Question

An organic compound on anlysis was found to contain 10.06% carbon,0.84%hydrogen and 89.10% chlorine.Calculate its empirical formula

Answer 1

Answer:

CHCl3

Explanation:

Carbon. 10.06/12 = 0.8383

hydrogen 0.84/1. = 0.84

chlorine. 89.1/35.5= 2.51

thus

simple ratios are 1:1:3

therefore CHCl3 is empirical formula

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Answer 2

The organic compound's empirical formula is $CHCl_{3}$.

Explanation:

Given:

An organic compound contains 10.06% carbon,0.84% hydrogen and 89.10% chlorine.

To Find:

The organic compound's empirical formula is to be found out.

Solution:

As given, an organic compound contains 10.06% carbon,0.84%hydrogen and 89.10% chlorine.

Carbon

% of  the carbon= 10.06

The atomic mass of C =12

The molar ratio $=\frac{10.06}{12} =0.8383$

The simpler molar ratio  $=\frac{0.84}{0.84}=1$

Hydrogen

% of  the hydrogen = 10.06

The atomic mass of H =1

The molar ratio $=\frac{0.84}{1} =0.84$.

The simpler molar ratio  $=\frac{0.84}{0.84}=1$

Chlorine

% of  the chlorine  = 89.10

The atomic mass of Cl = 35.5

The molar ratio $=\frac{89.10}{35.3} =2.5$

The simpler molar ratio  $=\frac{2.5}{0.84}=3$

Therefore, the simpler molar ratio C:H:Cl = 1:1:3

Thus, the organic compound empirical formula is $CHCl_{3}$.

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