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.
This is a problem with two equations and two unknowns and they are solved simultaneously.
Let X = mass Ag
and Y = mass Cu
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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
To calculate the mass percentage of silver metal in the starting mixture, we need to determine the amount of silver that was present in the mixture and then calculate its mass percentage.
First, let's calculate the moles of nitrate ions in the solution.
We are given that the total concentration of the nitrate ion (NO3-) in the solution is 0.650 M and the volume of the solution is 0.100 L. Using the equation:
Moles of nitrate ions (NO3-) = Concentration × Volume
Moles of nitrate ions (NO3-) = 0.650 M × 0.100 L = 0.065 moles
Since both silver(I) nitrate (AgNO3) and copper(II) nitrate (Cu(NO3)2) produce nitrate ions when dissolved, we can assume that the moles of nitrate ions in the solution are equal to the sum of the moles of silver nitrate and copper nitrate dissolved. Therefore:
Moles of silver nitrate (AgNO3) = Moles of nitrate ions (NO3-)
Moles of silver nitrate (AgNO3) = 0.065 moles
The molar mass of AgNO3 is calculated as follows:
Ag (silver) = 107.87 g/mol
N (nitrogen) = 14.01 g/mol
O (oxygen) = 16.00 g/mol
Therefore, molar mass of AgNO3 = (107.87 g/mol) + (14.01 g/mol × 3) = 169.88 g/mol
Now, we can calculate the mass of silver nitrate dissolved:
Mass of silver nitrate (AgNO3) = Moles of silver nitrate (AgNO3) × Molar mass of AgNO3
Mass of silver nitrate (AgNO3) = 0.065 moles × 169.88 g/mol = 11.04 grams
Since silver(I) nitrate (AgNO3) contains one mole of silver (Ag) per mole of silver nitrate (AgNO3), the mass of silver in the starting mixture is equal to the mass of silver nitrate (AgNO3).
Therefore, the mass percentage of silver in the starting mixture is given by:
Mass percentage of silver = (Mass of silver / Mass of starting mixture) × 100
Mass percentage of silver = (11.04 g / 3.00 g) × 100 = 368.0%
So, the mass percentage of silver metal in the starting mixture is 368.0%.