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

Percentage
`C=(9)/(13.5)xx100=66.6`
`H=(1)/(13.5)xx100=7.4`
`N=(3.5)/(13.5)xx100=26.0`
Atomic ration
`C=underset(=5.5)((66.6)/(12)),H=underset(=7.4)((7.4)/(1)),N=underset(=1.86)((26.0)/(14))`
Simplest ratio
`C = 3,H=4,N=1`
Empirical formula `= C_(3)H_(4)N`
Empirical formula mass `= 36 + 4 + 14 = 54`
`n=("Molecular mass")/("Empirical formula mass")`
`=(108)/(54)=2`
Molecular formula `= C_(6)H_(8)N_(2)`.