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15 October, 07:18

Complete combustion of 7.40 g of a hydrocarbon produced 23.8 g of co2 and 8.11 g of h2o. what is the empirical formula for the hydrocarbon?

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  1. 15 October, 09:55
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    You have to assume that the hydrocarbon contains only C, H and, possibly, O.

    Then, all the C in CO2 and the H in H2O come from the hydrocarbon.

    a) Number of moles of C in 23.8 g of CO2.

    molar mass of CO2 = 12.0 g + 2 * 16.0 g = 44.0 g / mol

    number of moles of C = mass in grams / molar mass = 23.8 g / 44.0 g/mol = 0.5409 mol

    b) number of moles of H in 8.11 g of H2O

    number of moles = mass in grams / molar mass = 8.11 grams / 18.0 g/mol = 0.4506 mol

    c) mass of O in the hydrocarbon, m of O:

    m of O = 7.40 g - mass of C - mass of H

    mass of C = number of moles * atomic mass = 0.5409 mol * 12 g/mol = 6.49 g

    m of H = 0.4506 mol * 1 g / mol = 0.45 g

    => grams of O = 7.40 g - 6.49 g - 0.45 g = 0.46 g of O.

    d) number of moles of O = mass / atomic mass

    => number of moles of O = 0.46 g / 16 g / mol = 0.02875 mol O.

    e) Empirical formula

    C: 0.5409 / 0.02875 = 18.8 ≈ 19

    H: 0.4506 / 0.02875 = 15.7 ≈ 16

    O: 0.02875 / 0.02875 = 1

    => Empirical formula = C19 H16 O
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