Chemistry

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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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	% 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

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

Hence, empirical formula is C₅H₅N

Empirical formula weight = (5 × 12) + (5 × 1) + (1 × 14) = 60 + 5 + 14 = 79 g/mol

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

(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.

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