written below it.\" width=\"878\" height=\"202\" \/> Figure\u00a01. The empirical formula of a compound can be derived from the masses of all elements in the sample.[\/caption]\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0<strong>Determining a Compound\u2019s Empirical Formula from the Masses of Its Elements<\/strong><\/h3>\r\nA sample of the black mineral hematite (Figure\u00a02), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"499\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211156\/CNX_Chem_03_03_hematite1.jpg\" alt=\"Two rounded, smooth black stones are shown.\" width=\"499\" height=\"307\" \/> Figure\u00a02. Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)[\/caption]\r\n\r\n[reveal-answer q=\"235849\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"235849\"]\r\n\r\nFor this problem, we are given the mass in grams of each element. Begin by finding the moles of each:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}\\\\ 34.97\\text{g Fe}\\left(\\frac{\\text{mol Fe}}{55.85\\text{g}}\\right)&amp; =&amp; 0.6261\\text{mol Fe}\\hfill \\\\ 15.03\\text{g O}\\left(\\frac{\\text{mol O}}{16.00\\text{g}}\\right)&amp; =&amp; 0.9394\\text{mol O}\\hfill \\end{array}[\/latex]<\/p>\r\nNext, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{0.6261}{0.6261}=1.000\\text{mol Fe}\\\\ \\frac{0.0394}{0.6261}=1.500\\text{mol O}\\end{array}[\/latex]<\/p>\r\nThe ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe<sub>1<\/sub>O<sub>1.5<\/sub>). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:\r\n<p style=\"text-align: center\">2(Fe<sub>1<\/sub>O<sub>1.5<\/sub>) = Fe<sub>2<\/sub>O<sub>3<\/sub><\/p>\r\nThe empirical formula is Fe<sub>2<\/sub>O<sub>3<\/sub>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\nFor additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip below.\r\n\r\nhttps:\/\/youtu.be\/mdNYDMoQ6As\r\n\r\n<\/div>\r\n<h3>Deriving Empirical Formulas from Percent Composition<\/h3>\r\nFinally, with regard to deriving empirical formulas, consider instances in which a compound\u2019s percent composition is available rather than the absolute masses of the compound\u2019s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0<strong>Determining an Empirical Formula from Percent Composition<\/strong><\/h3>\r\nThe bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure\u00a03). What is the empirical formula for this gas?\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"500\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211159\/CNX_Chem_03_03_BrewTank1.jpg\" alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\" width=\"500\" height=\"478\" \/> Figure\u00a03. An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \u201cDual Freq\u201d\/Wikipedia)[\/caption]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"93530\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"93530\"]\r\n\r\nSince the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is \u201cmost convenient\u201d because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element\u2019s mass percentage. This numerical equivalence results from the definition of the \u201cpercentage\u201d unit, whose name is derived from the Latin phrase <em>per centum<\/em> meaning \u201cby the hundred.\u201d Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}27.29\\%\\text{C}&amp; =\\hfill &amp; \\frac{27.29\\text{g C}}{100\\text{g compound}}\\hfill \\\\ 72.71\\%\\text{O}&amp; =\\hfill &amp; \\frac{72.71\\text{g O}}{100\\text{g compound}}\\hfill \\end{array}[\/latex]<\/p>\r\nThe molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element\u2019s mass by its molar mass:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}27.29\\text{g C}\\left(\\frac{\\text{mol C}}{12.01\\text{g}}\\right)&amp; =\\hfill &amp; 2.272\\text{mol C}\\hfill \\\\ 72.71\\text{g O}\\left(\\frac{\\text{mol O}}{16.00\\text{g}}\\right)&amp; =\\hfill &amp; 4.544\\text{mol O}\\hfill \\end{array}[\/latex]<\/p>\r\nCoefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{2.272\\text{ mol C}}{2.272}=1\\\\ \\frac{4.544\\text{ mol O}}{2.272}=2\\end{array}[\/latex]<\/p>\r\nSince the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO<sub>2<\/sub>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Derivation of Molecular Formulas<\/h2>\r\nRecall that empirical formulas are symbols representing the <em>relative<\/em> numbers of a compound\u2019s elements. Determining the <em>absolute<\/em> numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.\r\n\r\nMolecular formulas are derived by comparing the compound\u2019s molecular or molar mass to its <strong>empirical formula mass<\/strong>. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as <em>n<\/em>:\r\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{\\text{molecular or molar mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}{\\text{empirical formula mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}=n\\text{ formula units\/molecule}[\/latex]<\/p>\r\nThe molecular formula is then obtained by multiplying each subscript in the empirical formula by <em>n<\/em>, as shown below for the generic empirical formula A<sub>x<\/sub>B<sub>y<\/sub>:\r\n<p style=\"text-align: center\">[latex]{{\\text{(A}}_{\\text{x}}{\\text{B}}_{\\text{y}}\\text{)}}_{\\text{n}}={\\text{A}}_{\\text{nx}}{\\text{B}}_{\\text{nx}}[\/latex]<\/p>\r\nFor example, consider a covalent compound whose empirical formula is determined to be CH<sub>2<\/sub>O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound\u2019s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:\r\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{180\\text{amu\/molecule}}{30\\frac{\\text{amu}}{\\text{formula unit}}}=6\\text{ formula units\/molecule}[\/latex]<\/p>\r\nMolecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:\u00a0[latex]{\\text{(CH}}_{\\text{2}}{\\text{O)}}_{\\text{6}}={\\text{C}}_{\\text{6}}{\\text{H}}_{\\text{12}}{\\text{O}}_{\\text{6}}[\/latex]\r\n\r\nNote that this same approach may be used when the molar mass (g\/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5:\u00a0<strong>Determination of the Molecular Formula for Nicotine<\/strong><\/h3>\r\nNicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?\r\n\r\n[reveal-answer q=\"658532\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"658532\"]\r\n\r\nDetermining the molecular formula from the provided data will require comparison of the compound\u2019s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound\u2019s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}\\\\ \\left(74.02\\text{g C}\\right)\\left(\\frac{1\\text{mol C}}{12.01\\text{g C}}\\right)&amp; =\\hfill &amp; 6.163\\text{mol C}\\hfill \\\\ \\left(8.710\\text{g H}\\right)\\left(\\frac{1\\text{mol H}}{1.01\\text{g H}}\\right)&amp; =\\hfill &amp; 8.624\\text{mol H}\\hfill \\\\ \\left(17.27\\text{g N}\\right)\\left(\\frac{1\\text{mol N}}{14.01\\text{g N}}\\right)&amp; =\\hfill &amp; 1.233\\text{mol N}\\hfill \\end{array}[\/latex]<\/p>\r\nNext, we calculate the molar ratios of these elements.\r\n\r\nThe C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C<sub>5<\/sub>H<sub>7<\/sub>N. The empirical formula mass for this compound is therefore 81.13 amu\/formula unit, or 81.13 g\/mol formula unit.\r\n\r\nWe calculate the molar mass for nicotine from the given mass and molar amount of compound:\r\n<p style=\"text-align: center\">[latex]\\frac{40.57\\text{g nicotine}}{0.2500\\text{mol nicotine}}=\\frac{162.3\\text{g}}{\\text{mol}}[\/latex]<\/p>\r\nComparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:\r\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{162.3\\text{g\/mol}}{81.13\\frac{\\text{g}}{\\text{ formula unit}}}=2\\text{formula units\/molecule}[\/latex]<\/p>\r\nThus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:\u00a0[latex]{{\\text{(C}}_{\\text{5}}{\\text{H}}_{\\text{7}}\\text{N)}}_{\\text{6}}={\\text{C}}_{\\text{10}}{\\text{H}}_{\\text{14}}{\\text{N}}_{\\text{2}}[\/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\nThe chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound\u2019s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound\u2019s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound\u2019s molecular or molar mass to derive a molecular formula.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]\\%\\text{ X}=\\frac{\\text{mass X}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/li>\r\n \t<li>[latex]\\frac{\\text{molecular or molar mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}{\\text{empirical formula mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}=n\\text{ formula units\/molecule}[\/latex]<\/li>\r\n \t<li>(A<sub>x<\/sub>B<sub>y<\/sub>)<sub>n<\/sub> = A<sub>nx<\/sub>B<sub>ny<\/sub><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>empirical formula mass:\u00a0<\/strong>sum of average atomic masses for all atoms represented in an empirical formula\r\n\r\n<strong>percent composition:\u00a0<\/strong>percentage by mass of the various elements in a compound","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>Compute the percent composition of a compound<\/li>\n<li>Determine the empirical formula of a compound<\/li>\n<li>Determine the molecular formula of a compound<\/li>\n<\/ul>\n<\/div>\n<p>In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.<\/p>\n<h2>Percent Composition<\/h2>\n<p>The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound\u2019s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound\u2019s <strong>percent composition<\/strong>, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\%\\text{H}=\\frac{\\text{mass H}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\%\\text{C}=\\frac{\\text{mass C}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/p>\n<p>If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:<\/p>\n<p style=\"text-align: center\">[latex]\\%\\text{H}=\\frac{2.5\\text{g H}}{10.0\\text{g compound}}\\times 100\\%=25\\%[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\%\\text{C}=\\frac{7.5\\text{g C}}{10.0\\text{g compound}}\\times 100\\%=75\\%[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0<strong>Calculation of Percent Composition<\/strong><\/h3>\n<p>Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q82650\">Show Answer<\/span><\/p>\n<div id=\"q82650\" class=\"hidden-answer\" style=\"display: none\">\n<p>To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\\\ \\%\\text{C}=\\frac{7.34\\text{g C}}{12.04\\text{g compound}}\\times 100\\%=61.0\\%\\\\ \\%\\text{H}=\\frac{1.85\\text{g H}}{12.04\\text{g compound}}\\times 100\\%=15.4\\%\\\\ \\%\\text{N}=\\frac{2.85\\text{g N}}{12.04\\text{g compound}}\\times 100\\%=23.7\\%\\end{array}[\/latex]<\/p>\n<p>The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Determining Percent Composition from Formula Mass<\/h3>\n<p>Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH<sub>3<\/sub>), ammonium nitrate (NH<sub>4<\/sub>NO<sub>3<\/sub>), and urea (CH<sub>4<\/sub>N<sub>2<\/sub>O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH<sub>3<\/sub> contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3\u00a0\u00d7\u00a01.008\u00a0amu)\u00a0=\u00a03.024\u00a0amu The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\\\ \\%\\text{N}=\\frac{14.01\\text{amu N}}{17.03\\text{amu}{\\text{NH}}_{3}}\\times 100\\%=82.27\\%\\\\ \\%\\text{H}=\\frac{3.024\\text{amu N}}{17.03\\text{amu}{\\text{NH}}_{3}}\\times 100\\%=17.76\\%\\end{array}[\/latex]<\/p>\n<p>This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated in the example problem below. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0<strong>Determining Percent Composition from a Molecular Formula<\/strong><\/h3>\n<p>Aspirin is a compound with the molecular formula C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub>. What is its percent composition?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q97551\">Show Answer<\/span><\/p>\n<div id=\"q97551\" class=\"hidden-answer\" style=\"display: none\">\n<p>To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub>. It is convenient to consider 1 mol of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> and use its molar mass (180.159 g\/mole, determined from the chemical formula) to calculate the percentages of each of its elements:<\/p>\n<p style=\"text-align: left\">[latex]\\%\\text{C}=\\frac{9\\text{ mol C}\\times \\text{ molar mass C}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{9\\times 12.01\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{108.09\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]<\/p>\n<p style=\"text-align: left\">%C =60.00% C<\/p>\n<p>[latex]\\%\\text{H}=\\frac{8\\text{ mol H }\\times \\text{ molar mass H}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{8\\times 1.008\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{8.064\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]<\/p>\n<p>%H = 4.476% H<\/p>\n<p>[latex]\\%\\text{O}=\\frac{4\\text{mol O}\\times \\text{molar mass O}}{\\text{molar mass }{\\text{C}}_{9}{\\text{H}}_{18}{\\text{O}}_{4}}\\times 100=\\frac{4\\times 16.00\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100=\\frac{64.00\\text{g\/mol}}{180.159\\text{g\/mol}}\\times 100[\/latex]<\/p>\n<p>%O = 35.52% O<\/p>\n<p>Note that these percentages sum to equal 100.00% when appropriately rounded.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Determination of Empirical Formulas<\/h2>\n<p>As previously mentioned, the most common approach to determining a compound\u2019s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative <em>numbers<\/em>, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}1.17\\text{g C}\\times \\frac{1\\text{mol C}}{12.01\\text{g C}}=0.142\\text{mol C}\\\\ 0.287\\text{g H}\\times \\frac{1\\text{mol H}}{1.008\\text{g H}}=0.284\\text{mol H}\\end{array}[\/latex]<\/p>\n<p>Thus, we can accurately represent this compound with the formula C<sub>0.142<\/sub>H<sub>0.248<\/sub>. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle{\\text{C}}_{\\frac{0.142}{0.142}}{\\text{H}}_{\\frac{0.248}{0.142}}\\text{ or }{\\text{CH}}_{2}[\/latex]<\/p>\n<p>(Recall that subscripts of \u201c1\u201d are not written but rather assumed if no other number is present.)<\/p>\n<p>The empirical formula for this compound is thus CH<sub>2<\/sub>. This may or not be the compound\u2019s <em>molecular formula<\/em> as well; however, we would need additional information to make that determination (as discussed later in this section).<\/p>\n<p>Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:<\/p>\n<p style=\"text-align: center\">[latex]{\\text{C1}}_{0.150}{\\text{O}}_{0.525}={\\text{Cl}}_{\\frac{0.150}{0.150}}{\\text{O}}_{\\frac{0.525}{0.150}}={\\text{ClO}}_{3.5}[\/latex]<\/p>\n<p>In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl<sub>2<\/sub>O<sub>7<\/sub> as the final empirical formula.<\/p>\n<p>In summary, empirical formulas are derived from experimentally measured element masses by:<\/p>\n<ol>\n<li>Deriving the number of moles of each element from its mass<\/li>\n<li>Dividing each element\u2019s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula<\/li>\n<li>Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained<\/li>\n<\/ol>\n<p>Figure\u00a01\u00a0outlines this procedure in flow chart fashion for a substance containing elements A and X.<\/p>\n<div style=\"width: 888px\" class=\"wp-caption alignnone\"><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\/23211155\/CNX_Chem_03_03_empform1.jpg\" alt=\"A flow chart is shown that is composed of six boxes, two of which are connected together by a right facing arrow and located above two more that are also connected by a right-facing arrow. These two rows of boxes are connected vertically by a line that leads to a right-facing arrow and the last two boxes, connected by a final right facing arrow. The first two upper boxes have the phrases, \u201cMass of A atoms\u201d and \u201cMoles of A atoms\u201d respectively, while the arrow that connects them has the phrase, \u201cDivide by molar mass,\u201d written below it. The second two bottom boxes have the phrases, \u201cMass of X atoms\u201d and \u201cMoles of X atoms\u201d respectively, while the arrow that connects them has the phrase, \u201cDivide by molar mass\u201d written below it. The arrow that connects the upper and lower boxes to the last two boxes has the phrase \u201cDivide by lowest number of moles\u201d written below it. The last two boxes have the phrases, \u201cA to X mole ratio\u201d and \u201cEmpirical formula\u201d respectively, while the arrow that connects them has the phrase, \u201cConvert ratio to lowest whole numbers\u201d written below it.\" width=\"878\" height=\"202\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a01. The empirical formula of a compound can be derived from the masses of all elements in the sample.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0<strong>Determining a Compound\u2019s Empirical Formula from the Masses of Its Elements<\/strong><\/h3>\n<p>A sample of the black mineral hematite (Figure\u00a02), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?<\/p>\n<div style=\"width: 509px\" class=\"wp-caption alignnone\"><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\/23211156\/CNX_Chem_03_03_hematite1.jpg\" alt=\"Two rounded, smooth black stones are shown.\" width=\"499\" height=\"307\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a02. Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q235849\">Show Answer<\/span><\/p>\n<div id=\"q235849\" class=\"hidden-answer\" style=\"display: none\">\n<p>For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}\\\\ 34.97\\text{g Fe}\\left(\\frac{\\text{mol Fe}}{55.85\\text{g}}\\right)& =& 0.6261\\text{mol Fe}\\hfill \\\\ 15.03\\text{g O}\\left(\\frac{\\text{mol O}}{16.00\\text{g}}\\right)& =& 0.9394\\text{mol O}\\hfill \\end{array}[\/latex]<\/p>\n<p>Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{0.6261}{0.6261}=1.000\\text{mol Fe}\\\\ \\frac{0.0394}{0.6261}=1.500\\text{mol O}\\end{array}[\/latex]<\/p>\n<p>The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe<sub>1<\/sub>O<sub>1.5<\/sub>). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:<\/p>\n<p style=\"text-align: center\">2(Fe<sub>1<\/sub>O<sub>1.5<\/sub>) = Fe<sub>2<\/sub>O<sub>3<\/sub><\/p>\n<p>The empirical formula is Fe<sub>2<\/sub>O<sub>3<\/sub>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p>For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Calculating Percent Composition and Empirical Formulas\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mdNYDMoQ6As?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<h3>Deriving Empirical Formulas from Percent Composition<\/h3>\n<p>Finally, with regard to deriving empirical formulas, consider instances in which a compound\u2019s percent composition is available rather than the absolute masses of the compound\u2019s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0<strong>Determining an Empirical Formula from Percent Composition<\/strong><\/h3>\n<p>The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure\u00a03). What is the empirical formula for this gas?<\/p>\n<div style=\"width: 510px\" class=\"wp-caption alignnone\"><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\/23211159\/CNX_Chem_03_03_BrewTank1.jpg\" alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\" width=\"500\" height=\"478\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a03. An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \u201cDual Freq\u201d\/Wikipedia)<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q93530\">Show Answer<\/span><\/p>\n<div id=\"q93530\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is \u201cmost convenient\u201d because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element\u2019s mass percentage. This numerical equivalence results from the definition of the \u201cpercentage\u201d unit, whose name is derived from the Latin phrase <em>per centum<\/em> meaning \u201cby the hundred.\u201d Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}27.29\\%\\text{C}& =\\hfill & \\frac{27.29\\text{g C}}{100\\text{g compound}}\\hfill \\\\ 72.71\\%\\text{O}& =\\hfill & \\frac{72.71\\text{g O}}{100\\text{g compound}}\\hfill \\end{array}[\/latex]<\/p>\n<p>The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element\u2019s mass by its molar mass:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}27.29\\text{g C}\\left(\\frac{\\text{mol C}}{12.01\\text{g}}\\right)& =\\hfill & 2.272\\text{mol C}\\hfill \\\\ 72.71\\text{g O}\\left(\\frac{\\text{mol O}}{16.00\\text{g}}\\right)& =\\hfill & 4.544\\text{mol O}\\hfill \\end{array}[\/latex]<\/p>\n<p>Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{2.272\\text{ mol C}}{2.272}=1\\\\ \\frac{4.544\\text{ mol O}}{2.272}=2\\end{array}[\/latex]<\/p>\n<p>Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO<sub>2<\/sub>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Derivation of Molecular Formulas<\/h2>\n<p>Recall that empirical formulas are symbols representing the <em>relative<\/em> numbers of a compound\u2019s elements. Determining the <em>absolute<\/em> numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.<\/p>\n<p>Molecular formulas are derived by comparing the compound\u2019s molecular or molar mass to its <strong>empirical formula mass<\/strong>. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as <em>n<\/em>:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{\\text{molecular or molar mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}{\\text{empirical formula mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}=n\\text{ formula units\/molecule}[\/latex]<\/p>\n<p>The molecular formula is then obtained by multiplying each subscript in the empirical formula by <em>n<\/em>, as shown below for the generic empirical formula A<sub>x<\/sub>B<sub>y<\/sub>:<\/p>\n<p style=\"text-align: center\">[latex]{{\\text{(A}}_{\\text{x}}{\\text{B}}_{\\text{y}}\\text{)}}_{\\text{n}}={\\text{A}}_{\\text{nx}}{\\text{B}}_{\\text{nx}}[\/latex]<\/p>\n<p>For example, consider a covalent compound whose empirical formula is determined to be CH<sub>2<\/sub>O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound\u2019s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{180\\text{amu\/molecule}}{30\\frac{\\text{amu}}{\\text{formula unit}}}=6\\text{ formula units\/molecule}[\/latex]<\/p>\n<p>Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:\u00a0[latex]{\\text{(CH}}_{\\text{2}}{\\text{O)}}_{\\text{6}}={\\text{C}}_{\\text{6}}{\\text{H}}_{\\text{12}}{\\text{O}}_{\\text{6}}[\/latex]<\/p>\n<p>Note that this same approach may be used when the molar mass (g\/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5:\u00a0<strong>Determination of the Molecular Formula for Nicotine<\/strong><\/h3>\n<p>Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q658532\">Show Answer<\/span><\/p>\n<div id=\"q658532\" class=\"hidden-answer\" style=\"display: none\">\n<p>Determining the molecular formula from the provided data will require comparison of the compound\u2019s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound\u2019s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{lll}\\\\ \\left(74.02\\text{g C}\\right)\\left(\\frac{1\\text{mol C}}{12.01\\text{g C}}\\right)& =\\hfill & 6.163\\text{mol C}\\hfill \\\\ \\left(8.710\\text{g H}\\right)\\left(\\frac{1\\text{mol H}}{1.01\\text{g H}}\\right)& =\\hfill & 8.624\\text{mol H}\\hfill \\\\ \\left(17.27\\text{g N}\\right)\\left(\\frac{1\\text{mol N}}{14.01\\text{g N}}\\right)& =\\hfill & 1.233\\text{mol N}\\hfill \\end{array}[\/latex]<\/p>\n<p>Next, we calculate the molar ratios of these elements.<\/p>\n<p>The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C<sub>5<\/sub>H<sub>7<\/sub>N. The empirical formula mass for this compound is therefore 81.13 amu\/formula unit, or 81.13 g\/mol formula unit.<\/p>\n<p>We calculate the molar mass for nicotine from the given mass and molar amount of compound:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{40.57\\text{g nicotine}}{0.2500\\text{mol nicotine}}=\\frac{162.3\\text{g}}{\\text{mol}}[\/latex]<\/p>\n<p>Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle\\frac{162.3\\text{g\/mol}}{81.13\\frac{\\text{g}}{\\text{ formula unit}}}=2\\text{formula units\/molecule}[\/latex]<\/p>\n<p>Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:\u00a0[latex]{{\\text{(C}}_{\\text{5}}{\\text{H}}_{\\text{7}}\\text{N)}}_{\\text{6}}={\\text{C}}_{\\text{10}}{\\text{H}}_{\\text{14}}{\\text{N}}_{\\text{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound\u2019s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound\u2019s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound\u2019s molecular or molar mass to derive a molecular formula.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]\\%\\text{ X}=\\frac{\\text{mass X}}{\\text{mass compound}}\\times 100\\%[\/latex]<\/li>\n<li>[latex]\\frac{\\text{molecular or molar mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}{\\text{empirical formula mass}\\left(\\text{amu or}\\frac{\\text{g}}{\\text{mol}}\\right)}=n\\text{ formula units\/molecule}[\/latex]<\/li>\n<li>(A<sub>x<\/sub>B<sub>y<\/sub>)<sub>n<\/sub> = A<sub>nx<\/sub>B<sub>ny<\/sub><\/li>\n<\/ul>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>empirical formula mass:\u00a0<\/strong>sum of average atomic masses for all atoms represented in an empirical formula<\/p>\n<p><strong>percent composition:\u00a0<\/strong>percentage by mass of the various elements in a compound<\/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-615\">\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>: OpenStax College. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>How to Calculate Percent Composition, Empirical Formulas and Molecular Formulas - Chemistry. <strong>Authored by<\/strong>: Mr. Causey. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/mdNYDMoQ6As\">https:\/\/youtu.be\/mdNYDMoQ6As<\/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":141992,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"How to Calculate Percent Composition, Empirical Formulas and Molecular Formulas - Chemistry\",\"author\":\"Mr. Causey\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/mdNYDMoQ6As\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-615","chapter","type-chapter","status-publish","hentry"],"part":799,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/615","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/users\/141992"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/615\/revisions"}],"predecessor-version":[{"id":1234,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/615\/revisions\/1234"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/parts\/799"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/615\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/media?parent=615"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapter-type?post=615"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/contributor?post=615"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/license?post=615"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}