In a compound `C, H, N` atoms are present in `9:1:3.5` by weight. Molecular weight of compound is `108`. Its molecular formula is:

Options

A. `C_(2)H_(6)N_(2)`
B. `C_(3)H_(4)N`
C. `C_(6)H_(8)N_(2)`
D. `C_(9)H_(12)N_(3)`

Answer

C

Explanation

Ratio of masses :C:H:N =9:1:3.5.
Ratio of atoms `:C:H:N=9/12:1/1:3.5/14`

`1:1/1xx12/9:3.5/14xx12/9`
=1:1.33: 0.33= 3:4:1
E.F. of the compound `=C_3H_4N`
E.F mass=`12xx3+4xx1+14xx1`
=36+4+14=54
But mol.mass =108
`n=("Mol.mass ")/("E.F. mass ")=108/54=2`
M.F.=`(E.F)_(n)=(C_(3)H_(4)N)_(2)=C_(6)H_(8)N_(2)`