Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is, Given : Atomic masses of C, H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 g mol - 1.

Some Basic Concepts of Chemistry PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q1 with Solution

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Correct answer: D

Assume 100 g of compound

Then the masses of elements are:
- Carbon = $74\,$ g
- Hydrogen = $8.7\,$ g
- Nitrogen = $17.3\,$ g
Convert masses into moles

Using atomic masses $C=12$ , $H=1$ , $N=14$ :

$n_C=\frac{74}{12}=6.1667$ $n_H=\frac{8.7}{1}=8.7$ $n_N=\frac{17.3}{14}=1.2357$
Find simplest mole ratio

Divide all by the smallest value $1.2357$ :

$\frac{n_C}{1.2357}=\frac{6.1667}{1.2357}\approx 4.99 \approx 5$ $\frac{n_H}{1.2357}=\frac{8.7}{1.2357}\approx 7.04 \approx 7$ $\frac{n_N}{1.2357}=\frac{1.2357}{1.2357}=1$

So the empirical formula is:

$\boxed{C_5H_7N}$
Calculate empirical formula mass

$5(12)+7(1)+1(14)=60+7+14=81$

Empirical formula mass $=81\,$ g mol $^{-1}$ .
Use molecular molar mass

Given molecular molar mass $=162\,$ g mol $^{-1}$ .

$\text{Multiplier} = \frac{162}{81}=2$

Therefore molecular formula is:

$\boxed{(C_5H_7N)_2=C_{10}H_{14}N_2}$
Check options
- A: $C_4H_6N_2$ → not correct
- B: $C_2H_3N$ → not correct
- C: $C_5H_7N$ → empirical formula only, not molecular formula
- D: $C_{10}H_{14}N_2$ → correct

Hence, the molecular formula is:

$\boxed{C_{10}H_{14}N_2}$