An organic compound on analysis gave `C=48 g,H=8g` and `N=56 g.` Volume of `1.0 g` of the compound was foud to be `200ml` at `NTP.` Molecular formula of the compound is

Options

A. `C_(4)H_(8)N_(4)`
B. `C_(2)H_(4)N_(2)`
C. `C_(12)H_(24)N_(12)`
D. `C_(16)H_(32)N_(16)`

Answer

a

Explanation

`"Element % No. of mol es Simple ratio"`
`C" "48" "48//12=4" "1`
`H" "8" "8//1=8" "2`
`N" "56" "56//14=4" "1`
Empirical formula `CH_(2)N`
Emirical formula mass`=28`
Now, `200ml` of compound `=1g`
`22400ml` of compound `(1)/(200)xx22400=112`
`n=("Mol. mass")/("Emp formula mass")=(112)/(28)=4`
Therefore, Molecular formula `=9CH_(2)N_(4))=C_(4)H_(8)N_(4)`.