Why is "Subtracting the masses the wrong way round to find the oxygen" a common mistake in Determining the formula of a metal oxide?

Why it happens: Students forget whether the solid gained or lost mass in their route. How to avoid it: Oxidation: oxygen = oxide − metal (mass GAINED). Reduction: oxygen = oxide − metal left (mass LOST). The oxide is always the bigger mass, so oxygen is positive. Source: Edexcel Chemistry examiner reports — recurring error

Why is "Putting the raw masses straight into a ratio without converting to moles" a common mistake in Determining the formula of a metal oxide?

Why it happens: Students think the mass ratio is the formula. How to avoid it: ALWAYS divide each mass by its Aᵣ first. A formula is a ratio of MOLES (atoms), not of masses. Source: Edexcel Chemistry examiner reports

Why is "Rounding a 0.5 in the ratio instead of multiplying up (e.g. writing MO₂ when the ratio is 1 : 1.5)" a common mistake in Determining the formula of a metal oxide?

Why it happens: Students round 1.5 to the nearest whole number. How to avoid it: If a ratio contains a 0.5, multiply EVERY number by 2 to clear it (1 : 1.5 → 2 : 3 → M₂O₃). Never round a half away. Source: Edexcel Chemistry examiner reports

Why is "Writing the subscripts the wrong way round (e.g. CuO₂ instead of Cu₂O)" a common mistake in Determining the formula of a metal oxide?

Why it happens: Students attach the larger number to the wrong element. How to avoid it: The element with MORE moles gets the larger subscript. Cu : O = 2 : 1 means two coppers per oxygen → Cu₂O. Source: Edexcel Chemistry examiner reports

Why is "Not heating to constant mass, so the reaction is incomplete" a common mistake in Determining the formula of a metal oxide?

Why it happens: Students stop heating once it 'looks done'. How to avoid it: Reheat, cool and reweigh until two readings agree. Unreacted oxide or metal makes the masses — and the ratio — wrong. Source: Edexcel Chemistry examiner reports

Why is "Forgetting to flush air out before lighting hydrogen (and keeping gas flowing while cooling)" a common mistake in Determining the formula of a metal oxide?

Why it happens: Safety steps feel separate from the calculation, so they are skipped. How to avoid it: State that hydrogen + air is explosive, so flush the air out FIRST; keep hydrogen flowing during cooling so hot copper does not re-oxidise. Source: Edexcel Chemistry examiner reports

Chemical formulae, equations and calculationsEdexcel IGCSE Science(Double Award)-Chemistry (4WSD0-1C)

Determining the formula of a metal oxide

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Determining the Formula of a Metal Oxide — Edexcel International GCSE Science (Double Award) Chemistry (4WSD0-1C) Study Notes

The core practical that turns mass data into a chemical formula. Two routes: heating a metal in oxygen so the mass GAINED is the oxygen combined (oxidation), or reducing a copper oxide so the mass LOST is the oxygen removed (reduction). Convert masses → moles (÷ Aᵣ) → simplest ratio → formula. Covers Double Award Chemistry spec 1.39, assessed on Chemistry Paper 1 (4WSD0/1C).

What you’ll learn

Mapped to the Edexcel IGCSE 4WSD0-1C syllabus (2026 onwards).

1.39 — Determine the formula of a metal oxide from experimental mass data (oxidation or reduction).

The big idea — mass change tells you the oxygen (spec 1.39)

Find the mass of metal and the mass of oxygen, then turn both into moles.

A metal oxide is just a metal joined to oxygen. To find its formula by experiment you need two masses: the mass of the metal and the mass of the oxygen combined with it. You then convert each mass to moles and read off the simplest ratio — that ratio is the formula.

The clever part is getting the mass of oxygen. You never weigh oxygen directly. Instead you use the change in mass:

Oxidation (heat the metal in oxygen): the solid gets heavier. The extra mass is the oxygen that has joined on. So mass of oxygen = final mass − starting mass.
Reduction (remove oxygen from the oxide): the solid gets lighter. The mass lost is the oxygen that has been taken away. So mass of oxygen = starting mass − final mass.

Either way you end up with two masses — metal and oxygen — ready to turn into moles.

Once you have both masses, the rest is identical for every question: divide each mass by its Aᵣ to get moles, then divide both by the smaller number.

You need the mass of metal AND the mass of oxygen.
Oxygen mass comes from the change in mass, never weighed directly.
Oxidation = mass gained; reduction = mass lost.
Both routes end with: mass → moles → simplest ratio.

See the full worked example for determining the formula of a metal oxide →

Route 1 — oxidising a metal (magnesium → MgO)

Heat the metal in a crucible; the mass GAINED is the oxygen.

The classic example is burning magnesium in air. Magnesium reacts with oxygen to form magnesium oxide:

2Mg + O₂ → 2MgO

Method.

Weigh a clean crucible and lid (empty). Add a coil of magnesium ribbon and reweigh — the difference is the mass of magnesium.
Heat the crucible strongly. Lift the lid briefly and regularly so air (oxygen) gets in, but not so much that white smoke escapes (that smoke is MgO — losing it ruins the result).
When the magnesium has fully reacted, let it cool and reweigh the crucible + contents (now magnesium oxide).
Heat to constant mass — reheat, cool and reweigh until two readings agree. This proves all the magnesium has reacted.

Getting the two masses.

Mass of magnesium = (crucible + Mg) − (empty crucible).
Mass of magnesium oxide = (crucible + MgO) − (empty crucible).
Mass of oxygen = mass of oxide − mass of magnesium (the gain in mass).

Sources of error in this route:

White MgO smoke escapes when the lid is lifted → too little oxide → ratio wrong.
Some magnesium reacts with nitrogen in the air to form magnesium nitride (a small impurity).
Not heating to constant mass → some magnesium still unreacted.

2Mg + O₂ → 2MgO; the metal gains mass.
Lift the lid to let oxygen in, but keep the white smoke (MgO) inside.
Mass of oxygen = mass of oxide − mass of metal.
Heat to constant mass before recording.

See the full worked example for determining the formula of a metal oxide →

Route 2 — reducing a copper oxide (find CuO vs Cu₂O)

Pass hydrogen over the heated oxide; the mass LOST is the oxygen removed.

Copper has two oxides — copper(II) oxide CuO (black) and copper(I) oxide Cu₂O (red). The reduction experiment tells them apart by finding the Cu : O ratio. A reducing gas (hydrogen, or methane) takes the oxygen away, leaving pure copper:

CuO + H₂ → Cu + H₂O

Hydrogen flows over the heated oxide; oxygen leaves as water vapour, so the solid loses mass.

Method.

Weigh an empty reduction tube / boat; add the copper oxide and reweigh → mass of oxide.
Pass hydrogen through the tube and let it flush for a while FIRST to push all the air out (hydrogen + air explodes when lit).
Light the excess hydrogen at the small hole at the far end, then heat the oxide strongly.
When all the black oxide has turned to pink/brown copper, stop heating but keep the hydrogen flowing while it cools (so air does not re-oxidise the hot copper).
Reweigh → mass of copper. Heat to constant mass.

Getting the masses.

Mass of copper = mass of solid left at the end.
Mass of oxygen = mass of oxide − mass of copper (the loss in mass).

Sources of error:

Air not flushed out → risk of explosion / hot copper re-oxidises on cooling → too much "copper" mass.
Not heating to constant mass → some oxide unreacted → answer wrong.
Copper oxide "spitting" out of the boat → solid lost.

CuO + H₂ → Cu + H₂O; the solid loses mass.
Flush air out BEFORE lighting (safety) and keep gas flowing while cooling.
Mass of oxygen = mass of oxide − mass of copper.
Cu : O = 1 : 1 → CuO; Cu : O = 2 : 1 → Cu₂O.

See the full worked example for determining the formula of a metal oxide →

Turning the masses into a formula

mass → moles (÷ Aᵣ) → divide by smallest → whole-number ratio.

Every metal-oxide question — oxidation or reduction — finishes with the same calculation. Set it out in a small table; examiners love clearly laid-out working.

The simplest whole-number ratio of moles IS the formula of the oxide.

Worked example — copper oxide. A sample of copper oxide is reduced. It contains 3.18 g of copper and 0.80 g of oxygen. (Aᵣ: Cu = 63.5, O = 16.)

	Copper	Oxygen
Mass / g	3.18	0.80
÷ Aᵣ → moles	3.18 ÷ 63.5 = 0.0501	0.80 ÷ 16 = 0.0500
÷ smallest (0.0500)	1.00	1.00

Ratio Cu : O = 1 : 1, so the formula is CuO.

If the ratio comes out near a half, multiply to clear it. For example a ratio of 2 : 1 (Cu : O) gives Cu₂O; a ratio 1 : 1.5 (metal : O) would be multiplied by 2 to give 2 : 3, e.g. M₂O₃.

Tip: if you are only given the mass of oxide and the mass of metal, find oxygen by subtraction first, then start the table.

Lay it out as a table: mass, ÷ Aᵣ, ÷ smallest.
Use Aᵣ(Cu) = 63.5 and Aᵣ(O) = 16 unless told otherwise.
Round sensible values to the nearest whole number (1.00, 2.00).
If a 0.5 appears in the ratio, multiply everything to clear it.

See the full worked example for determining the formula of a metal oxide →

How it’s examined

Double Award Chemistry is assessed on one untiered paper — Chemistry Paper 1 (4WSD0/1C), 2 hours, 110 marks, 33.3% of the Double Award (calculator allowed). This practical appears as a structured calculation: you are given masses (of metal and oxide, or of oxide and the metal left) and asked to work out the formula, usually 4–5 marks. Method and safety marks also feature — describing how to heat to constant mass, how to let oxygen in without losing the oxide, or why the air must be flushed out before burning hydrogen. Show every step (mass → moles → ratio) to bank the working marks even if the final ratio slips.

Sources: Pearson Edexcel International GCSE Science (Double Award) 4SD0 specification — Issue 4 (valid for 2026 onwards); Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4 (shared Chemistry Paper 1 content); Pearson Edexcel 4SD0 / 4CH1 examiner reports 2022–2024. Last reviewed 2026-06-07.

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Step-by-step worked examples — Determining the formula of a metal oxide

Step-by-step solutions to past-paper-style questions on determining the formula of a metal oxide, written exactly the way a tutor would explain them at the board.

1Find the mass of oxygen from the mass change (Getting started)
Getting started• 4WSD0/1C style — short answer• metal-oxide, spec-1.39
Question
A piece of magnesium has a mass of 2.4 g. After heating in air it forms 4.0 g of magnesium oxide. Calculate the mass of oxygen that combined with the magnesium. (1 mark)
Step-by-step solution
1. Step 1
  The oxide is heavier than the metal because oxygen has joined on. The extra mass IS the oxygen.
2. Step 2
  Mass of oxygen = mass of oxide − mass of metal = 4.0 − 2.4 = 1.6 g.
Answer
Mass of oxygen = 4.0 − 2.4 = 1.6 g.
Examiner tip
In an oxidation experiment the mass GAINED is the oxygen. Subtract the wrong way round (metal − oxide) and you get a negative — an instant giveaway that you have muddled it.
2State the formula from a given mole ratio (Getting started)
Getting started• 4WSD0/1C style — short answer• metal-oxide, formula, spec-1.39
Question
An experiment on a copper oxide gives 0.040 mol of copper atoms and 0.020 mol of oxygen atoms. Write the formula of this copper oxide. (2 marks)
Step-by-step solution
1. Step 1
  Divide both by the smaller number (0.020). Cu = 0.040 ÷ 0.020 = 2; O = 0.020 ÷ 0.020 = 1.
2. Step 2
  Ratio Cu : O = 2 : 1, so the formula is Cu₂O (copper(I) oxide).
Answer
Cu : O = 2 : 1, so the formula is Cu₂O.
Examiner tip
The simplest whole-number mole ratio IS the formula. A 2 : 1 ratio of copper to oxygen gives Cu₂O, not CuO₂ — the bigger number belongs to the element you have more moles of (copper).
3Reduction of a copper oxide — find the formula (Building confidence)
Building confidence• 4WSD0/1C style — structured• reduction, metal-oxide, spec-1.39
Question
3.18 g of a copper oxide is reduced by heating in a stream of hydrogen. 2.54 g of copper remains. Use the data to find the formula of the copper oxide. (Aᵣ: Cu = 63.5, O = 16.) (4 marks)
Step-by-step solution
1. Step 1
  Mass of oxygen removed (the mass lost) = mass of oxide − mass of copper = 3.18 − 2.54 = 0.64 g.
2. Step 2
  Moles of copper = 2.54 ÷ 63.5 = 0.0400 mol.
3. Step 3
  Moles of oxygen = 0.64 ÷ 16 = 0.0400 mol.
4. Step 4
  Divide both by the smaller (0.0400): Cu = 1, O = 1. Ratio Cu : O = 1 : 1 → formula CuO.
Answer
Oxygen = 0.64 g; n(Cu) = 0.0400, n(O) = 0.0400; ratio 1 : 1 → CuO.
Examiner tip
In reduction the mass LOST is the oxygen. Note Aᵣ(Cu) = 63.5 — using 64 still rounds to a 1 : 1 ratio here, but always use the value on the data sheet.
4Oxidation of magnesium — find the formula (Building confidence)
Building confidence• 4WSD0/1C style — structured• oxidation, metal-oxide, spec-1.39
Question
A coil of magnesium of mass 2.40 g is heated in a crucible until it is fully oxidised. The magnesium oxide formed has a mass of 4.00 g. Find the formula of the oxide. (Aᵣ: Mg = 24, O = 16.) (4 marks)
Step-by-step solution
1. Step 1
  Mass of oxygen gained = mass of oxide − mass of magnesium = 4.00 − 2.40 = 1.60 g.
2. Step 2
  Moles of magnesium = 2.40 ÷ 24 = 0.100 mol.
3. Step 3
  Moles of oxygen = 1.60 ÷ 16 = 0.100 mol.
4. Step 4
  Divide both by 0.100: Mg = 1, O = 1. Ratio Mg : O = 1 : 1 → formula MgO.
Answer
Oxygen = 1.60 g; n(Mg) = 0.100, n(O) = 0.100; ratio 1 : 1 → MgO.
Examiner tip
A complete oxidation answer: oxygen by subtraction, both moles by ÷ Aᵣ, then ÷ smallest. If white MgO smoke escapes during the experiment the recorded oxide mass is too low and the ratio is thrown out.
5Unknown metal oxide from raw masses (Stretch)
Stretch• 4WSD0/1C style — extended• metal-oxide, formula, spec-1.39, data
Question
A metal M (Aᵣ = 56) is heated in oxygen. 1.12 g of M combines with oxygen to form 1.60 g of the oxide. Determine the formula of the oxide of M. (Aᵣ: O = 16.) (4 marks)
Step-by-step solution
1. Step 1
  Mass of oxygen = mass of oxide − mass of M = 1.60 − 1.12 = 0.48 g.
2. Step 2
  Moles of M = 1.12 ÷ 56 = 0.0200 mol.
3. Step 3
  Moles of oxygen = 0.48 ÷ 16 = 0.0300 mol.
4. Step 4
  Divide both by the smaller (0.0200): M = 1.00, O = 1.50. A 1.5 appears — multiply both by 2 → M = 2, O = 3.
5. Step 5
  Ratio M : O = 2 : 3 → formula M₂O₃ (M is iron, so this is Fe₂O₃).
Answer
Oxygen = 0.48 g; n(M) = 0.0200, n(O) = 0.0300; ratio 1 : 1.5 → ×2 → 2 : 3 → M₂O₃.
Examiner tip
When a ratio lands on a 0.5, you MUST multiply up to whole numbers — do not round 1.5 to 2. Students who write 'MO₂' here lose the formula mark by rounding instead of scaling.
6Explain a step in the reduction method (Stretch)
Stretch• 4WSD0/1C style — extended• reduction, method, safety, spec-1.39
Question
In a reduction experiment, copper oxide is heated in a stream of hydrogen. (a) Explain why hydrogen is passed through the apparatus BEFORE the gas is lit. (b) Explain why the experiment is reheated, cooled and reweighed several times. (4 marks)
Step-by-step solution
1. Step 1
  (a) Safety (1 + 1). Hydrogen mixed with air is explosive. Passing hydrogen through first flushes all the air out of the apparatus, so when the excess gas is lit there is no air–hydrogen mixture to explode.
2. Step 2
  (b) Constant mass (1 + 1). Reheating until the mass stops changing shows the reaction is complete — all the oxide has been reduced to copper. If the mass still falls, some oxide remains and the data would be wrong.
Answer
(a) To flush out the air, because hydrogen + air is explosive when lit. (b) To heat to constant mass, proving all the oxide has reacted so the masses used in the calculation are correct.
Examiner tip
These method/safety marks are easy to bank if you use the exact phrases 'flush out the air / hydrogen is explosive with air' and 'heat to constant mass / reaction complete'. Vague answers like 'for safety' score nothing.

Model Answers — Determining the formula of a metal oxide

High-scoring sample answers for determining the formula of a metal oxide on the Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4WSD0/1C style2 marks
In an experiment, 3.20 g of copper is heated in air and forms 4.00 g of copper oxide. Calculate the mass of oxygen that combined with the copper. (2 marks)
Model answer
The oxide is heavier than the metal because oxygen has joined on.

Mass of oxygen = mass of oxide − mass of copper = 4.00 − 3.20 = 0.80 g.
Why this scores
1 mark for the correct subtraction (oxide − metal), 1 mark for the answer 0.80 g. In oxidation the mass gained is the oxygen.
Question 2
4WSD0/1C style2 marks
An experiment gives a metal oxide containing 0.030 mol of magnesium atoms and 0.030 mol of oxygen atoms. State the simplest ratio of Mg : O and write the formula of the oxide. (2 marks)
Model answer
Divide both by the smaller value (0.030): Mg = 1, O = 1.

Ratio Mg : O = 1 : 1.

Formula = MgO.
Why this scores
1 mark for the 1 : 1 ratio, 1 mark for MgO. The simplest whole-number mole ratio is the formula — no balancing needed.
Question 3
4WSD0/1C style4 marks
2.16 g of a copper oxide is reduced in a stream of methane. 1.92 g of copper remains. Determine the formula of the copper oxide. (Aᵣ: Cu = 64, O = 16.) (4 marks)
Model answer
Mass of oxygen (lost) = 2.16 − 1.92 = 0.24 g. (1)

Moles of copper = 1.92 ÷ 64 = 0.0300 mol. (1)

Moles of oxygen = 0.24 ÷ 16 = 0.0150 mol. (1)

Divide both by the smaller (0.0150): Cu = 2, O = 1 → ratio Cu : O = 2 : 1 → formula Cu₂O. (1)
Why this scores
Mass lost = oxygen removed (reduction). The marks are for: oxygen mass, both mole values, and the final formula. Cu : O = 2 : 1 gives Cu₂O (copper(I) oxide), not CuO.
Question 4
4WSD0/1C style4 marks
Describe how a student could find the mass of oxygen that combines when magnesium is burned in a crucible. Include how they would make sure all the magnesium has reacted. (4 marks)
Model answer
Weigh the empty crucible and lid, then add magnesium and reweigh to find the mass of magnesium. (1)

Heat strongly, lifting the lid regularly so oxygen can get in, while keeping the white smoke (magnesium oxide) inside. (1)

Reheat, cool and reweigh until the mass is constant — this shows all the magnesium has reacted. (1)

Mass of oxygen = final mass (oxide) − starting mass (magnesium). (1)
Why this scores
Marks for: weighing to get the metal mass; letting oxygen in without losing oxide; heating to constant mass; and oxygen by subtraction. 'Heat to constant mass' is the phrase examiners look for to award the completeness mark.
Question 5
4WSD0/1C style — extended6 marks
A metal X forms an oxide. In an experiment, 1.40 g of X is fully oxidised to give 2.20 g of the oxide. The relative atomic mass of X is 56. Determine the formula of the oxide and name the type of reaction taking place. (Aᵣ: O = 16.) (6 marks)
Model answer
Step 1 — mass of oxygen. Mass of oxygen = mass of oxide − mass of X = 2.20 − 1.40 = 0.80 g. (1)

Step 2 — moles of each element.

Moles of X = 1.40 ÷ 56 = 0.0250 mol. (1)

Moles of O = 0.80 ÷ 16 = 0.0500 mol. (1)

Step 3 — simplest ratio. Divide both by the smaller (0.0250): X = 1, O = 2 → ratio X : O = 1 : 2. (1)

Step 4 — formula. Formula = XO₂. (1)

Step 5 — type of reaction. The metal gains oxygen, so this is oxidation. (1)
Why this scores
Full working banks the marks even if the final formula slips. Note this ratio is a clean 1 : 2 (XO₂) — but always check whether a 0.5 appears and scale up if it does. 'Oxidation = gain of oxygen' secures the last mark.
Question 6
4WSD0/1C style — extended8 marks
A student reduces 7.95 g of black copper oxide in a stream of hydrogen until only pink-brown copper remains, with a mass of 6.35 g. (a) Write a balanced equation for the reaction. (b) Determine the formula of the copper oxide. (c) Explain why the hydrogen supply is kept flowing while the apparatus cools. (Aᵣ: Cu = 63.5, O = 16.) (8 marks)
Model answer
(a) Equation (2).

CuO + H₂ → Cu + H₂O

(1 for correct formulae, 1 for balancing.)

(b) Formula (4).

Mass of oxygen removed = 7.95 − 6.35 = 1.60 g. (1)

Moles of copper = 6.35 ÷ 63.5 = 0.100 mol. (1)

Moles of oxygen = 1.60 ÷ 16 = 0.100 mol. (1)

Divide both by 0.100: Cu = 1, O = 1 → ratio 1 : 1 → formula CuO. (1)

(c) Cooling under hydrogen (2). The copper is hot and would react with oxygen in the air again (re-oxidise), turning back to copper oxide and giving a mass that is too high. Keeping hydrogen flowing keeps air away until the copper has cooled. (2)
Why this scores
The discriminating part is (c): hot copper re-oxidises in air, which would raise the final mass and ruin the ratio. State that the hydrogen keeps air/oxygen away from the hot copper. The equation must balance — CuO + H₂ → Cu + H₂O is already 1 : 1.

Key Definitions and Keywords — Determining the formula of a metal oxide

Definitions to memorise and the exact keywords mark schemes credit for determining the formula of a metal oxide answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) sitting.

Metal oxide
Examiner keyword
A compound formed when a metal combines with oxygen (e.g. MgO, CuO, Cu₂O, Fe₂O₃).
Oxidation
Examiner keyword
A reaction in which a substance gains oxygen. Heating a metal in air/oxygen to form its oxide is an oxidation.
Reduction
Examiner keyword
A reaction in which a substance loses oxygen. Removing oxygen from a metal oxide (e.g. with hydrogen or carbon) is a reduction.
Reducing agent
Examiner keyword
A substance that removes oxygen from another substance. Hydrogen, methane and carbon are used to reduce copper oxide.
Relative atomic mass (Aᵣ)
Examiner keyword
The average mass of an atom of an element compared with 1/12 the mass of a carbon-12 atom. Used to convert mass to moles.
Mole
The amount of substance that contains 6.02 × 10²³ particles. moles = mass ÷ Aᵣ for an element.
Simplest (empirical) ratio
Examiner keyword
The smallest whole-number ratio of atoms of each element in a compound; for a metal oxide this ratio is its formula.
Heating to constant mass
Examiner keyword
Repeatedly heating, cooling and reweighing until the mass stops changing — evidence that the reaction is complete.
Mass gained (oxidation)
The increase in mass when a metal is oxidised; it equals the mass of oxygen that combined with the metal.
Mass lost (reduction)
The decrease in mass when a metal oxide is reduced; it equals the mass of oxygen removed from the oxide.

Common Mistakes and Misconceptions — Determining the formula of a metal oxide

The traps other students keep falling into on determining the formula of a metal oxide questions — taken from recent Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) examiner reports and mark schemes — and how to avoid them.

✕Subtracting the masses the wrong way round to find the oxygen
Edexcel Chemistry examiner reports — recurring error
Why it happens
Students forget whether the solid gained or lost mass in their route.
How to avoid it
Oxidation: oxygen = oxide − metal (mass GAINED). Reduction: oxygen = oxide − metal left (mass LOST). The oxide is always the bigger mass, so oxygen is positive.
✕Putting the raw masses straight into a ratio without converting to moles
Edexcel Chemistry examiner reports
Why it happens
Students think the mass ratio is the formula.
How to avoid it
ALWAYS divide each mass by its Aᵣ first. A formula is a ratio of MOLES (atoms), not of masses.
✕Rounding a 0.5 in the ratio instead of multiplying up (e.g. writing MO₂ when the ratio is 1 : 1.5)
Edexcel Chemistry examiner reports
Why it happens
Students round 1.5 to the nearest whole number.
How to avoid it
If a ratio contains a 0.5, multiply EVERY number by 2 to clear it (1 : 1.5 → 2 : 3 → M₂O₃). Never round a half away.
✕Writing the subscripts the wrong way round (e.g. CuO₂ instead of Cu₂O)
Edexcel Chemistry examiner reports
Why it happens
Students attach the larger number to the wrong element.
How to avoid it
The element with MORE moles gets the larger subscript. Cu : O = 2 : 1 means two coppers per oxygen → Cu₂O.
✕Not heating to constant mass, so the reaction is incomplete
Edexcel Chemistry examiner reports
Why it happens
Students stop heating once it 'looks done'.
How to avoid it
Reheat, cool and reweigh until two readings agree. Unreacted oxide or metal makes the masses — and the ratio — wrong.
✕Forgetting to flush air out before lighting hydrogen (and keeping gas flowing while cooling)
Edexcel Chemistry examiner reports
Why it happens
Safety steps feel separate from the calculation, so they are skipped.
How to avoid it
State that hydrogen + air is explosive, so flush the air out FIRST; keep hydrogen flowing during cooling so hot copper does not re-oxidise.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Determining the formula of a metal oxide

Detailed Notes

What you’ll learn

How it’s examined

1Find the mass of oxygen from the mass change (Getting started)

2State the formula from a given mole ratio (Getting started)

3Reduction of a copper oxide — find the formula (Building confidence)

4Oxidation of magnesium — find the formula (Building confidence)

5Unknown metal oxide from raw masses (Stretch)

6Explain a step in the reduction method (Stretch)

Metal oxide

Oxidation

Reduction

Reducing agent

Relative atomic mass (Aᵣ)

Mole

Simplest (empirical) ratio

Heating to constant mass

Mass gained (oxidation)

Mass lost (reduction)

✕Subtracting the masses the wrong way round to find the oxygen

✕Putting the raw masses straight into a ratio without converting to moles

✕Rounding a 0.5 in the ratio instead of multiplying up (e.g. writing MO₂ when the ratio is 1 : 1.5)

✕Writing the subscripts the wrong way round (e.g. CuO₂ instead of Cu₂O)

✕Not heating to constant mass, so the reaction is incomplete

✕Forgetting to flush air out before lighting hydrogen (and keeping gas flowing while cooling)

Past paper style quiz