An organic compound having molecular mass 60 | vedclass

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Question

(C) The percentage of oxygen is $100 - (20 + 6.67 + 46.67) = 26.66\%$. The ratio of the number of gram atoms among $C$,$H$,$N$,and $O$ is: $C : H : N

Vedclass · Accepted Answer

$(NH_2)_2CO$

Vedclass · Answer

(C) The percentage of oxygen is $100 - (20 + 6.67 + 46.67) = 26.66\%$. The ratio of the number of gram atoms among $C$,$H$,$N$,and $O$ is: $C : H : N : O = \frac{20}{12} : \frac{6.67}{1} : \frac{46.67}{14} : \frac{26.66}{16} = 1.66 : 6.67 : 3.33 : 1.66$. Dividing by the smallest value $(1.66)$: $C : H : N : O = 1 : 4 : 2 : 1$. The empirical formula is $CH_4N_2O$. The empirical formula weight is $12 + 4 + 28 + 16 = 60 \ g/mol$. Since the molecular mass is $60$,the molecular formula is also $CH_4N_2O$,which is urea,$(NH_2)_2CO$. Urea on heating gives biuret as a solid residue,which gives a violet color with alkaline copper sulphate (biuret test).

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

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