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IGCSE Chemistry Recap – Atoms, Stoichiometry and Electrochemistry | Ho Him (2026)

Atoms, Stoichiometry and Electrochemistry

IGCSE Chemistry

Ho Him (2026)

For use on Sunday, 6 September 2026

01

Atomic Structure

Atoms, Elements and Compounds

A sodium atom is represented by 2311Na. Some information about four other substances is shown in the table.

Substance Formula
Helium He
Oxygen O2
Water H2O
Aluminium oxide Al2O3
  1. State the number of protons, neutrons and electrons in one neutral atom of 2311Na.
  2. Write the electronic structure of a sodium atom.
  3. Define the term element.
  4. Explain why oxygen is an element even though each oxygen molecule contains two atoms.
  5. Explain why water is a compound rather than an element.
  6. State the total number of atoms in one formula unit of aluminium oxide, Al2O3.
  7. State the ratio of aluminium atoms to oxygen atoms in aluminium oxide.
  8. Explain the difference between an atom and a molecule.
12 marks
02

Stoichiometry

Magnesium and Oxygen

Magnesium burns in oxygen to form magnesium oxide. A student completely burns 4.8 g of magnesium in excess oxygen.

Relative atomic masses: Mg = 24, O = 16
  1. Write the balanced chemical equation for the reaction between magnesium and oxygen.
  2. Calculate the relative formula mass, Mr, of magnesium oxide.
  3. Calculate the number of moles of magnesium used.
  4. Use the balanced equation to determine the number of moles of magnesium oxide formed.
  5. Calculate the mass of magnesium oxide formed.
  6. Calculate the mass of oxygen that reacts with the magnesium.
  7. State why the mass of magnesium oxide is greater than the original mass of magnesium.
12 marks
03

Stoichiometry

Calcium Carbonate and Hydrochloric Acid

Calcium carbonate reacts with hydrochloric acid according to the equation below.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

A student reacts 5.0 g of calcium carbonate completely with hydrochloric acid.

Relative atomic masses: Ca = 40, C = 12, O = 16
One mole of any gas occupies 24 dm3 at room temperature and pressure.
  1. Calculate the relative formula mass, Mr, of calcium carbonate.
  2. Calculate the number of moles of calcium carbonate used.
  3. Determine the number of moles of carbon dioxide formed.
  4. Calculate the volume of carbon dioxide formed at room temperature and pressure, in dm3.
  5. Convert your answer in part (d) into cm3.
  6. Determine the number of moles of hydrochloric acid needed for complete reaction.
  7. The concentration of the hydrochloric acid is 1.0 mol/dm3. Calculate the minimum volume of this acid required, in cm3.
14 marks
04

Electrochemistry

Electrolysis of Molten Lead(II) Bromide

Molten lead(II) bromide, PbBr2, is electrolysed using two graphite electrodes connected to a direct current power supply. Lead(II) bromide contains Pb2+ ions and Br− ions.

D.C. power supply + − Molten PbBr₂ Positive electrode Negative electrode Pb²⁺ Br⁻

Simplified apparatus for the electrolysis of molten lead(II) bromide.

  1. Explain why lead(II) bromide must be molten before it can conduct electricity and undergo electrolysis.
  2. State the name of the negative electrode.
  3. State which ion moves towards the negative electrode.
  4. Name the product formed at the negative electrode and write the ionic half-equation for its formation.
  5. Name the product formed at the positive electrode and write the ionic half-equation for its formation.
  6. State the type of chemical change occurring at each electrode: oxidation or reduction.
  7. Explain why graphite is suitable for use as the electrodes.
14 marks
05

Electrochemistry

Electrolysis of Aqueous Copper(II) Sulfate

A student electrolyses aqueous copper(II) sulfate using two different pairs of electrodes. The observations are shown below.

Experiment Electrodes used Observation at cathode Observation at anode
1 Graphite Pink-brown solid deposited Bubbles of a colourless gas
2 Copper Pink-brown solid deposited Copper anode becomes smaller
  1. Name the pink-brown solid deposited at the cathode in both experiments.
  2. Write the ionic half-equation for the reaction at the cathode.
  3. In Experiment 1, name the colourless gas formed at the graphite anode.
  4. Describe a test for this gas and state the positive result.
  5. State what happens to the intensity of the blue colour of the solution during Experiment 1. Explain your answer.
  6. Write the ionic half-equation for the reaction at the copper anode in Experiment 2.
  7. Explain why the blue colour of the solution remains approximately unchanged during Experiment 2.
  8. State how the mass of each copper electrode changes during Experiment 2.
  9. Explain how the apparatus in Experiment 2 could be used to electroplate a metal object with copper.
16 marks

Question 1

  1. For 2311Na:
    • 11 protons
    • 12 neutrons
    • 11 electrons
    Number of neutrons = 23 − 11 = 12.
  2. The electronic structure of sodium is 2,8,1.
  3. An element is a pure substance containing only one type of atom.
  4. Oxygen is an element because each O2 molecule contains only oxygen atoms. The two atoms in the molecule are atoms of the same element.
  5. Water is a compound because it contains hydrogen and oxygen atoms chemically bonded together. These are atoms of two different elements combined in a fixed ratio.
  6. Al2O3 contains two aluminium atoms and three oxygen atoms. Total number of atoms = 5.
  7. The ratio of aluminium atoms to oxygen atoms is 2 : 3.
  8. An atom is the smallest particle of an element that retains the chemical properties of that element. A molecule consists of two or more atoms chemically bonded together.
Key idea: A molecule may contain atoms of the same element, such as O2, or atoms of different elements, such as H2O.

Question 2

  1. 2Mg + O2 → 2MgO
  2. Mr of MgO = 24 + 16
    Mr of MgO = 40
  3. Moles = Mass / Mr
    Moles of Mg = 4.8 / 24 = 0.20 mol
    Number of moles of magnesium = 0.20 mol.
  4. The equation shows a 2 : 2 ratio between Mg and MgO. Therefore, the mole ratio of Mg to MgO is 1 : 1. Number of moles of MgO = 0.20 mol.
  5. Mass = Moles × Mr
    Mass of MgO = 0.20 × 40 = 8.0 g
    Mass of magnesium oxide = 8.0 g.
  6. Mass of oxygen = 8.0 − 4.8 = 3.2 g
    Mass of oxygen reacted = 3.2 g.
  7. The mass increases because magnesium combines with oxygen from the air. The mass of the product includes the mass of both the magnesium and the oxygen.
Stoichiometry method: Convert the known mass into moles, use the mole ratio from the balanced equation, and then convert the required number of moles into mass.

Question 3

  1. Mr of CaCO3 = 40 + 12 + (3 × 16)
    Mr of CaCO3 = 40 + 12 + 48
    Mr of CaCO3 = 100
  2. Moles = Mass / Mr
    Moles of CaCO3 = 5.0 / 100 = 0.050 mol
    Number of moles of calcium carbonate = 0.050 mol.
  3. The equation shows a 1 : 1 ratio between CaCO3 and CO2. Number of moles of carbon dioxide = 0.050 mol.
  4. Gas volume = Moles × 24 dm3
    Volume of CO2 = 0.050 × 24 = 1.2 dm3
    Volume of carbon dioxide = 1.2 dm3.
  5. 1 dm3 = 1000 cm3
    1.2 × 1000 = 1200 cm3
    Volume of carbon dioxide = 1200 cm3.
  6. The equation shows that one mole of CaCO3 reacts with two moles of HCl.
    Moles of HCl = 0.050 × 2 = 0.10 mol
    Number of moles of hydrochloric acid = 0.10 mol.
  7. Volume = Moles / Concentration
    Volume = 0.10 / 1.0 = 0.10 dm3
    0.10 dm3 × 1000 = 100 cm3
    Minimum volume of hydrochloric acid = 100 cm3.
Unit reminder: When concentration is given in mol/dm3, the volume used in the concentration formula must be measured in dm3.

Question 4

  1. In solid lead(II) bromide, the ions are held in fixed positions and cannot move. When it is molten, the Pb2+ and Br− ions are free to move and carry electric charge.
  2. The negative electrode is the cathode.
  3. The Pb2+ ion moves towards the negative electrode because it is positively charged.
  4. The product at the cathode is lead.
    Pb2+ + 2e− → Pb
  5. The product at the anode is bromine.
    2Br− → Br2 + 2e−
  6. At the cathode, Pb2+ ions gain electrons. This is reduction. At the anode, Br− ions lose electrons. This is oxidation.
  7. Graphite is suitable because it conducts electricity and is generally chemically unreactive under these conditions.
Electron rule: Reduction is the gain of electrons at the cathode. Oxidation is the loss of electrons at the anode.

Question 5

  1. The pink-brown solid is copper.
  2. Copper(II) ions gain electrons at the cathode.
    Cu2+ + 2e− → Cu
  3. The colourless gas formed at the graphite anode is oxygen.
  4. Insert a glowing splint into the collected gas. A positive result is that the glowing splint relights.
  5. The blue colour becomes paler. Cu2+ ions are removed from the solution and deposited as copper at the cathode. With graphite electrodes, Cu2+ ions are not replaced at the anode.
  6. At the copper anode, copper atoms lose electrons and enter the solution as copper(II) ions.
    Cu → Cu2+ + 2e−
  7. Cu2+ ions are removed from the solution at the cathode, but new Cu2+ ions are formed at the copper anode. The concentration of Cu2+ ions therefore remains approximately constant, so the blue colour remains approximately unchanged.
  8. The copper cathode gains mass because copper is deposited on it. The copper anode loses mass because copper atoms form Cu2+ ions and enter the solution.
  9. To electroplate an object with copper:
    • clean the metal object;
    • connect the object to the negative terminal so that it becomes the cathode;
    • use copper as the positive electrode or anode;
    • place both electrodes in aqueous copper(II) sulfate;
    • pass a direct electric current through the solution.
    Copper is deposited onto the surface of the object.
Copper electrode comparison: Copper is transferred from the anode to the cathode. The anode loses mass while the cathode gains mass.
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