Determine the molecular formula of an organic compound containing 49.3% carbon 6.84% hydrogen and its vapour density is 73 .

Question

TestHub · Accepted Answer

Element Atomic mass Simple ratio C 12 49.3 12 = 4.1 4.1 2.7 = 1.5 ; 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 Empirical Formula wt. = 12 × 3 + 1 × 5 + 16 × 2 = 73 Molecular wt = V . D . × 2 = 73 × 2 = 146 n = M . w t E . F . w t = 146 73 = 2 Molecular formula = E . F n = ( C 3 H 5 O 2 ) 2 = C 6 H 10 O 4 .

Element	Atomic mass	Simple ratio
C	12	$\frac{49.3}{12} = 4.1$	$\frac{4.1}{2.7} = 1.5; 1.5 \times 2 = 3$
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 - Nomenclature Question with Solution | TestHub

Options:

Doubts & Discussion