CBSE Class 11-science Answered
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 | 21 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 | 21 May, 2013, 09:54: PM
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