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

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

CH2CI

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

Answer 2

Given,

• Hydrogen — 4.07%
• Carbon–24.27%
• Chlorine—71.65%
• Molar Mass Is 98.96G

To Calculate Empirical Formulae,

☆ First, we have to divide the given percentages of atoms by their molecular masses.

Hydrogen — $\sf{\frac{4.07 }{1}}$= 4.07

Carbon–$\sf{\frac{24.27 }{12}}$= 2.02

Chlorine—$\sf{\frac{71.65 }{35.5}}$ = 2.01

☆ Now,

Hydrogen — $\sf{\frac{4.07 }{2.01}}$= 2

Carbon–$\sf{\frac{2.02 }{2.01}}$= 1

Chlorine—$\sf{\frac{2.01 }{2.01}}$ = 1

Therefore, the empirical formula is $\bf{CH_{2}Cl}$

To Calculate Molecular Formulae, we need to find the weight of the empirical formula $\bf{CH_{2}Cl}$

=> 12 + 2×1 + 35.5

=> 49.5

The molecular weight- 98.96 is double of empirical weight. Therefore molecular formula is $\bf{C_{2}H_{4}Cl_{2}}$.