chemistry

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Combustion analysis of a hydrocarbon produced 33.01 g {\rm CO}_2 and 27.04 g {\rm H}_2{\rm O}. Calculate the empirical formula of the hydrocarbon.
Express your answer as a chemical formula.

  • chemistry -

    33.01 g CO2
    27.04 g H2O

    mols CO2 = 33.01/44 = ? = mols C.
    mols H2O = 27.04/18 = ? and that x 2 = mols H.
    Now find the ratio of C to H with C being no larger than 1.0

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