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Curriculum: Irish Leaving Certificate (ILC)
Level: Ordinary Level (OL)
Subject: Chemistry
Chapter 3: Stoichiometry, Formulas and Equations
Version: Complete syllabus-checked chapter
Prepared By: ExamsLogic Academic Team
Chapter 3 Master Overview
This compiled version brings together Parts 1–4 and also adds the Ordinary Level syllabus points that were missing during the quality check.
Core Syllabus Map
3.1 States of Matter
3.2 Gas Laws
3.3 The Mole
3.4 Chemical Formulas
3.5 Chemical Equations
Added in this complete file
States of matter and diffusion
Combined gas law and correction to s.t.p.
Structural formulas with glucose and urea examples
Step-by-step balancing of chemical equations
Mandatory Experiment 3.1: volatile liquid method
Use this as your final Chapter 3 master file. It keeps your original premium notes and adds the remaining OL syllabus content so that the chapter is ready for revision use.
3.1 States of Matter and Diffusion
The Ordinary Level syllabus begins this chapter with the particle model. Before mole work and calculations, students should be clear about how particles behave in solids, liquids and gases.
Motion of Particles in Solids, Liquids and Gases
State
Particle Arrangement
Particle Motion
Main Result
Solid
Particles packed closely in fixed positions
Particles vibrate about fixed points
Definite shape and volume
Liquid
Particles close together but not fixed
Particles move past each other
Definite volume but no fixed shape
Gas
Particles far apart
Particles move rapidly in random directions
No fixed shape or volume
Particle spacing and movement explain the visible properties of matter.
Diffusion
Diffusion is the spreading out of particles from a region of high concentration to a region of low concentration. Diffusion is fastest in gases because gas particles move quickly and are far apart.
Diffusion = spreading of particles due to random motion
Syllabus Example
What is Observed
What it Shows
NH₃ and HCl
White ring forms where the gases meet
Gases diffuse through air
Ink and water
Colour slowly spreads through the water
Particles in liquids are moving
Smoke and air
Smoke spreads through the room
Gas particles move randomly in all directions
Examiner Tip: If you are asked why diffusion is faster in gases than liquids, say that gas particles move faster and are much further apart.
Part 1: The Mole and Core Stoichiometry Foundations
Your original Part 1 premium notes, followed by the added mandatory experiment section.
Chapter 3 Part 1: The Mole
1. Learning Objectives
Explain why chemists use the mole to count particles.
State Avogadro’s constant as approximately 6 × 10²³ particles per mole.
Calculate relative molecular mass from relative atomic masses.
Use molar mass to convert between mass and moles.
Use the number of particles and Avogadro’s constant in simple calculations.
Apply mole calculations to common ILC-style questions.
Examiner Tip: In mole questions, write the formula first, substitute clearly, then include units. Chemistry marks are often lost by messy working, not by difficult science.
2. Why Chemists Use the Mole
Atoms and molecules are extremely small. A tiny sample of a substance contains a huge number of particles, so counting them one by one is impossible. Chemists use a special counting unit called the mole.
A mole is a counting unit, just like a dozen, but much larger.
1 mole = 6 × 10²³ particles
Common Mistake: A mole is not a small furry animal in chemistry exams. Sadly, no marks for wildlife knowledge here.
3. Avogadro’s Constant
Avogadro’s constant is the number of particles in one mole of any substance.
Avogadro’s constant = 6 × 10²³ mol⁻¹
The word particles can mean atoms, molecules, ions, or formula units depending on the substance.
Substance
Particles Counted
Example
Element made of atoms
Atoms
1 mole of helium contains 6 × 10²³ helium atoms.
Molecular substance
Molecules
1 mole of water contains 6 × 10²³ water molecules.
Ionic compound
Formula units
1 mole of sodium chloride contains 6 × 10²³ NaCl formula units.
Particles and Moles Formula
Number of particles = number of moles × 6 × 10²³
Number of moles = number of particles ÷ 6 × 10²³
Worked Example 1: How many molecules are in 2 moles of water? Particles = moles × Avogadro’s constant = 2 × 6 × 10²³ = 1.2 × 10²⁴ molecules
Worked Example 2: How many moles are in 3 × 10²³ atoms of carbon? Moles = particles ÷ 6 × 10²³ = 3 × 10²³ ÷ 6 × 10²³ = 0.5 mol
Examiner Trap: Check whether the question asks for atoms, molecules, ions, or moles. The calculation may be simple, but the wording catches students.
4. Relative Atomic Mass and Relative Molecular Mass
The relative atomic mass of an element is found from the periodic table. The relative molecular mass is found by adding the relative atomic masses of all atoms in the formula.
Formula
Calculation
Relative Molecular Mass
H₂O
(2 × 1) + 16
18
CO₂
12 + (2 × 16)
44
NH₃
14 + (3 × 1)
17
CaCO₃
40 + 12 + (3 × 16)
100
Read the small numbers in the formula carefully. CO₃ means three oxygen atoms.
Examiner Tip: Brackets multiply everything inside them. For Ca(OH)₂, you have 1 calcium, 2 oxygen atoms and 2 hydrogen atoms.
5. Molar Mass
The molar mass of a substance is the mass of one mole of that substance. It has units of g mol⁻¹.
Molar mass of H₂O = 18 g mol⁻¹
Molar mass of CO₂ = 44 g mol⁻¹
Mass, Moles and Molar Mass Formula
moles = mass ÷ molar mass
mass = moles × molar mass
The formula triangle helps, but understanding the units is even better.
Worked Example 3: Calculate the number of moles in 36 g of water. Mr of H₂O = 18 Moles = mass ÷ molar mass = 36 ÷ 18 = 2 mol
Worked Example 4: Calculate the mass of 0.25 mol of carbon dioxide. Mr of CO₂ = 44 Mass = moles × molar mass = 0.25 × 44 = 11 g
Common Mistake: Do not write “moles = molar mass ÷ mass”. That flips the calculation and gives a nonsense answer.
6. Ordinary Level Calculation Method
Use this clean four-step method for most mole questions.
Step
What to Do
Example
1
Write the formula needed.
moles = mass ÷ molar mass
2
Find Mr from the formula.
CO₂ = 12 + 32 = 44
3
Substitute values.
moles = 22 ÷ 44
4
Calculate and add units.
0.5 mol
Examiner Trap: If the answer is larger or smaller than expected, pause. For example, 2 g of a substance cannot usually be thousands of moles in OL questions.
Mandatory Experiment 3.1: Relative Molecular Mass of a Volatile Liquid
The syllabus requires the determination of the relative molecular mass of a volatile liquid. This may be carried out using a conical flask or a gas syringe. The key idea is to vaporise a known mass of liquid and use the volume of the vapour to work out its relative molecular mass.
Aim
To determine the relative molecular mass of a volatile liquid by measuring the mass and volume of its vapour.
Apparatus
Dry conical flask with foil cap and pin hole
Water bath or hot plate
Volatile liquid sample
Balance
Thermometer
Measuring cylinder or gas-syringe setup
Method Summary
Measure the mass of the empty dry flask with foil cap.
Add a few drops of the volatile liquid and heat the flask in a water bath.
The liquid evaporates and displaces the air from the flask.
When evaporation is complete, cool and reweigh the flask.
The increase in mass gives the mass of the vapour.
Use the flask volume, temperature and pressure conditions to calculate the relative molecular mass.
Mᵣ = mass of 22.4 L of vapour at s.t.p. ÷ 1 mole
At s.t.p., 1 mole of gas occupies 22.4 L
Examiner Trap: The volatile liquid must completely vaporise before the measurement is valid. Any liquid left inside gives the wrong result.
Common sources of error: wet flask, loss of vapour, inaccurate temperature reading, and incomplete displacement of air.
7. Interactive Simulators and Digital Practice
Simulator 1: PhET – Molecules and Light / Build a Molecule Task: Build simple molecules such as H₂O, CO₂ and NH₃. Count the number of each atom in the formula. Exam Link: Helps students understand why Mr calculations use subscripts such as H₂O and CO₂.
Simulator 2: ChemCollective – Stoichiometry Virtual Problems Task: Practise converting mass to moles and moles to mass using guided problems. Exam Link: Builds confidence for Section B calculation questions.
Simulator 3: Interactive Periodic Table Task: Look up relative atomic masses for common elements: H, C, O, N, Na, Cl, Ca, Mg. Exam Link: Students must use periodic table data quickly and accurately.
Student Activity: Choose five substances from your textbook and calculate their Mr. Then check your answer using an online molar mass calculator.
8. Examiner Secrets, Mistakes and Traps
Examiner Secret: The first mark in calculation questions is often for the correct formula or correct Mr. Even if the final answer is wrong, clear working can still earn marks.
Common Mistake: Forgetting to multiply atoms outside brackets. Mg(OH)₂ contains two oxygen atoms and two hydrogen atoms.
Common Mistake: Writing Mr with grams. Relative molecular mass has no unit, but molar mass has units g mol⁻¹.
Examiner Trap: Avogadro’s constant is particles per mole, not grams per mole.
9. Exam Practice Questions
Q1. Define the mole. [2 marks]
Q2. State Avogadro’s constant. [1 mark]
Q3. Calculate the relative molecular mass of H₂SO₄. [3 marks]
Q4. Calculate the number of moles in 24 g of magnesium. [2 marks]
Q5. Calculate the mass of 0.2 mol of sodium chloride, NaCl. [3 marks]
Q6. How many molecules are present in 0.5 mol of carbon dioxide? [2 marks]
MCQs with Explanations
1. One mole contains: A. 6 particles B. 6 × 10²³ particles C. 12 particles D. 1 gram
Answer: B. One mole contains Avogadro’s constant number of particles.
2. The Mr of CO₂ is: A. 28 B. 32 C. 44 D. 48
Answer: C. Carbon = 12 and oxygen = 16, so 12 + (2 × 16) = 44.
3. The number of moles in 10 g of calcium carbonate, CaCO₃, is: A. 0.1 mol B. 1 mol C. 10 mol D. 100 mol
Answer: A. Mr of CaCO₃ = 100. Moles = 10 ÷ 100 = 0.1 mol.
10. Mark Scheme
Q1. A mole is the amount of substance [1] that contains 6 × 10²³ particles / Avogadro’s constant number of particles [1].
Molar mass has the same numerical value as Mr but uses units g mol⁻¹.
moles = mass ÷ molar mass.
mass = moles × molar mass.
particles = moles × Avogadro’s constant.
Read brackets and small numbers in formulas carefully.
12. Self-Assessment Checklist
I can define the mole.
I can state Avogadro’s constant.
I can calculate relative molecular mass from a formula.
I can convert mass into moles.
I can convert moles into mass.
I can calculate the number of particles in a sample.
I can show clear working for OL calculation questions.
Part 2: Gas Laws and Gas Volumes
Your original gas law notes, plus the missing combined gas law and s.t.p. correction section.
Chapter 3 Part 2: Gas Laws and Gas Volumes
1. Learning Objectives
State Boyle's Law, Charles's Law, Gay-Lussac's Law and Avogadro's Law in OL language.
Use simple gas law relationships to predict changes in pressure, volume and temperature.
Convert temperature from degrees Celsius to kelvin.
Understand molar volume of gases at STP.
Use gas volume ratios from balanced chemical equations.
Solve ordinary level exam-style gas calculation questions step by step.
Examiner Tip: Gas law questions are usually not difficult, but students lose marks because they forget to convert Celsius to kelvin.
2. Key Gas Terms
Term
Meaning
Unit Usually Used
Pressure
Force of gas particles hitting the walls of a container.
kPa, Pa, atm
Volume
Space occupied by a gas.
cm³, dm³, L
Temperature
Measure of average kinetic energy of particles.
K for calculations
Amount of gas
Number of moles of gas particles present.
mol
Temperature in kelvin = temperature in °C + 273
Worked Example: Convert 25°C to kelvin. 25 + 273 = 298 K.
Examiner Trap: Never use 25 directly in Charles's Law calculations. Use 298 K.
3. Boyle's Law
Boyle's Law states that, for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume.
P₁V₁ = P₂V₂
Compressing a gas into a smaller volume makes particles hit the container walls more often.
Worked Example: A gas has volume 600 cm³ at 100 kPa. Its volume is reduced to 300 cm³ at constant temperature. Find the new pressure. P₁V₁ = P₂V₂ 100 × 600 = P₂ × 300 P₂ = 60000 ÷ 300 = 200 kPa.
Common Mistake: Thinking pressure halves when volume halves. Actually, pressure doubles if temperature stays constant.
4. Charles's Law
Charles's Law states that, for a fixed mass of gas at constant pressure, volume is directly proportional to temperature in kelvin.
V₁ / T₁ = V₂ / T₂
When gas particles gain energy, they move faster and occupy a larger volume if pressure is constant.
Worked Example: A gas has volume 200 cm³ at 27°C. What is its volume at 87°C at constant pressure? T₁ = 27 + 273 = 300 K T₂ = 87 + 273 = 360 K V₁/T₁ = V₂/T₂ 200/300 = V₂/360 V₂ = 200 × 360 ÷ 300 = 240 cm³.
Examiner Trap: The formula uses kelvin, not Celsius. This is the most common lost mark in Charles's Law questions.
5. Gay-Lussac's Law
Gay-Lussac's Law states that, for a fixed mass of gas at constant volume, pressure is directly proportional to temperature in kelvin.
P₁ / T₁ = P₂ / T₂
Worked Example: A sealed gas cylinder has pressure 120 kPa at 20°C. Find its pressure at 70°C if volume remains constant. T₁ = 20 + 273 = 293 K T₂ = 70 + 273 = 343 K 120/293 = P₂/343 P₂ = 120 × 343 ÷ 293 = 140.5 kPa approximately.
Examiner Tip: In a sealed rigid container, volume is constant. So pressure changes with temperature.
6. Avogadro's Law and Molar Volume
Avogadro's Law states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules.
At STP: 1 mole of any gas occupies 22.4 litres
STP means standard temperature and pressure. At OL, you normally use the molar volume value given in the question or learned from class.
Amount of Gas
Volume at STP
1 mole
22.4 L
2 moles
44.8 L
0.5 mole
11.2 L
Worked Example: Find the volume occupied by 0.25 mol of oxygen gas at STP. Volume = moles × 22.4 L Volume = 0.25 × 22.4 = 5.6 L.
Common Mistake: Using 22.4 L for liquids or solids. Molar gas volume applies to gases only.
7. Gas Volume Ratios from Balanced Equations
When gases react at the same temperature and pressure, their volumes react in the same ratio as their mole ratio in the balanced equation.
Gas volume ratio = coefficient ratio in the balanced equation
The coefficients 2 : 1 : 2 become the gas volume ratio 2 : 1 : 2.
Worked Example: Nitrogen and hydrogen react: N₂ + 3H₂ → 2NH₃. What volume of hydrogen is needed to react with 10 L of nitrogen? Ratio N₂ : H₂ = 1 : 3 10 L nitrogen needs 3 × 10 = 30 L hydrogen.
Combined Gas Law and Correction of Gas Volumes to s.t.p.
The OL syllabus includes the combined gas law and simple correction of gas volumes to standard temperature and pressure.
Combined Gas Law
P₁V₁ / T₁ = P₂V₂ / T₂
This law is used when pressure, volume and temperature may all change together for a fixed mass of gas.
Worked Example:
A gas occupies 240 cm³ at 100 kPa and 300 K. What volume will it occupy at 120 kPa and 273 K?
To compare gas volumes fairly, chemists often correct them to s.t.p. where the temperature and pressure are standard. In Ordinary Level questions, keep units consistent and use Pa, cm³ and K where required.
Use the combined gas law to convert an observed gas volume to s.t.p.
Quantity
Common OL Unit
Pressure
Pa or kPa
Volume
cm³
Temperature
K
Examiner Tip: Do not mix Celsius with Kelvin in the same formula. Convert temperature first, then substitute.
8. Interactive Simulators / Virtual Labs
Simulator 1: Gas Properties Simulation Use: Change volume, temperature and number of particles. Watch how pressure changes. Student Task: Reduce the volume while keeping temperature constant. Record what happens to pressure. Exam Link: Boyle's Law.
Simulator 2: Balloons and Buoyancy / Gas Expansion Use: Heat a gas and observe expansion. Student Task: Explain why a balloon expands when warmed. Exam Link: Charles's Law and particle motion.
Simulator 3: Mole and Gas Volume Practice Use: Convert between moles and gas volume at STP. Student Task: Calculate the volume for 0.1 mol, 0.5 mol and 2 mol of gas at STP.
Teacher Note: Recommended simulator sources: PhET gas simulations, ChemCollective calculation practice, and classroom molar volume animations.
9. Examiner Secrets, Mistakes and Traps
Examiner Secret: Before using any gas law formula, write down what is constant: temperature, pressure or volume.
Common Mistake: Mixing cm³ and litres in the same calculation. Keep units consistent.
Common Mistake: Forgetting that 1 dm³ = 1 L = 1000 cm³.
Examiner Trap: In gas volume ratio questions, only gases follow the volume ratio directly. Do not use this shortcut for solids or liquids.
10. Exam Practice Questions
Q1. State Boyle's Law. [3 marks]
Q2. A gas has volume 400 cm³ at 150 kPa. Its pressure changes to 300 kPa at constant temperature. Calculate the new volume. [4 marks]
Q3. Convert 37°C to kelvin. [1 mark]
Q4. A gas has volume 250 cm³ at 300 K. Find its volume at 360 K at constant pressure. [4 marks]
Q5. Calculate the volume occupied by 1.5 mol of carbon dioxide at STP. [3 marks]
Q6. In the reaction 2CO + O₂ → 2CO₂, what volume of oxygen is required to react with 60 cm³ of carbon monoxide? [3 marks]
MCQs with Explanations
1. Which unit must be used for temperature in gas law calculations? A. °C B. K C. cm³ D. kPa
Answer: B. Gas law calculations use kelvin.
2. If volume is halved at constant temperature, pressure will: A. halve B. double C. stay the same D. become zero
Answer: B. Boyle's Law shows pressure and volume are inversely proportional.
3. At STP, 0.5 mol of gas occupies: A. 22.4 L B. 44.8 L C. 11.2 L D. 5.6 L
Answer: C. 0.5 × 22.4 = 11.2 L.
11. Mark Scheme
Q1. For a fixed mass of gas [1], at constant temperature [1], pressure is inversely proportional to volume / P₁V₁ = P₂V₂ [1].
Q6. Ratio CO : O₂ = 2 : 1 [1]; oxygen volume is half carbon monoxide volume [1]; 30 cm³ [1].
12. Last-Minute Revision Sheet
Boyle's Law: pressure and volume are inversely proportional at constant temperature.
Charles's Law: volume and temperature are directly proportional at constant pressure.
Gay-Lussac's Law: pressure and temperature are directly proportional at constant volume.
Always convert °C to K using K = °C + 273.
At STP, 1 mole of gas occupies 22.4 L.
Gas volume ratios follow balanced equation coefficients for gases at the same temperature and pressure.
Keep units consistent: do not mix cm³ and L without converting.
13. Self-Assessment Checklist
I can state Boyle's Law.
I can use P₁V₁ = P₂V₂.
I can convert Celsius to kelvin.
I can use Charles's Law for volume and temperature questions.
I can explain Gay-Lussac's Law in a sealed container.
I can calculate gas volume at STP from moles.
I can use balanced equations to solve gas volume ratio questions.
I can avoid common unit mistakes in gas calculations.
Part 3: Formulas, Equations and Chemical Calculations
Your original calculation notes, with added structural formulas, glucose and urea examples, and explicit equation-balancing coverage.
Structural Formulas, Glucose and Urea
The OL syllabus asks for more than empirical and molecular formulas. Students should also recognise structural formulas, which show how atoms are linked together in a molecule.
Structural Formulas
Type
What it Shows
Example
Molecular formula
Total number of each atom
C₂H₆O
Structural formula
How atoms are connected
CH₃CH₂OH
Empirical formula
Simplest whole-number ratio
C₂H₆O → C₂H₆O (already simplest)
Glucose
Molecular formula: C₆H₁₂O₆
Empirical formula: CH₂O
This is a simple biological example named directly in the syllabus.
Urea
Molecular formula: CH₄N₂O
Empirical formula: CH₄N₂O
Another simple biological example named in the syllabus.
Examiner Tip: If two substances have the same empirical formula, that does not mean they have the same molecular formula.
Chemical Equations and Balancing Equations
The stoichiometry sections use balanced equations all the time, so it helps to include one direct section on what balancing actually means.
What a Chemical Equation Shows
A chemical equation shows the reactants, the products, and the ratio in which particles react.
Reactants → Products
How to Balance an Equation
Write the correct formulas first.
Count the atoms of each element on both sides.
Change only the large coefficients in front of formulas.
Do not change small formula subscripts.
Check the final atom count again.
Unbalanced
Balanced
Why
H₂ + O₂ → H₂O
2H₂ + O₂ → 2H₂O
Now H atoms = 4 on both sides and O atoms = 2 on both sides
Mg + O₂ → MgO
2Mg + O₂ → 2MgO
Now Mg atoms = 2 and O atoms = 2 on both sides
Al + O₂ → Al₂O₃
4Al + 3O₂ → 2Al₂O₃
Same number of Al and O atoms on both sides
Examiner Trap: Never change a subscript to balance an equation. That changes the substance itself.
Chapter 3 Part 3: Chemical Calculations
1. Learning Objectives
Calculate percentage composition by mass.
Use percentage composition to find empirical formulas.
Use empirical formula and molecular mass to find molecular formulas.
Use balanced equations to calculate reacting masses.
Identify the limiting reactant in simple reactions.
Calculate theoretical yield, actual yield and percentage yield.
Show clear working with units, mole ratios and balanced equations.
Big Idea: Stoichiometry is chemistry counting. The balanced equation gives the ratio; the mole converts mass into number of particles.
2. Percentage Composition
Percentage composition tells us what percentage of a compound's mass comes from each element.
% by mass = (mass of element in compound ÷ molar mass of compound) × 100
In water, most of the mass comes from oxygen, even though there are two hydrogen atoms.
Worked Example 1: Percentage composition of CO₂
C = 12, O = 16, so Mᵣ of CO₂ = 12 + 32 = 44.
% carbon = 12 ÷ 44 × 100 = 27.3%
% oxygen = 32 ÷ 44 × 100 = 72.7%
Common Mistake: Students often forget to multiply oxygen by 2 in CO₂. Always count the atoms from the formula first.
3. Empirical Formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
Mass or % → moles → divide by smallest → whole-number ratio
The trick is to change masses into moles before comparing atoms.
Worked Example 2: Find the empirical formula
A compound contains 24 g carbon and 8 g hydrogen.
C: 24 ÷ 12 = 2 mol
H: 8 ÷ 1 = 8 mol
Divide by smallest: C = 2 ÷ 2 = 1, H = 8 ÷ 2 = 4
Empirical formula = CH₄
Worked Example 3: Using percentages
A compound is 40% carbon, 6.7% hydrogen and 53.3% oxygen.
Assume 100 g: C = 40 g, H = 6.7 g, O = 53.3 g.
C: 40 ÷ 12 = 3.33
H: 6.7 ÷ 1 = 6.7
O: 53.3 ÷ 16 = 3.33
Divide by 3.33: C = 1, H ≈ 2, O = 1
Empirical formula = CH₂O
Examiner Trap: A percentage formula question does not mean the formula must contain percentages. Treat percentages as masses out of 100 g.
4. Molecular Formula
The molecular formula shows the actual number of atoms in one molecule. It may be the same as the empirical formula, or it may be a multiple of it.
Multiplier = molecular mass ÷ empirical formula mass
Empirical Formula
Empirical Formula Mass
Molecular Mass
Molecular Formula
CH₂O
30
180
C₆H₁₂O₆
CH
13
78
C₆H₆
HO
17
34
H₂O₂
Worked Example 4: Molecular formula
The empirical formula of a compound is CH₂. Its molecular mass is 42.
Empirical formula mass = 12 + 2 = 14.
Multiplier = 42 ÷ 14 = 3.
Molecular formula = C₃H₆.
Examiner Tip: If your multiplier is not a whole number, check your empirical formula working. Molecular formulas require whole-number atom ratios.
5. Stoichiometric Calculations from Equations
A balanced equation gives the mole ratio of reactants and products.
balanced equation → mole ratio → convert moles to mass
Never jump straight from mass to mass without using the mole ratio.
Worked Example 5: Mass from balanced equation
What mass of magnesium oxide forms when 12 g of magnesium reacts completely with oxygen?
Balanced equation: 2Mg + O₂ → 2MgO
Moles of Mg = 12 ÷ 24 = 0.5 mol
Ratio Mg : MgO = 2 : 2, so 0.5 mol Mg gives 0.5 mol MgO
Mᵣ of MgO = 24 + 16 = 40
Mass MgO = 0.5 × 40 = 20 g
Worked Example 6: Reacting mass
What mass of oxygen is needed to react with 48 g of magnesium?
2Mg + O₂ → 2MgO
Moles Mg = 48 ÷ 24 = 2 mol
Ratio Mg : O₂ = 2 : 1, so moles O₂ = 1 mol
Mᵣ of O₂ = 32
Mass O₂ = 1 × 32 = 32 g
6. Limiting Reactant
The limiting reactant is the reactant that is used up first. It limits how much product can be made.
The limiting reactant is like the ingredient that runs out first.
Worked Example 7: Identify the limiting reactant
2H₂ + O₂ → 2H₂O
Suppose 4 mol H₂ reacts with 1 mol O₂.
The equation says 2 mol H₂ need 1 mol O₂.
So 4 mol H₂ would need 2 mol O₂, but only 1 mol O₂ is available.
Therefore, oxygen is the limiting reactant.
Examiner Trap: The reactant with the smaller mass is not always the limiting reactant. You must compare moles using the balanced equation.
7. Theoretical, Actual and Percentage Yield
The theoretical yield is the maximum amount of product expected from calculations. The actual yield is the amount collected in the experiment.
Worked Example 8: Percentage yield
A reaction should produce 10.0 g of product, but only 8.0 g is collected.
% yield = 8.0 ÷ 10.0 × 100 = 80%
Why is actual yield often lower?
Some product may be lost during filtration or transfer.
The reaction may not go to completion.
Side reactions may form unwanted products.
The product may be impure or difficult to dry fully.
Examiner Tip: When explaining low percentage yield, use practical reasons such as product loss, incomplete reaction or side reactions.
8. Interactive Simulators and Virtual Practice
Simulator 1: Reactants, Products and Leftovers
Use for: Limiting reactants and leftover reactants.
Student task: Change the number of reactant particles and observe which reactant runs out first.
Exam link: Explains why a balanced equation controls the amount of product formed.
Simulator 2: Balancing Chemical Equations
Use for: Practising balanced equations before mole ratio calculations.
Student task: Balance simple equations, then write the mole ratio between reactants and products.
Exam link: Stoichiometry questions always begin with a correct balanced equation.
Simulator 3: Mole and Formula Calculation Practice
Use for: Mass, moles, molar mass, empirical formula and molecular formula practice.
Student task: Complete 5 calculations and check whether the final answer has correct units.
Exam link: Helps students avoid the classic error of mixing grams and moles in the same line.
9. Examiner Secrets, Mistakes and Traps
Examiner Secret: In calculation questions, marks are often given for method. Even if your final answer is wrong, clear mole steps can still earn marks.
Examiner Tip: Write the balanced equation first, then write the mole ratio directly underneath it.
Common Mistake: Using the mass ratio from the equation instead of the mole ratio. Coefficients in equations are mole ratios, not gram ratios.
Common Mistake: Forgetting to multiply atoms inside brackets, for example Ca(OH)₂ contains two oxygen atoms and two hydrogen atoms.
Examiner Trap: Empirical formula is the simplest ratio. Molecular formula is the actual formula. They are not always the same.
Examiner Trap: Percentage yield should normally be less than or equal to 100%. If your answer is above 100%, check which value is actual and which is theoretical.
10. Exam Practice Questions
Q1. Calculate the percentage by mass of oxygen in magnesium oxide, MgO. [3 marks]
Q2. A compound contains 36 g carbon and 6 g hydrogen. Find its empirical formula. [4 marks]
Q3. A compound has empirical formula CH₂O and molecular mass 60. Find its molecular formula. [3 marks]
Q4. Calculate the mass of carbon dioxide formed when 12 g of carbon burns completely in oxygen. Equation: C + O₂ → CO₂. [4 marks]
Q5. In the reaction 2H₂ + O₂ → 2H₂O, 2 mol H₂ reacts with 0.5 mol O₂. Identify the limiting reactant and explain your answer. [4 marks]
Q6. A reaction has a theoretical yield of 25 g. The actual yield is 20 g. Calculate the percentage yield. [3 marks]
MCQs with Explanations
1. What is the empirical formula of C₆H₁₂O₆? A. C₆H₁₂O₆ B. CH₂O C. C₂H₄O₂ D. CHO
Answer: B. Divide all subscripts by 6 to get the simplest ratio, CH₂O.
2. Which step is needed first in most reacting mass calculations? A. Find percentage yield B. Write a balanced equation C. Draw a molecule D. Measure pH
Answer: B. The balanced equation gives the mole ratio needed for stoichiometry.
3. If actual yield is 6 g and theoretical yield is 8 g, the percentage yield is: A. 25% B. 50% C. 75% D. 133%
Answer: C. % yield = 6 ÷ 8 × 100 = 75%.
11. Last-Minute Revision Sheet
Percentage composition shows the percentage mass of each element in a compound.
Empirical formula is the simplest whole-number ratio of atoms.
To find empirical formula: divide by relative atomic mass, then divide by the smallest mole value.
Molecular formula is the actual number of atoms in a molecule.
Multiplier = molecular mass ÷ empirical formula mass.
Balanced equation coefficients give mole ratios.
Mass ÷ molar mass = moles.
Moles × molar mass = mass.
Limiting reactant is used up first.
% yield = actual yield ÷ theoretical yield × 100.
12. Self-Assessment Checklist
I can calculate percentage composition by mass.
I can find empirical formulas from masses.
I can find empirical formulas from percentages.
I can calculate molecular formulas from empirical formulas.
I can use balanced equations to find mole ratios.
I can calculate reacting masses.
I can identify a limiting reactant.
I can calculate percentage yield.
I can explain why actual yield may be lower than theoretical yield.
Example 2: Limiting Reactant 2H₂ + O₂ → 2H₂O If 4 mol H₂ reacts with 1 mol O₂, which reactant is limiting?
1 mol O₂ needs 2 mol H₂. There are 4 mol H₂ available, so oxygen runs out first. Limiting reactant = O₂
Example 3: Percentage Yield The theoretical yield is 10 g. The actual yield is 8 g. % yield = 8 ÷ 10 × 100 = 80%
5. Practical Skills Link
Stoichiometry is not only for written calculations. It is used when chemists plan reactions and analyse practical results.
Titration
Balanced equations help calculate unknown acid or base concentration.
Gas Collection
Gas volume can be converted into moles at STP.
Preparation of Salts
Yield calculations compare the expected product with the real product.
Common Mistake: Students often forget that practical yield is usually lower than theoretical yield because product may be lost during filtration, evaporation, transfer or drying.
6. Interactive Simulators / Virtual Labs
Simulator 1: PhET Reactants, Products and Leftovers Open simulator Task: Change the amount of each reactant and identify the limiting reactant. Record what is left over after the reaction.
Simulator 2: ChemCollective Stoichiometry Virtual Lab Open virtual lab collection Task: Choose a stoichiometry or solution preparation activity and practise converting between mass, moles and concentration.
Simulator 3: PhET Balancing Chemical Equations Open simulator Task: Balance each equation before attempting mole-ratio calculations.
Simulator Exam Link: After using a simulator, write one sentence explaining which reactant limits the reaction and one sentence explaining why balancing equations matters.
7. Examiner Secrets, Mistakes and Traps
Examiner Secret: Examiners reward clear working. Even if your final answer is wrong, correct mole steps can earn marks.
Common Mistake: Using formula masses before balancing the equation. Balance first, then use mole ratios.
Common Mistake: Writing only a number without units. Ordinary Level mark schemes often expect g, mol, L or cm³.
Examiner Trap: In percentage yield, the actual yield goes on top. Do not reverse the fraction.
8. Exam Practice Questions
Q1. Calculate the number of moles in 5.6 g of iron. Relative atomic mass of Fe = 56. [2 marks]
Q2. Calculate the mass of 0.25 mol of carbon dioxide, CO₂. [3 marks]
Q3. In the reaction 2Mg + O₂ → 2MgO, calculate the mass of MgO formed from 4.8 g Mg. [5 marks]
Q4. A reaction has a theoretical yield of 12 g but gives 9 g in the laboratory. Calculate the percentage yield. [3 marks]
Q5. Explain two reasons why the actual yield may be lower than the theoretical yield. [4 marks]
MCQs with Explanations
1. Which equation is used to calculate moles from mass? A. n = Mᵣ ÷ mass B. n = mass ÷ Mᵣ C. mass = n ÷ Mᵣ D. n = volume × 22.4
Answer: B. Moles = mass divided by molar mass.
2. What is the percentage yield if actual yield is 6 g and theoretical yield is 8 g? A. 25% B. 50% C. 75% D. 125%
Answer: C. 6 ÷ 8 × 100 = 75%.
3. Why must equations be balanced before stoichiometry calculations? A. To make products heavier B. To show correct mole ratios C. To change the reactants D. To remove gases
Answer: B. Balanced equations show the reacting mole ratio.
9. Mark Scheme
Q1. n = mass ÷ Ar [1]; n = 5.6 ÷ 56 = 0.10 mol [1].
Q2. Mr CO₂ = 12 + 16 + 16 = 44 [1]; mass = n × Mr [1]; mass = 0.25 × 44 = 11 g [1].
Q3. moles Mg = 4.8 ÷ 24 = 0.20 mol [1]; ratio Mg:MgO = 2:2 or 1:1 [1]; moles MgO = 0.20 mol [1]; Mr MgO = 40 [1]; mass = 0.20 × 40 = 8.0 g [1].
Q5. Product lost during transfer/filtration/evaporation [2]; reaction may not go to completion or side reactions may occur [2].
10. Last-Minute Revision Sheet
Always balance the equation before using mole ratios.
Moles = mass ÷ molar mass.
Mass = moles × molar mass.
At STP, 1 mole of gas occupies 22.4 L.
Avogadro's constant = 6 × 10²³ particles per mole.
Percentage yield = actual yield ÷ theoretical yield × 100.
The limiting reactant is used up first.
Actual yield is usually less than theoretical yield.
Show working clearly and include units.
11. Self-Assessment Checklist
I can calculate moles from mass.
I can calculate mass from moles.
I can use balanced equations to find mole ratios.
I can solve reacting mass calculations.
I can identify a limiting reactant in simple examples.
I can calculate percentage yield.
I can explain why actual yield may be lower than theoretical yield.
I can connect stoichiometry to titration, gas collection and product preparation.
12. Chapter 3 Complete
ExamsLogic Final Note: Chapter 3 is calculation-heavy. The secret is not memorising every question type. The secret is using the same pathway every time: balance equation → convert to moles → use ratio → convert to the required answer.
Chapter 3 Syllabus Completion Check
This compiled version now includes the core Ordinary Level Chapter 3 content:
3.1 States of Matter, including diffusion examples
3.2 Gas Laws, including the combined gas law and correction to s.t.p.
3.3 The Mole and Mandatory Experiment 3.1
3.4 Chemical Formulas, including structural formulas, glucose and urea
3.5 Chemical Equations, balancing and stoichiometric calculations
Use this file as your complete Chapter 3 master version. It compiles Parts 1–4 and includes the missing Ordinary Level syllabus points that were added during quality checking.