Question

A 0.4272 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgClby the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.1598 g.

What is the mass percentage of chlorine in the original compound?

Answer 1

molar mass of AgCl = 143.3 g/mol
number of mol of AgCl = (mass of AgCl)/(molar mass of AgCl)
= 1.1598/143.3
= (8.09*10^-3) mol

precipitation reation is
Ag + Cl --> AgCl
according to reaction
1 mol of AgCl is obtained from 1 mol of Cl
(8.09*10^-3) mol of AgCl is obtained from (8.09*10^-3) mol of Cl
so,
number of mol of Cl = (8.09*10^-3) mol

molar mass of Cl = 35.5 g/mol
mass of AgCl = (number of mol of Cl)*(molar mass of Cl)
= (8.09*10^-3)*35.5
= 0.287 g

mass % of Cl = {(mass of Cl)/(mass of sample)}*100
= (0.287/0.4272)*100
= 67.18 %

Answer: 67.18 %