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 $N{{H}_{3}}$ along with a solid residue. The solid residue gives violet colour with the alkaline copper sulphate solution, the compound is:
A. $C{{H}_{3}}NCO$
B. $C{{H}_{3}}CON{{H}_{2}}$
C. ${{(N{{H}_{2}})}_{2}}CO$
D. $C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}$

Answer 1

Hint: It is a colourless, crystalline substance which decomposes before boiling. It is used as a fertilizer and feed supplement, as well as a starting material for the manufacture of plastic and drugs. It is the highest nitrogen containing fertilizer.

Complete answer:
Firstly, from the given question, we can write the ratio of the number of gram atoms among C, H, N and O as-
C : H : N : O = $\dfrac{20}{12}$ : $\dfrac{6.67}{1}$ : $\dfrac{46.67}{14}$ : $\dfrac{26.66}{16}$ = 1 : 4 :2 :1
We know that the atomic numbers of the atoms are: C = 6, H = 1, N = 7 and O = 8.
Now, we will calculate their percentages to their atomic masses.
Empirical formula = $C{{H}_{4}}{{N}_{2}}O$
Therefore, Empirical weight = 60g
Given, Molecular weight = 60g
Molecular formula = ${{(Empiricalformula)}_{n}}$
$n=\dfrac{Molecular Weight}{Empiricalweight}$ = $\dfrac{60}{60}$ = 1
So, molecular formula is ${{(N{{H}_{2}})}_{2}}CO$
When gently heated, urea loses ammonia to form Biuret.
${{H}_{2}}NCO-N{{H}_{2}}+NHCON{{H}_{2}}\xrightarrow{-
N{{H}_{3}}}{{H}_{2}}NCONHCON{{H}_{2}}$ (Biuret)
When an aqueous biuret solution is treated with sodium hydroxide solution and a drop of copper sulphate solution, a violet colour is produced. This is known as the Biuret.
Now, let’s know more about the compound.
Urea is a nitrogenous compound containing a carbonyl group attached to two amine groups with osmotic diuretic activity. Urea is a highly soluble organic compound formed in the liver from ammonia used by the deamination of amino acids. Urea appears as solid odourless white crystals.
Therefore, the compound which formed is ${{(N{{H}_{2}})}_{2}}CO$

Note: Biuret test, also known as Piotrowski’s test, is a chemical test used for detecting the presence of peptide bonds. In the presence of peptides, a copper (II) ion forms mauve-coloured coordination complexes in an alkaline solution.