What will be the molecular formula of the compound of an organic compound contains 49.3 % carbon 6.84 % hydrogen and rest is oxygen and its vapour density is 73 .

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Element No.of moles Simple ratio C = 12 49.3 12 = 4.1 4 . 1 2.7 = 1.5 × 2 = 3 H = 1 6.84 1 = 6.84 6.84 2.7 = 2.5 × 2 = 5 O = 16 43.86 16 = 2.7 2.7 2.7 = 1 × 2 = 2 Empirical formula = C 3 H 5 O 2 E . F . wt = 12 · 3 + 1 · 5 + 16 · 2 = 73 M o l e c u l a r w t = V . D . 2 = 73 × 2 = 146 n = M . wt E . F . wt = 146 73 = 2 Molecular formula = E .F n = C 3 H 5 O 2 = C 6 H 10 O 4

Element	No.of moles	Simple ratio
$C = 12$	$\frac{49.3}{12} = 4.1$	$\frac{4.1}{2.7} = 1.5 \times 2 = 3$
$H = 1$	$\frac{6.84}{1} = 6.84$	$\frac{6.84}{2.7} = 2.5 \times 2 = 5$
$O = 16$	$\frac{43.86}{16} = 2.7$	$\frac{2.7}{2.7} = 1 \times 2 = 2$

Chemistry - Mole Concept Question with Solution | TestHub

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