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In an organic compound of molar mass $108 g m m o l e^{- 1} C$ , $H$ and N are present in 9:1:3.5 by mass. Molecular formula can be

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$C_{5} H_{6} N_{3}$

$C_{4} H_{18} N_{3}$

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

$C_{7} H_{10} N$

answer is A.

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Detailed Solution

C H N
Moles : $\frac{9}{12}$ : $\frac{1}{1}$ : $\frac{3.5}{14}$ $n = \frac{108}{54} = 2$

$\frac{3}{4}$ : 1 : $\frac{1}{4}$

3 : 4 : 1

E.M $= C_{3} H_{4} N = M . F = {(C_{3} H_{4} N)}_{n}$
MF $= {(C_{3} H_{4} N)}_{2} = C_{6} H_{8} N_{2}$

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