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ICSE Class 10 Chemistry Question 57 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 57

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Question

Question 49(b-i)

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. (Atomic mass : C = 12, H = 1, Br = 80)

Answer
Element% compositionAt. wt.Relative no. of atomsSimplest ratio
C12.671212.6712\dfrac{12.67}{12} = 1.051.051.05\dfrac{1.05}{1.05} = 1
H2.1312.131\dfrac{2.13}{1} = 2.132.131.05\dfrac{2.13}{1.05} = 2.02 = 2
Br85.118085.1180\dfrac{85.11}{80} = 1.061.061.05\dfrac{1.06}{1.05} = 1

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

Hence, empirical formula is CH2Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

n=Molecular weightEmpirical formula weight=2×9494=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 94}{94} = 2

∴ Molecular formula = n[E.F.] = 2[CH2Br] = C2H4Br2