Question

A mixture of 3.00 g of silver and copper metals was dissolved in excess nitric acid.
Resulting salts, silver(I) nitrate and copper(II) nitrate, were isolated and dissolved in enough
water to make 0.100 L of a solution with a total concentration of the nitrate ion equal to 0.650 M.
Calculate the mass percentage of silver metal in the starting mixture.

Answer 1

This is a problem with two equations and two unknowns and they are solved simultaneously.
Let X = mass Ag
and Y = mass Cu
-----------------
equation 1 is X + Y = 3.00

To get the second equation note that mols NO3^- = M x L = 0.650 M x 0.1L = 0.0650 mols. Therefore, mols NO3^- from Ag + mols NO3^- from Cu = 0.0650. To save typing I'll let am stand for atomic mass. Therefore, equation 2 is
(X/am Ag) + (2Y/am Cu) = 0.0650

Solve those two equations for X and Y simultaneously, then
%Ag = (mass Ag/mass sample)* 100 = (X/3.0)*100 = % Ag (mass percent)

Sol