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_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_2H`
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`