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_(2)`

Answer

C

Explanation

`{:("Element","%composition","At.wt"),("R,N.A",,"S.R"),(C,9,12),(,(9)/(12)=0.75,(0.75)/(0.25)=3),(H,1,1),(,(1)/(1)=1,(1.0)/(0.25)=4),(N,3.5,14),(,(3.5)/(14)=0.25,(0.25)/(0.25)=1):}`
`therefore E.F=C_(3)H_(4)N`
`therefore E.F.wt =3xx12+4xx1+1xx14=36+4+14=54`
`"M.wt"=108`
`n=(108)/(54)=2`
`M.F=(E.F)_(n)=(C_(3)H_(4)N)_(2)`
`=C_(6)H_(8)N_(2)`