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)xx100=64.98`
`"% of Hydrogen "=(2)/(18)xx(0.337)/(0.277)xx100=13.51`
`"% of Oxygen "=100-(64.98+13.51)=21.51`

Empirical formula of the compound `=C_(4)H_(10)O`
Empirical formula weight `=(4xx12)+(10xx1)+(1xx16)=74`
Molecular weight `= 2 xx V.D = 2 xx 37 = 74`
`n=("Molecular weight")/("Empirical formula weight")=(2xx37)/(74)=1`
`"Molecular formula" = "(Empirical formula)" xx=n`
`=(C_(4)H_(10)O)_(1)=C_(4)H_(10)O`