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 1.97 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

moles of carbon=.972/44=.021 moles H2=.3795/18=.021 moles of S=.0102 total weight=.021*12+.021*2+.01*32=.639 ratio=C:H:S::1:1:.5 therefore empirical formula is C2H2S Mass=59 Mass Ar=39 Rate(comp)/Rate(Ar)=M(Ar)/M(comp) n=1 molecular formula is C2nH2nSn=C2H2S

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