A hydrocarbon contains `10.5 g` of carbon per gram of hydrogen. `1 L` of vapour of the hydrocarbon at `127^(@)C` and 1 atm pressure weighs `2.8 g`. Find the molecular formula of the hydrocarbon.

Explanation

Calculation of percentage of carbon and hydrogen:
Since the given compound is a hydrocarbon, it contains only carbon and hydrogen.
Percentage of carbon `= (10.5 )/( 10.5 +1) xx 100 = 91.3`
Percentage of hydrogen = 100 - 91.3 = 8.7
Calculation of empirical formula:

`therefore `Empirical formula `= C_7 H_8`
Calculation of molecular mass :
P = 1 atm, V = 1L, T = 127 + 273 = 400 K, w = 2.8 g
According to the ideal gas equation,
`PV =n RT `
or ` PV = (w)/(m ) RT `
or ` m= (wRT)/(PV ) = ( 2.8 xx 0.0821 xx 400 )/(1 xx 1) = 91.95 = 92`
` therefore `Molecular mass of the compound = 92
Calculation of molecular formula
Empirical formula mass `= (7 xx 12) + (8 xx 1) = 92`
` therefore n= ("Molecular mass ")/(" Empirical formula mass ")= (92)/(92) =1`
Molecular formula `= n xx` Empirical formula
`= 1 xx C_7 H_8 = C_7 H_8 `
Hence, the molecular formula of the given compound is `C_7 H_8`.