A compound of carbon, hydrogen and nitrogen contains three elements in the respective weight ratio of 9 : 1 : 3. Calculate its empirical formula. If its molecular weight is 108, what is its molecular formula?

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Solution

$∴$ Emepricial formula = $C_{3} H_{4} N$

Emperical $w t = 54 g$

$∴$ n = $\frac{M o l . w t}{e m p . w t} = \frac{108}{54} = 2$
$∴$ Molecular formula = ( $C_{3} H_{4} N)_{2} = C_{6} H_{8} N_{2}$
( Note :- % ratio by wt. Nitorgen should be given 3.5 not 3)

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A compound of carbon, hydrogen and nitrogen contains three elements in the respective weight ratio of 9 : 1 : 3. Calculate its empirical formula. If its molecular weight is 108, what is its molecular formula?

∴ Emepricial formula = C3H4NEmperical wt=54g∴ n = Mol. wtemp. wt=10854=2∴ Molecular formula = (C3H4N)2=C6H8N2 ( Note :- % ratio by wt. Nitorgen should be given 3.5 not 3)

$∴$ Emepricial formula = $C_{3} H_{4} N$

Emperical $w t = 54 g$

$∴$ n = $\frac{M o l . w t}{e m p . w t} = \frac{108}{54} = 2$
$∴$ Molecular formula = ( $C_{3} H_{4} N)_{2} = C_{6} H_{8} N_{2}$
( Note :- % ratio by wt. Nitorgen should be given 3.5 not 3)