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

(D) $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}$

(A) $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{~N}_{2}$

(B) $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{~N}$

(C) $\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{~N}$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Some Basic Concepts of Chemistry

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}

(A) $C_{4} H_{6} N_{2}$
(B) $C_{2} H_{3} N$
(C) $C_{5} H_{7} N$
(D) $C_{10} H_{14} N_{2}$ ✓ Correct

Answer: (D)

Solution

To find the molecular formula, first determine the empirical formula from percentage composition. Convert percentages to moles: C: 74/12 = 6.16, H: 8.7/1 = 8.7, N: 17.3/14 = 1.23. Divide by the smallest (1.23) to get the simplest ratio: C:H:N = 5:7:1. The empirical formula is

C_{5} H_{7} N

with empirical mass = 5(12) + 7(1) + 14 = 81 g/mol. The multiplying factor = 162/81 = 2. Therefore, the molecular formula is

C_{10} H_{14} N_{2}

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Some Basic Concepts of Chemistry

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