In a compound C, H and N are present in 9:1:3.5 by weight. If molecular weight of compound is 108, the molecular formula of compound is

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MARKS Content Team · Accepted Answer

Correct answer: (C) C_6H_8 N_2. Element Ratio (by weight) Atomic weight Mol Ratio Simplest Mol Ratio C 9 12 9/12=0.75 0.75/0.25=3 H 1 1 1/1=1.00 1.00/0.25=4 N 3.5 14 3.5/14=0.25 0.25/0.25=1 ∴ Empirical formula =C_3H_4 N The empirical formula mass =3×12+4×1+14=54 g Given, molecular mass =108 g Thus, n=108/54=2 Hence, molecular formula = (Empirical formula) ×2 C_3H_4 N×2=C_6H_8 N_2

Element	Ratio (by weight)	Atomic weight	Mol Ratio	Simplest Mol Ratio
$C$	$9$	$12$	$\frac{9}{12} = 0.75$	$\frac{0.75}{0.25} = 3$
$H$	$1$	$1$	$\frac{1}{1} = 1.00$	$\frac{1.00}{0.25} = 4$
$N$	$3.5$	$14$	$\frac{3.5}{14} = 0.25$	$\frac{0.25}{0.25} = 1$

In a compound $C, H$ and $N$ are present in $9 : 1 : 3.5$ by weight. If molecular weight of compound is $108$ , the molecular formula of compound is

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