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20 July, 12:33

The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol. Determine the molecular formula of this compound.

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  1. 20 July, 12:47
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    Molecular formula = C20H30

    Explanation:

    NB 440mg = 0.44g, 135mg = 0.135g

    From the question, moles of CO2 = 0.44/44 = 0.01mol

    Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

    Also from the question, moles of H2O = 0.135/18 = 0.0075mole

    Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075*2 = 0.015 mol of H

    To get the empirical formula, divide by smallest number of mole

    Mol of C = 0.01/0.01=1

    Mol of H = 0.015/0.01 = 1.5

    Multiply both by 2 to obtain a whole number

    Mol of C = 1*2 = 2

    Mol of H = 1.5*2 = 3

    Empirical formula = C2H3

    [C2H3] not = 270

    [ (2*12) + 3]n = 270

    27n = 270

    n=10

    Molecular formula = [C2H3]10 = C20H30
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