IGCSE Chemistry Formula Practice
Worked Examples for Key Calculations | Cambridge IGCSE Chemistry (0620)
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Calculating Moles from Mass
Moles = Mass ÷ Molar Mass
How to Use This Guide
- Read the formula and understand what each variable means
- Study the example - note how values are substituted
- Try the practice problem before looking at the answer
- Check your working - write the equation, the substitution, then the answer with its unit
- Items marked (Extended) are Supplement content. Core candidates (Papers 1 and 3) are not tested on them
1. Mole Calculations (Extended)
The mole is chemistry's counting unit - it converts between mass, particles, and gas volumes. Master these before attempting any other calculation.
Calculating Moles from Mass
A step in most mole calculations. Converts grams of a substance to moles using its molar mass M in g/mol, which has the same number as its relative formula mass (Mr).
Moles = Mass ÷ Molar Mass
n = m / M
Worked Example:
Calculate the number of moles in 36 g of water (H₂O). (Ar: H = 1, O = 16)
Step 1: Calculate Mr of H₂O
Mr = (2 × 1) + 16 = 18
Step 2: Apply the formula
n = m / M = 36 / 18
n = 2 mol
Practice Problem:
Calculate the number of moles in 11 g of carbon dioxide (CO₂). (Ar: C = 12, O = 16)
Show Answer
Mr of CO₂ = 12 + (2 × 16) = 44
n = 11 / 44 = 0.25 mol
Calculating Mass from Moles
The reverse calculation - find the mass when you know how many moles you have. Useful for predicting product masses.
Mass = Moles × Molar Mass
m = n × M
Worked Example:
Calculate the mass of 0.5 mol of sodium chloride (NaCl). (Ar: Na = 23, Cl = 35.5)
Step 1: Calculate Mr of NaCl
Mr = 23 + 35.5 = 58.5
Step 2: Apply the formula
m = n × M = 0.5 × 58.5
m = 29.25 g
Practice Problem:
Calculate the mass of 2.5 mol of calcium carbonate (CaCO₃). (Ar: Ca = 40, C = 12, O = 16)
Show Answer
Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100
m = 2.5 × 100 = 250 g
Calculating Moles of Gas at RTP
At room temperature and pressure, 1 mole of any gas occupies 24 dm³. This lets you convert between gas volume and moles.
Moles of gas = Volume ÷ Molar Gas Volume
n = V / 24
At RTP (room temperature and pressure): 1 mol of any gas = 24 dm³
Worked Example:
Calculate the number of moles in 60 dm³ of oxygen gas at RTP.
n = V / 24
n = 60 / 24
n = 2.5 mol
Practice Problem:
Calculate the volume of 0.2 mol of carbon dioxide gas at RTP.
Show Answer
V = n × 24 = 0.2 × 24 = 4.8 dm³
Calculating Relative Atomic Mass from Isotopes
Most elements have multiple isotopes. The Ar is a weighted average based on each isotope's abundance in nature.
Ar = Σ(isotope mass × abundance) ÷ 100
Ar = (m₁×%₁ + m₂×%₂) / 100
Worked Example:
Chlorine has two isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Calculate the relative atomic mass.
Ar = (35 × 75 + 37 × 25) / 100
Ar = (2625 + 925) / 100
Ar = 3550 / 100
Ar = 35.5
Practice Problem:
Copper has two isotopes: ⁶³Cu (69%) and ⁶⁵Cu (31%). Calculate the relative atomic mass.
Show Answer
Ar = (63 × 69 + 65 × 31) / 100
Ar = (4347 + 2015) / 100 = 6362 / 100
Ar = 63.62
2. Concentration (Extended)
Concentration tells you how much solute is dissolved in a given volume of solution. Essential for any solution-based calculation including titrations.
Concentration in mol/dm³
Measures how many moles of solute are in each dm³ of solution.
Concentration = Moles ÷ Volume
c = n / V
Volume must be in dm³ (1 dm³ = 1000 cm³)
Worked Example:
0.2 mol of sodium hydroxide is dissolved in water to make 500 cm³ of solution. Calculate the concentration.
Step 1: Convert cm³ to dm³
V = 500 / 1000 = 0.5 dm³
Step 2: Apply the formula
c = n / V = 0.2 / 0.5
c = 0.4 mol/dm³
Practice Problem:
A solution contains 0.05 mol of HCl in 250 cm³. Calculate the concentration in mol/dm³.
Show Answer
V = 250 / 1000 = 0.25 dm³
c = 0.05 / 0.25 = 0.2 mol/dm³
Concentration in g/dm³
Measures how many grams of solute are in each dm³ of solution. Useful when mass is easier to measure than moles.
Concentration = Mass ÷ Volume
c = m / V
Worked Example:
8 g of sodium hydroxide is dissolved in water to make 2 dm³ of solution. Calculate the concentration in g/dm³.
c = m / V = 8 / 2
c = 4 g/dm³
Practice Problem:
A solution with concentration 20 g/dm³ has a volume of 500 cm³. Calculate the mass of solute.
Show Answer
V = 500 / 1000 = 0.5 dm³
m = c × V = 20 × 0.5 = 10 g
Converting Between mol/dm³ and g/dm³
Convert between the two concentration units using the molar mass M of the solute, in g/mol (the same number as its Mr).
g/dm³ = mol/dm³ × M
c (g/dm³) = c (mol/dm³) × M
Worked Example:
Convert 0.5 mol/dm³ NaOH to g/dm³. (Ar: Na = 23, O = 16, H = 1)
Step 1: Calculate Mr of NaOH
Mr = 23 + 16 + 1 = 40
Step 2: Convert
c (g/dm³) = 0.5 × 40
c = 20 g/dm³
Practice Problem:
Convert 7.3 g/dm³ HCl to mol/dm³. (Ar: H = 1, Cl = 35.5)
Show Answer
Mr of HCl = 1 + 35.5 = 36.5
c (mol/dm³) = 7.3 / 36.5 = 0.2 mol/dm³
3. Percentage Calculations (Extended)
Percentage calculations show how much product a reaction gave compared with the maximum possible, how pure a sample is, and how much of a compound's mass comes from each element.
Percentage Yield
Compares what you actually made to what you should have made. Usually less than 100% in practice due to side reactions and losses.
% Yield = (Actual Yield ÷ Theoretical Yield) × 100
% yield = (actual / theoretical) × 100
Worked Example:
A reaction should produce 50 g of product (theoretical yield). In the experiment, 42 g was obtained. Calculate the percentage yield.
% yield = (actual / theoretical) × 100
% yield = (42 / 50) × 100
% yield = 84%
Practice Problem:
The theoretical yield of a reaction is 120 g. If the percentage yield is 75%, calculate the actual yield.
Show Answer
Actual yield = (% yield / 100) × theoretical
Actual yield = (75 / 100) × 120 = 90 g
Percentage Purity
Measures how much of a sample is the desired substance vs. impurities. Used for ores, minerals, and industrial chemicals.
% Purity = (Mass of Pure Substance ÷ Total Mass) × 100
% purity = (pure / total) × 100
Worked Example:
A 25 g sample of limestone contains 22 g of pure calcium carbonate. Calculate the percentage purity.
% purity = (pure / total) × 100
% purity = (22 / 25) × 100
% purity = 88%
Practice Problem:
A 200 g sample of iron ore is 95% pure iron(III) oxide by mass. Calculate the mass of iron(III) oxide in the sample.
Show Answer
Mass of iron(III) oxide = (95 / 100) × 200 = 190 g
Percentage Composition by Mass
Calculates what fraction of a compound's mass comes from each element. Useful, for example, for comparing the nitrogen content of fertilisers.
% by mass = (Mass of Element ÷ Mass of Compound) × 100
% = (Ar × atoms / Mr) × 100
Worked Example:
Calculate the percentage of nitrogen in ammonium nitrate (NH₄NO₃). (Ar: N = 14, H = 1, O = 16)
Step 1: Calculate Mr
Mr of NH₄NO₃ = 14 + (4 × 1) + 14 + (3 × 16) = 80
Step 2: Calculate mass of N
Mass of N = 2 × 14 = 28
Step 3: Calculate percentage
% N = (28 / 80) × 100
% N = 35%
Practice Problem:
Calculate the percentage of oxygen in sulfuric acid (H₂SO₄). (Ar: H = 1, S = 32, O = 16)
Show Answer
Mr of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98
Mass of O = 4 × 16 = 64
% O = (64 / 98) × 100 = 65.3%
4. Energy & Enthalpy (Extended)
Energy calculations predict whether reactions release or absorb heat. Negative ΔH = exothermic (releases heat), positive ΔH = endothermic (absorbs heat).
Calculating Enthalpy Change from Bond Energies
Breaking bonds requires energy (endothermic), making bonds releases energy (exothermic). The difference gives the overall energy change.
ΔH = Energy to Break Bonds - Energy Released Making Bonds
ΔH = Σ(bonds broken) - Σ(bonds made)
+ΔH = endothermic, -ΔH = exothermic
Worked Example:
Calculate ΔH for: H₂ + Cl₂ → 2HCl
Bond energies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol
Step 1: Bonds broken (reactants)
1 × H-H + 1 × Cl-Cl = 436 + 242 = 678 kJ
Step 2: Bonds made (products)
2 × H-Cl = 2 × 431 = 862 kJ
Step 3: Calculate ΔH
ΔH = 678 - 862
ΔH = -184 kJ/mol (exothermic)
Practice Problem:
Calculate ΔH for: CH₄ + 2O₂ → CO₂ + 2H₂O
Bond energies: C-H = 413, O=O = 498, C=O = 805, O-H = 464 kJ/mol
Show Answer
Broken: 4(C-H) + 2(O=O) = 4(413) + 2(498) = 2648 kJ
Made: 2(C=O) + 4(O-H) = 2(805) + 4(464) = 3466 kJ
ΔH = 2648 - 3466 = -818 kJ/mol
Reading Energy from Diagrams
Energy profile diagrams show reactants, products, and activation energy. Learn to read values directly from these graphs.
ΔH = Products Energy - Reactants Energy
Activation energy (Ea) = Peak - Reactants
Worked Example:
An energy diagram shows: Reactants at 100 kJ, Products at 40 kJ, Peak at 150 kJ. Calculate ΔH and Ea.
Enthalpy change:
ΔH = 40 - 100 = -60 kJ (exothermic)
Activation energy:
Ea = 150 - 100 = 50 kJ
Practice Problem:
A reaction has ΔH = +80 kJ and Ea = 120 kJ. If the reactants are at 50 kJ, what are the energy levels of the products and the peak?
Show Answer
Products = Reactants + ΔH = 50 + 80 = 130 kJ
Peak = Reactants + Ea = 50 + 120 = 170 kJ
5. Chromatography (Extended)
Chromatography separates mixtures based on how substances travel through a medium. Rf values help identify unknown substances.
Calculating Rf Value
A substance has the same Rf value whenever the same solvent and conditions are used, so comparing Rf values with those of known substances helps identify unknown substances.
Rf = Distance moved by substance ÷ Distance moved by solvent
Rf = d(substance) / d(solvent)
Rf is always between 0 and 1 (no units)
Worked Example:
In a chromatography experiment, the solvent front moved 8 cm from the baseline. A substance moved 5.6 cm. Calculate its Rf value.
Rf = d(substance) / d(solvent)
Rf = 5.6 / 8
Rf = 0.7
Practice Problem:
A food dye has an Rf value of 0.45. If the solvent front moved 12 cm, how far did the dye travel?
Show Answer
d(substance) = Rf × d(solvent)
d = 0.45 × 12 = 5.4 cm
6. Reacting Masses (Extended)
Predicting how much product you can make from given reactants, using moles and the balanced equation. Core candidates do simpler reacting-mass calculations by proportion, without moles: for example, if 100 g of calcium carbonate gives 44 g of carbon dioxide, then 10 g gives 4.4 g.
Calculating Mass of Product from Reactant
Follow the chain: mass → moles → use ratio → moles → mass. The balanced equation gives you the mole ratio.
Use mole ratio from balanced equation
mass → moles → use ratio → moles → mass
Worked Example:
Calculate the mass of CO₂ produced when 10 g of CaCO₃ decomposes.
CaCO₃ → CaO + CO₂ (Ar: Ca = 40, C = 12, O = 16)
Step 1: Calculate moles of CaCO₃
Mr of CaCO₃ = 40 + 12 + 48 = 100
n = 10 / 100 = 0.1 mol
Step 2: Use mole ratio (1:1:1)
Moles of CO₂ = 0.1 mol
Step 3: Calculate mass of CO₂
Mr of CO₂ = 12 + 32 = 44
m = 0.1 × 44
m = 4.4 g
Practice Problem:
Calculate the mass of magnesium oxide formed when 4.8 g of magnesium burns.
2Mg + O₂ → 2MgO (Ar: Mg = 24, O = 16)
Show Answer
n(Mg) = 4.8 / 24 = 0.2 mol
Ratio Mg:MgO = 2:2 = 1:1
n(MgO) = 0.2 mol
Mr(MgO) = 24 + 16 = 40
m(MgO) = 0.2 × 40 = 8 g
Finding the Limiting Reactant
When reactants aren't in perfect proportions, one runs out first and stops the reaction. This is the limiting reactant.
Compare mole ratios to find which reactant runs out first
The limiting reactant determines how much product forms
Worked Example:
3 g of magnesium is added to hydrochloric acid containing 10 g of hydrogen chloride (HCl). Which is the limiting reactant?
Mg + 2HCl → MgCl₂ + H₂ (Ar: Mg = 24, H = 1, Cl = 35.5)
Calculate moles of each reactant:
n(Mg) = 3 / 24 = 0.125 mol
Mr(HCl) = 1 + 35.5 = 36.5
n(HCl) = 10 / 36.5 = 0.274 mol
Check ratio (need 2 mol HCl per 1 mol Mg):
HCl needed = 0.125 × 2 = 0.25 mol
HCl available = 0.274 mol (enough)
Mg is the limiting reactant
Practice Problem:
4 g of H₂ reacts with 48 g of O₂. Which is limiting?
2H₂ + O₂ → 2H₂O (Ar: H = 1, O = 16)
Show Answer
n(H₂) = 4 / 2 = 2 mol
n(O₂) = 48 / 32 = 1.5 mol
Ratio: Need 2:1, so 2 mol H₂ needs 1 mol O₂
O₂ available = 1.5 mol (more than needed)
H₂ is limiting
7. Titration Calculations (Extended)
Titrations use a solution of known concentration to find the concentration of an unknown. In this syllabus they are acid-base titrations.
Finding Unknown Concentration
Use the known solution to calculate moles, then use the equation ratio to find moles of the unknown, then calculate its concentration.
Moles = Concentration × Volume (in dm³)
n = c × V, then use the mole ratio from the equation
Worked Example:
25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.1 mol/dm³ HCl. Calculate the concentration of NaOH.
NaOH + HCl → NaCl + H₂O
Step 1: Calculate moles of HCl
n(HCl) = c × V = 0.1 × (20/1000) = 0.002 mol
Step 2: Use mole ratio (1:1)
n(NaOH) = 0.002 mol
Step 3: Calculate concentration of NaOH
c = n / V = 0.002 / (25/1000) = 0.002 / 0.025
c(NaOH) = 0.08 mol/dm³
Practice Problem:
25.0 cm³ of H₂SO₄ is neutralised by 40.0 cm³ of 0.2 mol/dm³ NaOH. Calculate the concentration of H₂SO₄.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Show Answer
n(NaOH) = 0.2 × (40/1000) = 0.008 mol
Ratio: 2 NaOH : 1 H₂SO₄
n(H₂SO₄) = 0.008 / 2 = 0.004 mol
c(H₂SO₄) = 0.004 / (25/1000) = 0.16 mol/dm³
Finding Molar Mass from Titration
If you know the mass of an unknown acid or base, titration can find its moles, allowing you to calculate its molar mass and so its Mr.
Use titration to find moles, then M = m/n
M is in g/mol; Mr is the same number with no units
Worked Example:
2.0 g of an acid, HX, is dissolved in water. The whole solution needs 25.0 cm³ of 0.8 mol/dm³ NaOH to neutralise it. Calculate the Mr of HX.
HX + NaOH → NaX + H₂O
Step 1: Calculate moles of NaOH
n(NaOH) = 0.8 × (25/1000) = 0.02 mol
Step 2: Use mole ratio (1:1)
n(HX) = 0.02 mol
Step 3: Calculate molar mass, then Mr
M = m / n = 2.0 / 0.02 = 100 g/mol
Mr = 100
Practice Problem:
3.0 g of an acid, H₂A, needs 50.0 cm³ of 0.5 mol/dm³ NaOH to fully neutralise it. Calculate the Mr of H₂A.
H₂A + 2NaOH → Na₂A + 2H₂O
Show Answer
n(NaOH) = 0.5 × (50/1000) = 0.025 mol
Ratio: 2 NaOH : 1 H₂A
n(H₂A) = 0.025 / 2 = 0.0125 mol
M = 3.0 / 0.0125 = 240 g/mol, so Mr = 240
8. Electrolysis
Electrolysis uses electricity to decompose compounds. Predicting products requires knowing the reactivity series and ion discharge rules.
Core: know the products for molten binary compounds (such as lead(II) bromide), concentrated aqueous sodium chloride and dilute sulfuric acid with inert electrodes. Extended: predict the products for any halide solution, dilute or concentrated.
Predicting Electrolysis Products (Extended)
For aqueous solutions with inert electrodes. At the cathode (−): metals below hydrogen deposit, otherwise H₂ forms. At the anode (+): a concentrated halide solution gives the halogen; a dilute one, or a solution with no halide ions, gives O₂.
At the Cathode (negative):
- • Metal deposited if below hydrogen in reactivity series (Cu, Ag)
- • Hydrogen gas (H₂) if metal is above hydrogen in reactivity series
At the Anode (positive):
- • Halogen (chlorine, bromine or iodine) if the halide solution is concentrated
- • Oxygen gas (O₂) if the halide solution is dilute, or if there are no halide ions (from water)
Worked Example:
Predict the products of electrolysis of dilute sodium chloride solution using inert electrodes.
Ions present: Na⁺, H⁺ (from water), Cl⁻, OH⁻ (from water)
At cathode: Na⁺ vs H⁺
Na is above H in reactivity series → H₂ gas
At anode: Cl⁻ vs OH⁻
Solution is dilute, so OH⁻ is oxidised → O₂ gas
Practice Problem:
Predict the products of electrolysis of concentrated copper(II) chloride solution using inert electrodes.
Show Answer
Ions: Cu²⁺, Cl⁻, H⁺, OH⁻
Cathode: Cu below H → Copper metal
Anode: Concentrated, so Cl⁻ → Chlorine gas (Cl₂)
More Resources
LumiExams.com | IGCSE Chemistry Formula Practice | Cambridge IGCSE 0620
Frequently Asked Questions
How do I calculate moles in IGCSE Chemistry?
Use the formula n = m/M where n is moles, m is mass in grams, and M is the molar mass in g/mol (the same number as Mr). For gases at RTP, use n = V/24 where V is volume in dm³. Always show your working step by step. Mole calculations are Extended (Supplement) content; Core candidates work out reacting masses by simple proportion, without moles.
What is the difference between percentage yield and percentage purity?
Percentage yield compares the mass of product you actually obtained with the maximum mass the reaction could give (actual ÷ theoretical × 100). Percentage purity compares the mass of the pure substance with the mass of the whole sample (mass of pure substance ÷ mass of sample × 100). Both are calculated from data you are given, and both are Extended (Supplement) calculations.
How do I convert between mol/dm³ and g/dm³?
This conversion is Extended (Supplement) content. To convert mol/dm³ to g/dm³, multiply by the molar mass M in g/mol (the same number as Mr). To convert g/dm³ to mol/dm³, divide by M. For example, 0.5 mol/dm³ NaCl (M = 58.5 g/mol) = 0.5 × 58.5 = 29.25 g/dm³.
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