Question

1.The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.50 mg of ethyl butyrate produces 5.69 mg of CO2 and 2.32 mg of H2O. What is the empirical formula of the compound?

2.Nicotine, a component of tobacco, is composed of C, H, and N. A 3.150-mg sample of nicotine was combusted, producing 8.5452 mg of CO2 and 2.450 mg of H2O. What is the empirical formula for nicotine?

Answer 1

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5.69 x103g - Ethyl butyrate + O₂ → CO + H₂O - No of males of (og → Mass = 5.69 mg - Molar mass 44.01glmal (As m= 103) 44.01 glmed = 0.000129 mol 1 mole (o contains 1 male c o No of moles of G in original compund = 0.000 12 9 mol y Mass of C in Original compound = No of molesx Molar mass = 0.000129 molx 12.01 g/mol = 0.00 1559 No of males of H₂O = 2 32x10 3g = 0.000 129 mal 18.015 glmal As I male ho contains 2 mole H . No of moles of H in original compound = 2x00.000/29 = 0.000 258 mol & Mass of H in original compound = 0.000 258 mod x 4.01 g/mol = 0.000 2600g . Mass.op O in Original compounda Total mass op Composand) ' (Mass of 6) +(Mass of 10) - = 42.50 g)-((0.001553) +(0.0002606.;)) = 0.000 6894 g ée No of moles of O in Original comparand = 0.00068948 - 15.999 glmol = 0.0000431 mol • Molar Ratio of C: H: 0 = 0.000 129 mel: 0.000258:0.0000421 Dividing by 0.0000431 and converting to whole no. Ratio V CHOO = 3:6:1 e Empirical formula of Ethyl Butyrate= C₂H6 Os = (₃ H60