Question

A compound on analysis found to contain following percentage composition C
= 11.3 %, and 0 = 45.3%. Determine the empirical and molecular formulae of this
compound. The relative molecular mass of the compound is 106. (A
wiecular mass of the compound is 106. (Atomic Mass of Na = 23,
C= 12,0 = 16)
​

Answer 1

Answer:

Here is your answer

Explanation:

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_2ClCH

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_2H_4Cl_2C

2

H

4

Cl

2

.