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1 November, 15:06

Complete combustion of 7.60 g of a hydrocarbon produced 23.4 g of CO2 and 10.8 g of H2O. What is the empirical formula for the hydrocarbon?

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  1. 1 November, 15:37
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    Mass of hydrocarbon = 7.60 g

    All the C gets converted to 23.4 g of CO2

    All the H gets converted to 10.8 g of H2O

    Now:

    Molecular mass of CO2 = 48 g/mol

    Molecular mass of H2O = 18 g/mol

    # moles of CO2 = 23.4 g/48 gmol-1 = 0.4875 = moles of C

    # moles of H2O = 10.8/18 = 0.6 moles = moles of H

    Now:

    Mass of C = 0.4875 moles * 12 g/mole = 5.85 g

    Mass of H = 0.6 moles * 1 g/mole = 0.6 g

    Total mass = C + H = 5.85 + 0.6 = 6.45 g. Ideally, this must be equal to the given mass of the hydrocarbon, 7.60 g.

    Nevertheless, in order to find the empirical formula, divide the # moles of C and H by the smallest value:

    C = 0.4875/0.4875 = 1

    H = 0.6/0.4875 = 1.23 (approximately = 1)

    Empirical formula = CH
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