Chemistry

On analysis, a substance was found to contain
C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44, calculate;
(a) it's empirical formula, and
(b) it's molecular formula

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	54.54	12	$\dfrac{54.54}{12}$ = 4.545	$\dfrac{ 4.545 }{2.275 }$ = 1.99 = 2
Hydrogen	9.09	1	$\dfrac{9.09 }{1}$ = 9.09	$\dfrac{9.09 }{ 2.275 }$ = 3.99 = 4
Oxygen	36.36	16	$\dfrac{36.36}{16}$ = 2.275	$\dfrac{2.275 }{ 2.275 }$ = 1

Simplest ratio of whole numbers = C : H : O = 2 : 4 : 1

Hence, empirical formula is C₂H₄O

Empirical formula weight = 2(12) + 4(1) + 16 = 24 + 4 + 16 = 44

V.D. = 44

Molecular weight = 2 x V.D. = 2 x 44 = 88

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{88}{44} = 2$

So, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

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