Question

An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives $NH_3$ alongwith?? solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is :

• A. $CH_3CH_2CONH_2$
• B. $(NH_2)_2CO$
• C. $CH_3CONH_2$
• D. $CH_3NCO$

Answer 1

Answer: (B)

$(NH_2)_2CO$

Answer 2

Empirical formula $= CH_4N_2O$ Empirical formula weight $= 12+ \left(4\times1\right)+ \left(2\times14\right)+ 16$ $= 60$ $\therefore n = \frac{Mol\,.\, formula \,weight}{Emp\,. \,formula\, weight} = \frac{60}{60} = 1$ $\therefore$ Molecular formula $= CH_{4}N_{2}O$ Given compound gives biuret test. Thus given compound is urea $\left(NH_{2}\right)_{2}CO.$ ${NH2CONH2 + HNHCONH2 ->[\Delta]}$ $\underset{\text{biuret}}{ {NH2CONHCONH2 }}$ ${+ NH3 ->[CuSO_4] }$ Violet colour