SCH3U
7.6 QUANTITATIVE ANALYSIS
Solution stoichiometry is often used in quantitative analysis, which involves
determining a quantity (how much) of a substance is present in a sample.
•
Recall that in Stoichiometry the mole ratio provides a necessary conversion factor:
molar mass of x
molar mass of y
grams x   moles of x  MOLE RATIO  moles of y   grams of y
We can do something similar for solutions:
molar concentration of x
molar concentration of y
volume x   moles of x  MOLE RATIO  moles of y   volume of y
•
In solution stoichiometry, known volumes and concentrations of reactants or
products are used to determine the volumes, concentrations, or masses of other
reactants or products.
Recall the following equations:
Converting grams to moles
n
Converting moles to grams
m  nM
Converting concentration to moles
n  cV
Converting moles to concentration
c
m
M
n
V
Examples
Example 1 – Finding the Minimum Volume to Precipitate
In this type of problem, you are trying to find the amount of one reactant needed to
combine with a specific amount of another reactant in order to create a precipitate.
A solution of sodium chloride (concentration = 0.0823 mol/L) is added to 21.40 mL of
silver nitrate (concentration = 0.962 mol/L), resulting in a precipitate of silver chloride.
Calculate the minimum volume of the sodium chloride solution required to precipitate all
of the silver ions.
Solution
First, write out the equation for this reaction, along with your knowns.
NaCl(aq)
+
AgNO3(aq)

AgCl(s) +
NaNO3(aq)
c = 0.0823 mol/L
V=?
c = 0.962 mol/L
V = 21.40 mL
= 0.02140 L
Calculate the number of moles of AgNO3, using the concentration and volume given.
n
= cV
= (0.962 mol/L)(0.02140 L)
= 0.02059 mol AgNO3
Use a mole ratio to determine the number of moles of NaCl needed. In this case, the ratio
of AgNO3:NaCl is 1:1.
nNaCl = nAgNO3 = 0.02059 mol NaCl
Calculate the volume of NaCl required, using the given concentration and the number of
moles.
V
=
=
=
n
C
0.02059 mol
0.0823 mol/L
0.250 L
(250 mL)
250 mL of 0.0823 mol/L NaCl solution is required to precipitate all the silver ions.
TRY THIS ONE:
What volume of 0.100 mol/L Na3PO4 is required to precipitate all the lead(II) ions from
150.0 mL of 0.250 mol/L Pb(NO3)2?
Example 2 – Finding the Mass of a Precipitated Compound
Given the concentration and volume of a particular reactant, you can calculate the mass
of precipitate that will form.
If excess barium chloride is added to 50.0 mL of 0.469 mol/L potassium chromate
solution, what mass of barium chromate will precipitate?
Solution
First, write out the equation for this reaction, along with your knowns.
BaCl2(aq)
+
K2CrO4(aq) 
V = 50.0 mL
C = 0.469 mol/L
BaCrO4(s)
+
m=?
M = 253.33 g/mol
2KCl(aq)
Using the given c and V, calculate the number of moles of potassium chromate.
n
= cV
= (0.469 mol/L)(0.050 L)
= 0.02345 mol
Use a mole ratio to determine the number of moles of BaCrO4 that will be formed. In this
case, the ratio of K2CrO4:BaCrO4 is 1:1.
nBaCrO4 = nK2CrO4 = 0.02345 mol
Calculate the mass of precipitate by multiplying the number of moles by the molar mass
of BaCrO4.
m
= (0.02345 mol)(253.33 g/mol)\
= 5.94 g
TRY THIS ONE:
How many grams of silver chloride will precipitate if 100.0 mL of
0.20 mol/L silver nitrate is combined with excess calcium chloride solution?
Example 3 – Solutions and Limiting Reactants
If you are given the amount of two reactants, and are asked to find the amount of product
formed, you must first find the limiting reactant.
Consider the following reaction:
Pb(CH3COO)2(aq) + Na2S(aq)  PbS(s) + 2NaCH3COO(aq)
If 257.8 mL of a 0.0468 mol/L solution of lead(II) acetate is added to 156.0 mL of a
0.095 mol/L solution of sodium sulfide, what mass of solid lead(II) sulfide will be
formed?
Solution
First, find the limiting reactant: calculate the number of moles of each reactant, and
divide by the coefficient from the balanced equation.
Pb(CH3COO)2(aq)
+ Na2S(aq)  PbS(s) + 2NaCH3COO(aq)
n = cV
= (0.0468 mol/L)(0.2578 L)
= 0.01207 mol/1
= 0.01207 mol
n = cV
= (0.095 mol/L)(0.156 L)
= 0.01482 mol/1
= 0.01482 mol
*limiting reactant
Use a mole ratio to determine the number of moles of PbS that will be formed. In this
case, the ratio of Pb(CH3COO)2: PbS is 1:1.
nPbS = nPb(CH3COO)2 = 0.01207 mol
Using the molar mass of PbS, find the mass of the precipitate.
m
= (0.01207 mol)(239.27 g/mol)
= 2.89 g PbS
TRY THIS ONE:
When aqueous solutions of silver nitrate and sodium chloride are mixed, silver chloride
is precipitated. What mass of silver chloride would be formed by combining 75.00 mL of
3.17 mol/L NaCl with 128 mL of 2.44 mol/L AgNO3?