Question 7(iii)An organic compound contains: H = 6.32 %, N = 17.76%. In the vapour state, this compound is 39.5 times as heavy as the same volume of hydrogen.(a) Find the molecular formula of the compound. (At wt: H = 1 N = 14 )(b) Calculate the number of hydrogen atoms in one mole of this compound.

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Accepted Answer

Given:Percentage of H = 6.32%Percentage of N = 17.76%Percentage of carbon = 100 - (6.32 + 17.76) = 75.92 %Atomic weights: H = 1, N = 14(a)Element% compositionAt. wt.Relative no. of atomsSimplest ratioHydrogen6.3216.321\dfrac{6.32 }{1}16.32​ = 6.326.321.27\dfrac{6.32 }{1.27}1.276.32​ = 5Nitrogen17.761417.7614\dfrac{ 17.76}{14}1417.76​ = 1.271.271.27\dfrac{1.27}{1.27 }1.271.27​ = 1Carbon75.921275.9212\dfrac{75.92 }{12}1275.92​ = 6.326.321.27\dfrac{ 6.32 }{1.27}1.276.32​ = 5Simplest ratio of whole numbers = H : N : C = 5 : 1 : 5Hence, empirical formula is C5H5NEmpirical formula weight = (5 × 12) + (5 × 1) + (1 × 14) = 60 + 5 + 14 = 79 g/molV.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(b) From the formula C5H5N, there are 5 hydrogen atoms per molecule.In 1 mole of the compound the number of molecules = 6.022 × 1023∴ Total hydrogen atoms = 5 × (6.022 × 1023) = 3.011 × 1024 hydrogen atoms.

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	6.32	1	$\dfrac{6.32 }{1}$ = 6.32	$\dfrac{6.32 }{1.27}$ = 5
Nitrogen	17.76	14	$\dfrac{ 17.76}{14}$ = 1.27	$\dfrac{1.27}{1.27 }$ = 1
Carbon	75.92	12	$\dfrac{75.92 }{12}$ = 6.32	$\dfrac{ 6.32 }{1.27}$ = 5

Model Paper 2 — Question 19

Question 7(iii)