What is a chemical formula and why is it important in Chemistry?

A chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule. It is important because it provides essential information about the types and numbers of atoms involved, which is crucial for understanding chemical reactions and properties. In the context of Cambridge International A Levels, mastering chemical formulae is fundamental for solving stoichiometry problems and predicting reaction outcomes.

How do you determine the empirical formula of a compound?

To determine the empirical formula, you need to find the simplest whole-number ratio of atoms in a compound. This involves converting the mass of each element to moles, dividing each by the smallest number of moles calculated, and adjusting to the nearest whole number if necessary. Understanding empirical formulae is essential for Cambridge International A Levels students as it lays the groundwork for more complex stoichiometric calculations.

What is the difference between an empirical formula and a molecular formula?

The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of each type of atom in a molecule. For example, the empirical formula of glucose is CH2O, whereas its molecular formula is C6H12O6. Recognizing this distinction is crucial for students studying Chemistry at the Cambridge International A Levels, as it affects how substances are represented and understood in chemical equations.

How can you calculate the molecular formula from the empirical formula?

To calculate the molecular formula from the empirical formula, you need the molar mass of the compound. Divide the molar mass by the empirical formula mass to find a multiplier, which is then used to convert the empirical formula to the molecular formula. This process is a key skill for Cambridge International A Levels students, as it is often required in stoichiometry problems and chemical analysis.

Why is balancing chemical equations important in stoichiometry?

Balancing chemical equations is crucial because it ensures that the law of conservation of mass is obeyed, meaning the same number of each type of atom is present on both sides of the equation. This is essential for accurately calculating reactants and products in stoichiometry. For Cambridge International A Levels students, mastering this skill is fundamental for success in both theoretical and practical chemistry assessments.

Why is "Changing a polyatomic ion's formula to balance charges." a common mistake in Formulae?

Why it happens: Treating SO₄ like separate S and O. How to avoid it: Keep the ion intact and use brackets: (SO₄)₃ etc. Source: 9701 Examiner Reports

Why is "Balancing an equation by changing formulae (subscripts)." a common mistake in Formulae?

Why it happens: Trying to force the atoms to match. How to avoid it: Only change the big numbers in front (coefficients); never alter a correct formula like H₂O to H₂O₂.

Why is "Omitting state symbols when asked for them." a common mistake in Formulae?

Why it happens: Rushing. How to avoid it: Add (s), (l), (g), (aq) to every species when required.

Why is "Ionic equation where the charges don't balance." a common mistake in Formulae?

Why it happens: Only checking atoms. How to avoid it: Check total charge is equal on both sides as well as atoms.

Why is "Rounding a 1 : 1.5 empirical ratio to 1 : 2." a common mistake in Formulae?

Why it happens: Wanting whole numbers immediately. How to avoid it: Multiply the whole ratio to clear fractions (×2, ×3).

Why is "Giving the empirical formula when the molecular formula is asked." a common mistake in Formulae?

Why it happens: Stopping after the ratio step. How to avoid it: If Mr is given, finish: divide Mr by the empirical formula mass and scale up. Source: 9701 Examiner Reports 2022-2024

Atoms, molecules and stoichiometry(AS-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

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Formulae, Equations and Empirical Formulae — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Build formulae of ionic compounds by balancing charges, write balanced full and ionic equations with state symbols, and calculate empirical and molecular formulae — including water of crystallisation.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

2.3.1 — Write formulae of ionic compounds from ionic charges and oxidation numbers, including predicting ionic charge from Periodic Table position and recalling the listed ions.
2.3.2 — Write and construct balanced equations, including ionic equations (no spectator ions) and using state symbols.
2.3.3 — Define and use the terms empirical and molecular formula.
2.3.4 — Understand and use the terms anhydrous, hydrated and water of crystallisation.
2.3.5 — Calculate empirical and molecular formulae using given data.

Writing formulae of ionic compounds

Find each ion's charge, then combine so the positives and negatives cancel.

Step 1 — predict the ion charge from the Group:

Group	1	2	13	15	16	17
Charge	$+1$	$+2$	$+3$	$-3$	$-2$	$-1$

Transition metals have variable charges, shown by a Roman numeral: iron(II) = $\text{Fe}^{2+}$ , iron(III) = $\text{Fe}^{3+}$ , copper(II) = $\text{Cu}^{2+}$ .

Step 2 — recall the common polyatomic and other ions (you must memorise these):

Ion	Formula	Ion	Formula
nitrate	$\text{NO}_3^-$	hydroxide	$\text{OH}^-$
carbonate	$\text{CO}_3^{2-}$	hydrogencarbonate	$\text{HCO}_3^-$
sulfate	$\text{SO}_4^{2-}$	phosphate	$\text{PO}_4^{3-}$
ammonium	$\text{NH}_4^+$	zinc	$\text{Zn}^{2+}$
silver	$\text{Ag}^+$

Step 3 — balance the charges so the compound is neutral. Use brackets for more than one polyatomic ion.

Worked. Aluminium sulfate: $\text{Al}^{3+}$ and $\text{SO}_4^{2-}$ . Cross the charges → two $\text{Al}^{3+}$ ( $+6$ ) balance three $\text{SO}_4^{2-}$ ( $-6$ ): $\text{Al}_2(\text{SO}_4)_3$

Worked. Ammonium phosphate: $\text{NH}_4^+$ and $\text{PO}_4^{3-}$ → $(\text{NH}_4)_3\text{PO}_4$ .

Charge from Group; transition metals use Roman numerals.
Combine so total + = total −.
Brackets around a polyatomic ion when more than one is needed.

Balanced and ionic equations

Balance atoms (and charge) on both sides; for ionic equations cancel the spectator ions.

Balanced symbol equation. The number of atoms of each element must be equal on both sides; only change the big numbers (coefficients), never the formulae.

$\text{2Na} + \text{2H}_2\text{O} \rightarrow \text{2NaOH} + \text{H}_2$

Add state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous.

Ionic equations show only the species that actually change. Spectator ions (present but unchanged on both sides) are cancelled.

Worked — neutralisation. Full: $\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ .

Spectators: $\text{Na}^+$ and $\text{Cl}^-$ .
Ionic equation: $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$ .

Worked — precipitation. $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$ .

Check both atoms AND charge balance in ionic equations: the total charge must be equal on each side.

Balance by changing coefficients only.
Include state symbols (s, l, g, aq).
Ionic equation: cancel spectator ions; balance atoms AND charge.

Empirical and molecular formulae; hydrates

Empirical = simplest ratio; molecular = real numbers. Hydrated salts carry water of crystallisation.

Empirical formula — the simplest whole-number ratio of atoms of each element in a compound.

Molecular formula — the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula.

Examples.

Ethane $\text{C}_2\text{H}_6$ : empirical $\text{CH}_3$ , molecular $\text{C}_2\text{H}_6$ .
Glucose $\text{C}_6\text{H}_{12}\text{O}_6$ : empirical $\text{CH}_2\text{O}$ .
Water $\text{H}_2\text{O}$ : empirical = molecular.

Hydration terms.

Hydrated — a salt whose crystals contain water of crystallisation (water chemically built into the lattice), e.g. $\text{CuSO}_4{\cdot}5\text{H}_2\text{O}$ (blue).
Anhydrous — the same salt with the water removed, e.g. $\text{CuSO}_4$ (white).
Water of crystallisation — the fixed number of water molecules per formula unit in a hydrated salt.

Empirical = simplest ratio; molecular = actual numbers.
Molecular formula = (empirical)ₙ.
Hydrated (with water of crystallisation) vs anhydrous.

Calculating empirical and molecular formulae

Divide each element's mass (or %) by its Ar, then by the smallest result; scale to whole numbers.

The method (works for masses or percentages):

Write the mass (or %) of each element.
Divide by $A_r$ → moles of each.
Divide by the smallest of those numbers.
If needed, scale up to the nearest whole-number ratio.

Worked — empirical formula. A compound is 40.0% C, 6.7% H, 53.3% O.

	C	H	O
% (≈ mass in 100 g)	40.0	6.7	53.3
÷ $A_r$	40.0/12 = 3.33	6.7/1 = 6.7	53.3/16 = 3.33
÷ smallest (3.33)	1	2	1

Empirical formula: $\text{CH}_2\text{O}$ .

Worked — molecular formula. If that compound has $M_r = 180$ :

Empirical mass of $\text{CH}_2\text{O} = 30$ .
$180 / 30 = 6$ → molecular formula $= (\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$ (glucose).

Worked — water of crystallisation. Finding $x$ in a hydrate is the same idea: find moles of anhydrous salt and moles of water, then take the ratio (see the worked examples for $\text{MgSO}_4{\cdot}x\text{H}_2\text{O}$ ).

Mass/% → ÷ $A_r$ → ÷ smallest → scale.
Molecular formula: $M_r ÷$ empirical mass = multiplier.
Hydrate: ratio of moles of salt to moles of water gives $x$ .

How it’s examined

Formula-writing and balancing run through every paper; ionic equations and empirical-formula calculations are reliable Paper 2 and Paper 5 earners. Examiner reports flag: altering polyatomic-ion formulae to balance, leaving out state symbols, charge not balancing in ionic equations, and rounding 1.5-type ratios instead of scaling.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 2.3; 9701 Examiner Reports 2022-2024; 9701/21 & 9701/51 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Formulae

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

1Formula of an ionic compound (2 marks)
Getting started• ionic formula, brackets
Question
Write the formula of iron(III) sulfate.
Step-by-step solution
1. Step 1
  Identify the ions. Iron(III) = $\text{Fe}^{3+}$ ; sulfate = $\text{SO}_4^{2-}$ .
2. Step 2
  Balance charges. Lowest common multiple of 3 and 2 is 6 → two $\text{Fe}^{3+}$ ( $+6$ ) and three $\text{SO}_4^{2-}$ ( $-6$ ).
  $\text{Fe}_2(\text{SO}_4)_3$
Answer
$\text{Fe}_2(\text{SO}_4)_3$ .
Examiner tip
Brackets are essential: 3 sulfate ions is $(\text{SO}_4)_3$ , not $\text{S}_3\text{O}_{12}$ .
2Balancing a combustion equation (2 marks)
Getting started• balancing, combustion
Question
Balance: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ .
Step-by-step solution
1. Step 1
  Balance C, then H. 3 C → 3 CO₂; 8 H → 4 H₂O.
  $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
2. Step 2
  Balance O last. Right side has 6 + 4 = 10 O → 5 O₂.
  $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
Answer
$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$ .
Examiner tip
Always balance oxygen last in a combustion equation — it appears in two products.
3Writing an ionic equation (3 marks)
Building confidence• ionic equation
Question
Barium chloride solution reacts with sodium sulfate solution to give a white precipitate of barium sulfate. Write the ionic equation, with state symbols.
Step-by-step solution
1. Step 1
  Full equation: $\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$ .
2. Step 2
  Identify spectators: $\text{Na}^+$ and $\text{Cl}^-$ are unchanged → cancel.
3. Step 3
  Ionic equation (atoms and charge balance).
  $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$
Answer
$\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$ .
Examiner tip
Only species that change state or form are kept; check both atoms AND charges balance.
4Empirical formula from masses (3 marks)
Building confidence• empirical formula
Question
A 2.80 g sample of iron combines with 1.20 g of oxygen. Find the empirical formula. ( $A_r$ : Fe 56.0, O 16.0)
Step-by-step solution
1. Step 1
  Moles of each.
  $n(\text{Fe}) = \frac{2.80}{56.0} = 0.0500;\quad n(\text{O}) = \frac{1.20}{16.0} = 0.0750$
2. Step 2
  Divide by the smallest (0.0500).
  $\text{Fe}:1\quad \text{O}:1.5$
3. Step 3
  Scale ×2 to clear the 0.5.
  $\text{Fe}_2\text{O}_3$
Answer
$\text{Fe}_2\text{O}_3$ .
Examiner tip
A ratio of 1 : 1.5 must be scaled (×2), not rounded to 1 : 2.
5Molecular formula from % composition + Mr (4 marks)
Stretch• empirical formula, molecular formula
Question
A hydrocarbon is 85.7% carbon and 14.3% hydrogen by mass and has $M_r = 56.0$ . Find its molecular formula. ( $A_r$ : C 12.0, H 1.0)
Step-by-step solution
1. Step 1
  Treat % as grams; find moles.
  $n(\text{C}) = \frac{85.7}{12.0} = 7.14;\quad n(\text{H}) = \frac{14.3}{1.0} = 14.3$
2. Step 2
  Divide by the smallest → empirical formula.
  $\text{C}:1\quad \text{H}:2 \;\Rightarrow\; \text{CH}_2$
3. Step 3
  Empirical mass = 14.0; divide Mr by it.
  $\frac{56.0}{14.0} = 4$
4. Step 4
  Multiply the empirical formula by 4.
  $\text{C}_4\text{H}_8$
Answer
Molecular formula $= \text{C}_4\text{H}_8$ .
Examiner tip
Empirical gives the ratio; the molecular formula needs Mr ÷ empirical formula mass to find the multiplier.
6Finding water of crystallisation (4 marks)
Stretch• hydrate
Question
4.92 g of hydrated magnesium sulfate $\text{MgSO}_4{\cdot}x\text{H}_2\text{O}$ is heated to give 2.40 g of anhydrous $\text{MgSO}_4$ . Find $x$ . ( $M_r$ : MgSO₄ 120.4, H₂O 18.0)
Step-by-step solution
1. Step 1
  Mass of water lost.
  $4.92 - 2.40 = 2.52\,\text{g}$
2. Step 2
  Moles of each.
  $n(\text{MgSO}_4) = \frac{2.40}{120.4} = 0.01993;\quad n(\text{H}_2\text{O}) = \frac{2.52}{18.0} = 0.1400$
3. Step 3
  Ratio H₂O : MgSO₄.
  $\frac{0.1400}{0.01993} \approx 7$
Answer
$x = 7$ , i.e. $\text{MgSO}_4{\cdot}7\text{H}_2\text{O}$ .
Examiner tip
Use the mass of water = (hydrated − anhydrous). Round x to the nearest whole number.

Key Formulae — Formulae

The formulae you need to memorise for formulae on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Empirical formula (mole ratio)
$\text{ratio} = \frac{n_{\text{A}}}{n_{\text{smallest}}} : \frac{n_{\text{B}}}{n_{\text{smallest}}} : \ldots$
$n$
moles of each element (= mass ÷ Ar, or % ÷ Ar)
When to use
Finding the simplest whole-number ratio from masses or percentage composition; scale up to clear fractions.
Molecular formula from empirical formula
$\text{multiplier} = \frac{M_r}{\text{empirical formula mass}}$
$M_r$
relative molecular mass of the compound
$empirical formula mass$
mass of one empirical unit
When to use
Turning an empirical formula into the molecular formula once you know Mr.
Example
CH₂ (mass 14) with Mr 56 → multiplier 4 → C₄H₈.

Key Definitions and Keywords — Formulae

Definitions to memorise and the exact keywords mark schemes credit for formulae answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Empirical formula
Examiner keyword
The simplest whole-number ratio of the atoms of each element in a compound.
Molecular formula
Examiner keyword
The actual number of atoms of each element in a molecule (a whole-number multiple of the empirical formula).
Balanced (stoichiometric) equation
Examiner keyword
An equation with equal numbers of each type of atom on both sides; the balancing numbers give the reacting mole ratio.
Ionic equation
Examiner keyword
An equation showing only the ions and species that take part, with spectator ions removed; atoms AND charges must balance.
Spectator ion
Examiner keyword
An ion present in the mixture but unchanged on both sides of the equation, so it is omitted from the ionic equation.
State symbols
Examiner keyword
(s) solid, (l) liquid, (g) gas, (aq) aqueous — added to each species to show its physical state.
Water of crystallisation
Examiner keyword
Water molecules that are part of the crystal structure of a hydrated salt, in a fixed ratio per formula unit.
Anhydrous / hydrated
Anhydrous = containing no water of crystallisation; hydrated = containing water of crystallisation.

Common Mistakes and Misconceptions — Formulae

The traps other students keep falling into on formulae questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Changing a polyatomic ion's formula to balance charges.
9701 Examiner Reports
Why it happens
Treating SO₄ like separate S and O.
How to avoid it
Keep the ion intact and use brackets: (SO₄)₃ etc.
✕Balancing an equation by changing formulae (subscripts).
Why it happens
Trying to force the atoms to match.
How to avoid it
Only change the big numbers in front (coefficients); never alter a correct formula like H₂O to H₂O₂.
✕Omitting state symbols when asked for them.
Why it happens
Rushing.
How to avoid it
Add (s), (l), (g), (aq) to every species when required.
✕Ionic equation where the charges don't balance.
Why it happens
Only checking atoms.
How to avoid it
Check total charge is equal on both sides as well as atoms.
✕Rounding a 1 : 1.5 empirical ratio to 1 : 2.
Why it happens
Wanting whole numbers immediately.
How to avoid it
Multiply the whole ratio to clear fractions (×2, ×3).
✕Giving the empirical formula when the molecular formula is asked.
9701 Examiner Reports 2022-2024
Why it happens
Stopping after the ratio step.
How to avoid it
If Mr is given, finish: divide Mr by the empirical formula mass and scale up.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Formulae — frequently asked questions

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Formulae

Short Study Notes in the form of Slides

Short Study Notes — Formulae

Detailed Notes

What you’ll learn

How it’s examined

1Formula of an ionic compound (2 marks)

2Balancing a combustion equation (2 marks)

3Writing an ionic equation (3 marks)

4Empirical formula from masses (3 marks)

5Molecular formula from % composition + Mr (4 marks)

6Finding water of crystallisation (4 marks)

Empirical formula (mole ratio)

Molecular formula from empirical formula

Empirical formula

Molecular formula

Balanced (stoichiometric) equation

Ionic equation

Spectator ion

State symbols

Water of crystallisation

Anhydrous / hydrated

✕Changing a polyatomic ion's formula to balance charges.

✕Balancing an equation by changing formulae (subscripts).

✕Omitting state symbols when asked for them.

✕Ionic equation where the charges don't balance.

✕Rounding a 1 : 1.5 empirical ratio to 1 : 2.

✕Giving the empirical formula when the molecular formula is asked.

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Formulae — frequently asked questions