Why is "Rounding a ratio like 1.5 or 2.5 to the nearest whole number" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Students treat the .5 as a rounding error rather than a signal to scale. How to avoid it: A value ending in .5 means **multiply everything by 2**; .33/.67 means ×3. Only round values very close to whole (e.g. 1.99 → 2). Source: Edexcel Chemistry examiner reports — recurring error

Why is "Dividing the percentages by 100 before finding moles" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Students think a percentage must be converted to a decimal fraction first. How to avoid it: Treat a percentage as a **mass in 100 g** of compound — 40% C is simply 40 g of C. Divide that by Ar directly. Source: Edexcel Chemistry examiner reports

Why is "Multiplying mass by Ar instead of dividing" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Confusion over the direction of the moles formula n = m ÷ Ar. How to avoid it: To get **moles**, always **divide mass by Ar**. Bigger atoms (large Ar) give fewer moles per gram. Source: Edexcel Chemistry examiner reports

Why is "Multiplying only one subscript by n when finding the molecular formula" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Students apply the multiplier to the first element and forget the rest. How to avoid it: Multiply **every** subscript in the empirical formula by n. Then check: Mr of your molecular formula must equal the given Mr. Source: Edexcel Chemistry examiner reports

Why is "Using the whole mass of CO₂ or H₂O as the mass of carbon or hydrogen" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Students forget that CO₂ and H₂O also contain oxygen. How to avoid it: Only part of each came from the compound: mass C = mass CO₂ × **12/44**; mass H = mass H₂O × **2/18**. Source: Edexcel Chemistry examiner reports

Why is "Giving the empirical formula when the question asks for the molecular formula (or vice versa)" a common mistake in Empirical & Molecular Formulae and calculation?

Why it happens: Students stop at the simplest ratio without reading the final command word. How to avoid it: If an Mr is given and the word is 'molecular', do the extra n = Mr(molecular) ÷ Mr(empirical) step. If no Mr, the answer is the empirical formula. Source: Edexcel Chemistry examiner reports

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

Empirical & Molecular Formulae and calculation

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Empirical & Molecular Formulae and calculation — Edexcel International GCSE Science (Double Award) Chemistry (4WSD0-1C) Study Notes

Two formulae, one method. The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of atoms in a molecule. Turn masses or percentages into moles (÷ Ar), divide by the smallest, scale up if needed — then use Mr to find the molecular formula. Covers Double Award Chemistry spec 1.39–1.40, assessed on Chemistry Paper 1 (4WSD0/1C).

What you’ll learn

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

1.39 — Calculate the empirical formula of a compound from masses or percentage composition.
1.40 — Deduce the molecular formula of a compound from its empirical formula and relative formula mass.

Empirical vs molecular formula — the difference (spec 1.39–1.40)

Empirical = simplest ratio; molecular = actual atom count. Know which one a question wants.

Every compound has an empirical formula and a molecular formula — and they are not always the same.

The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound.
The molecular formula is the actual number of atoms of each element in one molecule.

The molecular formula is always the empirical formula multiplied by a whole number (1, 2, 3, …).

Glucose: C₆H₁₂O₆ (molecular) divides down to CH₂O (empirical). Both describe the same compound.

For ionic compounds (e.g. NaCl, MgCl₂) the formula given is the empirical formula — there are no separate molecules.
For many molecular compounds the two differ: ethane is C₂H₆ (molecular) but CH₃ (empirical); benzene is C₆H₆ (molecular) but CH (empirical).

Read the command word. If the question gives masses or percentages and asks for "the empirical formula", stop at the simplest ratio. If it also gives an Mr and asks for "the molecular formula", you must do the extra step.

Empirical = simplest whole-number ratio of atoms.
Molecular = actual number of atoms in a molecule.
Molecular is always a whole-number multiple of empirical.
Glucose: molecular C₆H₁₂O₆, empirical CH₂O.

See the full worked example for empirical & molecular formulae and calculation →

Finding the empirical formula from masses or % (spec 1.39)

Same four-step routine every time: ÷ Ar, ÷ smallest, scale up, write the ratio.

Whether you are given masses or percentages, the routine is identical. (A percentage is just a mass in 100 g of compound: 53.3% O means 53.3 g of O.)

The four steps.

Write the mass (or %) of each element under its symbol.
Divide each by its Ar → this gives the number of moles of each element.
Divide every value by the smallest of them → this gives the simplest ratio.
If any value is not a whole number (e.g. 1.5, 2.5, 1.33), multiply them all up until they are whole.

Every empirical-formula question follows the same chain — set it out in columns and you cannot get lost.

Worked example (from percentages). A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Find the empirical formula.

	C	H	O
mass in 100 g	40.0	6.7	53.3
÷ Ar	40.0 ÷ 12 = 3.33	6.7 ÷ 1 = 6.7	53.3 ÷ 16 = 3.33
÷ smallest (3.33)	1	2.01 ≈ 2	1

Ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.

Tip — laying it out. Always keep the three numbers in line under their symbols. The smallest of the moles is the one you divide by; here it is 3.33 (carbon and oxygen tie). The hydrogen value 6.7 ÷ 3.33 ≈ 2.0, which is fine to round.

% → treat as grams in 100 g of compound.
Step 2: ÷ Ar to get moles of each element.
Step 3: ÷ the smallest mole value.
Step 4: scale up any 0.5 / 0.33 values to whole numbers.

See the full worked example for empirical & molecular formulae and calculation →

When the ratio is not whole — scaling up (spec 1.39)

1 : 1.5 → × 2; 1 : 1.33 → × 3. Never round a clear half or third.

Step 3 does not always give whole numbers. Never round 1.5 down to 1 or up to 2 — that changes the compound. Instead, multiply all the numbers in the ratio by a small whole number until they become whole.

A value ending in .5 → multiply everything by 2.
A value ending in .33 or .67 → multiply everything by 3.
A value ending in .25 or .75 → multiply everything by 4.

Worked example. A compound contains 27.3% C and 72.7% O by mass. Find the empirical formula.

	C	O
mass in 100 g	27.3	72.7
÷ Ar	27.3 ÷ 12 = 2.275	72.7 ÷ 16 = 4.544
÷ smallest (2.275)	1	1.997 ≈ 2

Ratio C : O = 1 : 2 → empirical formula CO₂.

Worked example needing × 2. A compound is 69.6% Ba and 30.4% Cl by mass (Ar: Ba = 137, Cl = 35.5).

	Ba	Cl
÷ Ar	69.6 ÷ 137 = 0.508	30.4 ÷ 35.5 = 0.856
÷ smallest (0.508)	1	1.685 ≈ 1.5

The 1.5 is a clear half, so multiply both by 2: Ba : Cl = 2 : 3 → empirical formula Ba₂Cl₃. (Always double-check by scaling — rounding 1.685 to 2 would have given the wrong formula.)

The judgement call. Examiners allow small rounding (e.g. 1.99 → 2, 3.02 → 3) because Ar values and data carry rounding. But a value sitting near .5 or .33 is meant to be scaled, not rounded.

.5 → multiply all by 2; .33/.67 → multiply all by 3.
.25/.75 → multiply all by 4.
Round 1.99 → 2 (ok); never round 1.5 → 2.
Scaling keeps the ratio — it does not change the compound.

See the full worked example for empirical & molecular formulae and calculation →

From empirical to molecular formula (spec 1.40)

n = Mr(molecular) ÷ Mr(empirical), then multiply every subscript by n.

Once you have the empirical formula, you need one extra piece of data — the relative formula mass (Mr) of the actual molecule — to find the molecular formula.

The three steps.

Work out Mr of the empirical formula (add up the Ar values of its atoms).
Find the multiplier: $n = \dfrac{M_r(\text{molecular})}{M_r(\text{empirical})}$
Multiply every subscript in the empirical formula by n.

Multiplier n = Mr(molecular) ÷ Mr(empirical). Here 180 ÷ 30 = 6, so CH₂O × 6 = C₆H₁₂O₆.

Worked example. A compound has empirical formula CH₂O and an Mr of 180. Find its molecular formula.

Mr(CH₂O) = 12 + (2 × 1) + 16 = 30.
n = 180 ÷ 30 = 6.
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (glucose).

Check it. Mr(C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 ✓ — equal to the given Mr, so the answer is right.

Watch out for n = 1. If Mr(molecular) = Mr(empirical), then n = 1 and the molecular formula is the empirical formula. That is a valid answer, not a mistake.

Step 1: Mr of the empirical formula.
Step 2: n = Mr(molecular) ÷ Mr(empirical).
Step 3: multiply every subscript by n.
Check: Mr of your molecular formula must equal the given Mr.

See the full worked example for empirical & molecular formulae and calculation →

Working back from combustion data (spec 1.39)

Mass of CO₂ → mass of C; mass of H₂O → mass of H; then the usual empirical routine.

A favourite stretch question gives you the masses of carbon dioxide and water produced when a compound burns, and asks for the empirical formula. The trick is that all the carbon ends up in the CO₂ and all the hydrogen ends up in the H₂O.

The extra first steps.

Mass of C = mass of CO₂ × (12 ÷ 44). (Because Mr(CO₂) = 44 and 12 of that is carbon.)
Mass of H = mass of H₂O × (2 ÷ 18). (Because Mr(H₂O) = 18 and 2 of that is hydrogen.)

Then carry on with the normal empirical routine (÷ Ar → ÷ smallest → scale up).

Worked example. Burning a hydrocarbon produces 8.8 g of CO₂ and 5.4 g of H₂O. Find its empirical formula.

Mass of C = 8.8 × (12 ÷ 44) = 8.8 × 0.2727 = 2.4 g.
Mass of H = 5.4 × (2 ÷ 18) = 5.4 × 0.1111 = 0.6 g.

Now the usual routine:

	C	H
mass (g)	2.4	0.6
÷ Ar	2.4 ÷ 12 = 0.20	0.6 ÷ 1 = 0.60
÷ smallest (0.20)	1	3

Ratio C : H = 1 : 3 → empirical formula CH₃.

A common trap. Do not use the whole mass of CO₂ or H₂O as if it were the mass of carbon or hydrogen — only the C part of the CO₂, and only the H part of the H₂O, came from the original compound.

All C ends up in CO₂; all H ends up in H₂O.
Mass C = mass CO₂ × 12/44.
Mass H = mass H₂O × 2/18.
Then use the normal ÷ Ar → ÷ smallest routine.

See the full worked example for empirical & molecular formulae and calculation →

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). Empirical and molecular formulae appear as a 4–6 mark calculation: deduce an empirical formula from given masses or percentages, then deduce the molecular formula using a supplied Mr. Method marks are awarded line by line (÷ Ar, ÷ smallest, correct ratio), so always show every step in columns — even a wrong final formula can score most of the marks if the working is clear.

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 — Empirical & Molecular Formulae and calculation

Step-by-step solutions to past-paper-style questions on empirical & molecular formulae and calculation, written exactly the way a tutor would explain them at the board.

1Classify empirical vs molecular formulae (Getting started)
Getting started• 4WSD0/1C style — short answer• empirical, molecular, spec-1.40
Question
Ethane has the molecular formula C₂H₆. (a) What is meant by the term empirical formula? (b) State the empirical formula of ethane. (2 marks)
Step-by-step solution
1. Step 1
  (a) The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound. (1)
2. Step 2
  (b) C₂H₆ has a C : H ratio of 2 : 6. Divide both by 2 → 1 : 3, so the empirical formula is CH₃. (1)
Answer
(a) the simplest whole-number ratio of atoms of each element; (b) CH₃.
Examiner tip
The definition mark needs the words 'simplest whole-number ratio'. Just saying 'the ratio of atoms' is not enough — it must be the simplest one.
2Empirical formula from a mass ratio (Getting started)
Getting started• 4WSD0/1C style — structured• empirical, spec-1.39
Question
3.0 g of magnesium combines with 2.0 g of oxygen to form an oxide. Find the empirical formula of the oxide. (Ar: Mg = 24, O = 16) (3 marks)
Step-by-step solution
1. Step 1
  Moles (1). n(Mg) = 3.0 ÷ 24 = 0.125 mol; n(O) = 2.0 ÷ 16 = 0.125 mol.
2. Step 2
  Divide by smallest (1). Both are 0.125, so divide both by 0.125 → Mg : O = 1 : 1.
3. Step 3
  Formula (1). The ratio is already whole, so the empirical formula is MgO.
Answer
MgO.
Examiner tip
Set the working out in columns. The mark scheme gives a mark for the moles, a mark for dividing by the smallest, and a mark for the final formula — show every line.
3Empirical formula from percentages (Building confidence)
Building confidence• 4WSD0/1C style — structured• empirical, percentage, spec-1.39
Question
An oxide of sodium contains 74.2% sodium and 25.8% oxygen by mass. Find its empirical formula. (Ar: Na = 23, O = 16) (3 marks)
Step-by-step solution
1. Step 1
  Treat % as mass in 100 g (1). 74.2 g Na and 25.8 g O.
2. Step 2
  Moles. n(Na) = 74.2 ÷ 23 = 3.23 mol; n(O) = 25.8 ÷ 16 = 1.61 mol.
3. Step 3
  Divide by smallest (1.61) (1). Na = 3.23 ÷ 1.61 = 2.0; O = 1.61 ÷ 1.61 = 1.
4. Step 4
  Formula (1). Ratio Na : O = 2 : 1 → empirical formula Na₂O.
Answer
Na₂O.
Examiner tip
A percentage is a mass in 100 g — never divide the percentage by 100 first. The 3.23 ÷ 1.61 = 2.0 is the step students rush; keep two decimal places until the final ratio.
4Empirical formula needing ×2 scaling (Building confidence)
Building confidence• 4WSD0/1C style — structured• empirical, scaling, spec-1.39
Question
An oxide of iron contains 70.0% iron and 30.0% oxygen by mass. Find its empirical formula. (Ar: Fe = 56, O = 16) (4 marks)
Step-by-step solution
1. Step 1
  Masses (1). In 100 g: 70.0 g Fe and 30.0 g O.
2. Step 2
  Moles (1). n(Fe) = 70.0 ÷ 56 = 1.25 mol; n(O) = 30.0 ÷ 16 = 1.875 mol.
3. Step 3
  Divide by smallest (1.25) (1). Fe = 1.25 ÷ 1.25 = 1; O = 1.875 ÷ 1.25 = 1.5.
4. Step 4
  Scale up (1). 1.5 is a clear half, so multiply both by 2 → Fe : O = 2 : 3 → empirical formula Fe₂O₃.
Answer
Fe₂O₃.
Examiner tip
The 1.5 must be scaled (×2), not rounded to 2. Rounding gives FeO₂, which does not exist — this is the single most common error in this question type.
5Molecular formula from empirical formula and Mr (Stretch)
Stretch• 4WSD0/1C style — extended• molecular, Mr, spec-1.40
Question
A hydrocarbon has the empirical formula CH₂ and a relative formula mass of 56. Deduce its molecular formula. (Ar: C = 12, H = 1) (3 marks)
Step-by-step solution
1. Step 1
  Mr of empirical formula (1). Mr(CH₂) = 12 + (2 × 1) = 14.
2. Step 2
  Find the multiplier (1). n = Mr(molecular) ÷ Mr(empirical) = 56 ÷ 14 = 4.
3. Step 3
  Multiply every subscript (1). (CH₂) × 4 = C₄H₈.
Answer
C₄H₈.
Examiner tip
Multiply BOTH subscripts by n — a frequent slip is to multiply only the carbon, giving C₄H₂. Check: Mr(C₄H₈) = 48 + 8 = 56, matching the data.
6Empirical formula from combustion data (Stretch)
Stretch• 4WSD0/1C style — extended• empirical, combustion, spec-1.39
Question
A hydrocarbon burns completely to produce 17.6 g of carbon dioxide and 10.8 g of water. Find the empirical formula of the hydrocarbon. (Ar: C = 12, H = 1, O = 16) (5 marks)
Step-by-step solution
1. Step 1
  Mass of C (1). All the C is in the CO₂. Mr(CO₂) = 44, of which 12 is C. Mass C = 17.6 × (12 ÷ 44) = 4.8 g.
2. Step 2
  Mass of H (1). All the H is in the H₂O. Mr(H₂O) = 18, of which 2 is H. Mass H = 10.8 × (2 ÷ 18) = 1.2 g.
3. Step 3
  Moles (1). n(C) = 4.8 ÷ 12 = 0.40 mol; n(H) = 1.2 ÷ 1 = 1.2 mol.
4. Step 4
  Divide by smallest (0.40) (1). C = 0.40 ÷ 0.40 = 1; H = 1.2 ÷ 0.40 = 3.
5. Step 5
  Formula (1). Ratio C : H = 1 : 3 → empirical formula CH₃.
Answer
CH₃.
Examiner tip
Only the carbon part of the CO₂ (× 12/44) and the hydrogen part of the H₂O (× 2/18) came from the compound. Using the whole 17.6 g or 10.8 g as the element mass is a guaranteed lost mark.

Model Answers — Empirical & Molecular Formulae and calculation

High-scoring sample answers for empirical & molecular formulae and calculation 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
Define (a) empirical formula and (b) molecular formula. (2 marks)
Model answer
(a) Empirical formula — the simplest whole-number ratio of the atoms of each element in a compound.

(b) Molecular formula — the actual number of atoms of each element in one molecule of the compound.
Why this scores
1 mark each. 'Simplest whole-number ratio' and 'actual number of atoms' are the keyword phrases the mark scheme looks for.
Question 2
4WSD0/1C style3 marks
1.2 g of carbon reacts with 0.4 g of hydrogen. Find the empirical formula of the compound formed. (Ar: C = 12, H = 1) (3 marks)
Model answer
Moles: n(C) = 1.2 ÷ 12 = 0.10 mol; n(H) = 0.4 ÷ 1 = 0.40 mol.

Divide by smallest (0.10): C = 0.10 ÷ 0.10 = 1; H = 0.40 ÷ 0.10 = 4.

Ratio C : H = 1 : 4 → empirical formula CH₄.
Why this scores
Marks for: moles of each element, dividing by the smallest, and the final formula. Lay the numbers out in columns so each step is clear.
Question 3
4WSD0/1C style4 marks
A compound contains 40.0% calcium, 12.0% carbon and 48.0% oxygen by mass. Find its empirical formula. (Ar: Ca = 40, C = 12, O = 16) (4 marks)
Model answer
Treat the percentages as masses in 100 g:

Moles: n(Ca) = 40.0 ÷ 40 = 1.0; n(C) = 12.0 ÷ 12 = 1.0; n(O) = 48.0 ÷ 16 = 3.0.

Divide by smallest (1.0): Ca = 1; C = 1; O = 3.

Ratio Ca : C : O = 1 : 1 : 3.

Empirical formula = CaCO₃ (calcium carbonate).
Why this scores
1 mark per correct mole value (Ca, C, O), 1 mark for the formula. A percentage is a mass in 100 g — do not divide the percentages by 100.
Question 4
4WSD0/1C style4 marks
An oxide of phosphorus contains 43.7% phosphorus and 56.3% oxygen by mass. Find its empirical formula. (Ar: P = 31, O = 16) (4 marks)
Model answer
In 100 g: 43.7 g P and 56.3 g O.

Moles: n(P) = 43.7 ÷ 31 = 1.41; n(O) = 56.3 ÷ 16 = 3.52.

Divide by smallest (1.41): P = 1; O = 3.52 ÷ 1.41 = 2.5.

Scale up: 2.5 is a clear half, so multiply both by 2 → P : O = 2 : 5.

Empirical formula = P₂O₅.
Why this scores
The discriminator is the 2.5 → ×2 step. Students who round 2.5 to 2 or 3 get PO₂ or PO₃ and lose the scaling mark. Always scale a clear half, never round it.
Question 5
4WSD0/1C style — extended6 marks
A compound contains 1.2 g of carbon, 0.2 g of hydrogen and 1.6 g of oxygen. Its relative formula mass is 60. (a) Find the empirical formula. (b) Deduce the molecular formula. (Ar: C = 12, H = 1, O = 16) (6 marks)
Model answer
(a) Empirical formula.

Moles: n(C) = 1.2 ÷ 12 = 0.10; n(H) = 0.2 ÷ 1 = 0.20; n(O) = 1.6 ÷ 16 = 0.10. (1)

Divide by smallest (0.10): C = 1; H = 2; O = 1. (1)

Empirical formula = CH₂O. (1)

(b) Molecular formula.

Mr(CH₂O) = 12 + (2 × 1) + 16 = 30. (1)

Multiplier n = 60 ÷ 30 = 2. (1)

Molecular formula = (CH₂O) × 2 = C₂H₄O₂ (ethanoic acid). (1)
Why this scores
Part (a) and (b) are marked separately, so even a wrong empirical formula can score the (b) method marks if the multiplier is applied correctly. Multiply EVERY subscript by n. Check: Mr(C₂H₄O₂) = 24 + 4 + 32 = 60. ✓
Question 6
4WSD0/1C style — extended7 marks
A hydrocarbon burns completely to give 22.0 g of carbon dioxide and 4.5 g of water. The relative formula mass of the hydrocarbon is 78. (a) Find the empirical formula. (b) Deduce the molecular formula. (Ar: C = 12, H = 1, O = 16) (7 marks)
Model answer
(a) Empirical formula.

Mass of C = 22.0 × (12 ÷ 44) = 6.0 g (all the C comes from the CO₂). (1)

Mass of H = 4.5 × (2 ÷ 18) = 0.5 g (all the H comes from the H₂O). (1)

Moles: n(C) = 6.0 ÷ 12 = 0.50; n(H) = 0.5 ÷ 1 = 0.50. (1)

Divide by smallest (0.50): C = 1; H = 1 → ratio C : H = 1 : 1, empirical formula CH. (1)

(b) Molecular formula.

Mr(CH) = 12 + 1 = 13. (1)

Multiplier n = 78 ÷ 13 = 6. (1)

Molecular formula = (CH) × 6 = C₆H₆ (benzene). (1)
Why this scores
Only the carbon in the CO₂ (× 12/44) and the hydrogen in the H₂O (× 2/18) came from the compound. Check: Mr(C₆H₆) = 72 + 6 = 78. ✓ Empirical and molecular can differ by a large multiple — here n = 6.

Key Definitions and Keywords — Empirical & Molecular Formulae and calculation

Definitions to memorise and the exact keywords mark schemes credit for empirical & molecular formulae and calculation answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) sitting.

Empirical formula
Examiner keyword
The simplest whole-number ratio of the atoms of each element present in a compound, e.g. CH₂O for glucose.
Molecular formula
Examiner keyword
The actual number of atoms of each element in one molecule of a compound, e.g. C₆H₁₂O₆ for glucose.
Relative atomic mass (Ar)
The average mass of an atom of an element compared with 1/12 of the mass of a carbon-12 atom; used to convert mass to moles.
Relative formula mass (Mr)
Examiner keyword
The sum of the relative atomic masses of all the atoms shown in a chemical formula.
Mole
Examiner keyword
The amount of a substance that contains 6.02 × 10²³ particles; the mass of one mole in grams is numerically equal to the Mr.
Percentage composition (by mass)
Examiner keyword
The percentage of the total mass of a compound made up by each element; treated as grams in 100 g of compound.
Mole ratio
The ratio of the number of moles of each element, obtained by dividing each mole value by the smallest.
Scaling up
Examiner keyword
Multiplying every number in a ratio by a small whole number (×2, ×3, ×4) so that a value such as 1.5 or 1.33 becomes a whole number.
Multiplier (n)
Examiner keyword
The whole number found from Mr(molecular) ÷ Mr(empirical); every subscript in the empirical formula is multiplied by it to give the molecular formula.
Combustion analysis
Finding an empirical formula from the masses of CO₂ and H₂O made when a compound burns; mass C = mass CO₂ × 12/44 and mass H = mass H₂O × 2/18.

Common Mistakes and Misconceptions — Empirical & Molecular Formulae and calculation

The traps other students keep falling into on empirical & molecular formulae and calculation questions — taken from recent Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) examiner reports and mark schemes — and how to avoid them.

✕Rounding a ratio like 1.5 or 2.5 to the nearest whole number
Edexcel Chemistry examiner reports — recurring error
Why it happens
Students treat the .5 as a rounding error rather than a signal to scale.
How to avoid it
A value ending in .5 means multiply everything by 2; .33/.67 means ×3. Only round values very close to whole (e.g. 1.99 → 2).
✕Dividing the percentages by 100 before finding moles
Edexcel Chemistry examiner reports
Why it happens
Students think a percentage must be converted to a decimal fraction first.
How to avoid it
Treat a percentage as a mass in 100 g of compound — 40% C is simply 40 g of C. Divide that by Ar directly.
✕Multiplying mass by Ar instead of dividing
Edexcel Chemistry examiner reports
Why it happens
Confusion over the direction of the moles formula n = m ÷ Ar.
How to avoid it
To get moles, always divide mass by Ar. Bigger atoms (large Ar) give fewer moles per gram.
✕Multiplying only one subscript by n when finding the molecular formula
Edexcel Chemistry examiner reports
Why it happens
Students apply the multiplier to the first element and forget the rest.
How to avoid it
Multiply every subscript in the empirical formula by n. Then check: Mr of your molecular formula must equal the given Mr.
✕Using the whole mass of CO₂ or H₂O as the mass of carbon or hydrogen
Edexcel Chemistry examiner reports
Why it happens
Students forget that CO₂ and H₂O also contain oxygen.
How to avoid it
Only part of each came from the compound: mass C = mass CO₂ × 12/44; mass H = mass H₂O × 2/18.
✕Giving the empirical formula when the question asks for the molecular formula (or vice versa)
Edexcel Chemistry examiner reports
Why it happens
Students stop at the simplest ratio without reading the final command word.
How to avoid it
If an Mr is given and the word is 'molecular', do the extra n = Mr(molecular) ÷ Mr(empirical) step. If no Mr, the answer is the empirical formula.

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Empirical & Molecular Formulae and calculation

Detailed Notes

What you’ll learn

How it’s examined

1Classify empirical vs molecular formulae (Getting started)

2Empirical formula from a mass ratio (Getting started)

3Empirical formula from percentages (Building confidence)

4Empirical formula needing ×2 scaling (Building confidence)

5Molecular formula from empirical formula and Mr (Stretch)

6Empirical formula from combustion data (Stretch)

Empirical formula

Molecular formula

Relative atomic mass (Ar)

Relative formula mass (Mr)

Mole

Percentage composition (by mass)

Mole ratio

Scaling up

Multiplier (n)

Combustion analysis

✕Rounding a ratio like 1.5 or 2.5 to the nearest whole number

✕Dividing the percentages by 100 before finding moles

✕Multiplying mass by Ar instead of dividing

✕Multiplying only one subscript by n when finding the molecular formula

✕Using the whole mass of CO₂ or H₂O as the mass of carbon or hydrogen

✕Giving the empirical formula when the question asks for the molecular formula (or vice versa)

Past paper style quiz