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An organic compound, whose vapour density is 45, has the following percentage composition, H = 2.22%, O = 71.19% and remaining carbon. Calculate its empirical formula. - Chemistry

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Question

An organic compound, whose vapour density is 45, has the following percentage composition,

H = 2.22%, O = 71.19% and remaining carbon. Calculate its empirical formula.

Numerical

Solution

As we know, molecular weight is 2 x vapour density.

∴ Molecular weight of compound = 2 x 45 = 90

∵ H2 + O2 = 2.22 + 71.19 = 73.41 < 100

The third element of organic compound C = 100 − 73.41 = 26.59%

Element Percentage Atomic Weight Ratio Simplest Ratio
C 26.59 12 `26.59/12` = 2.22 `2.22/2.22` = 1
H 2.22 1 `2.22/1` = 2.22 `2.22/2.22` = 1
O 71.19 16 `71.19/16` = 4.44 `4.44/2.22` = 2

∴ The empirical formula of the given organic compound = CHO2

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Determination of Empirical Formula
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Chapter 5: Mole concept and Stoichiometry - Exercise 5C [Page 90]

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Selina Concise Chemistry [English] Class 10 ICSE
Chapter 5 Mole concept and Stoichiometry
Exercise 5C | Q 12 | Page 90

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