Chapter 4
Solution Stoichiometry
Dr. Sapna Gupta
Concentrations of Solutions
• A solution is solute dissolved in a solvent.
• To quantify and know exactly how much of a solute is present in a
certain amount of solvent, one will need to calculate concentrations.
• Concentrations are given in
• percent solutions
• mass/mass % - (g of solute/g of solution) x 100%
• mass/volume % - (g of solute/mL of solution) x 100%
• volume/volume % - (mL of solute/mL of solution) x 100%
• Molarity (mol/L of solution)- used more in Chemistry
• Molality (mol/Kg of solution) – this is used more in Biology
Dr. Sapna Gupta/Solution Stoichiometry
2
Molar Concentration, Molarity (M)
In this chapter we will study Molarity – which is moles in a L of solution.
• Molarity is represented by M and the formula is given below
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Molarity =
𝐿 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
• Moles are converted to grams in order to make the solution in the lab.
• To prepare a solution, add the measured amount of solute to a volumetric
flask, then add water to bring the solution to the mark on the flask.
• A 3M solution of NaCl means there are 3 moles of NaCl in the solution.
• If you have a 200 mL of 2 M HCl – that means that there are 2 mols of HCl in 1
L solution. If you want to know how many grams of HCl you have in 200 mL
then you will have to calculate the amount of moles in 200 mL of that
solution using the Molarity equation; then you can calculate the grams from
those moles.
Dr. Sapna Gupta/Solution Stoichiometry
3
Example: You place a 1.52−g of potassium dichromate, K2Cr2O7, into a
50.0−mL volumetric flask. You then add water to bring the solution up to
the mark on the neck of the flask. What is the molarity of K2Cr2O7 in the
solution?
Molar mass of K2Cr2O7 is 294 g/mol
1 mol
294 g
 0.103 M
3
50.0  10 L
1.52 g
Example: A solution of sodium chloride used for intravenous transfusion
(physiological saline solution) has a concentration of 0.154 M NaCl. How
many moles of NaCl are contained in 500.−mL of physiological saline?
How many grams of NaCl are in the 500.−mL of solution?
M=
𝑚𝑜𝑙
𝐿
mol  M  L
Molar mass NaCl  58.4 g
58.4 g
1 mol
 0.154 M  0.500 L
0.0770 mol
 0.0770 mol NaCl
 4.50 g NaCl
Dr. Sapna Gupta/Solution Stoichiometry
4
Example: Calculate the molarity of a solution prepared by dissolving
45.00 grams of KI into a total volume of 500.0 mL.
45.00 g KI 1 mol KI 1000 mL


 0.5422 M
500.0 mL 166.0 g KI
1L
Example: How many milliliters of 3.50 M NaOH can be prepared from 75.00
grams of the solid?
75.00 g NaOH 
1 mol NaOH
1L
1000 mL


 536 mL
40.00 g NaOH 3.50 mol NaOH
1L
Dr. Sapna Gupta/Solution Stoichiometry
5
Dilution
• When a higher concentration solution is used to make a less−concentration
solution, the moles of solute are determined by the amount of the
higher−concentration solution.
• The number of moles of solute remains constant when more water is added.
MiVi = MfVf
Note:
The units on Vi and Vf must match.
Dr. Sapna Gupta/Solution Stoichiometry
6
Diluting a solution
quantitatively requires
specific glassware.
The photo at the right
shows a volumetric
flask used in dilution.
Dr. Sapna Gupta/Solution Stoichiometry
7
Example: A saturated stock solution of NaCl is 6.00 M. How much of this stock
solution is needed to prepare 1.00−L of physiological saline solution (0.154 M)?
M iVi  M fVf
Vi 
(0.154 M )(1.00 L)
6.00 M
Vi  0.0257 L or 25.7 mL
Vi 
M f Vf
Mi
Example: For the next experiment the class will need 250. mL of 0.10 M
CuCl2. There is a bottle of 2.0 M CuCl2. Describe how to prepare this solution.
How much of the 2.0 M solution do we need?
Concentrated: 2.0 M use ? mL (Vc)
Diluted: 250. mL of 0.10 M
McVc = MdVd
(2.0 M) (Vc) = (0.10 M) (250.mL)
Vc = 12.5 mL
12.5 mL of the concentrated solution are needed; add enough distilled water to
prepare 250. mL of the solution.
Dr. Sapna Gupta/Solution Stoichiometry
8
Solution Stoichiometry
• In solution stoichiometry you have to presume that soluble ionic
compounds dissociate completely in solution.
• Then using mole ratios we can calculate the concentration of all species
in solution.
• There are three common types of stoichiometric calculations
• Quantitative Analysis: The determination of the amount of a substance or
species present in a material.
• Volumetric Analysis: A type of quantitative analysis based on titration.
• Gravimetric Analysis: A type of quantitative analysis in which the amount
of a species in a material is determined by converting the species to a
product that can be isolated completely and weighed.
Dr. Sapna Gupta/Solution Stoichiometry
9
Volumetric Analysis - Titrations
A procedure for determining the amount of substance A by adding a carefully
measured volume with a known concentration of B until the reaction of A and B is
just complete. This can be for precipitation, neutralization or redox reactions.
• Standardization is the determination of the exact concentration of a solution.
• Equivalence point represents completion of the reaction.
• Endpoint is where the titration is stopped.
• An indicator is used to signal the endpoint.
Dr. Sapna Gupta/Solution Stoichiometry
10
Gravimetric Analysis
• In gravimetric analysis precipitation reactions are carried out.
• After the reaction the product is precipitated and collected in a crucible
or filter paper.
• The precipitate is weighed and then using mole ratios we can calculate
the concentration of all species in original solution.
The figure below shows the
reaction of Ba(NO3)2 with K2CrO4
forming the yellow BaCrO4
precipitate.
The BaCrO4 precipitate is being
filtered. It can then be dried and
weighed. Then concentration of
Ba2+ ions can be calculated
Dr. Sapna Gupta/Solution Stoichiometry
11
Example: Find the concentration of all species in a 0.25 M solution of MgCl2
MgCl2  Mg2+ + 2Cl
Given: MgCl2 = 0.25 M
[Mg2+ ] = 0.25 M (1:1 ratio)
[Cl ] = 0.50 M (1:2 ratio)
Example: A soluble silver compound was analyzed for the percentage of
silver by adding sodium chloride solution to precipitate the silver ion as
silver chloride. If 1.583 g of silver compound gave 1.788 g of silver
chloride, what is the mass percent of silver in the compound?
Strategy: g AgCl -> mol AgCl -> mol Ag -> g Ag -> % Ag
Molar mass of silver chloride (AgCl) = 143.32 g
1.788 g AgCl 
1mol AgCl
1mol Ag 107.9 g Ag


143.32 g AgCl 1mol AgCl 1mol Ag
1.346 g Ag
100%
1.583 g silver compound
= 1.346 g Ag in the compound
= 85.03% Ag
Dr. Sapna Gupta/Solution Stoichiometry
12
Example: Zinc sulfide reacts with hydrochloric acid to produce hydrogen
sulfide gas:
ZnS(s) + 2HCl(aq)  ZnCl2(aq) + H2S(g)
How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?
Strategy: g ZnS -> mol ZnS -> mol HCl (mol ratio from eq) -> vol HCl (using Molarity)
Molar mass of ZnS = 97.45 g
0.392 g ZnS 
1mol ZnS
2 mol HCl
1L solution


97.45 g ZnS 1mol ZnS 0.0512 mol HCl
= 0.157 L = 157 mL HCl solution
Dr. Sapna Gupta/Solution Stoichiometry
13
Example: A dilute solution of hydrogen peroxide is sold in drugstores as a
mild antiseptic. A typical solution was analyzed for the percentage of hydrogen
peroxide by titrating it with potassium permanganate:
5H2O2(aq) + 2KMnO4(aq) + 6H+(aq)  8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)
What is the mass percent of H2O2 in a solution if 57.5 g of solution required
38.9 mL of 0.534 M KMnO4 for its titration?
Strategy: mols KMnO4 -> mols H2O2 -> mass H2O2 -> % H2O2
Molar mass of H2O2 = 34.01 g
38.9  103 L 
0.534 mol KMnO4
5 mol H2O2
34.01g H2O2


= 1.77 g H2O2
1L
2 mol KMnO4
1mol H2O2
1.77 g H2 O 2
100%
57.5 g solution
= 3.07% H2O2
Dr. Sapna Gupta/Solution Stoichiometry
14
Example: A student measured exactly 15.0 mL of an unknown monoprotic
acidic solution and placed in an Erlenmeyer flask. An indicator was added to
the flask. At the end of the titration the student had used 35.0 mL of 0.12 M
NaOH to neutralize the acid. Calculate the molarity of the acid.
Strategy: mols NaOH -> mols acid (from eq) -> Molarity of acid
0.035 𝑀 𝑁𝑎𝑂𝐻 𝑥
0.12 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑
𝑥
= 0.0042 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑
1𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
0.0042 𝑚𝑜𝑙
= 0.28 M acid
0.015 𝐿
Example: Calculate the molarity of 25.0 mL of a monoprotic acid if it took
45.50 mL of 0.25 M KOH to neutralize the acid.
0.25 mol KOH
1 mol acid
 0.04550 L 
 0.01338 mol acid
L
1 mol KOH
0.01338 mol acid
 0.455 M
0.0250 L
Dr. Sapna Gupta/Solution Stoichiometry
15
Key Words/Concepts
• Solutions
o Solvent
o Solute
• Molarity (mol/L)
• Dilutions (MiVi = MfVf)
• Solution stoichiometry
o Volumetric analysis (titration)
o End point
o Equivalence point
o Gravimetric analysis
Dr. Sapna Gupta/Solution Stoichiometry
16