Question 6(c)Find the empirical formula and the molecular formula of an organic compound from the data given below: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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Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon75.921275.9212\dfrac{75.92}{12}1275.92​ = 6.326.321.26\dfrac{ 6.32 }{ 1.26}1.266.32​ = 5Hydrogen6.3216.321\dfrac{6.32}{1}16.32​ = 6.326.321.26\dfrac{6.32 }{ 1.26}1.266.32​ = 5Nitrogen17.761417.7614\dfrac{ 17.76}{14}1417.76​ = 1.261.261.26\dfrac{ 1.26}{ 1.26}1.261.26​ = 1Simplest ratio of whole numbers = C : H : N = 5 : 5 : 1Hence, empirical formula is C5H5NEmpirical formula weight = 5(12) + 5(1) + 14 = 79V.D. = 39.5Molecular weight = 2 x V.D. = 2 x 39.5 = 79n=Molecular weightEmpirical formula weight=7979=1\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{79}{79} = 1n=Empirical formula weightMolecular weight​=7979​=1∴ Molecular formula = n[E.F.] = 1[C5H5N] = C5H5N

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

Solved 2019 Question Paper ICSE Class 10 Chemistry — Question 15

Question 6(c)