Question

5. Nicotine, an alkaloid that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. Its molecular mass was found to be 162.28 g/mol. What is the empirical and molecular formulas of this compound? Show all your calculations

Answer 1

Molar Mass of Molar Mass of Molar Mass of carbon = 12.01 g molt Hydrogen = 1.008 g molt Nitrogen = 14.007 g molt Molar Mass of Nicotine = 162.28 g molt Amount of carbon present = 74.02% of 162.28 g molt = 74.02 & 162.28 = 120.11 g mot No. of carbon atoms present molar mass of carbon present Molar Mass of one carbon atom 120.11 g molt = 10 12.01 g molt Amount of Hydrogen present = 8.710% of 162.28 g mol- = 14.13 g molt No. of H-atoms present = 14.13 g molt = = 14 1.008g molt Amount of Nitrogen present = 17.27% of 162.28 g molt = 28.02 g molt No. of N- atoms present = 28.02 g molt - I = 7 2 14.007 g molt so molecular formula = I (Actual Ratio) lohiyN, Empirical formulae (smallest Whole Number Ratio) = C HyN!