An organic compound on analysis gave C = 42.8%, H = 7.2%, and N = 50%. Volume of 1 gm of the compound was found to be 200 ml at STP. Molecular formula of the compound is

Options

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

Answer

A

Explanation

`{:("C",:,"H",:,"N"),((42.8)/(12),:,(7.2)/(1),:,(50)/(14)),(3.56,:,7.2,:,3.57),(1,:,2,:,1):}`
`EF=CH_(2)N`
200 ml = 1 gm, `22400ml=(22400)/(200)=112gm`
M.W. = 112 gm
E.F.W. = `CH_(2)N=12+2+14=28`
`n=("M.W.")/("E.F.W.")=(112)/(28)=4, MF=C_(4)H_(8)N_(4)`