Question

A 24.117 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 42.963 mg of carbon dioxide and 17.587 mg of water. In another experiment, 32.443 mg of the compound is reacted with excess oxygen to produce 14.022 mg of sulfur dioxide. Add subscripts below to correctly identify the empirical formula of this compound (use this order of elements: CHSO).

Answer 1

moles of CO2 = 42.963 x 10^-3 / 44 = 9.764 x 10^-4

moles of C = 9.764 x 10^-4

mass of C = 9.764 x 10^-4 x 12 = 0.0117 g = 11.72 mg

moles of water = 17.587 x 10^-3 / 18 = 9.771 x 10^-4

mole of H = 2 x 9.771 x 10^-4 = 1.954 x 10^-3

mass of H = 1.954 x 10^-3 g = 1.954 mg

Mass of S = molar mass of S / molar mass of SO2 x mass of SO2

= 32 / 64 x 14.022 x 10^-3 x 24.117/32.443 = 5.212 mg

moles of S = 7.381 x 10^-3 / 32 = 1.629 x 10^-4

mass of O = 24.117 - total mass of C , H, S = 5.231 mg

mole of O = 0.163 x 10^-3 / 16 = 3.269 x 10^-4

C : H : S: O = 9.764 x 10^-4 : 1.954 x 10^-3 : 1.629 x 10^-4 : 3.269 x 10^-4

                     = 6 : 12 : 1 : 2

Empirical formula :   C₆ H₁₂SO₂