An organic compound on analysis gave C = 42.8%, H = 7.20%, and N = 50%. Volume of 1g of the compound was found to be 200 ml at STP. Molecular formula of the compound is:

Options

A. `C_4H_8N_4`
B. `C_(16)H_(32)N_16`
C. `C_(12)H_(24)N_12`
D. `C_2H_4N_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_2N`
`200 ml = 1gm, " " 22400 ml =(22400)/(200)=112 gm`
M.W. = 112 gm
E.F.W. = `CH_2N= 12+ 2 + 14 = 28`
`n=(M.W.)/(E.F.W.)=112/28=4, MF= C_4H_8N_4`