Question

If 0.9110 g of a sample of silver ore yielded 0.4162 g of AgCl in a gravimetric experiment, what is the percentage of Ag in the ore?

Answer 1

molar mass of AgCl = 143.3 g/mol
number of mol of AgCl = (mass of AgCl)/(molar mass of AgCl)
= 0.4162/143.3
= (2.91*10^-3) mol

reaction taking place is
Ag + Cl --> AgCl
according to reaction
1 mol of AgCl is obtained from 1 mol of Ag
(2.91*10^-3) mol of AgCl is obtained from (2.91*10^-3) mol of Ag
so,
number of mol of Ag required = (2.91*10^-3) mol

molar mass of Ag = 107.9 g/mol
mass of Ag = (number of mol of Ag required)*(molar mass of Ag)
= (2.91*10^-3)*107.9
= 0.314 g

mass % of Ag = {(mass of Ag)/(mass of sample)}*100
= (0.314/0.9110)*100
= 34.5 %

Answer: 34.5 %