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Chemistry

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

H = 2.22%, O = 71.19% and remaining carbon.

Calculate,

(a) it's empirical formula, and

(b) it's molecular formula

Mole Concept

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Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Hydrogen2.2212.221\dfrac{2.22}{1} = 2.222.222.21\dfrac{2.22 }{2.21} = 1
Oxygen71.191671.1916\dfrac{71.19}{16} = 4.444.442.21\dfrac{4.44}{2.21 } = 2
Carbon26.591226.5912\dfrac{26.59}{12} = 2.212.212.21\dfrac{ 2.21 }{2.21} = 1

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO2

Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

n=Molecular weightEmpirical formula weight=9045=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2

So, molecular formula = 2(CHO2) = C2H2O4

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