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Combustion analysis, Empirical and Molecular formulas, help!?

When 2.686 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.224 grams of CO2 and 1.511 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Enter the elements in the order presented in the qustion.

  • chemistry - ,

    empirical formula is the smallest whole number ratio

    molecular formula is the empirical doubled i believe.

  • chemistry - ,

    you would divide the grams of each molecule produced by the smallest ratio to find the subscripts.

  • chemistry - ,

    Start from a balanced equation

    CxHy + (x+y/4)O2 -> xCO2 + y/2 H2O

    so 1 mole of CxHy (molar mass 12x+y)produces
    x mole of CO2 and
    y/2 mole of H2O

    2.686 g/(12x+y) moles

    9.224/44 =x moles of CO2
    so x=0.2096

    and produces

    1.511/18=y/2 moles of H2O
    so y=0.1679

    so the ratio is
    or 5:4

    Thus the empirical formula or the simplest formula is
    for which molar mass is 64 g

    we are told in the question that the molar mass is 128.2 g so the molecular formula (ie. the formula for one mole) must be C10H8, that is twice the empirical formula in this case.

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