Question

An organic compound whose empirical and molecular formula
are same, contains 20% carbon, 6.7% hydrogen, 46.7%
nitrogen and the rest oxygen. On heating it yields ammonia,
leaving a solid residue. The solid residue gives a violet colour
with dilute solution of alkaline copper sulphate. The organic
compound is
(a) NH₂COONH₄ (b) HCOONH₄
(c) NH₂NHCHO (d) NH₂CONH₂

Answer 1

Explanation:

NH₂COONH₄ is the organic compound whose empirical and molecular formula

are same, contains 20% carbon, 6.7% hydrogen, 46.7%

nitrogen and the rest oxygen. On heating it yields ammonia,

leaving a solid residue. The solid residue gives a violet colour

with dilute solution of alkaline copper sulphate

Answer 2

The organic  compound is NH₂CONH₂

Explanation:

From the given data,

$\%$ C = 20, $\%$ H = 6.7, $\%$ N = 46.7

Remaining is oxygen. That is

$\%$ O = 100-(20+6.7+46.7)= 26.6

No. of moles of  C = $\frac{20}{12.011}=1.67$

No. of moles of H =$\frac{6.7}{1.00794}=6.61$  

No. of moles of N = $\frac{47.33}{14.01} =3.38$

No. of moles of O = $\frac{26.6}{15.999} =1.63$

To get the empirical formula, we have to divide all quantities with the smallest one,

Thus, the number of carbons = $\frac{1.67}{1.63} \approx 1$

the number of hydrogen's = $\frac{6.61}{1.63} \approx 4$  

the number of nitrogen's = $\frac{3.38}{1.63} \approx 2$

the number of oxygen's = $\frac{1.63}{1.63} \approx 1$

Therefore the empirical formula of the compound is $CH_4N_2O$

This is nothing but urea. i.e., $H_2NCONH_2$

Therefore , the correct answer is option (d)