An organic compound on analysis was found to contain 16.27% carbon, 0.67% hydrogen and 72.2% chlorine. Remaining is oxygen. The vapour density of the compound is equal to 73.75. Calculate the empirical formula and molecular formula of organic compound.

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Solution

Given: Carbon: $16.27 = \frac{16.27}{12} = 1.355$

Hydrogen: $0.67 = \frac{0.67}{1} = 0.67$

Chlorine: $72.2 = \frac{72.2}{35.5} = 2.03$

Oxygen: $10.86 = \frac{10.86}{16} = 0.67$

Mole ratio $C : H : C l : O = 2 : 1 : 3 : 1$

Emperical formula= $C_{2} H C l_{3} O$

E.F. weight= $24 + 1 + 106.5 + 16 = 147.5$

Given, vapour density= $73.75$

Molecular weight= $2 \times V . D$
$= 2 \times 73.75$
$= 147.5$ .
$∴$ Molecular Formula $C_{2} H C l_{3} O$

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