QJEEM22Ch908e: JEE Main 2022 Chemistry - Class 11 JEEM22Ch908e Solution: Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3%…

Mole ratio of H, C and N

$= {{8.7} \over 1}:{{74} \over {12}}:{{17.\,3} \over {14}}$

$= 8\,.7:6.167:1.23$

$= {{8.7} \over {1.23}}:{{6.167} \over {1.23}}:{{1.23} \over {1.23}}$

$= 7:5:1$

$\therefore$ Emperical formula = ${{C_5}{H_7}N}$

$\therefore$ Molecular formula $= {\left( {{C_5}{H_7}N} \right)_n}$

Given molecular mass = 162

Molecular mass of ${\left( {{C_5}{H_7}N} \right)_n}$

$= (5 \times 12 + 7 \times 1 + 14) \times n$

$= (81) \times n$

$\therefore$ $81 \times n = 162$

$\Rightarrow n = 2$

$\therefore$ Molecular formula = C₁₀H₁₄N₂