Chemistry

An organic compound contains 4.07% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96. Find its,
(a) Empirical formula
(b) Molecular formula

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	4.07	1	$\dfrac{4.07 }{1}$ = 4.07	$\dfrac{4.07 }{2.01}$ = 2
chlorine	71.65	35.5	$\dfrac{ 71.65}{35.5}$ = 2.01	$\dfrac{2.01}{2.01 }$ = 1
Carbon	24.28	12	$\dfrac{24.28 }{12}$ = 2.02	$\dfrac{ 2.02 }{2.01}$ = 1

Simplest ratio of whole numbers = H : Cl : C = 2 : 1 : 1

Hence, empirical formula is CH₂Cl

Empirical formula weight = 12 + (2 x 1) + 35.5 = 49.5

molar mass = 98.96

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{98.96}{49.5} = 1.99 = 2$

So, molecular formula = 2(CH₂Cl) = C₂H₄Cl₂

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