In a compound, $C$ , $H$ and $N$ atoms are present in ratio of $9 : 1 : 3.5$ by weight. Molecular weight of compound is $108$ . Molecular formula of compound is:

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

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C_{2} H_{6} N

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

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

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Solution

The correct option is C $C_{6} H_{8} N_{2}$
Ratio of atoms $=$ $C : H : N$ $= \frac{9}{12} : \frac{1}{1} : \frac{3.5}{14} = 3 : 4 : 1$

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

Since, $M o l e c u l a r F o r m u l a = (E m p i r i c a l F o r m u l a) \times n$
$n = \frac{M o l e c u l a r w e i g h t}{E m p i r i c a l w e i g h t} = \frac{108}{54} =$ 2

Therefore, molecular formula $= C_{6} H_{8} N_{2}$

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In an organic compound of molar mass 108 gram/mole C, H, and N atoms are present in 9:1:3.5 by weight find out the molecular formula of the compound

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