{"id":167,"date":"2021-07-23T09:19:09","date_gmt":"2021-07-23T13:19:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/determining-empirical-and-molecular-formulas\/"},"modified":"2022-06-22T09:42:07","modified_gmt":"2022-06-22T13:42:07","slug":"determining-empirical-and-molecular-formulas","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/determining-empirical-and-molecular-formulas\/","title":{"raw":"3.2 Determining Empirical and Molecular Formulas","rendered":"3.2 Determining Empirical and Molecular Formulas"},"content":{"raw":"<strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong>\r\n<div class=\"textbox textbox--learning-objectives\">\r\n\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\n<p id=\"fs-idp63066960\">The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains. Given the chemical formula of the substance, one may 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, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.<\/p>\r\n\r\n<div id=\"fs-idm175230352\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Percent Composition<\/strong><\/h3>\r\n<p id=\"fs-idm144417840\">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>\r\n\r\n<div id=\"fs-idp62101552\" data-type=\"equation\"><img class=\"wp-image-1078 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-300x140.png\" alt=\"\" width=\"251\" height=\"117\" \/><\/div>\r\n<p id=\"fs-idm144352192\">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>\r\n\r\n<div id=\"fs-idm95716912\" data-type=\"equation\"><img class=\"size-medium wp-image-1079 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-300x125.png\" alt=\"\" width=\"300\" height=\"125\" \/><\/div>\r\n<div id=\"fs-idm150394080\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm20386912\"><strong>Calculation of Percent Composition:<\/strong><\/p>\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<p id=\"fs-idm111369568\"><strong>Solution:<\/strong><\/p>\r\nTo calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:\r\n<div id=\"fs-idm181555088\" data-type=\"equation\"><img class=\"wp-image-1080 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-300x113.png\" alt=\"\" width=\"308\" height=\"116\" \/><\/div>\r\n<p id=\"fs-idm125592496\">The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.<\/p>\r\n<p id=\"fs-idp63575312\"><strong>Check Your Learning:<\/strong><\/p>\r\nA 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound\u2019s percent composition?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp63474576\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm146819008\">12.1% C, 16.1% O, 71.8% Cl<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm114973648\" class=\"bc-section section\" data-depth=\"2\">\r\n<h4 data-type=\"title\"><strong>Determining Percent Composition from Molecular or Empirical Formulas<\/strong><\/h4>\r\n<p id=\"fs-idm153839088\">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 u and three H atoms weighing a total of 3 \u00d7 1.008 u = 3.024 u. The formula mass of ammonia is therefore 14.01 u + 3.024 u = 17.03 u, and its percent composition is:<\/p>\r\n\r\n<div id=\"fs-idm150406144\" data-type=\"equation\"><img class=\"size-medium wp-image-1094 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-300x118.png\" alt=\"\" width=\"300\" height=\"118\" \/><\/div>\r\n<p id=\"fs-idm159904272\">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 <a class=\"autogenerated-content\" href=\"#fs-idm162294688\">(Figure)<\/a>. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements.<\/p>\r\n\r\n<div id=\"fs-idm162294688\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm28923440\"><strong>Determining Percent Composition from a Molecular Formula:<\/strong> 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>\r\n<p id=\"fs-idm77779888\"><strong>Solution:<\/strong><\/p>\r\nTo calculate the percent composition, the masses of C, H, and O in a known mass of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> are needed. It is convenient to consider one mole of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> and use its molar mass (180.159 g\/mol, determined from the chemical formula) to calculate the percentages of each of its elements:\r\n<div id=\"fs-idm106112\" data-type=\"equation\"><img class=\"wp-image-1082 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-300x141.png\" alt=\"\" width=\"491\" height=\"231\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm29762576\"><strong>Check Your Learning:<\/strong><\/p>\r\nTo three significant digits, what is the mass percentage of iron in the compound Fe<sub>2<\/sub>O<sub>3<\/sub>?\r\n\r\n<strong><span style=\"font-size: 1em\">Answer:<\/span><\/strong>\r\n<div id=\"fs-idm63137328\" data-type=\"note\">\r\n<p id=\"fs-idm72116192\">69.9% Fe<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm186235344\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Determination of Empirical Formulas<\/strong><\/h3>\r\n<p id=\"fs-idm181573632\">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, keep in mind that chemical formulas represent the relative <em data-effect=\"italics\">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. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios 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 amounts (in moles) are:<\/p>\r\n\r\n<div id=\"fs-idm182904000\" data-type=\"equation\"><img class=\"size-medium wp-image-1083 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-300x85.png\" alt=\"\" width=\"300\" height=\"85\" \/><\/div>\r\n<p id=\"fs-idm89096592\">Thus, this compound may be represented by the formula C<sub>0.142<\/sub>H<sub>0.248<\/sub>. Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:<\/p>\r\n\r\n<div id=\"fs-idm173344864\" data-type=\"equation\"><img class=\"wp-image-1084 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-300x69.png\" alt=\"\" width=\"183\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idm154418400\">(Recall that subscripts of \u201c1\u201d are not written but rather assumed if no other number is present.)<\/p>\r\n<p id=\"fs-idm83793840\">The empirical formula for this compound is thus CH<sub>2<\/sub>. This may or not be the compound\u2019s <em data-effect=\"italics\">molecular formula<\/em> as well; however, additional information is needed to make that determination (as discussed later in this section).<\/p>\r\n<p id=\"fs-idm172411952\">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>\r\n\r\n<div id=\"fs-idm153644640\" data-type=\"equation\"><img class=\"size-medium wp-image-1085 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-300x48.png\" alt=\"\" width=\"300\" height=\"48\" \/><\/div>\r\n<p id=\"fs-idm146854160\">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, 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>\r\n<p id=\"fs-idm99418064\">In summary, empirical formulas are derived from experimentally measured element masses by:<\/p>\r\n\r\n<ol id=\"fs-idm105983184\" type=\"1\">\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\n<p id=\"fs-idm116173824\"><a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_empform\">(Figure)<\/a> outlines this procedure in flow chart fashion for a substance containing elements A and X.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_03_03_empform\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">The empirical formula of a compound can be derived from the masses of all elements in the sample.<\/div>\r\n<span id=\"fs-idm114759824\" data-type=\"media\" data-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.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_empform-1.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.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idp70151968\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm148697040\"><strong>Determining a Compound\u2019s Empirical Formula from the Masses of Its Elements:<\/strong><\/p>\r\nA sample of the black mineral hematite (<a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_hematite\">(Figure)<\/a>), 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&nbsp;\r\n<div id=\"CNX_Chem_03_03_hematite\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)<\/div>\r\n<span id=\"fs-idm156718064\" data-type=\"media\" data-alt=\"Two rounded, smooth black stones are shown.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_hematite-1.jpg\" alt=\"Two rounded, smooth black stones are shown.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp9584080\"><strong>Solution:<\/strong><\/p>\r\nThis problem provides the mass in grams of each element. Begin by finding the moles of each:\r\n<div id=\"fs-idp81202080\" data-type=\"equation\"><img class=\"size-medium wp-image-1087 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-300x89.png\" alt=\"\" width=\"300\" height=\"89\" \/><\/div>\r\n<p id=\"fs-idp19490304\">Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:<\/p>\r\n\r\n<div id=\"fs-idm186572544\" data-type=\"equation\"><img class=\"wp-image-1095 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n-300x148.png\" alt=\"\" width=\"193\" height=\"95\" \/><\/div>\r\n<p id=\"fs-idm169565728\">The ratio is 1.000 iron to 1.500 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>\r\n\r\n<div id=\"fs-idm9646560\" data-type=\"equation\"><img class=\"wp-image-1089 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-300x50.png\" alt=\"\" width=\"222\" height=\"37\" \/><\/div>\r\n&nbsp;\r\n<p id=\"fs-idm180356384\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp52798448\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm112291760\">N<sub>2<\/sub>O<sub>5<\/sub><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm160004320\" class=\"chemistry link-to-learning\" data-type=\"note\">\r\n<p id=\"fs-idp89502368\">For additional worked examples illustrating the derivation of empirical formulas, watch the brief <a href=\"http:\/\/openstaxcollege.org\/l\/16empforms\">video<\/a> clip.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp48607392\" class=\"bc-section section\" data-depth=\"2\">\r\n<h4 data-type=\"title\"><strong>Deriving Empirical Formulas from Percent Composition<\/strong><\/h4>\r\n<p id=\"fs-idm130157248\">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>\r\n\r\n<div id=\"fs-idp11336208\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm16856224\"><strong>Determining an Empirical Formula from Percent Composition:<\/strong><\/p>\r\nThe bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (<a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_BrewTank\">(Figure)<\/a>). What is the empirical formula for this gas?\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_03_03_BrewTank\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \u201cDual Freq\u201d\/Wikimedia Commons)<\/div>\r\n<span id=\"fs-idm183208576\" data-type=\"media\" data-alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_BrewTank-1.jpg\" alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp8854528\"><strong>Solution:<\/strong><\/p>\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 data-effect=\"italics\">per centum<\/em> meaning \u201cby the hundred.\u201d Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:\r\n<div id=\"fs-idm50369216\" data-type=\"equation\"><img class=\"wp-image-1090 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-300x98.png\" alt=\"\" width=\"248\" height=\"81\" \/><\/div>\r\n<p id=\"fs-idp6456464\">The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element\u2019s mass by its molar mass:<\/p>\r\n\r\n<div id=\"fs-idm107059040\" data-type=\"equation\"><img class=\"wp-image-1091 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-300x81.png\" alt=\"\" width=\"281\" height=\"76\" \/><\/div>\r\n<p id=\"fs-idm174491744\">Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:<\/p>\r\n\r\n<div id=\"fs-idm112786736\" data-type=\"equation\"><img class=\"wp-image-1096 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2o.png\" alt=\"\" width=\"169\" height=\"101\" \/><\/div>\r\n<p id=\"fs-idm106276304\">Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO<sub>2<\/sub>.<\/p>\r\n<p id=\"fs-idm153398992\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?\r\n\r\n&nbsp;\r\n<div id=\"fs-idm61160608\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm49848016\">CH<sub>2<\/sub>O<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp23895104\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Derivation of Molecular Formulas<\/strong><\/h3>\r\n<p id=\"fs-idp60635424\">Recall that empirical formulas are symbols representing the <em data-effect=\"italics\">relative<\/em> numbers of a compound\u2019s elements. Determining the <em data-effect=\"italics\">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.<\/p>\r\n<p id=\"fs-idm152826480\">Molecular formulas are derived by comparing the compound\u2019s molar mass to its <span data-type=\"term\">empirical formula molar mass<\/span>. If the molar mass of the substance is known, it may be divided by the empirical formula molar mass to yield the number of empirical formula units per molecule (<em data-effect=\"italics\">n<\/em>):<\/p>\r\n\r\n<div id=\"fs-idm157027008\" data-type=\"equation\"><img class=\"size-full wp-image-1138 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2u.png\" alt=\"\" width=\"223\" height=\"52\" \/><\/div>\r\n<p id=\"fs-idp7653408\">The molecular formula is then obtained by multiplying each subscript in the empirical formula by <em data-effect=\"italics\">n<\/em>, as shown by the generic empirical formula A<sub>x<\/sub>B<sub>y<\/sub>:<\/p>\r\n\r\n<div id=\"fs-idp3369296\" style=\"text-align: center\" data-type=\"equation\">(A<sub>x<\/sub>B<sub>y<\/sub>)<sub>n<\/sub> = A<sub>nx<\/sub>B<sub>ny<\/sub><\/div>\r\n<p id=\"fs-idm149844688\">For example, consider a covalent compound whose empirical formula is determined to be CH<sub>2<\/sub>O. The empirical formula molar mass for this compound is approximately 30 g\/mol. If the compound\u2019s molar mass is determined to be 180 g\/mol, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:<\/p>\r\n\r\n<div id=\"fs-idm11480608\" data-type=\"equation\"><img class=\" wp-image-1098 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2q.png\" alt=\"\" width=\"117\" height=\"53\" \/><\/div>\r\n<p id=\"fs-idm156812016\">Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:<\/p>\r\n\r\n<div id=\"fs-idp29668368\" style=\"text-align: center\" data-type=\"equation\">(CH<sub>2<\/sub>O)<sub>6<\/sub> = C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/div>\r\n<div id=\"fs-idm153155296\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm56345360\"><strong>Determination of the Molecular Formula for Nicotine:<\/strong><\/p>\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<p id=\"fs-idm146789392\"><strong>Solution:<\/strong><\/p>\r\nDetermining the molecular formula from the provided data will require comparison of the compound\u2019s empirical formula molar 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<div id=\"fs-idm165807504\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1099 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-300x112.png\" alt=\"\" width=\"300\" height=\"112\" \/><\/div>\r\n<p id=\"fs-idm3598576\">Next, calculate the molar ratios of these elements relative to the least abundant element, N.<\/p>\r\n<p style=\"text-align: center\">6.163 mol C\/1.233 mol = 4.998 C<\/p>\r\n<p style=\"text-align: center\">8.624 mol H\/1.233 mol = 6.994 H<\/p>\r\n<p style=\"text-align: center\">1.233 mol N\/1.233 mol = 1.000 N<\/p>\r\n\r\n<div id=\"eip-272\" data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm152642880\">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 molar mass for this compound is therefore 81.13 g\/mol.<\/p>\r\n<p id=\"fs-idm172291248\">Calculate the molar mass for nicotine from the given mass and molar amount of compound:<\/p>\r\n\r\n<div id=\"fs-idp7634272\" data-type=\"equation\"><img class=\"wp-image-1101 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-300x64.png\" alt=\"\" width=\"220\" height=\"47\" \/><\/div>\r\n<p id=\"fs-idm113144656\">Comparing the molar mass and empirical formula molar mass indicates that each nicotine molecule contains two formula units:<\/p>\r\n\r\n<div id=\"fs-idp64453248\" data-type=\"equation\"><img class=\"wp-image-1102 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2t.png\" alt=\"\" width=\"159\" height=\"44\" \/><\/div>\r\n<p id=\"fs-idm217123120\">Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:<\/p>\r\n\r\n<div id=\"fs-idp60197184\" style=\"text-align: center\" data-type=\"equation\">(C<sub>5<\/sub>H<sub>7<\/sub>N)<sub>2<\/sub> = C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp8627344\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molar mass of 194.2 g\/mol?\r\n\r\n&nbsp;\r\n<div id=\"fs-idm149132352\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm152847216\">C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm130412048\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm171937152\">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 molar mass of a covalent compound may be compared to the compound\u2019s molecular or molar mass to derive a molecular formula.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm151713456\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm115920688\" data-type=\"exercise\">\r\n<div id=\"fs-idm175734160\" data-type=\"problem\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\r\n<dl id=\"fs-idp78260032\">\r\n \t<dt>percent composition<\/dt>\r\n \t<dd id=\"fs-idp78260672\">percentage by mass of the various elements in a compound<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong><\/p>\n<div class=\"textbox textbox--learning-objectives\">\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 id=\"fs-idp63066960\">The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains. Given the chemical formula of the substance, one may 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, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.<\/p>\n<div id=\"fs-idm175230352\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Percent Composition<\/strong><\/h3>\n<p id=\"fs-idm144417840\">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<div id=\"fs-idp62101552\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1078 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-300x140.png\" alt=\"\" width=\"251\" height=\"117\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-300x140.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-65x30.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-225x105.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-350x164.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a.png 761w\" sizes=\"auto, (max-width: 251px) 100vw, 251px\" \/><\/div>\n<p id=\"fs-idm144352192\">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<div id=\"fs-idm95716912\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1079 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-300x125.png\" alt=\"\" width=\"300\" height=\"125\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-300x125.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-768x321.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-65x27.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-225x94.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b-350x146.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2b.png 941w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div id=\"fs-idm150394080\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm20386912\"><strong>Calculation of Percent Composition:<\/strong><\/p>\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<p id=\"fs-idm111369568\"><strong>Solution:<\/strong><\/p>\n<p>To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:<\/p>\n<div id=\"fs-idm181555088\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1080 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-300x113.png\" alt=\"\" width=\"308\" height=\"116\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-300x113.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-768x289.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-65x24.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-225x85.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-350x132.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c.png 873w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><\/div>\n<p id=\"fs-idm125592496\">The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.<\/p>\n<p id=\"fs-idp63575312\"><strong>Check Your Learning:<\/strong><\/p>\n<p>A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound\u2019s percent composition?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp63474576\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm146819008\">12.1% C, 16.1% O, 71.8% Cl<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm114973648\" class=\"bc-section section\" data-depth=\"2\">\n<h4 data-type=\"title\"><strong>Determining Percent Composition from Molecular or Empirical Formulas<\/strong><\/h4>\n<p id=\"fs-idm153839088\">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 u and three H atoms weighing a total of 3 \u00d7 1.008 u = 3.024 u. The formula mass of ammonia is therefore 14.01 u + 3.024 u = 17.03 u, and its percent composition is:<\/p>\n<div id=\"fs-idm150406144\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1094 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-300x118.png\" alt=\"\" width=\"300\" height=\"118\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-300x118.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-65x26.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-225x89.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m-350x138.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2m.png 504w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm159904272\">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 <a class=\"autogenerated-content\" href=\"#fs-idm162294688\">(Figure)<\/a>. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound&#8217;s elements.<\/p>\n<div id=\"fs-idm162294688\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm28923440\"><strong>Determining Percent Composition from a Molecular Formula:<\/strong> 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<p id=\"fs-idm77779888\"><strong>Solution:<\/strong><\/p>\n<p>To calculate the percent composition, the masses of C, H, and O in a known mass of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> are needed. It is convenient to consider one mole of C<sub>9<\/sub>H<sub>8<\/sub>O<sub>4<\/sub> and use its molar mass (180.159 g\/mol, determined from the chemical formula) to calculate the percentages of each of its elements:<\/p>\n<div id=\"fs-idm106112\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1082 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-300x141.png\" alt=\"\" width=\"491\" height=\"231\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-300x141.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1024x482.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-768x361.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1536x723.png 1536w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-65x31.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-225x106.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-350x165.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e.png 1736w\" sizes=\"auto, (max-width: 491px) 100vw, 491px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm29762576\"><strong>Check Your Learning:<\/strong><\/p>\n<p>To three significant digits, what is the mass percentage of iron in the compound Fe<sub>2<\/sub>O<sub>3<\/sub>?<\/p>\n<p><strong><span style=\"font-size: 1em\">Answer:<\/span><\/strong><\/p>\n<div id=\"fs-idm63137328\" data-type=\"note\">\n<p id=\"fs-idm72116192\">69.9% Fe<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm186235344\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Determination of Empirical Formulas<\/strong><\/h3>\n<p id=\"fs-idm181573632\">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, keep in mind that chemical formulas represent the relative <em data-effect=\"italics\">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. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios 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 amounts (in moles) are:<\/p>\n<div id=\"fs-idm182904000\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1083 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-300x85.png\" alt=\"\" width=\"300\" height=\"85\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-300x85.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-768x216.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-65x18.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-225x63.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f-350x99.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2f.png 809w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm89096592\">Thus, this compound may be represented by the formula C<sub>0.142<\/sub>H<sub>0.248<\/sub>. Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:<\/p>\n<div id=\"fs-idm173344864\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1084 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-300x69.png\" alt=\"\" width=\"183\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-300x69.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-225x52.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g-350x80.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2g.png 463w\" sizes=\"auto, (max-width: 183px) 100vw, 183px\" \/><\/div>\n<p id=\"fs-idm154418400\">(Recall that subscripts of \u201c1\u201d are not written but rather assumed if no other number is present.)<\/p>\n<p id=\"fs-idm83793840\">The empirical formula for this compound is thus CH<sub>2<\/sub>. This may or not be the compound\u2019s <em data-effect=\"italics\">molecular formula<\/em> as well; however, additional information is needed to make that determination (as discussed later in this section).<\/p>\n<p id=\"fs-idm172411952\">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<div id=\"fs-idm153644640\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1085 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-300x48.png\" alt=\"\" width=\"300\" height=\"48\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-300x48.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-768x122.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-65x10.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-225x36.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h-350x56.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2h.png 853w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm146854160\">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, 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 id=\"fs-idm99418064\">In summary, empirical formulas are derived from experimentally measured element masses by:<\/p>\n<ol id=\"fs-idm105983184\" type=\"1\">\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 id=\"fs-idm116173824\"><a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_empform\">(Figure)<\/a> outlines this procedure in flow chart fashion for a substance containing elements A and X.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_03_03_empform\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">The empirical formula of a compound can be derived from the masses of all elements in the sample.<\/div>\n<p><span id=\"fs-idm114759824\" data-type=\"media\" data-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.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_empform-1.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.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idp70151968\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm148697040\"><strong>Determining a Compound\u2019s Empirical Formula from the Masses of Its Elements:<\/strong><\/p>\n<p>A sample of the black mineral hematite (<a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_hematite\">(Figure)<\/a>), 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<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_03_03_hematite\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)<\/div>\n<p><span id=\"fs-idm156718064\" data-type=\"media\" data-alt=\"Two rounded, smooth black stones are shown.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_hematite-1.jpg\" alt=\"Two rounded, smooth black stones are shown.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp9584080\"><strong>Solution:<\/strong><\/p>\n<p>This problem provides the mass in grams of each element. Begin by finding the moles of each:<\/p>\n<div id=\"fs-idp81202080\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1087 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-300x89.png\" alt=\"\" width=\"300\" height=\"89\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-300x89.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-768x228.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-225x67.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1-350x104.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2a-1.png 878w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp19490304\">Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:<\/p>\n<div id=\"fs-idm186572544\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1095 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n-300x148.png\" alt=\"\" width=\"193\" height=\"95\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n-300x148.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n-65x32.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n-225x111.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2n.png 328w\" sizes=\"auto, (max-width: 193px) 100vw, 193px\" \/><\/div>\n<p id=\"fs-idm169565728\">The ratio is 1.000 iron to 1.500 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<div id=\"fs-idm9646560\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1089 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-300x50.png\" alt=\"\" width=\"222\" height=\"37\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-300x50.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-65x11.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-225x38.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1-350x58.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2c-1.png 486w\" sizes=\"auto, (max-width: 222px) 100vw, 222px\" \/><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm180356384\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp52798448\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm112291760\">N<sub>2<\/sub>O<sub>5<\/sub><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm160004320\" class=\"chemistry link-to-learning\" data-type=\"note\">\n<p id=\"fs-idp89502368\">For additional worked examples illustrating the derivation of empirical formulas, watch the brief <a href=\"http:\/\/openstaxcollege.org\/l\/16empforms\">video<\/a> clip.<\/p>\n<\/div>\n<div id=\"fs-idp48607392\" class=\"bc-section section\" data-depth=\"2\">\n<h4 data-type=\"title\"><strong>Deriving Empirical Formulas from Percent Composition<\/strong><\/h4>\n<p id=\"fs-idm130157248\">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 id=\"fs-idp11336208\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm16856224\"><strong>Determining an Empirical Formula from Percent Composition:<\/strong><\/p>\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 (<a class=\"autogenerated-content\" href=\"#CNX_Chem_03_03_BrewTank\">(Figure)<\/a>). What is the empirical formula for this gas?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_03_03_BrewTank\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \u201cDual Freq\u201d\/Wikimedia Commons)<\/div>\n<p><span id=\"fs-idm183208576\" data-type=\"media\" data-alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_03_03_BrewTank-1.jpg\" alt=\"A picture is shown of four copper-colored industrial containers with a large pipe connecting to the top of each one.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp8854528\"><strong>Solution:<\/strong><\/p>\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 data-effect=\"italics\">per centum<\/em> meaning \u201cby the hundred.\u201d Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:<\/p>\n<div id=\"fs-idm50369216\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1090 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-300x98.png\" alt=\"\" width=\"248\" height=\"81\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-300x98.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-65x21.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-225x74.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1-350x114.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2d-1.png 627w\" sizes=\"auto, (max-width: 248px) 100vw, 248px\" \/><\/div>\n<p id=\"fs-idp6456464\">The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element\u2019s mass by its molar mass:<\/p>\n<div id=\"fs-idm107059040\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1091 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-300x81.png\" alt=\"\" width=\"281\" height=\"76\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-300x81.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-768x208.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-65x18.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-225x61.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1-350x95.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2e-1.png 815w\" sizes=\"auto, (max-width: 281px) 100vw, 281px\" \/><\/div>\n<p id=\"fs-idm174491744\">Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:<\/p>\n<div id=\"fs-idm112786736\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1096 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2o.png\" alt=\"\" width=\"169\" height=\"101\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2o.png 276w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2o-65x39.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2o-225x135.png 225w\" sizes=\"auto, (max-width: 169px) 100vw, 169px\" \/><\/div>\n<p id=\"fs-idm106276304\">Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO<sub>2<\/sub>.<\/p>\n<p id=\"fs-idm153398992\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm61160608\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm49848016\">CH<sub>2<\/sub>O<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp23895104\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Derivation of Molecular Formulas<\/strong><\/h3>\n<p id=\"fs-idp60635424\">Recall that empirical formulas are symbols representing the <em data-effect=\"italics\">relative<\/em> numbers of a compound\u2019s elements. Determining the <em data-effect=\"italics\">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.<\/p>\n<p id=\"fs-idm152826480\">Molecular formulas are derived by comparing the compound\u2019s molar mass to its <span data-type=\"term\">empirical formula molar mass<\/span>. If the molar mass of the substance is known, it may be divided by the empirical formula molar mass to yield the number of empirical formula units per molecule (<em data-effect=\"italics\">n<\/em>):<\/p>\n<div id=\"fs-idm157027008\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1138 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2u.png\" alt=\"\" width=\"223\" height=\"52\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2u.png 223w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2u-65x15.png 65w\" sizes=\"auto, (max-width: 223px) 100vw, 223px\" \/><\/div>\n<p id=\"fs-idp7653408\">The molecular formula is then obtained by multiplying each subscript in the empirical formula by <em data-effect=\"italics\">n<\/em>, as shown by the generic empirical formula A<sub>x<\/sub>B<sub>y<\/sub>:<\/p>\n<div id=\"fs-idp3369296\" style=\"text-align: center\" data-type=\"equation\">(A<sub>x<\/sub>B<sub>y<\/sub>)<sub>n<\/sub> = A<sub>nx<\/sub>B<sub>ny<\/sub><\/div>\n<p id=\"fs-idm149844688\">For example, consider a covalent compound whose empirical formula is determined to be CH<sub>2<\/sub>O. The empirical formula molar mass for this compound is approximately 30 g\/mol. If the compound\u2019s molar mass is determined to be 180 g\/mol, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:<\/p>\n<div id=\"fs-idm11480608\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1098 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2q.png\" alt=\"\" width=\"117\" height=\"53\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2q.png 183w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2q-65x29.png 65w\" sizes=\"auto, (max-width: 117px) 100vw, 117px\" \/><\/div>\n<p id=\"fs-idm156812016\">Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:<\/p>\n<div id=\"fs-idp29668368\" style=\"text-align: center\" data-type=\"equation\">(CH<sub>2<\/sub>O)<sub>6<\/sub> = C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/div>\n<div id=\"fs-idm153155296\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm56345360\"><strong>Determination of the Molecular Formula for Nicotine:<\/strong><\/p>\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<p id=\"fs-idm146789392\"><strong>Solution:<\/strong><\/p>\n<p>Determining the molecular formula from the provided data will require comparison of the compound\u2019s empirical formula molar 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<div id=\"fs-idm165807504\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1099 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-300x112.png\" alt=\"\" width=\"300\" height=\"112\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-300x112.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-768x286.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-65x24.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-225x84.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r-350x131.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2r.png 917w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm3598576\">Next, calculate the molar ratios of these elements relative to the least abundant element, N.<\/p>\n<p style=\"text-align: center\">6.163 mol C\/1.233 mol = 4.998 C<\/p>\n<p style=\"text-align: center\">8.624 mol H\/1.233 mol = 6.994 H<\/p>\n<p style=\"text-align: center\">1.233 mol N\/1.233 mol = 1.000 N<\/p>\n<div id=\"eip-272\" data-type=\"equation\"><\/div>\n<p id=\"fs-idm152642880\">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 molar mass for this compound is therefore 81.13 g\/mol.<\/p>\n<p id=\"fs-idm172291248\">Calculate the molar mass for nicotine from the given mass and molar amount of compound:<\/p>\n<div id=\"fs-idp7634272\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1101 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-300x64.png\" alt=\"\" width=\"220\" height=\"47\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-300x64.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-225x48.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s-350x75.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2s.png 695w\" sizes=\"auto, (max-width: 220px) 100vw, 220px\" \/><\/div>\n<p id=\"fs-idm113144656\">Comparing the molar mass and empirical formula molar mass indicates that each nicotine molecule contains two formula units:<\/p>\n<div id=\"fs-idp64453248\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1102 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2t.png\" alt=\"\" width=\"159\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2t.png 285w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2t-65x18.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/3.2t-225x62.png 225w\" sizes=\"auto, (max-width: 159px) 100vw, 159px\" \/><\/div>\n<p id=\"fs-idm217123120\">Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:<\/p>\n<div id=\"fs-idp60197184\" style=\"text-align: center\" data-type=\"equation\">(C<sub>5<\/sub>H<sub>7<\/sub>N)<sub>2<\/sub> = C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp8627344\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molar mass of 194.2 g\/mol?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm149132352\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm152847216\">C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm130412048\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm171937152\">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 molar mass of a covalent compound may be compared to the compound\u2019s molecular or molar mass to derive a molecular formula.<\/p>\n<\/div>\n<div id=\"fs-idm151713456\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm115920688\" data-type=\"exercise\">\n<div id=\"fs-idm175734160\" data-type=\"problem\"><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\n<dl id=\"fs-idp78260032\">\n<dt>percent composition<\/dt>\n<dd id=\"fs-idp78260672\">percentage by mass of the various elements in a compound<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-167","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":130,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/167","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":11,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/167\/revisions"}],"predecessor-version":[{"id":2114,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/167\/revisions\/2114"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/130"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/167\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=167"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=167"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=167"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}