An organic compound contains `49.3%` carbon `6.84%` hydrogen and its vapour density is 73 . Molecular formula of the compund is

Options

A. `C_(3)H_(8)O_(2)`
B. `C_(4)H_(10)O_(2)`
C. `C_(3)H_(10)O_(2)`
D. `C_(6)H_(10)O_(4)`

Answer

D

Explanation

`% "Oxygen" = (100%) - (%C + %H)`
`=(100%) - (49.3%) + 6.44%)`
`= 43.86%`
Considering `100g` of o .c.,
`m_(C ) = 49.3 g, m_(H) = 6.84 g`, and `m_(O) = 43.86`
`n_(C ) : n_(H) : n_(O) = (49.3)/(12):(6.84)/(1):(43.86)/(16)`
`= 4.11:6.84:2.74`
`= (4.11)/(2.74) : (6.84)/(2.74) : (2.74)/(2.74)`
`= 1.5 : 2.5 : 1 = 3 : 5 : 2`
Therefore, empirical formula `= C_(3)H_(5)O_(2)`
and empirical formula mass `= 73u`
Molecular mass `= 2 xx` Vapor density
`= 2 xx (73) = 146 u`
Molecular formula `= n xx` Empirical formula
`= ((146 u)/(73 u))(C_(3)H_(5)O_(2))`
`= C_(6)H_(10)O_(4)`
Shortcut method:
Molecular mass `= 2 xx` Vapor density `= 146 u`
Only `C_(6)H_(10)O_(4)(6 xx 12 + 10 xx 1 + 4 xx 16)` satisfies the given molecular mass.