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Q.

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

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a

$C_{2} H_{6} N_{2}$

b

$C_{3} H_{4} N_{2}$

c

$C_{6} H_{8} N_{2}$

d

$C_{9} H_{12} N_{3}$

answer is C.

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Detailed Solution

	C	H	N
	9	1	3.5
Molar ratio	$\frac{9}{12} = 0.75$	$\frac{1}{1} = 1$	$\frac{3.5}{14} = 0.25$
Simple molar ratio	$\frac{0.75}{0.25} = 3$	$\frac{1}{0.25} = 4$	$\frac{0.25}{0.25} = 1$

So, empirical formula = $C_{3} H_{4} N$

$Empirical formula weight = (3 \times 12 + 4 \times 1 + 14) = 54$

$n = \frac{108}{54} = 2$

$Molecular formula = {(C_{3} H_{4} N)}_{2}$

${=C}_{6} H_{8} N_{2}$

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