Show transcribed image text Example 6.7 Determination of the Molecular Formula for Nicotine 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? Solution Determining the molecular formula from the provided data will require comparison of the compound's Download for free at https llopenstax.org/details/books/che 18 Chapter 6 | Composition of Substances and Solutions empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound's empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements: (74.02 g C)(m )-6-163 mol C (8.710 g H)(H ) = 8.624 mol H (17.27 g N)(4쫌%%)= 1.233 mol N = 6.163 mol C 01 g Next, we calculate the molar ratios of these elements relative to the least abundant element, N 6.163 mol C / 1.233 mol N 5 8.264 mol H / 1.233 mol N = 7 1.233 mol N / 1.233 mol N = 1 1233 = 1.000 mol N 4.998 mol C 8.624 = 6.994 mol H 1.233 The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C,H,N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/ mol formula unit. We calculate the molar mass for nicotine from the given mass and molar amount of compound: 40.57 g nicotine_ 0.2500 mol nicotine 162.3 g mol Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units: 162.3 g/mol- 81.13 2 formula units/molecule = formula unit Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two: Check Your Learning 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 molecular mass of 194.2 amu? Answer: CaH1oN40