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?

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