Chemistry

Find the empirical and molecular formula of an organic compound from the data given: C = 75.92%, H = 6.32% and N = 17.76%. The vapour density of the compound is 39.5
[C = 12, H = 1, N = 14]

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	75.92	12	$\dfrac{75.92}{12}$ = 6.32	$\dfrac{ 6.32 }{ 1.26}$ = 5
Hydrogen	6.32	1	$\dfrac{6.32}{1}$ = 6.32	$\dfrac{6.32 }{ 1.26}$ = 5
Nitrogen	17.76	14	$\dfrac{ 17.76}{14}$ = 1.26	$\dfrac{ 1.26}{ 1.26}$ = 1

Simplest ratio of whole numbers = C : H : N = 5 : 5 : 1

Hence, empirical formula is C₅H₅N

Empirical formula weight = 5(12) + 5(1) + 14 = 79

V.D. = 39.5

Molecular weight = 2 x V.D. = 2 x 39.5 = 79

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{79}{79} = 1$

∴ Molecular formula = n[E.F.] = 1[C₅H₅N] = C₅H₅N

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