Question

A 29.137 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 51.905 mg of carbon dioxide and 21.248 mg of water. In another experiment, 31.321 mg of the compound is reacted with excess oxygen to produce 13.537 mg of sulfur dioxide. Add subscripts below to correctly identify the empirical formula of this compound (use this order of elements: CHSO).

Answer 1

:  mass of C in 51.905mg of CO2 = 51.905 x 12/ 44 = 14.156 mg of C
mass of H in 21.248 mg of H2O = 21.248 x 2 / 18 = 2.361 mg of H

mass of S in 13.537 mg of SO2 = 13.537 x 32 / 64 = 6.769 mg
mass of S in 29.137 g of sample = 6.769 x 29.137 / 31.321 = 6.297 mg of S

mass of O by difference = 29.137 - ( 14.156 + 2.361+6.297) = 6.323 mg of O

mmole of C = 14.156 / 12= 1.177 mmoles
mmoles of H = 2.361 / 1.0079 = 2.342 mmoles
mmoles of S = 6.297 / 32 = 0.197 mmoles
mmoles of O = 6.323 / 16 = 0.395 mmoles


molar ratio of C : H : S : O = 1.177: 2.342: 0.197: 0.395
if we divide by the smallest number we get the simple ratio

6:12:1:2