Question

A compound contains 4.07% hydrogen. 24.27% carbon and rest chlorine. Its molecular mass is 98.96. What are its empirical and molecular formula.



here, how to find "rest of chlorine" ?​

Answer 1

Answer:

Explanation:

Given data

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 — 4.07 / 1 = 4.07

Carbon–24.27 / 12 = 2.02

Chlorine—71.65 / 35.5 = 2.01

Now we have to divide all the values with the lowest obtained value.

Hydrogen — 4.07 / 2.01 = 2

Carbon–2.02 / 2.01 = 1

Chlorine— 2.01 / 2.01 = 1

Therefore, the empirical formula is CH2Cl

To Calculate Molecular Formulae, we need to find the weight of the empirical formula

CH2Cl

12 + 2*1 + 35.5

= 49.5

The molecular weight- 98.96 is double of empirical weight. Therefore molecular formula is

C2H4Cl2