Question

An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of the compound is :
(a) $C_3H_5O_2$
(b) $C_4H_{10}O_2$
(c) $C_6H_{10}O_4$
(d) $C_3H_{10}O_2$

Answer 1

Empirical formula =C3H5O2C3H5O2

n=Mol.massEmp.massn=Mol.massEmp.mass

=2×7373=2×7373

=2=2

Molecular formula =2×C3H5O2=2×C3H5O2

=C6H10O4=C6H10O4

Hence (C) is the correct answer.

Answer 2

$\textbf{Answer}$

$c) C_{6}H_{10}O_{2}$

$\textbf{Explanation}$

» Given :

% of C = 49.3

% of H = 6.84

% of O = 43.86

And vapour density = 73 (Molecular Mass)

° Refer the Attachment !!

(There %age = percentage

At. Mass = Atomic Mass

S.W.N.R = Simple While Number Ratio

S.R = Simple Ratio)

Empirical Formula = $C_{3}H_{5}O$

Empirical Formula Mass = 3 × 12 + 5 × 1 + 2 × 16

= 73

Now..

Empirical Mass × n = Molecular Mass

n = $\dfrac{Molecular\:Mass}{Empirical\:Mass}$

n = $\dfrac{2\:\times\:73}{73}$

n = 2

Molecular Formula = n × Empirical Formula

= 2 × $C_{3}H_{5}O$

= $C_{6}H_{10}O_{2}$

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