Chapter A3 – Chemical Reactions
Important Examples of
Chemical Change
Review: Chemical reactions
 Recall, in Chapter A1 we learned that all chemical
reactions share the same two features:
 1) A change in energy
Review: Chemical reactions

EXOthermic reactions
 chemical reactions that release
energy, usually in the form of:
 heat
 light
 electricity
 one common type of
exothermic reactions is
combustion = burning of
some kind of fuel
Review: Chemical reactions

ENDOthermic reactions
 chemical reactions that
absorb energy
 the reaction will not begin on
its own but rather, requires
electricity or some kind of
“activation” energy
 usually feel cold to the touch,
like instant cold packs
Showing an energy change in a
chemical reaction
 Sometimes, to illustrate whether a chemical reaction
is ENDO or EXOthermic, the word “energy” is
included in the equation
 Endothermic reactions: energy is shown as a
reactant

e.g. photosynthesis - a plant’s cells are able to capture the
sun’s light energy, and use it to convert carbon dioxide and
water into glucose (sugar) and oxygen
 carbon dioxide + water + energy  glucose + oxygen
Showing an energy change in a
chemical reaction
 Sometimes, to illustrate whether a chemical reaction
is ENDO or EXOthermic, the word “energy” is
included in the equation
 Exothermic reactions: energy is shown as a product

e.g. in cellular respiration, a plant or animal’s cells burn
the sugar to release the chemical energy stored in it
 glucose + oxygen  carbon dioxide + water + energy
 notice that the two reactions are the reverse of one another
Review: Chemical reactions
 2) One or more new substances are produced
 These new substances (the products) have different
properties that the chemicals that went into the
reaction (the reactants)
 The reaction could involve:
 the formation of a gas, as evidenced by the
production of bubbles
 the formation of a solid, as evidenced by the
solution becoming cloudy, and forming a
precipitate
Precipitates
 Precipitates sometimes form when two aqueous
solutions are mixed
 While both reactants were very soluble in water, one of
the products formed is only slightly soluble
 e.g. NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s)
Review of states
 ELEMENTS:
 their state is shown on the periodic table using a shading
system:
SOLID
LIQUID
GAS
 COMPOUNDS:
 ionic compounds are always solid, unless dissolved in
water and highly soluble, in which case, aqueous
 acids are always aqueous
 molecular compounds can be solid, liquid or gas, and
are usually given to you
Conservation of Mass
 in 1789, French chemist Antoine Lavoisier made an
important observation about chemical reactions
 because his theory could be tested and no exception
has been found to contradict it, his idea is accepted as
a scientific law
 The Law of Conservation of Mass says that:
 in a chemical reaction, matter cannot be created or
destroyed
 this means that, the total mass of the reactants is equal
to the total mass of the products
Conservation of Mass
 Antoine Lavoisier knew very little about the structure
of elements and compounds (recall, he was working
PRIOR to Dalton, Thomson, Rutherford or Bohr’s
model of the atom), he simply made a conclusion
based on empirical evidence (based on experiments)
 today, we are able to explain Lavoisier’s findings
theoretically
 the total number of atoms of each element in a chemical
reaction is equal before and after the reaction
 a chemical reaction is not a destruction or production of
atoms, but simply a rearrangement of them
Writing chemical equations
Chemical equations
 A symbolic representation of the process of chemical
change
 Chemical equations can be written in word form or using
chemical symbols, but all use the following format:
+

+
reactant one
reacts with
reactant two
to produce
product one
and
product two
 It does not matter in what order the reactants or listed or
what order the products are listed as long as the reactants
are on the left and the products on the right side of the
arrow
Example: the combustion of
methane
 The main gas in natural gas is methane, CH4(g)
 When the methane in your furnace burns in the
presence of oxygen, it produces carbon dioxide and
water vapour
 The reaction can be depicted as follows:
Word equations
 word equations are the simplest form of writing a
reaction equation
 rather than use formulas, word equations use the
names of the compounds and elements
 e.g. methane + oxygen  carbon dioxide + water
Practice problems
 Read the following descriptions of reactions and create
a word equation.
1. Solid magnesium metal reacts with hydrochloric
acid to produce aqueous magnesium chloride and
hydrogen gas.
2. An iron nail is placed in a solution of copper (II)
chloride. As a result, small amounts of copper metal
are produced in a solution of iron (II) chloride.
3. In a hydrogen fuel cell, hydrogen gas and oxygen gas
react to produce liquid water.
Practice problems (solutions):
 Read the following descriptions of reactions and create
a word equation.
1. Solid magnesium metal reacts with hydrochloric
acid to produce aqueous magnesium chloride and
hydrogen gas.
magnesium + hydrochloric acid  magnesium chloride + hydrogen
Practice problems (solutions):
2. An iron nail is placed in a solution of copper (II)
chloride. As a result, small amounts of copper metal
are produced in a solution of iron (II) chloride.
iron + copper (II) chloride  copper + iron (II) chloride
Practice problems (solutions):
3. In a hydrogen fuel cell, hydrogen gas and oxygen gas
react to produce liquid water.
hydrogen + oxygen  water
Skeleton equations
 A skeleton equation is a chemical reaction
equation written using chemical formulas
 It is not considered to be a finished equation
because it has yet to be balanced
 It includes the both the chemical’s formula
and its state
 e.g. CH4(g) + O2(g)  CO2(g) + H2O(g)
Skeleton equations
 Writing skeleton equations requires you to
be able to write up to four correct chemical
formulas
 Remember:
 ionic compounds must have balanced
charges
 acid formulas are balanced in the same way
as ionic compounds
 you must include states – look to the
equation description to guide you
Practice problems:
 Write skeleton equations for the three reactions
described above.
Practice problems (solutions):
1.
Solid magnesium metal reacts with hydrochloric
acid to produce aqueous magnesium chloride and
hydrogen gas.
magnesium + hydrochloric acid  magnesium chloride + hydrogen
Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
Practice problems (solutions):
2. An iron nail is placed in a solution of copper (II)
chloride. As a result, small amounts of copper metal
are produced in a solution of iron (II) chloride.
iron + copper (II) chloride  copper + iron (II) chloride
Fe(s) + CuCl2(s)  Cu(s) + FeCl2(aq)
Practice problems (solutions):
3. In a hydrogen fuel cell, hydrogen gas and oxygen gas
react to produce liquid water.
hydrogen + oxygen  water
H2(g) + O2(g)  H2O(l)
Balancing equations
 Recall, the Law of Conservation of Mass says that the
number of atoms of each element stays constant in a
reaction
 This means, you may need more than one “batch” of a
reactant to make all the products, or you may make
more than one batch of a product with the ingredients
you have in the reactants
 Consider the reaction of the combustion of methane
again
Balancing equations
 In order to use all the atoms in the methane, we need
two molecules of O2, and in turn produce one
molecule CO2 and two molecules of H2O
 This way, the total number of carbon atoms, oxygen
atoms and hydrogen atoms stays consistent
throughout
Balancing equations
 To illustrate this in a chemical equation, we use
coefficients – large numbers written before the
chemical
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Balancing equations
 All chemical equations are balanced in this way – by
adding coefficients in front of the chemical formula
 The coefficient applies to the entire formula
 e.g. 2H2O  means two water molecules, so four
hydrogens and two oxygens
Balancing equations
 To balance an equation:
 write the skeleton equation, including states
 once you have the skeleton equation written, you cannot
alter the chemical formulas
 count the number of atoms of each element in the
reactants and then in the products
 start adding coefficients, re-tallying as you go until the
number of atoms are balanced
Example:
CH4(g) + O2(g)  CO2(g) + H2O(g)
Elements
Number of atoms Number of atoms
in reactants
in products
C
1
1
H
4
2
O
2
2+1=3
 The carbons are balanced, but the hydrogens and
oxygens are not
 Balance the hydrogens by adding a coefficient before
the product containing hydrogen
Example:
CH4(g) + O2(g)  CO2(g) + 2H2O(g)
Elements
Number of atoms Number of atoms
in reactants
in products
C
1
1
H
4
2x2=4
O
2
2+2=4
 With the addition of the “2” in front of the water, we
balance the hydrogen but also increase the amount of
oxygen in the product side
Example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Elements
Number of atoms Number of atoms
in reactants
in products
C
1
1
H
4
4
O
2x2=4
4
 Now we are balanced!
Example:
 Aqueous iron (II) nitrate reacts with aqueous sodium
phosphate. The products are aqueous sodium nitrate
and solid iron (II) phosphate.
 Write the balanced chemical equation for this
reaction.
Example (solution)
 Aqueous iron (II) nitrate reacts with aqueous sodium
phosphate. The products are aqueous sodium nitrate
and solid iron (II) phosphate.
 Write the balanced chemical equation for this
reaction.
 Step #1 – Write the skeleton equation, including states
Fe(NO3)2(aq) + Na3PO4(aq)  NaNO3(aq) + Fe3(PO4)2(s)
Example (solution)
 Step #2 – Tally the number of atoms of each element
 *note – polyatomic ions can be kept as one unit, as
long as they appear on both sides of the equations
Fe(NO3)2(aq) + Na3PO4(aq)  NaNO3(aq) + Fe3(PO4)2(s)
Element / ion
# in reactants
# in products
Fe
1
3
NO3
2
1
Na
3
1
PO4
1
2
Example (solution)
 Step #3 – Start adding coefficients
3Fe(NO3)2(aq) + Na3PO4(aq)  NaNO3(aq) + Fe3(PO4)2(s)
Element / ion
# in reactants
# in products
Fe
1 3
3
NO3
2 6
1
Na
3
1
PO4
1
2
Example (solution)
 Step #3 – Start adding coefficients
3Fe(NO3)2(aq) + Na3PO4(aq)  6NaNO3(aq) + Fe3(PO4)2(s)
Element / ion
# in reactants
# in products
Fe
1 3
3
NO3
2 6
1 6
Na
3
1 6
PO4
1
2
Example (solution)
 Step #3 – Start adding coefficients
3Fe(NO3)2(aq) + 2Na3PO4(aq)  6NaNO3(aq) + Fe3(PO4)2(s)
Element / ion
# in reactants
# in products
Fe
1 3
3
NO3
2 6
1 6
Na
3 6
1 6
PO4
1 2
2
Practice problems:
 Balance the three reactions you wrote skeleton
equations for:
 Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
 Fe(s) + CuCl2(s)  Cu(s) + FeCl2(s)
 H2(g) + O2(g)  H2O(l)
Practice problems (solutions):
 Balance the three reactions you wrote skeleton equations
for:
 Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
 Fe(s) + CuCl2(s)  Cu(s) + FeCl2(s)  already balanced!
 2H2(g) + O2(g)  2H2O(l)
Homework:
 A3.2 – Check and Reflect (p. 90)
 #6 - 9
Five types of Chemical Reactions
Chemical reactions
 despite there being millions of different possible
chemical reactions, they follow certain patterns
 most reactions can be grouped into one of five
categories:
1.
2.
3.
4.
5.
Formation reactions (also called synthesis reactions)
Decomposition reactions
Hydrocarbon combustion
Single replacement reactions
Double replacement reactions
Overview of the 5 types of
chemical reactions
Reaction #1: Formation (Synthesis)
 Two elements combine to form a compound
 element + element  compound
 A + B  AB
 These reactions are exothermic - they occur without
the need of added energy
E.g. 2K(s) + Cl2(g)  2KCl(s)
Reaction #1: Formation (Synthesis)
 How you recognize this reaction:
 It’s the only reaction where both reactants are elements
(not compounds)
 What’s the challenge?
 watch for polyatomic elements (e.g. O2, H2. etc.)
 if the compound produced is an ionic compound,
remember to balance your charges
 if the compound includes a multivalent metal, unless
otherwise specified, assume the common ion charge
(the one listed first on the periodic table)
Practice problems:
 Magnesium oxide is a compound found in many
cosmetics and lotions. It is produced by reacting
magnesium metal with oxygen gas.
 a) Write the word equation for this reaction.
 b) Write the skeleton equation for this reaction.
 c) Balance the equation.
Practice problems (solution):
 Magnesium oxide is a compound found in many
cosmetics and lotions. It is produced by reacting
magnesium metal with oxygen gas.
 a) Write the word equation for this reaction.
magnesium + oxygen  magnesium oxide
 b) Write the skeleton equation for this reaction.
Mg(s) + O2(g)  MgO(s)
 c) Balance the equation.
2 Mg(s) + O2(g)  2 MgO(s)
Practice problems:
 Write the skeleton equation and balanced equation for
the formation of lithium oxide from its elements.
 Write the skeleton equation and balanced equation for
the formation of lead (IV) bromide from its elements.
Practice problems (solutions):
 Write the skeleton equation and balanced equation for
the formation of lithium oxide from its elements.
Li(s) + O2(g)  Li2O(s)
4Li(s) + O2(g)  2Li2O(s)
 Write the skeleton equation and balanced equation for
the formation of lead (IV) bromide from its elements.
Pb(s) + Br2(l)  PbBr4(s)
Pb(s) + 2Br2(l)  PbBr4(s)
Practice problems:
 Name the products in each of the following reactions
 potassium + iodine 
 magnesium + phosphorus 
 cesium + chlorine 
 Complete each equation below (the products are ionic
compounds)
 Na(s) + Br2(l) 
 Mg(s) + F2(g) 
 Al(s) + Cl2(g) 
Practice problems (solutions):
 Name the products in each of the following reactions
 potassium + iodine  potassium iodide
 magnesium + phosphorus  magnesium phosphide
 cesium + chlorine  cesium chloride
 Complete each equation below (the products are ionic
compounds)
 2Na(s) + Br2(l)  2NaBr(aq)
 Mg(s) + F2(g)  MgF2
 2Al(s) + 3Cl2(g)  2AlCl3(aq)
Reaction #2: Decomposition
 A compound breaks down into its
composite elements
 compound  element + element
 AB  A+ B
 These reactions are endothermic
 because the compound form is
more stable than the elements,
decomposition reactions require
energy to occur
 e.g. electrolysis of water
E.g. H2O(l)  O2(g) + H2(g)
Reaction #2: Decomposition
 How you recognize this reaction:
 It’s the only reaction where there is only one reactant
 What’s the challenge?
 watch for polyatomic elements (e.g. O2, H2. etc.)
Practice problems:
 Pure aluminium is not typically found in nature, but
can be produced by the decomposition of aluminium
chloride.
 a) Write the word equation for this reaction.
 b) Write the skeleton equation for this reaction.
 c) Balance the equation.
Practice problems (solution):
 Pure aluminium is not typically found in nature, but
can be produced by the decomposition of aluminium
chloride.
 a) Write the word equation for this reaction.
aluminium chloride  aluminium + chlorine
 b) Write the skeleton equation for this reaction.
AlCl3(aq)  Al(s) + Cl2(g)
 c) Balance the equation.
2AlCl3(aq)  2Al(s) + 3Cl2(g)
Practice problems:
 Write the balanced equations for these reactions:
 magnesium sulfide  magnesium + sulfur
 potassium iodide  potassium + iodine
 Complete the following decomposition reactions and
balance the equation:
 Al2O3(s) 
 NiCl2(s) 
 Na3P(s) 
Practice problems (solutions):
 Write the balanced equations for these reactions:
 magnesium sulfide  magnesium + sulfur
8MgS(s)  8Mg(s) + S8(s)
 potassium iodide  potassium + iodine
2KI(s)  2K(s) + I2(s)
 Complete the following decomposition reactions and
balance the equations:
 2Al2O3(s)  4Al(s) + 3O2(g)
 NiCl2(s)  Ni(s) + Cl2(g)
 4Na3P(s)  12Na(s) + P4(s)
Reaction #3: Hydrocarbon combustion
 Hydrocarbons are molecular compounds containing
hydrogen and carbon
 e.g. octane C8H18(l), glucose C6H12O6(s), methane CH4(g)
 “Combustion” means to burn
 Like all fuels, hydrocarbons require oxygen in order to burn
 Because they produce heat and light, combustion reactions
are always exothermic
Reaction #3: Hydrocarbon combustion
 Regardless of what hydrocarbon is burning, the
products are carbon dioxide and water vapour
 hydrocarbon fuel + oxygen  carbon dioxide + water
 CxHy + O2(g)  CO2(g) + H2O(g)
Reaction #3: Hydrocarbon combustion
 How you recognize this reaction:
 The reactants are a hydrocarbon and oxygen, or in the
description of the reaction, it says “combusts” or “burns”
 What’s the challenge?
 while these reactions always produce the same two
products, they are the hardest equations to balance

start by balancing the carbons, then the hydrogens, and leave
the O2 coefficient to last
Practice problems:
 Complete and balance each equation:
 CH4(g) + O2(g) 
 C2H6(g) + O2(g) 
 C3H8(g) + O2(g) 
 C6H6(l) + O2(g) 
Reaction #4: Single replacement
 An element reacts with a compound, and produces a
new element and a new compound
 element + compound  compound + element
 A + BC  AC + B
 If the reacting element is a metal, it will replace the
metal cation from the compound

e.g. Na(s) + KCl(s)  NaCl(s) + K(s)
 If the reacting element is a non-metal, it will replace
the non-metal anion from the compound

e.g. Br2(l) + MgCl2(s)  MgBr2(s) + Br2(l)
Reaction #4: Single replacement
 How you recognize this reaction:
 The reactants are a compound, and an element that is
NOT oxygen
 What’s the challenge?
 make sure you balance the charges in the new ionic
compound
 if the new element is a non-metal, watch for polyatomic
elements
 if the reaction occurs in solution, you will need to check
the solubility table for the solubility of the new
compound
Reaction #4: Single replacement
 E.g. Copper is added to a solution of
silver nitrate as shown on left:
 Cu(s) + AgNO3(aq) →
Reaction #4: Single replacement
 After reaction, two important
observations are made:
 1) The solution has turned blue
indicating the solid copper has
dissolved forming copper (II) ions
which are blue.
 2) A crystal precipitate has formed
on the outside of the copper wire,
and is silver in colour.
Practice problems:
 Write the products of this single replacement reaction, and
balance the reaction.
Practice problems (solution):
 Cu(s) + 2AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Practice problem:
 While copper compounds found in drinking water are
poisonous, pure copper is not. To remove copper
compounds such as copper (II) chloride from water,
aluminium reacts with the copper (II) chloride
solution.
 a) write the word equation for this reaction.
 b) write the skeleton equation.
 c) balance the equation.
Practice problem (solution):
 While copper compounds found in drinking water are
poisonous, pure copper is not. To remove copper
compounds such as copper (II) chloride from water,
solid aluminium reacts with the copper (II) chloride
solution.
 a) write the word equation for this reaction.
aluminium + copper (II) chloride  aluminium chloride + copper
 b) write the skeleton equation.
Al(s) + CuCl2(aq)  Al2Cl3(aq) + Cu(s)
 c) balance the equation.
4Al(s) + 3CuCl2(aq)  2Al2Cl3(aq) + 3Cu(s)
Reaction #5: Double replacement
 Two compounds exchange ions to form two new
compounds
 compound+ compound compound + compound
 AB + CD  AD + CB


“A” and “C” are both cations (+ ions) so they will always appear
first in the formula
“B” and “D” are both anions (- ions) so they will always appear
second in the formula
Reaction #5: Double replacement
 How you recognize this reaction:
 It’s the only reaction where both reactants are
compounds
 What’s the challenge?
 Because you will be dealing with two new ionic
compounds, you will have to balance the charges for
both compounds
 You will also have to check the solubility table for both
new compounds
Example:
 When aqueous lead (II) nitrate and aqueous sodium
iodide are mixed, a bright yellow precipitate of solid lead
(II) iodide forms, along with a second aqueous solution.
 a) Write the word equation.
 b) Write the skeleton equation, including states.
 c) Balance the equation.
Example (solution):
 When aqueous lead (II) nitrate and aqueous sodium
iodide are mixed, a bright yellow precipitate of solid lead
(II) iodide forms, along with a second aqueous solution.
 a) Write the word equation.
lead (II) nitrate + sodium iodide  lead (II) iodide + sodium nitrate
 b) Write the skeleton equation, including states.
Pb(NO3)2(aq) + NaI(aq)  PbI2(s) + NaNO3(aq)
 c) Balance the equation.
Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2 NaNO3(aq)
Practice problems:
 Write the word, skeleton and balanced equations for
the following reactions.
 a) When aqueous copper (I) nitrate and aqueous
potassium bromide are mixed, a precipitate of solid
copper (J) bromide forms, along with a second product.
 b) When aqueous aluminium chloride and aqueous
sodium hydroxide are mixed, a precipitate and another
aqueous solution form.
Practice problems (solutions):
 Write the word, skeleton and balanced equations for
the following reactions.
 a) When aqueous copper (I) nitrate and aqueous
potassium bromide are mixed, a precipitate of solid
copper (J) bromide forms, along with a second product.



copper (I) nitrate + potassium bromide 
copper (I) bromide + potassium nitrate
CuNO3(aq) + KBr(aq)  CuBr(s) + KNO3(aq)
already balanced!
Practice problems (solutions):
 Write the word, skeleton and balanced equations for
the following reactions.
 b) When aqueous aluminium chloride and aqueous
sodium hydroxide are mixed, a precipitate and another
aqueous solution form.



aluminium chloride + sodium hydroxide 
aluminium hydroxide + sodium chloride
AlCl3(aq) + NaOH(aq)  Al(OH)3(s) + NaCl(aq)
AlCl3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaCl(aq)
Summary of the five reactions:
Reaction type
Format
How to recognize:
Formation
A + B  AB
Reactants are both
elements
Decomposition
AB  A + B
Only one reactant
Hydrocarbon
combustion
CxHy + O2(g)  CO2(g) + H2O(g)
Fuel reacting with oxygen
Single
replacement
A + BC  AC + B
Reactants are a
compound and an
element (not oxygen)
Double
replacement
AB + CD  AD + CB
Reactants are both
compounds
Practice problems:
 Classify the reactions below as F, D, HC, SR or DR
 a) C6H12(l) + 9O2(g)  6CO2(g) + 6H2O(g)
 b) CaCl2(aq) + Na2SO4(aq)  CaSO4(s) + 2NaCl(aq)
 c) 2Fe(s) + O2(g)  2FeO(s)
 d) Be(s) + 2LiBr(aq)  BeBr2(aq) + 2Li(s)
 e) MnI4(aq)  Mn(s) + 2I2(s)
Practice problems (solutions):
 Classify the reactions below as F, D, HC, SR or DR
 a) C6H12(l) + 9O2(g)  6CO2(g) + 6H2O(g)
 b) CaCl2(aq) + Na2SO4(aq)  CaSO4(s) + 2NaCl(aq)
 c) 2Fe(s) + O2(g)  2FeO(s)
 d) Be(s) + 2LiBr(aq)  BeBr2(aq) + 2Li(s)
 e) MnI4(aq)  Mn(s) + 2I2(s)
HC
DR
F
SR
D
Predicting the products of chemical
reactions
 Once you become familiar with the five reaction types,
you will be expected to classify reactions without the
products, AND predict the products of the reaction.
 step #1 – classify the reaction
 step #2 – use the reaction type to predict the products
 step #3 – write the formula for the product(s)
 step #4 – include the states
 step #5 – balance the reaction
Example:
 Classify each of the following reactions, and complete
the reaction.
 a) PbBr4(aq) 
 b) Ni(NO3)3(aq) + Na2SO3(aq) 
 c) C8H18(l) + O2(g) 
 d) Cu(s) + Au(ClO3)3(aq) 
 e) Zn(s) + S8(s) 
Example:
 a) PbBr4(aq) 
 step #1 – classify the reaction

one reactant  decomposition
 step #2 – use the reaction type to predict the products

breaks down into its elements
 step #3 – write the formula for the product(s)
 step #4 – include the states

PbBr4(aq)  Pb(s) + Br2(l)
 step #5 – balance the reaction

PbBr4(aq)  Pb(s) + 2 Br2(l)
Example:
 b) Ni(NO3)3(aq) + Na2SO3(aq) 
 step #1 – classify the reaction

both reactants are compounds  double replacement
 step #2 – use the reaction type to predict the products

two cations switch places
 step #3 – write the formula for the product(s)
 step #4 – include the states

Ni(NO3)3(aq) + Na2SO3(aq)  NaNO3(aq) + Ni2(SO3)3(s)
 step #5 – balance the reaction

2 Ni(NO3)3(aq) + 3 Na2SO3(aq)  6 NaNO3(aq) + Ni2(SO3)3(s)
Example:
 c) C8H18(l) + O2(g) 
 step #1 – classify the reaction

a hydrocarbon reacting with oxygen  combustion
 step #2 – use the reaction type to predict the products

carbon dioxide and water vapour
 step #3 – write the formula for the product(s)
 step #4 – include the states

C8H18(l) + O2(g)  CO2(g) + H2O(g)
 step #5 – balance the reaction

2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Example:
 d) Cu(s) + Au(ClO3)3(aq) 
 step #1 – classify the reaction

an element (not O2) reacting with a compound  single
replacement
 step #2 – use the reaction type to predict the products
 the element is a metal  it will replace the other metal
 step #3 – write the formula for the product(s)
 step #4 – include the states
 Cu(s) + Au(ClO3)3(aq)  Cu(ClO3)2(aq) + Au(s)
 step #5 – balance the reaction
 3 Cu(s) + 2 Au(ClO3)3(aq)  3 Cu(ClO3)2(aq) + 2 Au(s)
Example:
 e) Zn(s) + S8(s) 
 step #1 – classify the reaction

both reactants are elements  formation
 step #2 – use the reaction type to predict the products

the product is the ionic compound formed by these elements
 step #3 – write the formula for the product(s)
 step #4 – include the states

Zn(s) + S8(s)  ZnS(s)
 step #5 – balance the reaction

8 Zn(s) + S8(s)  8 ZnS(s)
Homework:
 Predicting products worksheet
The Mole
Counting atoms
 How many carbon atoms are there in the tip of your
pencil? Is it possible to count them? Is there a way of
expressing the number of atoms without having to
count them one by one?
“Mass” as a method of counting
 Counting is useful when the number of things to be
counted is relatively small. In some cases it isn't. Consider
this example:
 Imagine that you have to make up a materials list for a deck
you are building. The list might look like this:
 20 pieces of 1" x 6" x 12' pressure treated lumber for decking
 10 pieces of 2" x 8" x 12' pressure treated lumber for framing
 2 kg of 3.5" deck screws
 You might ask: "Why buy the wood in individual pieces and
the screws by the kilogram?"
“Mass” as a method of counting
 The answer is convenience. It is convenient to count
out pieces of wood because relatively few pieces are
required; however, in the case of the deck screws, it is
easier to measure out a 2 kg amount than it is to count
out the large number of screws needed to fasten the
boards together.
Mass amounts
 Substituting a mass amount for a counted amount is a
common practice in chemistry because of the
impracticalities of counting things as small as atoms and
molecules.
 However, in chemistry (and in carpentry) there's a catch.
There's something you need to know before you can
substitute mass measurements for counted amounts.
 When you measure by mass, how do you know how many
items there actually are in a sample?
 Some sort of conversion factor is needed to convert a mass to
a number of items. (e.g. 1 kg = 57 screws)
 Can the same process be applied to atoms? YES!
The mole
 In the case of atoms, the solution is not as simple
because counting out enough atoms to make up a
kilogram is totally impractical, if not impossible.
 Atoms are so small that it takes an incredibly huge
number of them to make up a visible amount.
 An iron spike for example might contain more than 6.02
x 1023 atoms of iron.
 Dealing with such large numbers of atoms on a
consistent basis can get to be tiresome after a while.
That's where the mole comes in.
A mole is a “group” of atoms
 We use groupings of objects all the time to make
counting large amounts easier, for example:
 “a pair of shoes” means two shoes
 “a dozen eggs” means twelve eggs
 “a flat of pop” means twenty-four cans
A mole is a “group” of atoms
 “A mole” is similar – but it means a much larger
number – 602 000 000 000 000 000 000 000 atoms
 How big is a mole?
 one mole of peas is enough to cover Earth and 250 more
planets the same size as Earth one metre deep in green
peas
 a mole of marbles spread over the Earth would cover it
to a depth of 80 km
 if you own a mole of dollars and you spend a billion
dollars a day, then you could spend that amount per day
for over a trillion years before you run out of money
A mole is a “group” of atoms
 Where do we get this number?
 Avogadro was the chemist that first
suggested this amount – we now refer to
6.02 x 1023 as Avogadro’s number
 A mole is the number of atoms in exactly
12 grams of carbon-12.
 This number was determined by dividing
the mass of a single carbon-12 atom into a
12 g mass of carbon-12.
 Since the mass of a single carbon-12 atom
is tiny (1.99 x 10-23 g), the value of a mole
comes out to be a huge number: 6.02 x 1023
How do we measure moles?
 Back to our example with the screws – we don’t have a
device that counts screws, but because you know that a
bag of 57 screws has a mass of 1 kg, you can use a scale
 However, if you are dealing with different hardware (e.g.
57 nails instead of 57 screws) you will get a different
mass
How do we measure moles?
 Since there’s no measuring device to measure moles,
we use the mass of a mole of atoms
 Since each element has different sized atoms, the mass
of one mole of carbon is different than one mole of
sulfur, etc.
Atomic molar mass
 The mass of one mole of an
element is its atomic
molar mass
 it is measured in grams per
mole (g/mol)
 The atomic molar mass of
each element is listed on
the periodic table – you’ve
already been using it.
 carbon: 12.01 g/mol
 sulfur: 32.07 g/mol
Practice problem:
 What is the mass of the following elements?
Practice problem (solution):
55.85 g
63.55 g
118.69 g
126.90 g
How much is one mole?
 The pictures below represent ONE MOLE of
 the elements copper, aluminum, sulfur
 the compounds potassium dichromate, water, and
copper(II) chloride
How do we know the molar mass
of compounds?
 The periodic table only lists the molar mass of
elements, so how can you find the molar mass of
compounds?
 To find the molar mass of a compound, you have to:
 write the chemical formula,
 list the number of atoms of each element,
 multiply this number by the molar mass of the element.
Atomic molar mass
 Example: Find the molar mass of sodium chloride
 Formula: NaCl  1 atom of sodium and 1 of chlorine
 MNaCl
= (1 x MNa) + (1 x MCl)
= (1 x 22.99 g/mol) + (1 x 35.45 g/mol)
= 58.44 g/mol
 Example: Find the molar mass of water
 Formula: H2O  2 atoms of hydrogen and 1 of oxygen
 MH2O
= (2 x MH) + (1 x MO)
= (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 18.02 g/mol
Practice problems:
 What is the molar mass of the following compounds?
 methane (CH4)
 calcium chloride
 sucrose (C12H22O11)
Practice problems (solution):
 methane (CH4)

MCH4
= (1 x MC) + (4 x MH)
= (1 x 12.01 g/mol) + (4 x 1.01 g/mol)
= 16.05 g/mol
 calcium chloride

MCaCl2 = (1 x MCa) + (2 x MCl)
= (1 x 40.08 g/mol) + (2 x 35.45 g/mol)
= 110.98 g/mol
 sucrose (C12H22O11)

MC12H22O11 = (12 x MC) + (22 x MH) + (11 x MO)
= (12 x 12.01 g/mol) + (22 x 1.01 g/mol) + (11 x 16.00 g/mol)
= 342.34 g/mol
Moles, mass and molar mass
 The relationship between these three quantities can be
summed up in a formula:
n = m
where:
M
 n = number of moles, measured in moles (mol)
 m = mass, measured in grams (g)
 M = molar mass, measured in grams per mole (g/mol)
Using the formula
 You can use the formula n = m/M to find the number of
moles of a sample if you know:
 the mass of the sample,
 the identity of the compound  use its chemical formula to
calculate the molar mass
 Example: How many moles are in 100 g of carbon dioxide?
 m = 100g
 MCO2 = (1 x 12.01) + (2 x 16.00) = 44.01 g/mol
 n= m
M
=
100g
44.01g/mol
= 2.27 mol
Practice problems:
 How many moles of silicon are in a 56.18-g sample?
 How many moles of potassium fluoride are in a 25.0-g
sample?
Practice problems (solutions):
 How many moles of silicon are in a 56.18-g sample?
m = 56.18 g
MSi = 28.09 g/mol
n =
m =
M
56.18 g
= 2.00 mol
28.09 g/mol
 How many moles of potassium fluoride are in a 25.00-g
sample?
m = 25.00 g
MKF = (1 x 39.10) + (1 x 19.00) = 58.10 g/mol
n =
m =
M
25.00 g
58.10 g/mol
= 0.43 mol
Using the formula
 You can also use the formula to find the mass of a
sample, if you know:
 the number of moles in the sample
 the identity of the sample  allows you to find molar
mass
 Using the formula to find mass requires you to
rearrange the formula to solve for mass
 m = nM
Example:
 What is the mass of 10.0 mol of water?
n = 10.0 mol
MH2O = 18.02 g/mol (calculated in a previous question)
m = nM = (10.0 mol)(18.02 g/mol) = 180.2 g
Practice problems:
 What is the mass of 5.0 mol of NaOH?
 What is the mass of 4.3 mol of ammonia, NH3?
Practice problems (solutions):
 What is the mass of 5.0 mol of NaOH?
n = 5.0 mol
MNaOH = (1 x 22.99) + (1 x 16.00) + (1 x 1.01) = 40.00 g/mol
m = nM = (5.0 mol)(40.00 g/mol) = 200 g
 What is the mass of 4.3 mol of ammonia, NH3?
n = 4.3 mol
MNH3 = (1 x 14.01) + (3 x 1.01) = 17.04 g/mol
m = nM = (4.3 mol)(17.04 g/mol) = 73.27 g
Showing your work
 Calculation questions are usually worth several marks,
because in science, you are required to show the
process you went through to solve the question.
 Typically, a mole question will be worth 3 marks:
 1 mark for writing the formula you are using
 1 mark for substituting numbers into the formula,
including units
 1 mark for the correct answer, with units
The mole concept and the Law of
Conservation of Mass
 Recall, the Law of Conservation of Mass says that the
total mass of the reactants is equal to the total mass of
the products
 In balancing equations, you applied this Law
 the coefficient you add in front of a compound in a
balanced equation is the number of moles of that
substance
 e.g. In the reaction 4 Na(s) + O2(g)  2Na2O(s)
 4 moles of sodium reacted with one mole of oxygen to
produce 2 moles of sodium oxide
Practice problem:
 Using the coefficients as the number of moles,
calculate the mass of the reactants and the mass of the
products.
Practice problem (solutions):
 Using the coefficients as the number of moles, calculate the
mass of the reactants and the mass of the products.
4 Na(s) + O2(g)  2Na2O(s)
 sodium:
 m = nM = (4 mol)(22.99 g/mol) = 91.96 g
 oxygen
 m = nM = (1 mol)(2 x 16.00 g/mol) = 32.00g
 mass of reactants = 91.96 g + 32.00 g = 123.96 g
 sodium oxide
 m = nM = (2 mol)(2 x 22.99 + 1 x 16.00 g/mol) = 123.96 g
Work Time
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