20 cm3 of gaseous hydrocarbon-stichiometry question?

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  • 1 month ago
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    Since everything is at the same T and P, the volumes are directly proportional to the number of moles present.

    Every atom of carbon in the hydrocarbon makes a molecule of CO2.  So there must have been (100 cm^3 / 20 cm^3) = 5 atoms of C in each molecule of the hydrocarbon.

    Producing 100 cm^3 of CO2 required 100 cm^3 of O2.

    (200 cm^3 O2 total) - (40 cm^3 O2 excess) - (100 cm^3 O2 reacted to make CO2) =

    60 cm^3 O2 reacted to make H2O

    4 H + O2 → 2 H2O

    (60 cm^3 O2) x (4 mol H / 1 mol O2) = 240 cm^3 of H in theory

    (240 cm^3 H) / (20 cm^3 hydrocarbon) = 12 atoms H in each molecule of hydrocarbon

    So the empirical formula of the hydrocarbon is C5H12.  (And that is also likely the molecular formula, since it is difficult to imagine where the hydrogen atoms would be attached in any molecule with a multiple of the the empirical formula)

    C5H12 + 8 O2 → 5 CO2 + 6 H2O

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