An organic compound on analysis gave the following percentage composition: C=57.8% ; H=3.6% and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound.

Asked by Prachi Panwar | 21st May, 2013, 06:42: PM

Expert Answer:

Element

% amount

Relative number of atoms

Ratio of number of atoms

Simplest ratio

C

57.8

(57.8/12) = 4.82

(4.82/2.41) = 2

2x2 = 4

H

3.6

(3.6/1) = 3.6

(3.6/2.41) = 1.5

1.5x2 = 3

O

(100-57.8-3.6) = 38.6

(38.6/16) = 2.41

(2.41/2.41) = 1

1x2=2

 

So, we can deduce the empirical formula of the organic compound =  C4H3O2.

The empirical Formula mass = 4x12 + 3x1 + 2x16 = 83 u.

The molecular mass of compound can be calculated as:

Molecular mass = 2 x Vapour Density = 2 x 83 = 166 u.

n = Molecular mass / Empirical Formula Mass = 166/83 = 2.

Molecular Formula of the compound = n(Empirical Formula) = 2(C4H3O2) = C8H6O4.

Answered by  | 21st May, 2013, 09:54: PM

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