Pearson Edexcel International GCSE Chemistry 4CH1: Gases in the atmosphere – Study Guide

This guide teaches sub-topic 2(c), Gases in the atmosphere, of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). It covers specification points 2.9–2.14, including the required practical 2.14. The qualification is untiered, and none of these six points carries a “C” reference, so all of them can be examined on Paper 1 and Paper 2. One calculation step below uses molar volume, which comes from point 1.35C in Topic 1; that step is labelled Paper 2 only.

Useful links: the 4CH1 Chemistry course hub, the printable 4CH1 checklist, the condensed Gases in the atmosphere revision notes, and the wider Topic 2 pages: the Reactivity and group trends study guide, the Reactivity revision notes, the Reactivity practice questions and the Groups, acids, salts and tests practice set. To find your weak spots first, take the free 4CH1 all-topics diagnostic.

What this sub-topic covers

Spec point	What you must be able to do	Paper
2.9	Know the approximate percentages by volume of the four most abundant gases in dry air	1 and 2
2.10	Understand how to find the percentage of oxygen in air using reactions of metals (e.g. iron) and non-metals (e.g. phosphorus) with air	1 and 2
2.11	Describe the combustion of elements in oxygen, including magnesium, hydrogen and sulfur	1 and 2
2.12	Describe the formation of carbon dioxide by thermal decomposition of metal carbonates, including copper(II) carbonate	1 and 2
2.13	Know that carbon dioxide is a greenhouse gas and that increasing amounts may contribute to climate change	1 and 2
2.14	Practical: determine the approximate percentage by volume of oxygen in air using a metal or a non-metal	1 and 2

Paper 1 is a 2-hour paper (110 marks); Paper 2 lasts 1 hour 15 minutes (70 marks). A calculator may be used on both.

2.9 The composition of dry air

“Dry” means the water vapour has been left out. Water vapour varies from place to place and day to day, so the fixed percentages are always quoted for dry air.

Gas	Approximate % by volume
Nitrogen, N₂	78
Oxygen, O₂	21
Argon, Ar	0.9
Carbon dioxide, CO₂	0.04

The rest is tiny amounts of other noble gases and trace gases. Learn the order as well as the numbers: argon is more abundant than carbon dioxide. Oxygen is about one-fifth of air, so removing the oxygen from a sample should shrink its volume by about one-fifth.

Worked example 1: volumes in a sample

Question. Estimate the volume of each of the four main gases in 250 cm³ of dry air.

Working. Multiply the total volume by each percentage.

nitrogen:       250 × 78/100   = 195 cm³
oxygen:         250 × 21/100   = 52.5 cm³
argon:          250 × 0.9/100  = 2.25 cm³
carbon dioxide: 250 × 0.04/100 = 0.1 cm³

Answer. About 195 cm³ nitrogen, 52.5 cm³ oxygen, 2.25 cm³ argon and 0.1 cm³ carbon dioxide.

2.10 and 2.14 Measuring the percentage of oxygen in air

Every method traps a known volume of air and lets a substance react with its oxygen, forming a solid (or a product that dissolves). After returning to room temperature, the decrease in gas volume is the volume of oxygen.

volume of oxygen = starting volume of air − final volume of gas
% oxygen = (volume of oxygen ÷ starting volume of air) × 100

Always divide by the starting volume, never by the final volume.

Method A: iron (a metal)

1. Wet the inside of a measuring cylinder (or a graduated tube) and push a small plug of iron wool to the closed end.
2. Turn the cylinder upside down in a trough of water and read the volume of trapped air.
3. Leave it for about a week. The iron rusts: iron + oxygen + water → hydrated iron(III) oxide. Rusting needs both water and oxygen, which is why the iron wool is damp.
4. As oxygen is used up, water rises into the cylinder. When the level stops changing, read the final volume.

Iron wool has a large surface area, so it reacts faster than a nail. No heating is needed, but the method is slow.

Method B: phosphorus (a non-metal)

A small piece of phosphorus is placed in a dish floating on water inside a bell jar or tube of trapped air, and lit (usually with a hot wire). It burns with a bright flame and gives dense white smoke of phosphorus(V) oxide:

4P(s) + 5O₂(g) → P₄O₁₀(s)

The oxide dissolves in the water, so water rises to take the place of the oxygen that was used. Phosphorus is hazardous: this is normally a teacher demonstration. You must wait for the gas to cool to room temperature before reading the volume, because hot gas has expanded.

Method C: copper and gas syringes (another metal)

A known volume of air is pushed back and forth between two gas syringes over strongly heated copper. The pink-brown copper turns black: 2Cu(s) + O₂(g) → 2CuO(s). Keep passing the air over until the volume stops falling, so that all the oxygen reacts, then cool and read the final volume.

Worked example 2: iron wool result

Question. A student traps 94.0 cm³ of air over damp iron wool. After eight days the gas volume is 74.6 cm³ and no longer changing. Calculate the percentage by volume of oxygen in the air.

Working.

volume of oxygen = 94.0 − 74.6 = 19.4 cm³
% oxygen = 19.4 ÷ 94.0 × 100 = 20.6%

Answer. 20.6%, close to the accepted 21%.

Worked example 3: why reading too early gives a low answer

Question. In a phosphorus experiment, a tube holds 60.0 cm³ of air. Straight after burning, a student reads 50.5 cm³. After cooling to room temperature, the volume is 47.8 cm³. Calculate the percentage of oxygen from each reading and explain which is correct.

Working.

warm reading: 60.0 − 50.5 = 9.5 cm³  →  9.5 ÷ 60.0 × 100 = 15.8%
cool reading: 60.0 − 47.8 = 12.2 cm³ →  12.2 ÷ 60.0 × 100 = 20.3%

Answer. The cooled reading, 20.3%, is correct. The warm gas had expanded, so its volume was too large and the apparent decrease was too small.

Why results can come out low or high

• Low: not all the oxygen reacted (too little iron or copper, not enough time, too few passes over the copper); reading the volume while the gas is still warm; air leaking in at a joint.
• High: reading the final volume when the apparatus is colder than at the start.

2.11 Combustion of elements in oxygen

Combustion is a reaction with oxygen that releases heat and light. Elements burn far more brightly in pure oxygen than in air, because air is only about one-fifth oxygen. Each element forms its oxide.

Element	What you see	Product	Equation
Magnesium	Very bright white flame (don’t look straight at it); white powder left	Magnesium oxide	2Mg(s) + O₂(g) → 2MgO(s)
Hydrogen	Pale blue flame, hard to see; a mixture with air ignites with a squeaky pop	Water	2H₂(g) + O₂(g) → 2H₂O(l)
Sulfur	Melts, then burns with a blue flame	Sulfur dioxide, a colourless gas with a choking smell	S(s) + O₂(g) → SO₂(g)

Water forms as steam in the flame and condenses on a cold surface, so H₂O(l) is the usual state symbol at room temperature.

Worked example 4: describing a combustion

Question. Describe what you would see when a piece of magnesium ribbon is heated and lowered into a gas jar of oxygen. Write the balanced equation with state symbols.

Answer. A very bright white flame; a white powder is left.

2Mg(s) + O₂(g) → 2MgO(s)

Two marks need two separate observations. “It burns” alone is not an observation.

2.12 Carbon dioxide from metal carbonates

Thermal decomposition is breaking down a compound by heat alone. Metal carbonates decompose to the metal oxide and carbon dioxide:

metal carbonate → metal oxide + carbon dioxide
CuCO₃(s) → CuO(s) + CO₂(g)

Copper(II) carbonate is a green powder. On heating it turns black (copper(II) oxide) and gives off carbon dioxide. Other examples:

• ZnCO₃(s) → ZnO(s) + CO₂(g): white solid, yellow while hot, white again on cooling.
• CaCO₃(s) → CaO(s) + CO₂(g): needs strong heating.

How easily a carbonate decomposes varies. Copper(II) carbonate breaks down easily in a Bunsen flame; calcium carbonate needs much stronger heating; sodium and potassium carbonates do not decompose at Bunsen temperatures.

Apparatus. Heat the carbonate in a boiling tube fitted with a delivery tube that dips into limewater. Carbon dioxide turns the limewater milky (cloudy). When you stop, lift the delivery tube out of the limewater before you remove the flame, otherwise cold limewater can be sucked back into the hot tube and crack it.

Worked example 5: reacting masses

Question. 9.88 g of copper(II) carbonate is heated until there is no further change. Calculate (a) the mass of copper(II) oxide left and (b) the mass of carbon dioxide given off. (Ar: Cu = 63.5, C = 12, O = 16)

Working.

Mr CuCO₃ = 63.5 + 12 + (3 × 16) = 123.5
Mr CuO   = 63.5 + 16 = 79.5
Mr CO₂   = 12 + (2 × 16) = 44

moles CuCO₃ = 9.88 ÷ 123.5 = 0.0800 mol
1 mol CuCO₃ → 1 mol CuO + 1 mol CO₂

(a) mass CuO = 0.0800 × 79.5 = 6.36 g
(b) mass CO₂ = 0.0800 × 44 = 3.52 g

Answer. (a) 6.36 g (b) 3.52 g. Check: 6.36 + 3.52 = 9.88 g, so mass is conserved. In practice the mass loss on the balance is the mass of CO₂ that escaped.

(c) Paper 2 only (uses point 1.35C). Volume of CO₂ at rtp = 0.0800 × 24 = 1.92 dm³.

2.13 Carbon dioxide as a greenhouse gas

You need to know two things: carbon dioxide is a greenhouse gas, and increasing amounts of it in the atmosphere may contribute to climate change. A clear explanation runs in steps:

1. The Sun’s radiation passes through the atmosphere and warms the Earth’s surface.
2. The warm surface gives out infrared (heat) radiation.
3. Greenhouse gases such as carbon dioxide absorb some of this infrared and send some of it back towards the surface.
4. This keeps the lower atmosphere warmer than it would otherwise be.
5. Burning fossil fuels and cutting down forests increase the amount of carbon dioxide, so more heat is trapped. This may contribute to climate change, for example rising average temperatures, melting ice, rising sea levels and changes in weather patterns.

Use the specification’s wording: “may contribute to climate change”.

Common errors

• Swapping argon and carbon dioxide in the abundance order, or giving CO₂ as 4% instead of 0.04%.
• Dividing the volume decrease by the final volume instead of the starting volume.
• Reading the gas volume before the apparatus has cooled after burning phosphorus or heating copper.
• Forgetting that iron needs water as well as oxygen to rust in the iron-wool method.
• Calling the sulfur product “sulfur trioxide” or writing S + O₂ → SO₃.
• Describing copper(II) carbonate as “blue” (it is green) or its oxide as “brown” (it is black).
• Removing the flame before the delivery tube, causing suck-back.

Next steps

Condense this with the revision notes, then try questions 4 and 5 of the Groups, acids, salts and tests practice set and questions 2 and 5 of the Reactivity practice questions.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1), Specification, Issue 3, September 2024, published by Pearson Education Limited. Topic 2: Inorganic chemistry, sub-topic (c) Gases in the atmosphere, points 2.9–2.14.