Question

A compound containing 4.04% H , 24.47% C and 71.65% Cl .find the emperical formula.​

Answer 1

Answer:

CH2Cl

Explanation:

1st step) DIVIDE THESE PERCENTAGES WITH THEIR ATOMIC MASSES

H = 4.04÷1 =4.04

C = 24.47÷12 = 2.03

Cl =71.65÷35.5 = 2.01

2nd step) DIVIDE THR VALUES WITH THE LOWEST ONE OF THE THESE VALUES

IN THIS THE LOWEST VALUE IS OF Cl = 2.01 SO:

H = 4.04÷2.01 = 2.009

C = 2.03÷2.01 = 1.009

Cl = 2.01÷2.01 = 1

HERE COMES THE EMPIRICAL FORMULA WHICH BECOME "CH2Cl"

Hope it helps...