Question

A 23.113 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 41.174 mg of carbon dioxide and 16.855 mg of water. In another experiment, 20.101 mg of the compound is reacted with excess oxygen to produce 8.6877 mg of sulfur dioxide. Add subscripts below to correctly identify the empirical formula of this compound (use this order of elements: CHSO).

Answer 1

41.170 mg of carbon dioxide we have 41.170 x 10^-3 x 12 / 44 g of C = 0.01122 g
moles of C = 0.01122 / 12 = 0.00093 moles
all the carbon came from the compound so
% C in the compound = 0.01122 / .023113 = 48.54%

mass of H in the sample of water = 16.855 x 10^-3 x 2 / 18 = 0.00187 g
moles of H = 0.00187 moles
% H in the compound = 0.00187/ .023113 = 8.09%

mass of S in 8.6877 mg of SO2 = 8.6877 x 10^-3 x 32 / 64 = 0.00434
%S = 0.00434 / 20.101 x 10^-3 = 21.88%
moles of S = 0.000135 moles
moles in 23.113 g of sample = 0.000135x 23.113 / 20.101 = 0.0001552 moles

%O = 100 - ( 48.54% +8.09%+ 21.88%) = 21.49
mass in 23.113 mg of compound = 21.49% of 23.113 x 10^-3 = 0.004966 g of O
= 0.000310 moles of O

molar ratio of the elements in 23.113 mg of sample =
C:H:S:O = 0.00093 : 0.00187: 0.0001552: 0.000310

=6.0: 12: 1.0: 1.99

empirical formula is C6H12SO2