Question

A compound contains 4.07 % hydrogen,
24.27 % carbon and 71.65 % chlorine.
Its molar mass is 98.96 g. What are its
empirical and molecular formulas ?​

Answer 1

step 1

Divide the given percentages of atoms with their molecular masses

H---4.07/1 =4.07

C---24.27/12 =2.02

Cl--71.65/35.5=2.01

step 2

divide all values wstep 1

Divide the given percentages of atoms with their molecular masses

H---4.07/1 =4.07

C---24.27/12 =2.02

Cl--71.65/35.5=2.01

step 2

divide all values with the lowest value obtained.

H---4.07/2.01=2

C---2.02/2.01=1

Cl---2.01/2.01=1

therefore the empirical formula is

CH

2

Cl

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C

2

H

4

Cl

2

.ith the lowest value obtained.

H---4.07/2.01=2

C---2.02/2.01=1

Cl---2.01/2.01=1

therefore the empirical formula is

CH

2

Cl

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C

2

H

4

Cl

2

.

Answer 2

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step 1

Divide the given percentages of  atoms with their molecular masses

H---4.07/1  =4.07

C---24.27/12 =2.02

Cl--71.65/35.5=2.01

step 2

divide all values with the lowest value obtained.

 H---4.07/2.01=2

 C---2.02/2.01=1

 Cl---2.01/2.01=1

therefore the empirical formula is

CH2Cl

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C2H4Cl2.

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