Chemistry, asked by TAPASHMISHRA2104, 8 months ago

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Answers

Answered by shushritha
2

Answer:

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

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 C2H4Cl2

Answered by sonamazhar11
2

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

CH2Cl2

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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