Combustion of 0.277g of an organic compound gave 0.66g carbondioxide and 0.337g water. Vapour density of the compound is equal to 37. Calculate its molecular formula. 

Explanation

% of Carbon `=(12)/(44) xx (0.66)/(0.277) xx 100=64.98`
% of Hydrogen `=(2)/(18) xx (0.337)/(0.277)xx 100=13.51 `
% of Oxygen `=100-(64.98+13.51)=21.51`

Empirical formula of the compound `=C_(4)H_(10)O`
Empirical formula weight `=(4 xx 12) +(10 xx 1)+(1 xx 16)=74`
`n=("Molecular weight")/("Empirical formula weight")=(2 xx 37)/(74)=1`
Molecular formula =`("Empirical formula") xx n `
`=(C_(4)H_(10)O)_(1)=C_(4)H_(10)O`