A compound contains 4.07% hydrogen,24.27%carbon, 71.65%chlorine. its molecular mass is 98.96. What is its empirical and molecular formula?

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Solution

Element Percentage Atomic Mass Relative no. of atoms Simplest ratio Whole no. ratio

Carbon 24.27 12 $\frac{24.27}{12} = 2.022$ $\frac{2.022}{2.018} = 1$ 1

Hydrogen 4.07 1 $\frac{4.07}{1} = 4.07$ $\frac{4.07}{2.018}$ 2

Chlorine 71.65 35.5 $\frac{71.65}{35.5} = 2.018$ $\frac{2.018}{2.018}$ 1

Emperical formula = CH₂Cl
Molecular formula = (CH₂Cl)_n

$n = \frac{M o l e c u l a r M a s s}{E m p e r i c a l f o r m u l a m a s s} = \frac{98.96}{49.5} = 2$

Therefore, molecular formula = (CH₂Cl)₂= C₂H₄Cl₂

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A compound contains 4.07% hydrogen, 24.27% Carbon and 71.65% Chlorine. Its molar mass is 90.96g. What are it's empirical and molecular formula?

A compound 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its emperical formula and molecular formula?

Q. A compound contains

4.07

% hydrogen,

24.27

% carbon and

71.65

% chlorine. It molar mass is

98 g

. What are its empirical and molecular formulas? (Atomic mass of C=12,H=1,Cl=35)

An organic compound contains 4.7% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96% find its,
(a) Empirical formula (b) Molecular formula

Q. A carbon compound contains

4.07

24.27

71.6

. Its molar mass is

98.96 g {m o l}^{- 1}

, What are its empirical and molecular formulae ?

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Element	Percentage	Atomic Mass	Relative no. of atoms	Simplest ratio	Whole no. ratio
Carbon	24.27	12	$\frac{24.27}{12} = 2.022$	$\frac{2.022}{2.018} = 1$	1
Hydrogen	4.07	1	$\frac{4.07}{1} = 4.07$	$\frac{4.07}{2.018}$	2
Chlorine	71.65	35.5	$\frac{71.65}{35.5} = 2.018$	$\frac{2.018}{2.018}$	1