Question

21. A compound contains 4.07% H, 24.27% C and 71.65% Cl. What is its empirical
formula? (Atomic mass of H=1, C=12 and Cl=35.5)
Chemistry

Answer 1

Answer:

Explanation:

I can  explain it in 2 steps

$\underline{\mathfrak{step 1}}$

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

$\underline{\mathfrak{step 2}}$

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

So the empirical formula is

$CH_2Cl$