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7 December, 09:32

Complete combustion of 4.40 g of a hydrocarbon produced 14.1 g of co2 and 5.04 g of h2o. what is the empirical formula for the hydrocarbon?

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  1. 7 December, 11:46
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    For a hydrocarbon, the combustion reactions are the following:

    C + O₂ - - > CO₂

    H₂ + 1/2 O₂ - - > H₂O

    The molar mass of CO₂ is 44 g/mol while C is 12 g/mol. Let's solve for amount of C in hydrocarbon.

    Mass of C = (14.1 g CO₂) (1mol/44g) (1 mol C/1 mol CO₂) (12 g/mol) = 3.845 g C

    So, that means that the mass of hydrogen is:

    Mass of H = 4.4 - 3.845 = 0.555 g

    Moles C = 3.845/12 = 0.32042

    Moles H = 0.555/1 = 0.555

    Divide both by the smaller value, 0.32042.

    C: 0.32042/0.32042 = 1

    H: 0.555/0.32042 = 1.732

    We have to get an answer that is closest to a whole number. Let's try multiplying both with 4.

    C: 1*4 = 4

    H: 1.732*4 = 6.93≈7

    Thus, the empirical formula is C₄H₇.
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