The empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio. For a covalent substance, we are usually more interested in the molecular formula, which gives the actual number of atoms of each kind present per molecule. Without additional information, however, it is impossible to know whether the formula of penicillin G, for example, is C16H17N2NaO4S or an integral multiple, such as C32H34N4Na2O8S2, C48H51N6Na3O12S3, or (C16H17N2NaO4S)n, where n is an integer. (The actual structure of penicillin G is shown in Figure 3.4 "Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G".) Consider glucose, the sugar that circulates in our blood to provide fuel for our bodies and especially for our brains. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. If we assume that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, we divide the mass of each element by its molar mass: Once again, we find the subscripts of the elements in the empirical formula by dividing the number of moles of each element by the number of moles of the element present in the smallest amount: The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH2O, but what is its molecular formula? Many known compounds have the empirical formula CH2O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in our blood. At this point, we cannot know whether glucose is CH2O, C2H4O2, or any other (CH2O)n. We can, however, use the experimentally determined molar mass of glucose (180 g/mol) to resolve this dilemma. First, we calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose, This is much smaller than the observed molar mass of 180 g/mol. Second, we determine the number of formula units per mole. For glucose, we can calculate the number of (CH2O) units—that is, the n in (CH2O)n—by dividing the molar mass of glucose by the formula mass of CH2O:
Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol. Given: percent composition and molar mass Asked for: molecular formula Strategy: A Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine. B Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present. C Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula. Solution: A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine. moles C = 49 .18 g C × 1 mol C 12 .011 g C = 4 .095 mol C moles H = 5 .39 g H × 1 mol H 1 .0079 g H = 5 .35 mol H moles N = 28 .65 g N × 1 mol N 14 .0067 g N = 2 .045 mol N moles O = 16 .68 g O × 1 mol O 15 .9994 g O = 1 .043 mol OTo obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount: O: 1 .043 1 .043 = 1 .000 C: 4 .095 1 .043 = 3 .926 H: 5 .35 1 .043 = 5 .13 N: 2 .045 1 .043 = 1 .960These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C4H5N2O. B The molecular formula of caffeine could be C4H5N2O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows: Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives 196 g/mol 97 .096 g/C 4 H 5 N 2 O = 2.02 ≈ 2 C 4 H 5 N 2 O empirical formula unitsC There are two C4H5N2O formula units in caffeine, so the molecular formula must be (C4H5N2O)2 = C8H10N4O2. The structure of caffeine is as follows: Exercise Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer. Answer: C2Cl2F4
Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems.
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Can the empirical formula of a compound always be used to determine the compounds molar mass?
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