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14 November, 09:11

Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of co2 and 1.37 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

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  1. 14 November, 09:26
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    The general equation for a combustion reaction would be expressed as:

    CxHy + (x+y) O2 = xCO2 + y/2H2O

    To determine the empirical formula of the substance, toluene, we calculate the moles of C and H in the product. The only source of carbon would be from carbon dioxide and for hydrogen, from water. We do as follows:

    1.37 g H2O (1 mol / 18.02 g) (2 mol H / 1 mol H2O) = 0.1521 mol H

    5.86 g CO2 (1 mol / 44.01 g) (1 mol C / 1 mol CO2) = 0.1332 mol C

    So we have,

    C = 0.1332 C / 0.1332 = 1

    H = 0.1521 H / 0.1332 = 1

    Thus, the empirical formula woud be CH.
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