What does Edexcel GCSE Chemistry Topic 5 cover?

Topic 5 is separate-science only. It covers the transition metals and how they compare with Group 1, alloys and why they are harder than pure metals, corrosion and rusting and how to prevent it, and quantitative analysis including concentration in mol/dm3, the acid-alkali titration core practical and calculations, percentage yield, atom economy, and gas volume calculations.

Why is an alloy harder than a pure metal?

A pure metal has layers of identically sized atoms that can slide over one another, so it is soft. An alloy contains atoms of a different size, which distort the regular layers so they cannot slide as easily, making the alloy harder.

How do you calculate an unknown concentration from a titration?

Calculate the moles of the solution whose concentration you know using moles equals concentration times volume in dm3, use the mole ratio from the balanced equation to find the moles of the unknown, then divide by its volume in dm3 to get the concentration. Convert all volumes from cm3 to dm3 first.

What is the difference between percentage yield and atom economy?

Percentage yield compares the actual product obtained with the theoretical maximum, showing losses in practice. Atom economy compares the relative formula mass of the useful product with the total of all products, showing how much reactant mass becomes useful product. A sustainable process has both a high yield and a high atom economy.

What is the most common mistake in Topic 5?

Common mistakes are forgetting to convert cm3 to dm3, using a 1 to 1 ratio for sulfuric acid in titrations, inverting the percentage yield formula, dividing instead of multiplying when converting mol/dm3 to g/dm3, and saying rusting needs only water or only oxygen rather than both.

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Edexcel GCSE Chemistry Topic 5 Separate chemistry 1: a complete overview

A deep-dive Edexcel GCSE Chemistry guide to Topic 5 Separate chemistry 1. Covers the transition metals, alloys and corrosion, and the quantitative analysis content: moles and concentration, the titration core practical and calculations, percentage yield, atom economy, and gas volumes using the molar gas volume.

Generated by Claude Opus 4.815 min readTopic 5Updated 2026-06-02

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What Topic 5 actually demands
Transition metals, alloys and corrosion
Quantitative analysis
Yield, atom economy and gas volumes
How Topic 5 is examined
Check your knowledge

What Topic 5 actually demands

Separate chemistry 1 is sat only by students taking GCSE Chemistry as a separate science. It pairs the descriptive chemistry of the transition metals, alloys and corrosion with the quantitative analysis that pushes the calculations to their highest level. Edexcel tests confident, well-shown calculation and clear structure-and-property explanations.

This guide walks through the topic in specification order, then sets out the exam patterns Edexcel repeats. Each dot point has a matching page with practice questions; this overview ties them together.

Transition metals, alloys and corrosion

The transition metals (central block) are harder, stronger, denser and higher-melting than Group 1, react more slowly, form coloured compounds and act as catalysts. An alloy is harder than a pure metal because the different-sized atoms distort the layers so they cannot slide. Rusting needs both water and oxygen and forms hydrated iron(III) oxide; it is prevented by barriers (paint, oil, coating) or by sacrificial protection with a more reactive metal such as zinc (galvanising).

Quantitative analysis

The mole links mass and particles, and concentration in mol/dm $^3$ equals moles divided by volume in dm $^3$ . Convert mol/dm $^3$ to g/dm $^3$ by multiplying by the $M_r$ . The titration core practical uses a pipette and burette to find an unknown concentration, repeating to concordant titres. The calculation is moles of the known solution, mole ratio, then concentration of the unknown.

Yield, atom economy and gas volumes

Percentage yield is actual over theoretical, below 100 percent because of losses, incomplete or reversible reactions, impurities and side reactions. Atom economy is the $M_r$ of the useful product over the total $M_r$ of all products, and a high value means a sustainable, low-waste process. Gas volumes use the molar gas volume of $24$ dm $^3$ /mol at rtp, so volume equals moles times 24.

How Topic 5 is examined

A typical Edexcel profile for this topic:

Transition metals and alloys. Comparing properties and explaining alloy hardness.
Corrosion. The conditions for rusting and methods of prevention.
Titration calculations. Finding an unknown concentration from titration data.
Yield, atom economy and gas volumes. Multi-step calculation questions.

Check your knowledge

A mix of recall and calculation questions covering Topic 5. Attempt them under timed conditions, then check against the solutions.

Give two properties of transition metals that differ from Group 1 metals. (2 marks)
Explain why an alloy is harder than the pure metal. (2 marks)
State the two substances needed for iron to rust. (2 marks)
Calculate the moles in $20.0$ cm $^3$ of $0.150$ mol/dm $^3$ acid. (2 marks)
$25.0$ cm $^3$ of alkali is neutralised by $25.0$ cm $^3$ of $0.100$ mol/dm $^3$ acid in a $1:1$ reaction. Find the concentration of the alkali. (2 marks)
A reaction has a theoretical yield of $20$ g and an actual yield of $17$ g. Calculate the percentage yield. (2 marks)
Calculate the volume of $0.40$ mol of carbon dioxide at rtp. (2 marks)
Explain why a high atom economy is desirable. (2 marks)

Solutions

Q1: Transition metals are harder and stronger and have higher melting points (also denser, less reactive, form coloured compounds, act as catalysts).
Q2: The different-sized atoms in the alloy distort the regular layers so they cannot slide over one another as easily.
Q3: Water and oxygen.
Q4: $0.150 \times (20.0/1000) = 0.150 \times 0.0200 = 0.00300$ mol.
Q5: Moles of acid $= 0.100 \times 0.0250 = 0.00250$ mol; ratio $1:1$ , so moles of alkali $= 0.00250$ ; concentration $= 0.00250 / 0.0250 = 0.100$ mol/dm $^3$ .
Q6: $(17 / 20) \times 100 = 85\%$ .
Q7: $0.40 \times 24 = 9.6$ dm $^3$ .
Q8: More of the reactant mass becomes useful product, so there is less waste, making the process more sustainable and often more profitable.

Sources & how we know this

Pearson Edexcel GCSE (9-1) Chemistry (1CH0) specification — Pearson Edexcel (2016)