When a 0.6333-g sample of a compound of carbon, hydrogen and sulfur was burned i

Question

When a 0.6333-g sample of a compound of carbon, hydrogen and sulfur was burned in excess oxygen, 0.9272 g CO2, 0.3795 g H2O and 0.6749 g SO2 were obtained. In a separate experiment, it was found that, for fixed temperature and pressure, a 0.100 L sample of this compound takes 1.73 times longer to effuse than does a 0.100 L sample of Ar(g). What is the molecular formula of the compound? , draw Structures for the molecule (1 bonus mark for every acceptable structure, up to a maximum of 5 bonus marks).

Explanation / Answer

0.9272g CO2 / 44g per mole * 1 mole C per CO2= 0.021moles C ==>2 moles C
0.3795 g H2O / 18g per mole *2 mole H per H2O = 0.042 moles H==>4 moles H
0.6749g SO2 / 64 g per mole * 1 mole S per SO2 = 0.0105 moles S==>1 mole 1
Empirical formula is C2H4S Empirical mass is 60 g/mole

(Mx/Mar) = Var/Vx

(Mx/40) = (1.73)2

Mx = 3 * 40 = 120

120 is twice the empirical mass of 60

MOlecular formula is C4H8S2

I will post structures if you post a 2nd question asking for the "Structural isomers of C4H8S2."

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