Question

A compound gave the following data : C = 57.82%, O = 38.58% and the rest hydrogen. It’s vapour density is 83. Find it’s empirical and molecular formula. [C = 12, O = 16, H = 1]

August 2, 2024

Answer

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	57.82	12	57.12/12 = 4.81	4.81/2.41 = 2 × 2 = 4
Oxygen	38.58	16	38.58/16 = 2.41	2.41/2.41 = 1 × 2 = 2
Hydrogen	3.60	1	3.60/1 = 3.6	3.60/2.41 = 1.5 × 2 = 3

C : O : H = 4 : 2 : 3

Simplest ratio of whole numbers = 4 : 2 : 3

Hence, the empirical formula is C₄O₂H₃ or C₄H₃O₂

Empirical formula weight = (4 × 12) + (3 × 1) + (2 × 16)

= 48 + 3 + 32

= 83

Vapour density (V.D.) = 83

Molecular weight = 2 × V.D.

= 2 × 83 = 166

n = $Molecular\ weight\over Empirical\ formula\ weight$

= $166\over 83$ = 2

Molecular formula = (C₄H₃O₂)₂ = C₈H₆O₄

∴ Molecular formula = C₈H₆O₄

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Question

A compound gave the following data : C = 57.82%, O = 38.58% and the rest hydrogen. It’s vapour density is 83. Find it’s empirical and molecular formula. [C = 12, O = 16, H = 1]

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