Course Specification
Term 2 (AY 21-22)
Course Name:
Chemistry
Course Code:
CHM51
Grade:
10 Advanced
Aim:
The Grade 10 Chemistry course provides students with a college-level foundation to
support future advanced coursework in Chemistry. The course will give students the
opportunity to cultivate their understanding of Chemistry through inquiry-based
investigations as they explore topics as Structure & Properties of Matter, Chemical
Bonding & Reaction, Matter, Energy & Equilibrium, Organic & Nuclear Chemistry.
Course Outline:
Unit
Number of Periods
Term 2
Unit 2: Chemical Bonding and Reaction (Continued)
36
•
2.3 – Chemical Reactions
(Inspire Chemistry, Unit 2 – Module 8: Lesson 1 – 3)
10
•
2.4 – The Mole
(Inspire Chemistry, Unit 2 – Module 9: Lesson 1 – 4)
13
•
2.5 – Stoichiometry
(Inspire Chemistry, Unit 2 – Module 10: Lesson 1 – 4)
13
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 1 of 46
Unit 2: Chemical Bonding and Reaction
2.3 – Chemical Reactions
Inspire Chemistry – Module 8 – Lesson 1: Reactions and Equations
CHM.5.3.01.014.01 Distinguish between chemical reaction and chemical equation
A chemical reaction is a rearrangement of the atoms in one or more substances to form different
substances. It occurs when reactants are converted to products.
A chemical equation is a statement that uses chemical formulas to show the identities and relative
amounts of the substances involved in a chemical reaction.
CHM.5.3.01.014.02 List different observations (or physical evidences) that indicate that a chemical
reaction may be taking place
The different observations (Physical Evidences) include:
1) Release of energy as heat or light
2) Absorption of heat
3) Production of gas
4) Formation of a precipitate (solid)
5) Color change
6) Odor change
CHM.5.3.01.014.03
Explain the difference between reactants and products in a chemical equation
Reactants are the initial components (starting substances).
Products are the resultant components (substances formed during the reaction).
Example:
For the following reaction, identify the reactant(s) and product(s)
CaO(s) + CO2(g) → CaCO3(s)
Reactants: CaO and CO2
Product: CaCO3
CHM.5.3.01.014.04
Identify reactants and products in a chemical equation
Example:
For each of the following chemical reactions or equations, identify the reactant(s) and product(s)
Chemical Reaction or Equation
Solid aluminum, Al, reacts with oxygen gas, O2,
to form solid aluminum oxide, Al2O3
Reactant
Aluminum, Al
Oxygen, O2
Product
Aluminum oxide, Al2O3
Carbon tetrachloride, CCl4, is prepared in the
liquid form, by the reaction of chlorine gas, Cl2,
and methane gas, CH4. Hydrogen chloride is
also formed in the reaction.
Chlorine, Cl2
Methane, CH4
Carbon tetrachloride, CCl4
Hydrogen chloride, HCl
Ca + 2HCl → CaCl2 + H2
Ca
HCl
NH3
O2
CaCl2
H2
NO
H2O
4 NH3 + 5 O2 → 4 NO + 6 H2O
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.014.05
Describe what do the arrows and coefficients mean in chemical equations
Arrows: They separate reactants from products and specify direction
A coefficient is the number written in front of a reactant or a product that is used to balance
chemical equations.
Coefficients: They specify the relative amount of the components
CHM.5.3.01.014.06
Identify the meaning of different symbols used in equations
Symbols Used in Chemical Equations
Symbol
Explanation
+
Separates two or more reactants or products
→
“Yields,” separates reactants from products
⇌
Yields,” separates reactants from products
Indicates a reversible reaction
(s)
Designates a reactant or product in the solid state
It is placed after the formula
(l)
Designates a reactant or product in the liquid state
It is placed after the formula
(g)
Designates a reactant or product in the gaseous state
It is placed after the formula
(aq)
Designates an aqueous solution; the substance is dissolved in water
It is placed after the formula
∆
→
Indicates that heat is supplied to the reaction
X
A formula written above or below the yields sign indicates its use as a
catalyst
→
CHM.5.3.01.014.06
Describe word equation
A word equation is a statement to indicate the reactants and products of chemical reactions.
Example:
Write the word equation for the following chemical Reaction: N2(g) + 3H2(g) → 2NH3(g)
Nitrogen gas and hydrogen gas react together to produce ammonia gas.
CHM.5.3.01.014.08
Compare and contrast a skeleton equation and a chemical equation
The skeleton equation includes the formulas of reactants and products (without indicating their
amounts).
The chemical equation gives the relative amounts of reactants and products.
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.014.09 Write a skeleton equation for a word equation and vice versa
Example:
Write the skeleton equation for the following word equation:
“Iron reacts with oxygen to form iron (III) oxide”
Skeleton equation:
CHM.5.3.01.014.10
is conserved
Fe + O2 → Fe2O3
Explain why it is important to balance a chemical equation while identifying what
Because mass is neither created nor destroyed in chemical reactions (according to the Law of
conservation of mass), the number of atoms of all elements must be equal on both sides of the
reaction arrow
In a chemical equation, mass and number of atoms are conserved.
CHM.5.3.01.014.11 Explain why when balancing a chemical equation, the subscript of the formula is
not adjusted
When balancing a chemical equation, the subscript of the formula is not adjusted because this will
change the identity of the substance.
CHM.5.3.01.014.12
Identify the quantitative information revealed by a chemical equation
The quantitative information revealed by a chemical equation is the number of molecules, and
atoms of each element or compound composing the reactants and products
CHM.5.3.01.014.13
Describe the steps used in balancing an equation
The steps for balancing an equation:
1) Write the skeleton equation for the reaction
2) Count the atoms of the elements in the reactants
3) Count the atoms of the elements in the products
4) Change the coefficients to make the number of atoms of each element equal on both sides of
the equation, showing that atoms are conserved
5) Write the coefficients in their lowest possible ratio
6) Check you work
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.014.14
Balance different chemical equations
Example:
Balance each of the following chemical equations.
__Fe2 O3(g) + __H2 (g) → __ Fe(s) + __H2 O(g)
1, 3, 2, 3
__Cl2(g) + __CH4 (g) → __CCl4 (l) + __HCl(g)
4, 1, 1, 4
__Zn(s) + __CuCl2 (aq) → __ZnCl2 (aq) + __Cu(s)
1, 1, 1, 1
__F2(g) + __H2 O(l) → __HF(g) + __O2(g)
2, 2, 4, 1
CHM.5.3.01.015.01 Build, using molymods, a model for representing chemical reactions and chemical
compounds
Inspire Chemistry – Module 8 – Lesson 2: Classifying Chemical Reactions
CHM.5.3.01.016.01 Define a synthesis reaction while writing the general equation, particulate diagram
and some examples
Synthesis Reaction is a chemical reaction in which two or more substances react to produce a
single product
General Equation:
X + Y → XY
Particulate Diagram:
Examples:
Fe(s) + S(s) → FeS(s)
2Al(s) + 3Br2(g) → 2AlBr3(s)
2Cu(s) + O2(g) → 2CuO(s)
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.017.01
Write a balanced chemical equation for a synthesis reaction
Example:
Consider the following diagram that represents the reaction between iron and sulfur.
Identify the type of the reaction and write a balanced chemical equation for the reaction taking
place.
Type of the reaction: Synthesis Reaction
Balanced Chemical Equation: Fe(s) + S(s) → FeS(s)
or
5Fe(s) + 5S(s) → 5FeS
CHM.5.3.01.016.02 Define a combustion reaction while writing the general equation, particulate
diagram and some examples
Combustion reaction is a simple type of synthesis reaction, where a substance combines with
oxygen and releases energy in the form of heat and light.
Not all combustion reactions are synthesis reactions. In some combustion reaction, one substance
replaces another in the formation of products.
General Equation:
X + O2 → XO2
Particulate Diagram:
Examples:
2H2(g) + O2(g) → 2H2O(l)
C(s) + O2(g) → CO2(g)
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)
C3H6O(l) + 4 O2(g) → 3 CO2(g) + 3 H2O(l)
CHM.5.3.01.024.01
combustion
-
Compare between the products of complete combustion and incomplete
Complete Combustion: The products are carbon dioxide and water
Example: CH4 + 2O2 → CO2 + 2H2O
-
Incomplete Combustion: The products are carbon monoxide and water
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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Example: 2CH4 + 3O2 → 2CO + 4H2O
CHM.5.3.01.024.02
Describe the side effect of incomplete combustion
Incomplete combustion produces carbon monoxide; a toxic poisonous gas, that binds with
heamoglobin to form carboxyheamoglobin, and this reduces the ability of red blood cells to carry
oxygen and causes suffocation
CHM.5.3.01.017.02 Write balanced chemical equations, and/or word equations for the complete and
incomplete combustion of hydrocarbons (as methane and butane)
Examples:
Combustion of Methane, CH4
Complete combustion:
Combustion of Butane, C4H10
Complete combustion:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
Methane reacts with oxygen to produce carbon
dioxide and water
Butane reacts with oxygen to produce carbon
dioxide and water
Incomplete combustion:
Incomplete combustion:
2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(l)
2 C4H10(g) + 9 O2(g) → 8 CO(g) + 10 H2O(l)
Methane reacts with oxygen to produce carbon
monoxide and water
Butane reacts with oxygen to produce carbon
dioxide and water
CHM.5.3.01.025.01
Compare between complete and incomplete combustion through a lab activity
CHM.5.3.01.016.03 Define a decomposition reaction while writing the general equation, particulate
diagram and some examples
Decomposition reaction is a chemical reaction in which a single compound breaks into two or
more elements or new compounds
General Equation:
XY → X + Y
Particulate Diagram:
Examples:
2 NaCl → 2 Na + Cl2
∆
K2CO3(s) → K2O(s) + CO2(g)
CaCO3 → CaO + CO2
CHM.5.3.01.017.03
Write a balanced chemical equation for a decomposition reaction
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 7 of 46
Example:
Consider the following diagram that represents heating some solid
ammonium chloride, NH4Cl, in a test tube. Upon Heating, bubbles of
ammonia gas, NH3, and hydrogen chloride gas, HCl, are produced.
Identify the type of the reaction and write a balanced chemical
equation for the reaction taking place.
Type of Reaction: Decomposition Reaction
Balanced Chemical Equation: NH4Cl(s) → NH3(g) + HCl (g)
CHM.5.3.01.020.01 Define a single-replacement reaction while writing the general equation, particulate
diagram and some examples
Single-replacement reaction is a chemical reaction in which the atoms of one element replace the
atoms of another element in a compound
General Equation:
X + YZ → XZ + Y
Particulate Diagram:
Examples:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Cl2(g) + 2KBr(aq) ⟶ 2KCl(aq) + Br2(l)
MQ + Z → ZQ + M
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 8 of 46
CHM.5.3.01.020.02 Explain how an activity (reactivity) series of metals is organized
Reactivity is the ability to react
with another substance.
The activity series of metals
orders metals by their reactivity
with other metals.
The most active metals are on
the top of the list, and the least
active metals are at the bottom
of the list.
CHM.5.3.01.020.03 Explain; using activity (reactivity) series, whether or not a metal will always
replace another metal in a compound dissolved in water
A specific metal can replace any metal listed below it that is in a compound.
A specific metal cannot replace any metal listed above it that is in a compound.
CHM.5.3.01.020.04 Use the activity (reactivity) series of metals to predict if a metal can replace
hydrogen or another metal in a solution while writing the products of the reaction; if any
Example:
Using the reactivity series, predict whether each of the following reactions will occur or not.
If a reaction will occur, write (R). If a reaction does not occur write (NR).
Mg + HCl →
R
Cu + MgCl2 →
NR
Zn + AgNO3 →
R
Ag + NaNO3 →
NR
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.021.02 Perform a practical experiment to predict the relative reactivity of certain metallic
elements
CHM.5.3.01.020.05 Use the activity (reactivity) series of halogens to predict if a halogen can replace
another halogen in a solution while writing the products of the reaction; if any
Example:
Using the reactivity series, predict whether each of the following reactions will occur or not.
If a reaction will occur, write (R). If a reaction does not occur write (NR).
Br2 + HCl →
CHM.5.3.01.017.04
NR
Cl2 + 2 HBr →
R
Cl2 + 2 HI →
R
I2 + 2 HCl →
NR
Write a balanced chemical equation for a single-replacement reaction
Example:
A student carried out an experiment to compare the reactivity of magnesium, Mg, and copper, Cu.
a) In which of the two experiments did a reaction take place? Justify your answer.
Experiment 1.
The color of solution changed from blue to colorless.
Pieces of reddish brown copper are deposited.
b) For the reaction that took place: Identify its type and write a balanced chemical equation.
A single-replacement reaction
Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)
c) Which metal is more reactive Cu or Mg?
Magnesium, Mg, is more reactive than Cu.
Mg can replace copper in its solution but Cu cannot replace Mg in its solution.
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.016.05 Define a double-replacement reaction while writing the general equation,
particulate diagram and some examples
Double replacement is a chemical reaction that involves exchange of ions between two
compounds.
General Equation:
XY + WZ → XZ + WY
Particulate Diagram:
Examples:
Ca(OH)2(aq) + 2HCl(aq) ⟶ CaCl2(aq) + 2H2O(l)
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
2NaOH(aq) + FeCl2(aq) →Fe(OH)2(s) + 2NaCl(aq)
FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq)
CHM.5.3.01.016.06
Describe what happens to the anions in a double-replacement reaction
In a double replacement reaction, the anions change places. The anion of one compound bonds
with the cation of the second compound. The anion of the second compound bonds with the cation
of the first compound. Two new compounds are formed.
CHM.5.3.01.016.07
Describe the result of a double-replacement reaction
A double replacement reaction produces two different compounds (A solid precipitate, water or
gas)
CHM.5.3.01.017.05
Describe the guidelines for writing double-replacement reactions
The guidelines for writing double-replacement reactions:
1) Write the components of the reactants in a skeleton equation
2) Identify the cations and the anions in each compound
3) Pair up each cation with the anion from the other compound
4) Write the formulas for the products using the pairs from Step 3
5) Write the complete equation for the double-replacement reaction
6) Balance the equation
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.016.08
Compare and contrast single-replacement and double-replacement reactions
Single-replacement Reaction
Double-replacement Reaction
The atoms of one element replace atoms of
another element in a compound
Two compounds dissolved in water exchange
positive ions
Examples:
Examples:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)
Mg (s) + CuSO4 (aq) ⟶ MgSO4 (aq) + Cu(s)
K2CO3(aq) + BaCl2(aq) → 2KCl(aq) + BaCO3(s)
Cl2 (g) + 2NaI (aq) ⟶ 2NaCl (aq) + I2 (aq)
CHM.5.3.01.017.06
Write a balanced chemical equation for a double-replacement reaction
Example:
Aqueous solutions of sulfuric acid, H2SO4, and potassium hydroxide, KOH, react to form
potassium sulfate, K2SO4 and water, H2O.
Write the balanced equation for the above reaction.
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
CHM.5.3.01.016.04 Identify the reaction type when a chemical equation, word equation or a particulate
diagram is given and vice versa
Example:
Identify the type of the chemical reaction given in the table below.
Chemical equation
Type of chemical reaction
C2H4 (g) + 3O2 (g) ⟶ 2CO2 (g) + 2H2O (g)
Combustion
Zn (s) + CuSO4 (aq) ⟶ ZnSO4 (aq) + Cu(s)
Single-replacement
CaCO3(s) ⟶ CaO(s) + CO2(g)
Decomposition
H2(g) + F2(g) ⟶ 2 HF(g)
Synthesis
Ca(OH)2(aq) + 2HCl(aq) ⟶ CaCl2(aq) + 2H2O(l)
Double-replacement
Cl2(g) + 2KBr (aq) ⟶ 2KCl (aq) + Br2 (l)
Single-replacement
Decomposition
Synthesis
AX + BY → AY + BX
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Double replacement
Page 12 of 46
CHM.5.3.01.021.01 Conduct practical experiments or develop simulation software to compare the types
of reactions (synthesis, decomposition, single displacement, double displacement)
Inspire Chemistry – Module 8 – Lesson 3: Reactions in Aqueous Solutions
CHM.5.3.03.006.01 Define aqueous solution, solute and solvent
Aqueous solution is a solution in which the most plentiful substance is water.
Solute is the substance dissolved in a solution.
Solvent is the most plentiful substance in a solution.
In aqueous solutions, the solvent is always water.
CHM.5.3.03.006.02 Describe the difference in the behavior of molecular compounds and ionic
compounds as solutes in aqueous solutions
Molecular Compounds in Solution
Some solutes are molecular compounds that form ions when they dissolve in water.
Example: Hydrogen chloride, HCl:
+
HCl(aq) → H(aq)
+ Cl−
(aq)
Ionic Compounds in Solution
Ionic compounds consist of positive and negative ions held together by ionic bonds.
When some ionic compounds dissolve in water, their ions can separate (they dissociate).
Example: Sodium chloride, NaCl:
NaCl(aq) → Na+(aq) + Cl−
(aq)
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CHM.5.3.03.006.03 List three common types of products produced by reactions that occur in aqueous
solutions
The three common types of products are:
1) Solids or precipitates
2) Water
3) Gases
CHM.5.3.03.003.01
Compare chemical equations and complete ionic equations
Chemical equation: It is a statement using chemical formulae to describe the identities and relative
amounts of the reactants and products involved in the chemical reaction.
Complete ionic equation: It is chemical equation that shows all of the particles in a solution as they
really exist. It is a chemical equation where the electrolytes in an aqueous solution are written as
dissociated ions.
Note:
1. Only strong electrolytes are to be written as ions.
2. Compounds or elements in the solid, liquid or gaseous states remain intact and do not
dissociate.
Example:
Consider the reaction of aqueous solutions of sodium hydroxide, NaOH, and hydrochloric acid,
HCl, to form aqueous sodium chloride, NaCl, and water, H2O.
Write the balanced chemical equation and the complete ionic equation for the reaction taking
place.
Balanced Chemical Equation: NaOH(aq) + HCl(aq) ⟶ NaCl(aq) + H2O(l)
−
+
+
−
Complete Ionic Equation: Na+(aq) + OH(aq)
+ H(aq)
+ Cl−
(aq) → Na (aq) + Cl(aq) + H2 O(l)
CHM.5.3.03.003.02 Compare a net ionic equation and a complete ionic equation
Complete ionic equation: All dissolved ionic compounds and highly ionized molecular compounds
are shown as free ions.
Net ionic equation: It includes only the particles that take part in the reaction.
Example:
Consider the reaction of aqueous solutions of sodium hydroxide, NaOH, and hydrochloric acid,
HCl, to form aqueous sodium chloride, NaCl, and water, H2O.
Write the balanced chemical equation and the complete ionic equation for the reaction taking
place.
−
+
+
−
Complete Ionic Equation: Na+(aq) + OH(aq)
+ H(aq)
+ Cl−
(aq) → Na (aq) + Cl(aq) + H2 O(l)
+
−
Net Ionic Equation: H(aq)
+ OH(aq)
→ H2 O(l)
CHM.5.3.03.006.04 Define spectator ion
Spectator ions are ions that do not participate in a chemical reaction.
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.03.006.05 Identify spectator ions in a given chemical equation
Examples:
Identify the spectator ions for each of the following reactions.
Reaction
Spectator Ions
Na2S(aq) + 2HCl(aq) → 2NaCl(aq) + H2S(g)
Na+ and Cl‒
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
Na+ and NO3‒
CuCl2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaCl(aq)
Na+ and Cl‒
KHCO3(aq) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l)
K+ and NO3‒
CHM.5.3.03.006.06 Describe the different solubility rules for ionic compounds
Solubility Rules for Ionic Compounds
Compounds containing
the following ions
Solubility
Li+, Na+, K+ and
ammonium (NH4+ )
Soluble
Nitrates (NO−
3 ) and
acetate, (CH3 COO− )
Soluble
Chloride, (Cl‒), bromide,
(Br‒), and iodide, (I‒)
Sulfate salts (SO2−
4 )
Hydroxide, OH‒
sulfide, S2‒
Carbonates, (CO2−
3 ), and
phosphates, (PO3−
4 )
Exceptions
Soluble
When these ions pair with Ag+, or Pb2+,
the resulting compounds are
insoluble
Soluble
2+
2+
2+
When SO2−
4 pairs with Sr , Ca , Ba ,
Pb2+ and Ag+, the resulting compound
is insoluble
Insoluble
When these ions part with Li+, Na+, K+,
NH4+ , Ca2+, Sr2+, or Ba2+, the resulting
compounds are soluble
Insoluble
When these ions part with Li+, Na+, K+,
NH4+ , Ca2+, Sr2+, or Ba2+, the resulting
compounds are soluble
Insoluble
When these ions part with Li+, Na+, K+,
or NH4+ , the resulting compounds are
soluble
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CHM.5.3.03.003.03 Write a balanced chemical equation, complete ionic equation, net ionic equation
and word equation for reactions that form precipitates (using solubility rules)
Example:
Consider the reaction between copper (II) sulfate solution, CuSO4, and barium nitrate solution,
Ba(NO3)2.
a) Write the balanced chemical equation with state symbols for the above reaction.
CuSO4(aq) + Ba(NO3)2(aq) ⟶ Cu(NO3)2(aq) + BaSO4(s)
b) Write the complete ionic equation for the above reaction.
−
−
2+
2+
Cu2+ (aq) + SO2−
4 (aq) + Ba (aq) + 2NO3 (aq) ⟶ BaSO4(s) + Cu (aq) + 2NO3 (aq)
c) Write the net ionic equation for the above reaction
Ba2+ (aq) + SO2−
4 (aq) ⟶ BaSO4(s)
Example:
In a lab experiment, when solutions of lead (II) nitrate, Pb(NO3)2 (aq), and potassium iodide, KI(aq),
were mixed together, a yellow precipitate was formed.
a) Identify the type of product formed in the above reaction.
A precipitate
b) Write the word equation for the above reaction.
Lead (II) nitrate + Potassium iodide → Lead (II) iodide + Potassium nitrate
c) Write a balanced chemical equation with state symbols for the above reaction.
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
d) Write the net ionic equation for the above reaction.
Pb2+(aq) + 2 I―(aq) → PbI2(s)
e) Identify the spectator ion(s).
K+ and NO3―
CHM.5.3.03.006.07 Use solubility rules to predict the result of a double displacement reaction (using
solubility rules)
Example:
Using the solubility rules, predict whether each of the following reactions of aqueous solutions
result in the formation of a precipitate or not. If a precipitate is formed, write (P), if no precipitate
it formed write (NP).
NaOH + KCl →
CHM.5.3.03.006.08
NP
P
NaCl + AgNO3 →
NP
CaCl2 + KI →
P
Ca(NO3)2 + Na2SO4 →
Identify the type of reaction that occurs in aqueous solutions
The reaction that occurs in an aqueous solution is a double-replacement reaction
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CHM.5.3.03.006.09
Type of
Reaction
Describe the three reactions that take place in aqueous solutions
Description & Examples
Reactions that form precipitates -
It is a double-replacement reaction.
Dissolved substances are mixed.
When they react, a solid is formed, visible as a white or colored cloudiness in the
reaction mixture.
Example:
Consider the reaction of aqueous solutions of sodium chloride, NaCl, and silver nitrate,
AgNO3, to form solid silver chloride, AgCl, and aqueous sodium nitrate, NaNO3.
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction taking place.
Balanced Chemical Equation: NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
−
−
Complete Ionic Equation: Na+(aq) + Cl−(aq) + Ag +(aq) + NO3(aq)
→ Na+(aq) + NO3(aq)
+ AgCl(s)
−
Net Ionic Equation: Ag +
(aq) + Cl(aq) → AgCl(s)
Reactions
that form
water
-
It is a double-replacement reaction.
Evidence of the reaction may not be observable because water is colorless, odorless,
and already makes up most of the solution.
Example:
Consider the reaction of aqueous solutions of sodium hydroxide, NaOH, and
hydrochloric acid, HCl, to form aqueous sodium chloride, NaCl, and water, H2O.
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction taking place.
Balanced Chemical Equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
+
+
−
−
Complete Ionic Equation: Na+(aq) + OH(aq)
+ H(aq)
+ Cl−
(aq) → Na (aq) + Cl(aq) + H2 O(l)
−
+
Net Ionic Equation: OH(aq)
+ H(aq)
→ H2 O(l)
Reactions
that form
gases
-
It is a double-replacement reaction.
Gases as carbon dioxide, CO2, hydrogen, H2, sulfur dioxide, SO2, and hydrogen
sulfide, H2S, may be formed.
Bubbles are produced or seen as the reaction proceeds.
1. Acid + Metal → Salt + Hydrogen
✓ Metal is not an alkali metal as Na, K or Li or Ca because the reaction is highly
explosive
✓ Metal is not Cu or Ag because they are highly Unreactive
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between hydrochloric acid solution, HCl, and magnesium, Mg.
Balanced Chemical Equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
+
2+
−
Complete Ionic Equation: Mg (s) + 2H(aq)
+ 2Cl−
(aq) → Mg (aq) + 2Cl(aq) + H2 (g)
+
Net Ionic Equation: Mg (s) + 2H(aq)
→ Mg 2+
(aq) + H2 (g)
2. Acid + Metal Carbonate → Salt + Water + Carbon Dioxide
✓ This is a double replacement reaction and a decomposition reaction.
Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between hydrochloric acid solution, HCl, and sodium carbonate solution,
Na2CO3.
Balanced Chemical Equation: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
+
−
+
−
Complete Ionic Equation: 2Na+(aq) + CO2−
3(aq) + 2H(aq) + 2Cl(aq) → 2Na (aq) + 2Cl(aq) + CO2(g) + H2 O(l)
+
Net Ionic Equation: CO2−
3(aq) + 2H(aq) → CO2(g) + H2 O(l)
3.
Acid + Metal Bicarbonate (Metal hydrogen carbonate) → Salt + Water + Carbon Dioxide
Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between hydrochloric acid solution, HCl, and sodium bicarbonate solution,
NaHCO3.
Balanced Chemical Equation: NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
When carbonic acid, H2CO3, is formed, it decomposes to water, H2O(l), and carbon
dioxide gas, CO2(g).
−
+
+
−
Complete Ionic Equation: Na+(aq) + HCO3(aq)
+ H(aq)
+ Cl−
(aq) → Na (aq) + Cl(aq) + CO2(g) + H2 O(l)
+
Net Ionic Equation: HCO−
3(aq) + H(aq) → CO2(g) + H2 O(l)
4. Acid + Metal Sulfite → Salt + Water + Sulfur dioxide
Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between hydrochloric acid solution, HCl, and sodium sulfite solution,
Na2SO3.
Balanced Chemical Equation: Na2SO3(aq) + 2HCl(aq) → 2NaCl(aq) + SO2(g) + H2O(l)
+
−
+
−
Complete Ionic Equation: 2Na+(aq) + SO2−
3(aq) + 2H(aq) + 2Cl(aq) → 2Na (aq) + 2Cl(aq) + SO2(g) + H2 O(l)
+
Net Ionic Equation: SO2−
3(aq) + 2H(aq) → SO2(g) + H2 O(l)
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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5. Acid + Metal Sulfide → Salt + Hydrogen Sulfide
Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between hydrochloric acid solution, HCl, and sodium sulfide solution, Na2S.
Balanced Chemical Equation: Na2S(aq) + 2HCl(aq) → 2NaCl(aq) + H2S(g)
2−
+
+
−
Complete Ionic Equation: 2Na+(aq) + S(aq)
+ 2H(aq)
+ 2Cl−
(aq) → 2Na (aq) + 2Cl(aq) + H2 S(g)
2−
+
Net Ionic Equation: S(aq)
+ 2H(aq)
→ H2 S(g)
6. Base + Ammonium Salt → Salt + Water + Ammonia
Example:
Write the balanced chemical equation, complete ionic equation and net ionic equation for
the reaction between ammonium chloride solution, NH4Cl, and sodium hydroxide
solution, NaOH.
Balanced Chemical Equation: NH4Cl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) + NH3(g)
+
+
−
+
−
Complete Ionic Equation: NH4(aq)
+ Cl−
(aq) + Na (aq) + OH(aq) → Na (aq) + Cl(aq) + H2 O(l) + NH3 (g)
+
−
Net Ionic Equation: NH4(aq)
+ OH(aq)
→ NH3 (g) + H2 O(l)
CHM.5.3.03.003.04 Write a balanced chemical equation, complete ionic equation, net ionic equation
and word equation for reactions that form water (Reaction of a strong acid with a strong base)
Example: Mixing sodium hydroxide with hydrochloric acid
o Balanced chemical equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
o Complete ionic equation: Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) → Na+(aq) + Cl–(aq) + H2O (l)
o Net ionic equation: H+(aq) + OH–(aq) → H2O (l)
o Word equation: Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and
water
CHM.5.3.03.003.05 Write a balanced chemical equation, complete ionic equation, net ionic equation
and word equation for reactions that form gases (Reaction of acid with metal, metal carbonate, metal
bicarbonate, metal sulfite, metal sulfide, and reaction of a base with an ammonium salt)
Please refer to KPI 2.3.47 (Reaction that form gases)
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.03.002.01
and H2S)
Describe how to test for the presence of different gases (NH3, CO2, H2, O2, SO2
Testing for Gases
Some common experiments that will most likely produce a gas are:
1) When a solid substance is heated
2) When an acid is added to a substance
3) When an alkali is added to a substance
The following table shows the different tests and observations the students should carry out and
have when identifying the listed gases.
Gas
Test & Result
Smell
Ammonia
NH3
-
Test: Approach a red litmus paper
Result: The gas turns red litmus paper blue
Carbon dioxide
CO2
-
Test: Bubble the gas through limewater for a short time
Result: Lime water turns milky
Odorless
Hydrogen
H2
-
Test: Approach a lighted splint
Result: The gas burns with a squeaky pop sound with a
lighted splint
Odorless
Oxygen
O2
-
Test: Approach a glowing splint
Result: The gas relights a glowing splint
Odorless
Sulfur dioxide
SO2
-
Test: Bubble the gas through an orange solution of
potassium dichromate (VI).
Result: The gas turns aqueous potassium dichromate
(VI) into green
Smells like burnt
matches
Test: Approach a damp lead (II) acetate paper
Test: The gas turns the damp lead (II) acetate paper
black
Smell of rotten eggs
Hydrogen sulfide
H2S
CHM.5.3.03.003.07
-
Pungent smelling gas
Describe an overall equation while listing the steps
An overall equation is an equation that combines two reactions.
The steps in writing an overall equation are:
1. Write the equation for the first reaction
2. Write the equation for the second reaction
3. Combine the equations with both reactants on one side and both products on the other side
4. Cross out any substances that are on both sides of the equation
CHM.5.3.03.003.08
Write the overall equation for different reactions
Example:
Consider the reaction between sodium bicarbonate, NaHCO3, and hydrochloric acid, HCl. Write
the overall equation for the reaction taking place.
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2CO3(aq)
H2CO3(aq) → CO2(g) + H2O(l)
______________________________________
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 20 of 46
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
CHM.5.3.01.022.01
Compare between the reaction of metallic and non-metallic oxides with water and
the types of obtained solutions
Metallic oxides are basic while non-metallic oxides are acidic.
A metallic oxide solution is basic, it turns red litmus paper blue and has a pH greater than 7.
A non-metallic oxide solution is acidic, it turns blue litmus paper red and has a pH less than 7.
CHM.5.3.01.022.02 Plan and conduct a demonstration (considering safety rules) to compare the
properties of nonmetallic and metallic oxides
CHM.5.3.03.006.10 Predict the nature of products (water, gas, a precipitate) of the chemical reactions
that take place in aqueous solutions
Type of Reaction
Synthesis
Combustion
Decomposition
Reactants
Suggested Products
Two or more
substances
One compound
✓ Metal and oxygen
✓ Non-metal and
oxygen
✓ A compound and
oxygen
✓ The oxide of the metal
✓ The oxide of the nonmetal
✓ Two or more oxides
One Compound
Two or more elements
and/or compound
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
General Equation
X + Y → XY
X + O2 → XO2
XY → X + Y
Page 21 of 46
Single-replacement
✓ A metal and a
compound
✓ A non-metal and a
compound
✓ A new compound and
the replaced metal
✓ A new compound and
the replaced non-metal
X + WZ → XZ + W
Double-replacement
Two compound
Two different compounds,
one of which is a solid,
water or a gas
XY + WZ → XZ + WY
2.4 – The Mole
Inspire Chemistry – Module 9 – Lesson 1: Measuring Matter
CHM.5.3.01.002.01 Define mole
Mole is the SI base unit to measure the amount of a substance.
Abbreviation: mol
A mole is the number of carbon atoms in exactly 12 g of pure carbon-12.
A mole of anything contains 6.022140857 × 1023 representative particles
CHM.5.3.01.002.02
Explain why chemists use the mole
Chemists use the mole because it is a convenient way of knowing how many representative
particles are in a sample. It allows chemists to accurately calculate the number of atoms, molecules
or formula units in a substance
CHM.5.3.01.002.03
Describe Avogadro's number
It is the number 6.022140857 × 1023, which is the number of representative particles in a mole.
CHM.5.3.01.003.01
State the mathematical relationship between Avogadro's number and one mole
One mole contains Avogadro's number (6.00×1023) of representative particles
Number of Representative Particles = Number of moles × 6.00×1023
Number of moles =
Number of Representative Particles
6.00 × 1023
CHM.5.3.01.003.02 Describe, using particulate diagrams, different types of representative particles
Representative particles can be atoms, ions, molecules, or formula unit.
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 22 of 46
CHM.5.3.01.003.03 List the conversion factors used to convert between particles and moles
6.00 × 1023 representaive particles
and
1 mol
6.00 ×
1023
1 mol
represntative particles
CHM.5.3.01.004.01 Calculate the number of representative particles present in given moles of an
element (atomic or molecular) or a compound and vice versa
Representative Particles and Moles
Substance
Chemical Formula
Representative
Particle
Representative
Particles in 1.00 mol
Copper
Cu
Atom
6.02 × 1023
Atomic Nitrogen
N
Atom
6.02 × 1023
Nitrogen gas
N2
Molecule
6.02 × 1023
Water
H2O
Molecule
6.02 × 1023
Calcium ion
Ca2+
Ion
6.02 × 1023
Calcium fluoride
CaF2
Formula unit
6.02 × 1023
Example:
Calculate the number of representative particles of each of the following:
(Use Avogadro’s number as 6.00 × 1023)
12.0 mol Ag
Number of Particles = 12.0 × 6.00 × 1023 = 7.20 × 1024 atoms of Ag
18.0 mol H2O
Number of Particles = 18.0 × 6.00 × 1023 = 1.08 × 1025 molecules of H2 O
0.150 mol NaCl
Number of Particles = 0.150 × 6.00 × 1023 = 9.00 × 1023 formula units of NaCl
1.20⨯10‒2 mol H2
Number of Particles = 1.20 × 10−2 × 6.00 × 1023 = 7.20 × 1021 molecules of H2
Example:
How many moles contain the given quantity below?
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 23 of 46
(Use Avogadro’s number as 6.00 × 1023)
1.2⨯1015 molecules of CO2
Number of moles =
Number of Particles
1.2 × 1015
=
= 2.0 × 10−9 mol
6.00 × 1023
6.00 × 1023
3.6⨯1021 formula units of NaCl
Number of moles =
Number of Particles
3.6 × 1021
=
= 6.0 × 10−3 mol
6.00 × 1023
6.00 × 1023
3.0⨯1027 formula units of KI
Number of moles =
Number of Particles
3.0 × 1027
=
= 5.0 × 103 mol
6.00 × 1023
6.00 × 1023
Example:
Which of the following contains larges number of moles?
3.00 ⨯ 1024 atoms of Ne or 6.00 ⨯ 1024 atoms of Xe
Number of Representative Particles
3.00 × 1024
Number of moles of Ne =
=
= 5.00 mol
6.00 × 1023
6.00 × 1023
Number of Representative Particles
6.00 × 1024
Number of moles of Xe =
=
= 10.0 mol
6.00 × 1023
6.00 × 1023
6.00 ⨯ 1024 atoms of Xe contains larger number of moles than 3.00 ⨯ 1024 atoms of Ne
Inspire Chemistry – Module 9 – Lesson 2: Mass and the Mole
CHM.5.3.01.003.04 Define molar mass
Molar mass is the mass in grams of one mole of any pure substance
Unit is g/mol
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 24 of 46
CHM.5.3.01.004.02
Calculate molar mass of an element
The molar mass of any element is numerically equal to its atomic mass and has a unit of g/mol.
Example:
As per the periodic table, an atom of copper, Cu, has an atomic mass of 63.55 amu.
The molar mass of Cu is 63.55 g/mol.
1 mole of Cu or 6.00 × 1023 atoms of Cu has a mass of 63.55 g.
CHM.5.3.01.003.05 List the conversion factor needed to convert between mass and moles of an atom
Mass = Number of Moles × Molar Mass
Mass
Conversion Factor: Molar Mass and
CHM.5.3.01.004.03
Molar Mass
Mass
Calculate the mass (in grams) of an element given its moles and vice versa
Example:
Calculate the mass of 4.50 moles of cobalt, Co.
m = n × M = 4.50 × 58.93 = 265 g
Example:
Calculate number of moles in 118 g of cobalt, Co.
Number of moles =
Mass
118
=
= 2.00 moles
Molar Mass
58.93
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 25 of 46
CHM.5.3.01.003.06
Define atomicity
Atomicity is the total number of atoms present in one mole of an element or compound
CHM.5.3.01.003.07
Identify the atomicity of a given species
Example:
Identify the atomicity of each of the following species.
Species
Atomicity
NH3
4
H2SO4
7
Al2(SO4)3
17
H2O
3
S8
8
H2
2
Na
1
CaI2
3
Mg(OH)2
5
C12H22O11
45
CHM.5.3.01.004.05 Calculate the number of atoms in an element or compound given moles and vice
versa
Number of atoms = Number of Moles × (6.00 × 1023) × atomicity
Example:
Calculate the number of nitrogen atoms present in 0.50 moles of dinitrogen trioxide, N2O3,
molecules.
Number of nitrogen atoms = 0.50 × (6.00×1023) × 2 = 6.0×1023 atoms
Example:
Calculate the number of oxygen atoms present in 2.00 moles of aspirin, C9H8O4, molecules.
Number of oxygen atoms = 2.00 × (6.00×1023) × 4 = 4.8×1024 atoms
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 26 of 46
Inspire Chemistry – Module 9 – Lesson 3: Moles of Compounds
CHM.5.3.01.003.08 Describe what a subscript in a chemical formula indicates
The subscripts in a chemical formula represent the number of moles of an element in a given
compound.
CHM.5.3.01.001.01
chemical formula
Describe the relationship between the mole information of a substance and its
The mole information of the individual elements shows the ratio of the elements in a compound.
These mole ratios become the subscripts of the chemical formula.
CHM.5.3.01.001.02 Identify the number of moles of atoms and ions for a given chemical formula of a
compound (ionic and molecular)
Example:
Consider one mole of CCl2F2. Identify the number of moles of each atom in the compound.
1 mole of CCl2F2 contains:
1 mole of C atoms
2 moles of Cl atoms
2 moles of F atoms
Example:
Consider one mole of Al2(SO4)3. Identify the number of moles of each ion in the compound.
3
1 mole of Al2(SO4)3 contains:
2 moles of Al3+ ions
moles of SO2−
4 ions
CHM.5.3.01.004.06
compound
Calculate the number of moles of atoms (elements) in a given moles of a
Example:
Calculate the number of moles of phosphorus atoms and oxygen atoms in 2.00 moles of P4O10.
1 mole of P4O10 contains 4 moles of P atoms and 10 moles of O atoms.
2.00 mole of P4O10 contains 8.00 moles of P atoms and 20.0 moles of O atoms.
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 27 of 46
CHM.5.3.01.004.07
Calculate the number of moles of ions in a given moles of a compound
Example:
Iron (III) sulfate, Fe2(SO4)3, is sometimes used in the process of water purification. Calculate the
number of moles of iron (III) ions and sulfate ions in 4.00 moles of Fe2(SO4)3.
1 mole of Fe2(SO4)3 contains 2 moles of Fe3+ ions and 3 moles of SO2−
4 ions.
3+
4.00moles of Fe2(SO4)3 contains 8.00 moles of Fe ions and 12.0 moles of SO2−
4 ions.
CHM.5.3.01.003.09 Describe the molar mass of a compound
The molar mass of a compound is the sum of the masses of all elements in that compound.
CHM.5.3.01.003.10 Describe how to determine the molar mass of a compound and how the molar
masses of the ions that make up this compound relate to the law of conservation of mass
Multiply the mass of one mole of each element by the number of moles of that element in the
chemical formula. Add the resulting masses. During the formation of a compound, the
calculated molar masses of the reactants are equal to the calculated molar masses of the
products. This shows that matter is conserved in a chemical reaction.
CHM.5.3.01.004.08
Calculate the molar mass of a compound (ionic and molecular)
Example:
Calculate the molar mass of CaSO4
Molar mass = (1 × 40.08) + (1 × 32.07) + (4 × 16.00) = 136.15 g/mol
Example:
Calculate the molar mass of CO2
Molar mass = (1 × 12.01) + (2 × 16.00) = 44.01 g/mol
CHM.5.3.01.004.09
Calculate the mass of a given moles of a compound and vice versa
Example:
Calculate the mass of 2.00 moles of CaSO4
m = n × M = 2.00 × 136.15 = 272 g
Example:
Calculate number of moles in 10.0 g of calcium carbonate, CaCO3.
Number of moles =
Mass
10.0
=
= 0.100 moles
Molar Mass
100.09
CHM.5.3.01.001.03 Explain how a chemical formula is used to determine the number of moles in a
given mass of compound
Convert the mass to moles, multiply the number of moles by the ratio of the number of atoms or
ions to one mole, and multiply by Avogadro’s number.
CHM.5.3.01.004.10 Calculate the number of representative particles present in given mass of a
compound (ionic and molecular) and vice versa
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 28 of 46
Example:
Consider a 10.0 g sample of calcium carbonate, CaCO3 (Ca = 40, C = 12, O = 16)
Answer the following questions:
a) Calculate the number of moles of the sample
Number of moles =
Mass
10.0
=
= 0.100 moles
Molar Mass 100
b) How many calcium ions, Ca2+, are present in the sample?
1 mole of CaCO3 contains 1 mol of Ca2+ ions
0.100 moles of CaCO3 contains 0.100 mol of Ca2+ ions
Number of Ca2+ ions = 0.100 × 6.00 × 1023 = 6.00 × 1022 Ca2+ ions
c) How many carbonate ions, CO2−
3 , are present in the sample?
1 mole of CaCO3 contains 1 mol of CO2−
3 ions
0.100 moles of CaCO3 contains 0.100 mol of CO2−
3 ions
2−
23
Number of CO3 ions = 0.100 × 6.00 × 10 = 6.00 × 1022 CO2−
3 ions
d) Calculate the mass, in grams, of one formula unit of CaCO3.
Mass of on Formula unit =
Molar Mass
100
=
= 1.67 × 10−22 g
23
23
6.00 × 10
6.00 × 10
Example:
Consider a 2.3 g sample of ethanol, C2H5OH (C = 12, H = 1, O = 16)
Answer the following questions:
a) Calculate the number of moles of the sample
Number of moles =
Mass
2.3
=
= 0.050 moles
Molar Mass 46
b) How many carbon atoms, C, are present in the sample?
1 mole of C2H5OH contains 2 moles of C atoms
0.050 moles of C2H5OH contains 0.10 mol of C atoms
Number of C atoms = 0.10 × 6.00 × 1023 = 6.0 × 1022 C atoms
c) How many hydrogen atoms, H, are present in the sample?
1 mole of C2H5OH contains 6 moles of H atoms
0.050 moles of C2H5OH contains 0.30 mol of H atoms
Number of H atoms = 0.30 × 6.00 × 1023 = 1.8 × 1023 H atoms
d) How many oxygen atoms, O, are present in the sample?
1 mole of C2H5OH contains 1 mole of O atoms
0.050 moles of C2H5OH contains 0.050 mol of O atoms
Number of O atoms = 0.050 × 6.00 × 1023 = 3.0 × 1022 H atoms
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 29 of 46
Inspire Chemistry – Module 9 – Lesson 4: Empirical and Molecular Formulas
CHM.5.3.01.006.01 Define percent composition
Percent composition is the percent by mass of each element in a compound.
CHM.5.3.01.009.01
experimental data
Calculate the percent composition (percent by mass of an element) from
If you know the relative mass of each element in a compound, you can calculate the percent
composition of the compound.
The percent by mass of an element in a compound is the mass in grams of the element divided by
the mass in grams of the compound, multiplied by 100%.
The percent composition of a compound is always the same.
The percent of all the elements in a compound adds up to 100%.
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐛𝐲 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 =
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
× 𝟏𝟎𝟎
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
Example:
What is the percent by mass of a sample of a compound that contains 85 g element X and 15 g
element Y?
Percent by mass of X =
85
× 100 = 85 %
100
Percent by mass of Y =
15
× 100 = 15 %
100
CHM.5.3.01.009.02
chemical formula
Calculate the percent composition (percent by mass of an element) from the
The subscripts in the formula are used to calculate the mass of each element in a mole of that
compound.
Using the individual masses of the elements and the molar mass, you can calculate the percent by
mass of each element.
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐛𝐲 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 =
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐢𝐧 𝟏 𝐦𝐨𝐥𝐞 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
× 𝟏𝟎𝟎
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
Example:
Calculate the percentage by mass of hydrogen in methane, CH4.
(C = 12, H = 1)
Percent by mass of H =
4×1
16
× 100 = 25 %
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
Page 30 of 46
CHM.5.3.01.006.02
Define empirical formula
Empirical formula is the formula with the smallest whole-number mole ratio of the elements.
CHM.5.3.01.006.03 Describe the experimental data that is needed to calculate percent composition of
an unknown compound
Experimental analyses are used to determine the mass of each element in a sample.
CHM.5.3.01.009.03 Determine the empirical formula of a compound given the mass of the elements
Example:
A sample of a compound contains 3.2 g of sulfur, S, and 4.8 g of oxygen, O. Find the empirical
formula of the compound. (S = 32, O = 16)
1. Write given quantity
2. Calculate number of moles
m
n=M
n=
O
3.2 g
4.8 g
3.2
= 0.10 mol of S
32
n=
0.10
=1
0.10
3. Divide moles by the smallest
value
4. Write the simplified ratio then
deduce the empirical formula
S
4.8
= 0.30 mol ofO
16
0.30
=3
0.10
SO3
CHM.5.3.01.009.04 Determine the empirical formula of a compound given the percent composition
Example:
A sample of a compound contains 50% sulfur, S, and 50% oxygen, O. Find the empirical formula
of the compound. (S = 32, O = 16)
1. Write given quantity
S
50%
O
50%
2. Write the quantity in grams
50 g
50 g
3. Calculate number of moles
m
n=
M
n=
50
= 1.5625 mol of S
32
4. Divide moles by the smallest
value
1.5625
=1
1.5625
n=
50
= 3.125 mol ofO
16
3.125
=2
1.5625
5. Write the simplified ratio then
SO2
deduce the empirical formula
CHM.5.3.01.006.04 Define molecular formula
Molecular formula specifies the actual number of atoms of each element in one molecule or
formula unit of a substance
CHM.5.3.01.007.01 Describe, using examples, the difference between empirical formula and molecular
formula
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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An empirical formula is the smallest-whole number ratio of elements that make up a compound,
ex The empirical formula of C6H6 is CH
A molecular formula specifies the actual number of atoms of each element in one molecule or
formula unit of the substance, ex C6H6
Example:
Use the following table to write the empirical formula for each given molecular formula.
Molecular formula
Empirical formula
CO
CO
C12H22O11
C12H22O11
C3H6O3
CH2O
P4O10
P2O5
NH3
NH3
C6H12O6
CH2O
C8H10N4O2
C4H5N2O
N2O4
NO2
CHM.5.3.01.009.05 Determine molecular formula of a compound given the experimentally determined
molar mass of the compound
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐞𝐦𝐩𝐢𝐫𝐢𝐜𝐚𝐥 𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐮𝐧𝐢𝐭𝐬 =
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐟𝐨𝐫𝐦𝐮𝐥𝐚
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐦𝐩𝐢𝐫𝐢𝐜𝐚𝐥 𝐟𝐨𝐫𝐦𝐮𝐥𝐚
Example:
A compound has a molar mass of 360 g/mol and its empirical formula is CH2O. What is the
molecular formula of the compound?
(C = 12, H = 1, O = 16)
Number of empirical formula units =
molar mass of molecular formula
360
=
= 12
molar mass of empirical formula
30
There are 12 units of CH2O
The molecular formula is C12H24O12
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Example:
A student analyzed a sample of a compound whose molecular formula has a molar mass of 32 g/mol. The
compound contained 1.4 g of nitrogen, N, and 0.20 g of hydrogen, H. Answer questions a and b.
a) What is the empirical formula of the compound?
(N = 14, H = 1)
1. Write given quantity
2. Calculate number of
m
moles n = M
n=
N
H
1.4 g
0.20 g
1.4
= 0.10 mol of N
14
3. Divide moles by the
smallest value
n=
0.10
=1
0.10
4. Write the simplified ratio
then deduce the empirical
formula
0.20
= 0.20 mol of H
1
0.20
=2
0.10
NH2
b) What is the molecular formula of the compound?
Number of empirical formula units =
Molar mass of the molecular formula
32
=
=2
Molar mass of the empirical formula
16
Molecular formula is N2H4
CHM.5.3.01.010.01 Plan and conduct a practical experiment, (considering safety rules) to determine the
empirical formula of a chemical compound (such as magnesium oxide)
CHM.5.3.01.008.01 Explain the relationship between the empirical formula and the molecular formula
through simulation software
CHM.5.3.01.008.02 Explain the relationship between the empirical formula and the molecular formula
using models and/or drawings
CHM.5.3.01.005.01 Determine the Percent of composition of a specific chemical compound through
practical experiment
3.3 – Stoichiometry
Inspire Chemistry – Module 10 – Lesson 1: Defining Stoichiometry
CHM.5.3.01.011.01 Define stoichiometry
Stoichiometry is the study of quantitative relationships between the amounts of reactants used and
amounts of products formed by a chemical reaction.
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CHM.5.3.01.011.02 Explain the importance of the law of conservation of mass in chemical reactions
The law of conservation of mass states that matter is neither created nor destroyed.
Bonds in a chemical reaction break and reform to produce products, but the amount of matter at
the end of the reaction is the same as it was at the beginning. This is seen in a chemical reaction
when the same number of each type of atom exists on both sides of the equation.
CHM.5.3.01.011.03 Interpret a balanced chemical equation in terms of moles, mass and representative
particles (atoms, molecules and formula units)
Example:
Interpret the following balanced equation in terms of moles, mass, representative particles and the
show that the law of conservation of mass is observed.
(Na = 23, Cl = 35.5, Pb = 207, N = 14, O = 16)
2 NaCl(aq)
In terms of
representative
particles
In terms of
moles
In terms of Mass
Law of
Conservation of
Mass
+
Pb(NO3)2(aq)
→
2NaNO3(aq)
+
PbCl2(s)
2 formula units of
NaCl
1 formula unit of
Pb(NO3)2
2 formula units of
NaNO3
1 formula unit of
PbCl2
2 moles of NaCl
1 mole of Pb(NO3)2
2 moles of NaNO3
1 mole of PbCl2
117 g of NaCl
331 g of Pb(NO3)2
170 g of NaNO3
278 g of PbCl2
117 g + 331 g = 448 g Reactants
170 g + 278 g = 448 g Products
Mass is Conserved
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Example:
Interpret the following balanced equation in terms of moles, mass, representative particles and the
show that the law of conservation of mass is observed.
(C = 12, O = 16, H = 1)
CH4(g)
+
2O2(g)
→
CO2(g)
+
2H2O(g)
In terms of
representative
particles
1 molecule of
CH4
2 molecules of O2
1 molecule of CO2
2 molecules of
H2O
In terms of
moles
1 mole of CH4
2 moles of O2
1 mole of CO2
2 moles of H2O
16 g of CH4
64 g of O2
44 g of CO2
36 g H2O
In terms of Mass
Law of
Conservation of
Mass
16 g + 64 g = 80 g Reactants
44 g + 36 g = 80 g Products
Mass is Conserved
CHM.5.3.01.011.04 Compare the mass of the reactants and the mass of the products in a chemical
reaction while explaining how these masses are related
The coefficients indicate the molar relationship between each pair of reactants and products.
The masses of reactants and products are equal.
CHM.5.3.01.011.05 Describe a mole ratio
A mole ratio is the ratio between the numbers of moles of any two substances in a balanced
chemical equation.
CHM.5.3.01.011.06 Identify the source from which a chemical reaction’s mole ratios are derived
A chemical reaction’s mole ratios are derived from the relationships between coefficients in a
balanced chemical equation.
The number of mole ratios that can be written for a chemical reaction that has a total number of n
substances is: (n)(n – 1).
Number of substances in a chemical reaction
(n)
3
4
5
6
Total number of mole ratio
(n)(n – 1)
6
12
20
30
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CHM.5.3.01.011.07 Write a balanced chemical equation for a word equation while determining the
possible mole ratios
Example:
Consider the thermal decomposition of solid potassium chlorate to form solid potassium chloride
and oxygen gas.
Write a balanced chemical equation and the possible mole ratios for the reaction.
Balanced Chemical Equation:
2 KClO3(s) → 2KCl(s) + 3O2(g)
The mole ratios that can be written for the above reaction are as follows:
2 mol KClO3
2 mol KClO3
and
2 mol KCl
3 mol O2
2 mol KCl
2 mol KCl
and
2 mol KClO3
3 mol O2
3 mol O2
3 mol O2
and
2 mol KClO3
2 mol KCl
Inspire Chemistry – Module 10 – Lesson 2: Stoichiometric Calculations
CHM.5.3.01.011.08 Explain why a balanced chemical equation is needed to solve a stoichiometric
problem
The coefficients in the balanced equation indicate the molar relationship between each pair of
reactants and products.
CHM.5.3.01.012.01
List the steps included in stoichiometric calculations
o Step 1 – Start with a balanced equation. Interpret the equation in terms of moles.
o Step 2 – Convert from grams to moles of the given substance. Use the inverse of the molar
mass as the conversion factor.
o Step 3 – Convert from moles of the given substance to moles of the unknown substance. Use
the appropriate mole ratio from the balanced chemical equation as the conversion factor.
o Step 4 – Convert from moles of unknown to grams of unknown. Use the molar mass as the
conversion factor.
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CHM.5.3.01.012.02 Calculate the number of moles of a reactant or a product given the number of
moles of another reactant or product
Example:
Consider the following reaction: N2(g) + 3 H2(g) → 2 NH3(g).
0.60 moles of nitrogen, N2, reacts with hydrogen, H2, to form ammonia, NH3.
a) Calculate number of moles of H2 needed to react with 0.60 moles of N2.
N2
1 mole
0.60 moles
3 H2
3 moles
?
Number of moles of H2 = 1.8 moles
b) Calculate number of moles of NH3 produced when 0.60 moles of N2 reacted.
N2
1 mole
0.60moles
2 NH3
2 moles
?
Number of moles of NH3 = 1.2 moles
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CHM.5.3.01.012.03 Calculate the mass of a reactant or a product given the number of moles of another
reactant or product and vice versa
Example:
Acetylene gas, C2H2, is produced by adding water to calcium carbide, CaC2, according to the
following equation: CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq).
Answer the following question.
a) Calculate the mass of C2H2 produced when 4.00 moles of CaC2 reacted. (C = 12, H = 1)
CaC2
1 mole
1 mole
4.00 moles
C2H2
1 mole
1 × 26 g
???? g
Mass of C2H2 = 104 g
b) Calculate the mass of H2O needed to react with 4.00 moles of CaC2. (C = 12, H = 1)
CaC2
1 mole
1 mole
4.00moles
2 H2O
2 mole
2 × 18 g
???? g
Mass of CaC2 = 144 g
CHM.5.3.01.012.04 Calculate the mass of a reactant or a product given the mass of another reactant or
product
Example:
Acetylene gas, C2H2, is produced by adding water to calcium carbide, CaC2, according to the
following equation: CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq).
Answer the following questions.
(Ca = 40, C = 12, H = 1, O = 16)
a) Calculate the mass of CaC2 needed to react with 3.60 g of H2O.
CaC2
1 mole
1 × 64 g
???? g
2 H2O
2 moles
2 × 18 g
3.60 g
Mass of CaC2 = 6.40 g
b) Calculate the mass of C2H2 produced when 3.60 g of H2O reacts with CaC2.
C2H2
1 mole
1 × 26 g
???? g
2 H2O
2 moles
2 × 18 g
3.60 g
Mass of C2H2 = 2.60 g
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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Inspire Chemistry – Module 10 – Lesson 3: Limiting Reactants
CHM.5.3.01.013.01 Describe the reason why a reaction between two substances comes to an end
One of the reactants is used up.
CHM.5.3.01.013.02 Distinguish between limiting reactant and excess reactant in a chemical reaction
Limiting reactant is the reactant that controls the amount of product formed in a chemical reaction
Excess reactant is the substance that is not used up completely in a reaction (It is only partially
consumed in the reaction)
CHM.5.3.01.033.01
Explain the importance of using an excess reactant in a chemical reaction
Using an excess of one reactant causes reactions to be driven to continue until all of the limiting
reactant is used up. This technique can also speed up a reaction.
Steps to identify limiting reactant:
1) Find number of moles of each reactant
2) Divide each number of moles by the reacting ratio
3) The smaller quantity is the limiting reactant
CHM.5.3.01.013.03 Identify limiting reactant and excess reactant in a chemical reaction given the
particulate diagram of reactants
Example:
Hydrogen gas reacts with oxygen gas to form water vapor according to the following balanced
chemical reaction: 2H2(g) + O2(g) → 2 H2O(g)
Consider the following particulate diagram for the reaction taking place answer questions a & b?
a) Identify the limiting reactant and the excess reactant.
The limiting Reactant is H2 and the excess reactant is O2.
Explanation:
H2
O2
Given
6 moles
5 moles
Divide by reacting ratio
6
=3
2
5
=5
1
Limiting reagent
H2 (The smaller amount)
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Hence, the 6 moles of H2, will completely react with 3 moles of O2 to form 6 moles of H2O
b) How many moles of the excess reactant remains when the reaction stops?
6 moles of H2 will react with 3 moles of O2
Amount of O2 remaining = 5 – 3 = 2 moles
CHM.5.3.01.013.04 Identify limiting reactant and excess reactant in a chemical reaction given the
number of moles of reactants
Example:
Methane,CH4, burns in oxygen, O2, to form carbon dioxide, CO2, and water vapor, H2O, according
to the following equation, 𝐂𝐇𝟒(𝐠) + 𝟐𝐎𝟐(𝐠) → 𝐂𝐎𝟐(𝐠) + 𝟐𝐇𝟐 𝐎(𝐠) .
A sample of 0.250 moles of CH4 was burnt in 1.25 moles of O2. Answer questions a & b.
a) Identify the limiting reactant and the excess reactant.
The limiting reactant is CH4 and the excess reactant is O2.
Number of moles
Divide by reacting ratio
Limiting reagent
CH4
O2
0.250 moles
1.25 moles
0.250
= 0.250
1
1.25
= 0.625
2
CH4 (The smaller amount)
b) Calculate the mass of the excess reactant that remains when the reaction stops.
0.250 moles of CH4 will react with 0.500 moles of O2
Amount of O2 remaining = 1.25 – 0.500 = 0.750 moles
Mass of excess reactant remaining = n × M = 0.750 × 32 = 24.0 g
CHM.5.3.01.013.05 Identify limiting reactant and excess reactant in a chemical reaction given the mass
of reactants
Example:
A 3.4 g sample of ammonia, NH3, reacts with 3.2 g of oxygen, O2, to form nitrogen monoxide,
NO, and water, H2O, according to the following balanced equation:
4 NH3 + 5 O2 → 4 NO + 6 H2O
(Molar mass NH3 = 17 g/mol, O2 = 32 g/mol, NO = 30 g/mol & H2O = 18 g/mol)
a) Identify the limiting reagent.
NH3
Number of moles
m
n=
M
Divide by reacting ratio
Limiting reagent
n=
3.4
= 0.20 moles
17
0.20
= 0.050
4
O2
n=
3.2
= 0.10 moles
32
0.10
= 0.020
5
O2 (The smaller amount)
b) Identify the excess reagent.
NH3
G10 Advanced Chemistry– CHM51 – Course Specification – Term 2 (Detailed KPIs)
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CHM.5.3.01.013.06 Calculate the mass of excess reactant that remains after the reaction is complete
Please refer to KPI 2.5.17
CHM.5.3.01.013.07 Calculate the number of moles of excess reactant that remains after the reaction is
complete
Please refer to KPI 2.5.16
CHM.5.3.01.013.08 Calculate the mass of product(s) produced in a chemical reaction given a particulate
diagram, mass or mole of the reactants
Please refer to the example below
CHM.5.3.01.013.09 Calculate number of moles of product(s) produced in a chemical reaction given a
particulate diagram, mass or mole of the reactants
Please refer to the example below
Example:
The particulate diagram of the reactants is given
Hydrogen gas reacts with oxygen gas to form water vapor according to the following balanced chemical
reaction: 2H2(g) + O2(g) → 2 H2O(g)
Consider the following particulate diagram for the reaction taking place answer questions a – d.
a) Identify the limiting reactant and the excess reactant.
The limiting Reactant is H2 and the excess reactant is O2.
Explanation:
H2
O2
Given
4 moles
3 moles
Divide by reacting ratio
4
=2
2
3
=3
1
Limiting reagent
H2 (The smaller amount)
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b) Calculate the amount of the excess reactant, in moles, that remain when the reaction stops.
2H2
2 moles
4 moles
:
O2
1 mole
?????
2 moles of O2 (out of 3 moles) will react with 4 moles of H2
Amount of O2 remaining = 3 – 2 = 1 mole
c) Calculate the amount of the product, in moles, that forms when the reaction stops.
2H2
2 moles
4 moles
:
2H2O
2 moles
?????
4 moles of H2O will form
d) Draw the amount of the product formed and the excess reactant remaining.
4 moles of H2 will completely react with 2 moles of O2 where 1 mole of O2 remains.
4 moles of H2O are formed
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Example:
The mass of the reactants is given
Consider the following information to answer questions a – g.
A 1.24 g sample of P4, reacts with 9.60 g of O2, to form P4O6, according to the following balanced
equation:
P4 + 3 O2 → P4O6
(Molar mass P4 = 124 g/mol, O2 = 32 g/mol & P4O6 = 220 g/mol)
a) Identify the limiting reagent.
P4
Number of moles
m
n=
M
n=
Divide by reacting ratio
O2
1.24
= 0.0100 moles
124
n=
0.0100
= 0.0100
1
Limiting reagent
9.60
= 0.300 moles
32
0.300
= 0.100
3
P4
b) Identify the excess reagent.
O2
c) Calculate the number of moles P4O6 formed.
P4
1 mole
1×124 g
1.24 g
P4O6
1 mole
1 mole
??
Number of moles of P4O6 = 0.0100 moles
d) Calculate the mass of P4O6 formed.
Mass of P4O6 = n × M = 0.0100 × 220 = 2.20 g
e) After the reaction is completed, calculate the number of moles of excess reactant that reacted.
P4
1 mole
1×124 g
1.24 g
3 O2
3 moles
3 moles
??
Number of moles of O2 reacting = 0.0300 moles
f) After the reaction is completed, calculate the number of moles of excess reactant REAMAINING.
Number of moles of O2 remaining = 0.300 – 0.0300 = 0.270 moles
g) Calculate the mass of the excess reactant remaining.
Mass = n × M = 0.270 × 32 = 8.64 g
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Example:
The number of moles of the reactants is given
Methane, CH4, burns in oxygen to form carbon dioxide, CO2, and water vapor, H2O, according to the
following balanced equation:
𝐂𝐇𝟒(𝐠) + 𝟐𝐎𝟐(𝐠) → 𝐂𝐎𝟐(𝐠) + 𝟐𝐇𝟐 𝐎(𝐠) .
A sample of 0.250 moles of methane was burnt in 1.25 moles of oxygen. Answer questions a – f.
a) Identify the limiting reactant.
Number of moles
m
n=
M
Divide by reacting ratio
Limiting reactant
CH4
O2
0.250 moles
1.25 moles
0.250
= 0.250
1
1.25
= 0.625
2
CH4 (The smaller amount)
b) Calculate the number of moles and mass of CO2 formed.
CH4
1 mole
0.250 mol
CO2
1 mole
?????
Number of moles of CO2 formed = 0.250 moles
Mass of CO2 formed = n × M = 0.250 × 44 = 11.0 g
c) Calculate the number of moles and mass of H2O produced.
CH4
1 mole
0.250 mol
2H2O
2 moles
?????
Number of moles of H2O produced = 0.500 moles
Mass of H2O produced = n × M = 0.500 × 18 = 9.00 g
d) Calculate the total number of moles of products formed.
Total moles of products = number of moles of CO2 + number of moles of H2O
= 0.250 + 0.500 = 0.750 moles
e) Calculate the total mass of products formed.
Total mass of products = mass of CO2 + mass of H2O = 11.0 + 9.00 = 20.0 g
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f) After the reaction is completed, calculate the mass of the excess reactant remaining.
CH4
1 mole
0.250 mol
2 O2
2 moles
???
Number of moles of O2 reacting = 0.500 moles
Number of moles of O2 remaining = 1.25 – 0.500 = 0.750 moles
Mass of O2 remaining = n × M = 0.750 × 32 = 24.0 g
CHM.5.3.01.013.10 Conduct an experiment to determine the limiting reactant, the quantity of the
products and of the reactant in excess in a chemical reaction
Inspire Chemistry – Module 10 – Lesson 4: Percent Yield
CHM.5.3.01.013.11 Distinguish between theoretical yield and actual yield in stoichiometric calculations
Theoretical yield is the maximum amount of product that can be produced from a given amount of
reactant
Actual yield is the measured amount obtained
CHM.5.3.01.013.12 Explain why the actual yield is usually less than the calculated theoretical yield
Due to one of the following:
1) Not all reactions often go to completion
2) Impure reactants produce less product than expected
3) Competing side reactions may consume some of the reactants
4) Some of the product may be lost when it is purified
CHM.5.3.01.013.13 Define the percent yield of a chemical reaction
Percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100
𝐏𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐲𝐢𝐞𝐥𝐝 =
𝐀𝐜𝐭𝐮𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
× 𝟏𝟎𝟎
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
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CHM.5.3.01.013.14 Calculate percent yield and theoretical in a chemical reaction
Example:
A 500. g sample of copper (II) sulfate pentahydrate, CuSO4.5H2O, is heated until complete
decomposition where anhydrous copper(II)sulfate, CuSO4, is formed according to the following
∆
reaction: CuSO4 . 5H2 O → CuSO4 + 5H2 O. Answer questions a and b.
a) Calculate the theoretical yield of CuSO4.
CuSO4.5H2O
1 mole
1 × 250 g
500. g
CuSO4
1 mole
1 × 160 g
???? g
Mass of CuSO4 = 320. g
b) Calculate the percent yield of the anhydrous salt, CuSO4 , if the actual mass of the CuSO4 is
160.g.
Percentage yield =
Actual Yield
Theoretical
160.
× 100 = 320. × 100 = 50.0 %
CHM.5.3.01.033.02 Identify which type of yield (theoretical yield, actual yield, or percent yield) is a
measure of the efficiency of a chemical reaction
The percent yield
This document was created for educational purposes. The document is not meant for publication or mass
printing or production. It is made to support the UAE Ministry of Education and Applied Technology
High School students for AY 2021 – 2022
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