A 05051 g sample of a pure soluble chloride compound is diss

A 0.5051 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.5204 g.

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

Solution

Molar mass of AgCl,
MM = 1*MM(Ag) + 1*MM(Cl)
= 1*107.9 + 1*35.45
= 143.35 g/mol


mass(AgCl)= 1.5204 g

use:
number of mol of AgCl,
n = mass of AgCl/molar mass of AgCl
=(1.5204 g)/(143.35 g/mol)
= 1.061*10^-2 mol
This is number of moles of AgCl

one mole of AgCl contains 1 moles of Cl
use:
number of moles of Cl = 1 * number of moles of AgCl
= 1 * 1.061*10^-2
= 1.061*10^-2

Molar mass of Cl = 35.45 g/mol

use:
mass of Cl,
m = number of mol * molar mass
= 1.061*10^-2 mol * 35.45 g/mol
= 0.376 g

Mass % of Cl = mass of Cl * 100 / mass of sample
= 0.376*100/0.5051
= 74.4 %
Answer: 74.4 %