The study of relative amounts – ratios – of
substances involved in chemical reactions is
known as stoichiometry.
 The word stoichiometry comes from two
Greek words meaning ‘element’ and
‘measure’
 Knowledge of stoichiometry is essential
whenever quantitative information about a
chemical reaction is required


Stoichiometry – on line explanations and
practice questions
What a balanced chemical equation
tells us
• In a chemical sense, if you want to say, ‘One
molecule of nitrogen gas reacts with three
molecules of hydrogen gas to form two
molecules of ammonia gas’, then you write
» N2(g) + 3H2(g) ---> 2NH3(g)
– It would be impossible and impractical to make
ammonia molecule by molecule. Chemists work with
larger quantities called moles (symbol mol).
– The moles of reactants and products taking part in the
reaction
» 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas
to produce 2 moles of ammonia gas
What a balanced chemical equation
tells us
• N2(g) + 3H2(g) ---> 2NH3(g)
• The equation actually indicates the mole ratio
of reactants and products.
• The beauty of the this method is that the ratios
never change.
• If 1 mole of nitrogen reacts with 3 moles of
hydrogen,
• then 2 moles of nitrogen will react with
• 6 moles of hydrogen,
• 0.5 moles of nitrogen will react with
• 1.5 moles of hydrogen.......etc
What a balanced chemical equation
tells us
• N2(g) + 3H2(g) ---> 2NH3(g)
• Assuming that there is plenty of nitrogen,
how much ammonia will be produced by 3
moles of hydrogen?
• The equation tells us 2 moles of ammonia
• 6 moles of hydrogen will produce?
• 4 moles of ammonia
What a balanced chemical equation
tells us
• An equation also tells us about the ratio
between the reactants and the products of
other quantities.
• It tells us about the ratio between the
number of particles, and the masses of the
reactants and products
What a balanced chemical equation
tells us
• When 1 mole of N2 gas reacts with 3 moles
of H2 gas to produce 2 moles of NH3 gas.
• We know that this can be restated as
• 28g of N2 gas (or 6.02 x 1023 N2 molecules)
reacts with
• 6g of H2 gas (or 1.806 x 1024 H2 molecules)
to produce
• 34g of NH3 gas (or 1.204 x 1024 NH3
molecules)
Writing and Balancing Equations
• Determine the reactants and products.
• Write the equation.
– Use the correct formulas for each compound watch for the diatomics.
• Balance the equation.
Balancing Examples
• Solid copper(II) oxide reacts with hydrogen
gas to form solid copper and liquid water.
• CuO(s) + H2(g) ---> Cu(s) + H2O(l)
• Aluminum metal reacts with oxygen gas to
form solid aluminum oxide.
• Al(s) + O2(g) ---> Al2O3(s)
Balancing Examples
• When balancing an equation, ONLY the
coefficients can be changed.
• NEVER change the subscripts.
• For example: 3H2O
• 3 is the coefficient.
2 and 1 are the subscripts.
• Changing the subscripts changes the
compound.
• H2O2
HELPFUL HINTS
• Keep polyatomic ions together as a unit if
they show up on both sides of the equation.
• NO3-1 , C2H3O2-1 , NH4+1 , ….
• Ca(NO3)2 + NaCl ---> CaCl2 + NaNO3
• Save H and O until last.
• They will tend to balance out together with
the H2O.
Example
Aluminum sulfide reacts with water to produce
aluminum hydroxide and hydrogen sulfide.
Al2S3 + H2O --> Al(OH)3 + H2S
What should we look for immediately?
Save O and H for last - there is water.
Al2S3 + H2O ---> 2 Al(OH)3 + H2S
Al2S3 + H2O ---> 2 Al(OH)3 + 3 H2S
The Al and S are balanced - now - H & O.
Al2S3 + 6 H2O ------> 2 Al(OH)3 + 3 H2S
Balance The Following Equations
•
•
•
•
•
•
•
•
•
•
•
•
Al + S8
Al2S3
16 Al + 3 S8
8 Al2S3
CO + O2
CO2
2 CO + O2
2 CO2
C2H4 + O2
CO2 + H2O
C2H4 + 3 O2
2 CO2 + 2 H2O
Pb(NO3)2 + NaCl
PbCl2 + NaNO3
Pb(NO3)2 + 2 NaCl
PbCl2 + 2 NaNO3
Mg2C3 + H2O
Mg(OH)2 + C3H4
Mg2C3 + 4 H2O
2 Mg(OH)2 + C3H4
Ca(OH)2 + HBr
CaBr2 + H2O
Ca(OH)2 + 2 HBr
CaBr2 + 2 H2O
Practice
Predict the products. Balance the equation
1. HCl(aq) + AgNO3(aq) 
2. CaCl2(aq) + Na3PO4(aq) 
3. Pb(NO3)2(aq) + BaCl2(aq) 
4. FeCl3(aq) + NaOH(aq) 
5. H2SO4(aq) + NaOH(aq) 
6. KOH(aq) + CuSO4(aq) 
What a chemical equation does not tell us
• An equation does not give us information
about the rate of reaction
• It might be fast (eg the explosive
oxidation of hydrogen to form water)
• It might be slow (eg the oxidation or
rusting of iron)
What a chemical equation does not tell us
• An equation does not tell us whether a
reaction requires heat or gives off heat, or
what temperature or pressure is needed
• An equation does not tell us how the
individual atoms or molecules are
transformed from reactants to products.
• This information is essential for full
understanding of how a chemical reaction
takes place
Review
• Work through the sample problem on
page 337
• Complete the revision questions 1, 2 page
338
Mass – Mass stoichiometry
• Mass – mass stoichiometry involves
solving a problem in which the mass of
a reactant or product is given.
• You are then asked to calculate the
mass of another reactant or product.
• Mass – mass stoichiometry requires the
conversion of masses of substances to
moles, or moles of substances to
masses.
• Mass (m) = moles (n) x molar mass (M)
Mass – Mass stoichiometry
1.
2.
3.
4.
Mass – mass stoichiometry problems can be
solved in four steps
Write a balanced chemical equation for the
reaction, identifying the known (given) and
unknown (required) quantities of substance.
Calculate the number of moles of the known
quantity of substance present
From the equation, find the molar ratio that
states the proportion of known to unknown
quantities in the reaction and use it to
calculate the number of moles of the required
substance
Calculate the quantity (mass) of the required
substance
Mass – Mass stoichiometry
• Work through the sample problem steps
1 – 4 on page 339
• Complete the revision questions 3 – 6
pages 339, 340
Mass – Mass stoichiometry
• Limited reactant calculations
– When the amounts of reactants provided
for a reaction are not in the mole ratio,
amounts of one or more reactants will be
left over once the reaction has taken place
– The amounts of products will be limited by
the amount of the reactant that is
completely used up in the reaction
– In the previous example
– N2(g) + 3H2(g) ---> 2NH3(g)
Mass – Mass stoichiometry
• Limited reactant calculations
– A balanced equation indicates the mole ratios
in which the reactants are used up and the
products are formed
– N2(g) + 3H2(g) ---> 2NH3(g)
– According to this equation – one mole of
nitrogen gas reacts with three moles of
hydrogen gas to produce two moles of
ammonia gas
– If one mole of nitrogen and four moles of
hydrogen are mixed and allowed to react, then
all the nitrogen and three moles hydrogen are
used up.
– This means that one mole of hydrogen is left
over
Mass – Mass stoichiometry
• Limited reactant calculations
– N2(g) + 3H2(g) ---> 2NH3(g)
– The reactant that is completely used up (in this
case nitrogen) is called the limiting reactant.
– Any unreacted reactants (hydrogen in this
example) are called excess reactants.
– The amount of product formed by the reaction
is limited by the amount of the limiting
reactant.
– In this case the addition of more hydrogen to
the mixture will have no effect on the amount
of product formed, as there is no more
nitrogen available to react with it.
Atoms or
Molecules
Flowchart
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by
atomic/molar mass
from periodic table
Divide by atomic/molar
mass from periodic
table
Mass
(grams)
Practice
• Calculate the Molar Mass of calcium
phosphate
• Formula =
Ca3(PO4)2
• Masses elements:
• Molar Mass =
Review
• Complete the sample problem page 341
• Complete the revision questions 8, 9
page 342
Concentrating on Solutions
• Many chemical reactions take place
between substances that have been
dissolved in a liquid, most commonly
water, to form a solution.
• The liquid in which the substances are
dissolved is called the solvent; the
dissolved substance is called the solute.
Concentrating on Solutions
• When working with solutions, the most
easily measured quantity is their
volume.
• However, for the volume of a solution to
provide useful information, the solution’s
concentration – that is, the amount of
solute in a given volume – must be
known
Calculating concentrations of solutions
• The concentration of a solution is the
number of moles that it contains in each
litre.
• The concentration of a solution is
calculated using the formula
n
• C = --V
• This can be transposed to find the
number of moles, n = cV
Calculating concentrations of
solutions
• If 1 mol of solute is dissolved in a total
volume of 1L of water, the concentration of
solution is 1mol L -1.
• If 0.5 mol of solute is dissolved in a total
volume of 1L of water, the concentration of
the solution is 0.5 mol L -1.
• If, however, 0.5 mol of solute is dissolved
in a total volume of 0.5L of water, the
concentration of the solution is 1 mol L -1
Calculating concentrations of
solutions
quantity of solute (moles)
• Concentration (c) = -----------------------------volume of solution (L)
moles
c = ------------ (in mol L -1)
volume
Calculating concentrations of
solutions
• The formula for concentration can be
manipulated to calculate the amount,
and hence the mass, of the solute
• n = c x V (volume must be in litres)
• m = n x molar mass
Calculating concentrations of
solutions
• In order to prepare a particular volume of solution of
known concentration, the following five steps should be
followed
1. Calculate the number of moles of solute that are
needed to obtain the correct concentration of solution
for the volume of solvent to be used, according to the
formula n = c x V
2. Calculate the mass of the solute needed, using the
formula m = n x M
3. Partially fill a volumetric flask with water, and add the
correct mass of solute
4. Dissolve the solute
5. Add water to the required volume
Review
• Work through the sample problem on
page 343
• Complete the revision question 10 page
344
The concentration of water
• All pure liquids have a concentration, since
concentration can be defined as the number of
particles in a given volume.
• Given that it is possible to mix two liquids
together and get a smaller volume, it follows
that empty space exists between the liquid
molecules.
• At a molecular level there is always empty
space (vacuum) between the water molecules.
So liquid water has a concentration.
The concentration of water
• The concentration of water can be
expressed as grams per litre (density) or
moles per litre
• Given that the density of water is 1g mL-1,
it follows that 1000mL of water has a mass
of 1000g.
• This can be converted to moles using the
formula .
The concentration of water
• The molar mass of water (M) is 18 g mol -1
1000
• n (H2O) = --------18
= 55.5mol
• One litre of pure water therefore contains
55.5mol of water molecules
 Many
ionic substances are soluble.
 The process of dissolving involves the ionic
lattice breaking up and the anions and
cations dissociating from each other
 The concentrations of the resulting ions can
be calculated
 These concentrations are called ionic
concentrations and are designated by square
brackets
 [Na+] = 0.5 mol L -1 means that the ionic
concentration of Na+ ions in solution is
0.5mol L -1
Review
• Work through the sample problems page
344 -345
• Complete the revision questions 11 – 13
page 345


When a solution is diluted by the addition of
more solvent (eg water), the number of
moles of solute remains the same.
The addition of water to a concentrated
solution does not alter the number of moles
or the mass of the solute in that solution



After the solution has been diluted, the
number of moles of solute remains the same.
Therefore, if n1 represents the number of
moles in the initial or concentrated solution
and n2 represents the number of moles of the
final or dilute solution then we can say that:
n1 = n2





n1 = n2
Using the equation n = c x V, we can write
equations for n1 and n2 as follows:
n1 = c1 x V1 and n2 = c2 x V2
Since the values of n1 and n2 are equal, these
equations can be combined to form the
following formula:
c1 x V1 = c2 x V2
Review
• Work through the sample problem page 346
• Complete the revision questions 14, 15 page
346
Mass – Concentration Stoichiometry




Chemical reactions involving interactions between
solids and solutions
May involve mixing two solutions to form a
precipitate or
Solids may dissolve in some solutions to form new
products
Mass – concentration stoichiometry calculations
require the use of two formulae to calculate moles.
Mass – Concentration Stoichiometry

When dealing with solids – use the formula:

m
n = -----M

When dealing with solutions – use the formula:

n=cxV
Mass – Concentration Stoichiometry

1.
2.
3.
4.
Mass – concentration stoichiometry problems may be solved
in four steps:
Write a balanced chemical equation for the reaction,
identifying the known (given) and unknown (required)
quantities of substance
Calculate the number of moles of the known quantity of
substance present
From the equation, find the molar ratio that states the
proportion of known to unknown quantities in the reaction
and use it to calculate the number of moles of the required
substance
Calculate the quantity of the required substance
Review
• Work through the sample problems page
348
• Complete the revision questions 16 - 18




When some solutions are mixed, chemical
reactions may occur.
Solution stoichiometry, sometimes also called
‘concentration-concentration’ stoichiometry,
involves reactions in solution such as
precipitation reactions and neutralisation of an
acid and a base to form water and a salt.
Solution stoichiometry utilises the formula:
n = cV

1.
2.
3.
4.
Solution stoichiometry problems can be solved in four
steps
Write a balanced chemical equation for the reaction,
identifying the known (given) and unknown (required)
quantities of substance
Calculate the number of moles of the known quantity
of substance present using the formula: n = cV
From the equation, find the molar ratio of known to
unknown quantities in the reaction and use it to
calculate the number of moles of the required substance
Calculate the quantity (ie concentration) of the required
substance
Review
• Work through the sample problem page 350
• Complete the revision questions 19 – 23
pages 350, 351.
Acid – Base Titrations
• Acid – base titrations are based on neutralisation
reactions.
• They are a type of volumetric analysis where the
unknown concentration of a solution is determined
by reacting it with a solution of known
concentration.
• The solution of known concentration is called a
known concentration is called a standard solution
and is prepared by dissolving an accurate amount
of solute in water using a volumetric flask that is
calibrated to contain the specified volume.
Acid – Base Titrations
• The standard solution is placed in a burette,
which is used to deliver definite but variable
volumes of liquid.
• In a titration, the volume of liquid measured
by the burette is called a titre.
• The solution of unknown concentration is
added to a conical flask using a pipette.
• A pipette is used to deliver a known volume
of liquid which is then called an aliquot
Acid – Base Titrations
• A suitable indicator is added to the aliquot in the
conical flask
• The solution in the burette is slowly added to the
aliquot until the indicator changes colour
• This process is called titration
• The point at which chemically equivalent amounts
of acid and base (according to the equation) are
present is called the equivalence point
• The point at which the indicator changes colour is
called the end point and is usually about one drop
after the equivalence point
Acid – Base Titrations
• Titres within 0.05mL of each other are
called concordant titres
• Three concordant titres are needed to
calculate the average titre.
Review
• Work through the sample problem page 352
• Complete the revision questions 24, 25 page
352
CALCULATING PH VALUES
The pH scale can used to compare the acidity or
basicity of different solutions
 Acidity of aqueous solutions is due to the
presence of H30+, therefore the pH can be
calculated from the concentration of H30+.
 Mathematically, pH can be found by taking the
logarithm to base 10 of the hydrogen (or
hydronium) ion concentration and then adding a
negative sign.

CALCULATING PH VALUES


pH = -log[H30+] or [H30+] = 10-pH
Eg. – to find the pH of an aqueous solution in
which [H30+] is 0.01 mol L-1 first express [H30+]
as a power of 10 before calculating the logarithm,
then multiply by -1.
CALCULATING PH VALUES
[H30+] = 0.01 mol L -1
= 10-2 mol L -1
Log10 [H30+] = log10 [10-2]
= -2
pH = -1 x (-2)
=2

CALCULATING PH
This means that when the [H30+] in a solution is
expressed as a power of 10 the pH is equal to the
power multiplied by -1
 If [H30+] = 10-x, then pH = x


Therefore [H30+] = 10-pH
Review
• Work through the sample problems pages
353, 354
Pure water is neutral and has a pH of 7 at
25°C.
 Water molecules can react with
themselves to produce a small number
of H30+ and OH- ions.
 The concentration of these ions, when
multiplied together, always equals
10-14 (mol L -1)
 This is call the ionic product of water

There is a direct relationship between
levels of [H30+] and OH- ions in aqueous
solutions.
 At 25°C the product of their
concentrations equals 1 x 10-14.
 This product is called the ionic product of
water and is expressed by the equation.
 [H30+] x [OH-] = 10-14 (mol L-1)2 at 25°C

In a neutral solution, such as water,
[H30+] = [OH-]
 Substituting [H30+] for [OH-], the above
expression becomes
 [H30+] x [H30+] = 10-14 (mol L-1)2
 [H30+]2 = 10-14 (mol L-1)2
 [H30+] = 10-7 (mol L-1)2

Therefore both [H30+] and [OH-] are
equal to 10-7 mol L-1 in a neutral solution
at 25°C.
 It follows that it the amount of either
[H30+] or [OH-] in aqueous solution is
increased, the other quantity must
decrease by a corresponding factor.

The equation for the ionic product of
water has useful applications in pH
calculations, due to the fact that pH is
always measure in terms of [H30+] or [H+].
 When the concentration of a base is
given, we need to convert this
information to the concentration of H30+,
using the ionic product of water, before
we can determine the pH

Review
• Work through the sample problem page 355
• Complete the revision questions 26 – 31
page 356
Percentage Yield
When an equation is used to calculate the
amount of product that will form during a
reaction, then a value for the theoretical
yield is obtained.
 This is the maximum amount of product
that could be formed from a given amount
of reactant.
 In contrast, the amount of product that
actually forms when the reaction is carried
out in a laboratory is the actual yield.

Percentage Yield
The actual yield is often less than the theoretical
yield.
 The percentage yield is the ratio of the actual
yield to the theoretical yield.
 It measures the efficiency of the reaction
actual yield
 Percentage yield = ------------------------- x 100%
theoretical yield

Percentage Yield
A percentage yield should not be larger
than 100%
 Factors which may cause percentage
yield to be less than 100%

 Reactions do not always go to completion
 Impure reactants and competing side
reactions may cause other products to be
formed
 During purification, some of the product may
be lost
Review
• Work through the sample problem page 357
• Complete the revision question 32 page 357
• Complete the multiple choice questions 1 – 14
• Complete the review questions 1, 2, 3, 4, 5, 6, 7, 13, 16,
18, 20, 22, 24, 26, 27, 29, 30, 32, 35, 38, 41, 44, 48, 49,
52, 54, 57, 59, 62
• Complete the extended response question 1 page 365