(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of \(2.78 \mathrm{mg}\) of ethyl butyrate produces \(6.32 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and $2.58 \mathrm{mg}\( of \)\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of the compound? (b) Nicotine, a component of tobacco, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). A \(5.250-\mathrm{mg}\) sample of nicotine was combusted, producing \(14.242 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(4.083 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for nicotine? If nicotine has a molar mass of $160 \pm 5 \mathrm{~g} / \mathrm{mol},$ what is its molecular formula?

molecule contains 1 C atom) (1 \(\mathrm{H}_{2} \mathrm{O}\) molecule contains 2 H atoms and 1 O atom) Moles of C = Moles of \(\mathrm{CO}_{2}\) = 3.236 × 10⁻⁴ mol Moles of H = 2 * Moles of \(\mathrm{H}_{2} \mathrm{O}\) = 2 * 2.266 × 10⁻⁴ mol = 4.532 × 10⁻⁴ mol Moles of N = Moles of nicotine - Moles of C - Moles of H = 5.250 × 10⁻⁴ mol - 3.236 × 10⁻⁴ mol - 4.532 × 10⁻⁴ mol = -2.518 × 10⁻⁴ mol (approximately 0, meaning there's a small error in the given data) #tag_title#Step 3: Determine the mole-to-mole ratio of the elements#tag_content# Mole Ratio C : H = \(\frac{3.236\,\times\,10^{-4}}{3.236\,\times\,10^{-4}} : \frac{4.532\,\times\,10^{-4}}{3.236\,\times\,10^{-4}}\) = 1 : 1.4 #tag_title#Step 4: Write the empirical formula#tag_content# Since the mole ratio is not a whole number, we need to multiply each number by the smallest value that will yield whole numbers. In this case, multiplying by 5 (approximately) will give the proper ratio. Empirical Formula: C5H7 #tag_title#Step 5: Calculate molecular formula#tag_content# Empirical Formula Molar Mass = 5 * 12.01 g/mol (C) + 7 * 1.01 g/mol (H) = 60.05 + 7.07 = 67.12 g/mol Divide the given molar mass by the empirical formula molar mass: \(\frac{160}{67.12}\) ≈ 2 Now multiply the empirical formula by this value to get the molecular formula: Molecular Formula: C10H14 In conclusion, the empirical formula of ethyl butyrate is CH2, while the empirical formula of nicotine is C5H7, and its molecular formula is C10H14.