Question

Combustion analysis of toluene, a common organic solvent, gives 8.20 mg of co2 and 1.92 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

Answer 1

For the purpose of proper representation in this item, we let the number of moles of carbon in the compound be x, that of H is y. The equation of toluene now becomes,
                   CxHy
The combustion reaction is,
        CxHy + O2 --> CO2 + H2O

The equation presented above may not be balanced yet. Then, we determine the number of mmols of C, H, and O in the product using the given masses.
    
(1) 8.20 mg CO2
             (8.2 mg CO2)(1 mmol CO2/44 mg CO2) = 0.186 mmol CO2
     which means,
                     0.186 mmol C
                     0.373 mmol O

(2) 1.92 mg H2O
            (1.92 mg H2O)(1 mmol H2O/18 mg H2O) = 0.107 mmol H2O
    which means
                    0.2133 mmol H
                    0.107 mmol O

Thus, the equation for toluene is,
               C(0.186)H(0.2133)

Dividing the numbers by the lesser value,
               CH(8/7)
To eliminate the fraction, we multiply by the denominator. Thus, the final answer would be,
              C7H8