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?

Question

Parker Price

Chemistry

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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Answers (1)

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Zoe Chen · Answer 1

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.