Question

A compound on analysis gave the following percentage composition C=54.55% H=9.09% O=36.36%. Determine the empirical formula of the compound
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Answer 1

GIVEN :-

• Amount of carbon = 54.55 g

• Amount of hydrogen = 9.09 g

• Amount of oxygen = 36.36 g

TO FIND :-

• Empirical formula .

SOLUTION :-

  We know ,

  GAM of carbon = 12 g

  GAM of hydrogen = 1 g

  GAM of oxygen = 16 g

  $\bullet \sf \ No \ of \ moles \ of \ carbon = \dfrac{54.55}{12}=4.55$

  $\bullet \sf \ No \ of \ moles \ of \ hydrogen = \dfrac{9.09}{1}=9.09$

  $\bullet \sf \ No \ of \ moles \ of \ oxygen = \dfrac{36.36}{16}=2.27$

 

 Simplest ratio ,

  $\longrightarrow \sf 4.55:9.09 :2.27 \\\\\longrightarrow \dfrac{4.55}{2.27}:\dfrac{9.09}{2.27}:\dfrac{2.27}{2.27}\\\\\longrightarrow \bf 2: 4 : 1$

  Empirical formula of the compound = $\bf C_2H_4O$