What is a chemical formula and how is it used in Chemistry?

A chemical formula is a symbolic representation of a compound that indicates the elements present and their relative proportions. In Chemistry, particularly in the Edexcel IGCSE curriculum, chemical formulae are used to convey information about the types and numbers of atoms in a molecule, helping students understand the composition and structure of compounds.

How do you balance chemical equations in Edexcel IGCSE Chemistry?

Balancing chemical equations involves ensuring that the number of atoms for each element is equal on both sides of the equation. This is achieved by adjusting the coefficients (the numbers in front of molecules) without changing the chemical formulae. This process reflects the Law of Conservation of Mass, which is a key principle in Chemistry.

What are the steps to calculate the relative formula mass of a compound?

To calculate the relative formula mass (RFM) of a compound, sum the relative atomic masses of all the atoms in its chemical formula. For example, for water (H2O), the RFM is calculated as (2 x 1) + 16 = 18. This calculation is essential in Chemistry for understanding the mass relationships in chemical reactions.

Why is it important to understand moles and molar calculations in Chemistry?

Understanding moles and molar calculations is crucial because they allow chemists to quantify the amount of substance involved in chemical reactions. The mole concept links the macroscopic world of grams and liters to the microscopic world of atoms and molecules, which is a fundamental aspect of the Edexcel IGCSE Chemistry curriculum.

How do you determine the empirical formula from percentage composition?

To determine the empirical formula from percentage composition, convert the percentage of each element to grams, then to moles by dividing by their atomic masses. Next, divide each mole value by the smallest number of moles calculated. This gives the simplest whole number ratio of atoms in the compound, which is the empirical formula. This skill is vital for solving problems in Principles of Chemistry.

Why is "Changing subscripts inside a formula when balancing" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Trying to balance H by changing H₂O to H₃O. How to avoid it: NEVER change subscripts — they define the compound's identity. Only adjust the coefficients (the numbers IN FRONT). H₃O is a different species from H₂O! Source: 4CH1 Examiner Reports 2022-2024

Why is "Using cm³ in $c = n/V$ when concentration is in mol/dm³" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Forgetting to convert. How to avoid it: ALWAYS convert volume from cm³ to dm³ first: divide by 1000. 25 cm³ = 0.025 dm³. Then apply c = n/V. Source: 4CH1 Examiner Reports 2022-2024

Why is "Using a 1:1 mole ratio when the balanced equation says otherwise" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Rushing past the balanced equation. How to avoid it: Always WRITE the balanced equation first and identify the mole ratio explicitly. For 2Mg + O_2 arrow 2MgO, the ratio Mg : O₂ is 2 : 1.

Why is "Confusing atom economy with percentage yield" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Both are percentages relating to product. How to avoid it: **Yield** = actual / theoretical (depends on the actual experiment). **Atom economy** = Mr(desired product) / Mr(all products) (depends ONLY on the equation — same value every time). A reaction with atom economy 50% can still have yield 95%. Source: 4CH1 Examiner Reports 2022-2024

Why is "Rounding empirical-formula ratios too aggressively (e.g. 2.5 → 3)" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Assuming small deviations from whole numbers are experimental error. How to avoid it: If your ratio is 1 : 2.5, multiply ALL values by 2 to get 2 : 5 — a whole-number ratio. 2.5 should not be rounded to 3.

Why is "Forgetting state symbols" a common mistake in Chemical Formulae, Equations and Calculations?

Why it happens: Treating them as optional decoration. How to avoid it: Most Edexcel marks for 'balance and add state symbols' questions split the state-symbols mark separately. Always add (s), (l), (g) or (aq) — it's a free mark.

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Chemical Formulae, Equations and Calculations

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Chemical Formulae, Equations and Calculations — Pearson Edexcel International GCSE Chemistry 4CH1 Study Notes (2026 onwards, Spec Issue 4)

The hardest-marking subtopic of 4CH1. Writing chemical formulae from ion charges. Balancing equations with state symbols. The mole concept and Avogadro's constant. Eight calculation skills: $M_r$ , $n = m/M_r$ , reacting masses, gas volumes at rtp, concentration ( $c = n/V$ ), percentage composition, empirical and molecular formulae (Paper 2), percentage yield and atom economy (Paper 2).

What you’ll learn

Mapped to the Cambridge IGCSE 4CH1 syllabus (2026 onwards).

1.22 — Write word equations and balanced chemical equations (including state symbols) for reactions.
1.23 — Use chemical formulae and ion charges to deduce the formulae of compounds.
1.25 — Calculate relative formula masses ( $M_r$ ) from relative atomic masses ( $A_r$ ).
1.29 — Know that one mole of a substance contains $6.02 \times 10^{23}$ particles (Avogadro's constant).
1.30 — Calculate amount in moles using $n = m / M_r$ .
1.31 — Use reacting masses and mole ratios in calculations.
1.36 — Calculate gas volumes at room temperature and pressure ( $V_{\text{gas}} = n \times 24$ dm³).
1.38 — Calculate the concentration of solutions (c = n/V).
1.39P — Calculate the empirical formula of a compound from masses or percentages (Paper 2).
1.40P — Deduce molecular formula from empirical formula and Mr (Paper 2).
1.42P — Calculate percentage yield (Paper 2).
1.45P — Calculate atom economy (Paper 2).

Writing chemical formulae (spec 1.23)

Balance ion charges.

Compounds form when ions of opposite charge come together; the total POSITIVE charge MUST equal the total NEGATIVE charge so the compound is overall neutral. To write the formula:

Write the cation symbol with its charge, then the anion with its charge.
Swap and drop the charges (the magnitude becomes the subscript of the OTHER ion).
Simplify the subscripts to the smallest whole-number ratio.

Worked examples.

Sodium chloride: Na⁺ + Cl⁻ → NaCl (1:1).
Calcium chloride: Ca²⁺ + Cl⁻ → CaCl₂ (need 2 Cl⁻ to balance Ca²⁺).
Magnesium oxide: Mg²⁺ + O²⁻ → MgO (charges already cancel).
Aluminium chloride: Al³⁺ + Cl⁻ → AlCl₃.
Aluminium oxide: Al³⁺ + O²⁻ → Al₂O₃ (2 × 3 = 6, 3 × 2 = 6).
Sodium oxide: Na⁺ + O²⁻ → Na₂O.

The magnitude of each ion's charge becomes the subscript of the other ion.

Polyatomic ions to memorise. Wrap brackets around polyatomic ions when you need more than one:

Ion	Formula	Charge
Hydroxide	OH⁻	-1
Nitrate	NO₃⁻	-1
Carbonate	CO₃²⁻	-2
Sulfate	SO₄²⁻	-2
Ammonium	NH₄⁺	+1

Examples: calcium hydroxide Ca(OH)₂; aluminium hydroxide Al(OH)₃; ammonium sulfate (NH₄)₂SO₄; calcium nitrate Ca(NO₃)₂.

Total + charge = total − charge.
Swap the magnitudes of the charges to set subscripts.
Polyatomic ions need brackets when more than one is present.

Balancing chemical equations and state symbols (spec 1.22)

Atoms are conserved — coefficients only.

A balanced chemical equation has the SAME number of atoms of each element on both sides — required by the law of conservation of mass. To balance:

Write the unbalanced equation with correct chemical formulae for all reactants and products.
Balance atoms one element at a time, leaving H and O for last (they often appear in multiple species).
Only adjust coefficients (the numbers IN FRONT of formulae) — NEVER change subscripts.
Reduce to smallest whole-number ratio at the end.
Add state symbols in brackets after each formula.

State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

Worked example — combustion of propane.

Unbalanced: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$

Balance C: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}$
Balance H: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
Balance O: RHS = 6 + 4 = 10 → 5 O₂ on LHS: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
Add state symbols: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$

Common pitfall. Students sometimes try to "balance" by changing CO₂ to CO. This is illegal — CO and CO₂ are entirely different compounds. Adjust coefficients only.

Conserve atoms — same count each side.
Coefficients only — never alter subscripts.
State symbols: (s), (l), (g), (aq).

See the full worked example for chemical formulae, equations and calculations →

The mole and relative formula mass (spec 1.25, 1.29, 1.30)

$M_r$ + $n = m/M_r$ + Avogadro's constant.

Relative formula mass ( $M_r$ ) = sum of $A_r$ values of all the atoms in a chemical formula. For example:

$M_r$ (H₂O) = 2(1) + 16 = 18
$M_r$ (CO₂) = 12 + 2(16) = 44
$M_r$ (CaCO₃) = 40 + 12 + 3(16) = 100
$M_r$ (NaOH) = 23 + 16 + 1 = 40

The mole. One mole = the amount of substance containing $6.02 \times 10^{23}$ particles (Avogadro's constant, $L$ ).

Why $L$ is useful. A mole of any substance has a mass equal to its $M_r$ expressed in grams. So:

1 mole of water = 18 g of water.
1 mole of carbon dioxide = 44 g of CO₂.
1 mole of calcium carbonate = 100 g of CaCO₃.

Number of moles from mass.

$n = \dfrac{m}{M_r}$

(units: $n$ in mol, $m$ in g, $M_r$ unitless)

Cover the quantity you need: m = n × Mr, n = m ÷ Mr, Mr = m ÷ n.

Worked example. How many moles of NaOH are in 24.0 g? $M_r$ (NaOH) = 40 → $n = 24.0 / 40 = 0.60$ mol.

Reverse calculation. Mass from moles: $m = n \times M_r$ . E.g. how much CaCO₃ is in 0.5 mol? $m = 0.5 \times 100 = 50$ g.

Number of particles. Number of particles = moles × $L$ . E.g. number of water molecules in 0.50 mol of water = $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ .

$M_r$ = sum of $A_r$ values.
1 mol of substance has mass = $M_r$ in grams.
$n = m / M_r$ .
1 mol contains $6.02 \times 10^{23}$ particles.

See the full worked example for chemical formulae, equations and calculations →

Reacting masses — stoichiometry (spec 1.31)

Mass → moles → ratio → moles → mass.

Stoichiometry problems use the mole ratio from a balanced equation to convert between reactant and product amounts. Method:

Calculate moles of the substance you know ( $n = m / M_r$ ).
Use the balanced equation to find the mole ratio.
Use the ratio to find moles of the substance you want.
Convert moles back to mass: $m = n \times M_r$ (or to volume / concentration if needed).

Every reacting-mass problem follows the same chain: mass → moles → ratio → moles → mass.

Worked example. What mass of magnesium oxide forms when 6.0 g of magnesium burns in excess oxygen?

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$

Step 1. $n$ (Mg) = 6.0 / 24 = 0.25 mol.

Step 2. Mole ratio Mg : MgO = 2 : 2 = 1 : 1.

Step 3. $n$ (MgO) = 0.25 mol.

Step 4. $m$ (MgO) = 0.25 × 40 = 10 g.

Generalising to non-1:1 ratios. Aluminium + chlorine → aluminium chloride: $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$ . What mass of Cl₂ is needed to react with 5.4 g of Al? $n$ (Al) = 5.4/27 = 0.20 mol. Ratio Al : Cl₂ = 2 : 3, so $n$ (Cl₂) = 0.20 × (3/2) = 0.30 mol. $m$ (Cl₂) = 0.30 × 71 = 21.3 g.

Limiting reagent. When both reactant masses are given, the reactant that runs out first is the limiting reagent. Calculate moles of each, divide by their coefficient, and the SMALLER quotient is the limiting reagent.

ALWAYS write the balanced equation first.
Mass → moles → ratio → moles → mass.
Identify the limiting reagent if both masses given.

See the full worked example for chemical formulae, equations and calculations →

Gas volumes at rtp (spec 1.36)

$V = n \times 24$ dm³ at room temperature and pressure.

At room temperature and pressure (rtp, ~25 °C and 1 atm), one mole of ANY gas occupies 24 dm³ (= 24,000 cm³). This is the molar volume of a gas at rtp.

$V_{\text{gas (dm}^3\text{)}} = n \times 24 \quad \text{or} \quad V_{\text{gas (cm}^3\text{)}} = n \times 24000$

Worked example 1. How many moles in 480 cm³ of gas at rtp?

$n = \dfrac{V}{24000} = \dfrac{480}{24000} = 0.020\ \text{mol}$

Worked example 2. What volume (in dm³) of CO₂ at rtp is produced when 5.0 g of CaCO₃ decomposes? $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

$n$ (CaCO₃) = 5.0 / 100 = 0.050 mol. Mole ratio 1:1, so $n$ (CO₂) = 0.050 mol. $V$ = 0.050 × 24 = 1.2 dm³ (or 1200 cm³).

Why 24 dm³ regardless of the gas? Avogadro's hypothesis: equal volumes of any gas at the same T and P contain equal numbers of molecules. The IDENTITY of the gas (H₂, O₂, CO₂, etc.) doesn't matter at rtp — they all occupy the same volume per mole.

Conversion reminders. 1 dm³ = 1000 cm³ = 1 litre. So dm³ × 1000 = cm³, cm³ ÷ 1000 = dm³.

Molar volume = 24 dm³ at rtp.
$V$ = $n$ × 24 (in dm³) or $V$ = $n$ × 24 000 (in cm³).
Same volume regardless of gas identity.

See the full worked example for chemical formulae, equations and calculations →

Concentration of solutions (spec 1.38)

$c = n / V$ ; volume MUST be in dm³.

Concentration = the amount of dissolved solute per unit volume of solution. Units: mol/dm³ (also written mol dm⁻³ or M for molarity).

$c = \dfrac{n}{V_{\text{dm}^3}}$

Rearrangements:

Moles from concentration and volume: $n = c \times V$ (the form used in titration calculations).
Volume from moles and concentration: $V = n / c$ .

Critical unit conversion. Concentrations are quoted per dm³ but laboratory volumes are usually in cm³. ALWAYS convert before substituting:

$V_{\text{dm}^3} = \dfrac{V_{\text{cm}^3}}{1000}$

Worked example. 4.00 g of NaOH dissolved in water to make 250 cm³ of solution. What is the concentration?

$n$ (NaOH) = 4.00 / 40 = 0.100 mol.
$V$ = 250 / 1000 = 0.250 dm³.
$c = 0.100 / 0.250 =$ 0.400 mol/dm³.

Titration calculations. In a titration, you measure the volume of one solution that reacts exactly with a measured volume of another (the endpoint shown by an indicator). Use the mole ratio from the balanced equation to convert between the two:

$n$ (reagent 1) = $c_1 \times V_1$ (in dm³).
Use ratio from balanced equation → $n$ (reagent 2).
$c_2 = n_2 / V_2$ (in dm³).

$c = n / V$ with $V$ in dm³.
ALWAYS convert cm³ → dm³ (÷ 1000) before substituting.
Titrations: $n = cV$ + mole ratio + $c = n/V$ to find unknown concentration.

See the full worked example for chemical formulae, equations and calculations →

Empirical and molecular formulae (spec 1.39P, 1.40P — Paper 2)

% → moles → ratio → whole numbers.

These specifications are Paper 2 only (the "P" suffix). They appear regularly on 4CH1/2C.

Empirical formula = the simplest whole-number ratio of atoms of each element in a compound. Molecular formula = the actual number of atoms in one molecule (always a whole-number multiple of the empirical formula).

Method for empirical formula from percentage composition.

Treat percentages as masses in 100 g of compound. So 40% C in 100 g of compound = 40 g of C.
Divide each mass by $A_r$ to get moles.
Divide all moles by the smallest value → simplest ratio.
Multiply if needed to get whole numbers (e.g. 1 : 2.5 → multiply by 2 → 2 : 5).
Write the empirical formula.

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

$n$ (C) = 40.0 / 12 = 3.33; $n$ (H) = 6.7 / 1 = 6.7; $n$ (O) = 53.3 / 16 = 3.33.
Divide all by smallest (3.33): C = 1, H = 2.01, O = 1.
Ratio = 1 : 2 : 1.
Empirical formula = CH₂O.

Step from empirical to molecular formula. Given $M_r$ of the compound:

Calculate $M_r$ (empirical) by summing $A_r$ values.
Multiplier = $M_r$ (compound) / $M_r$ (empirical).
Multiply ALL subscripts in the empirical formula by this multiplier.

Worked example continued. The compound has $M_r$ = 180. $M_r$ (CH₂O) = 30. Multiplier = 180/30 = 6. Molecular formula = C₆H₁₂O₆ (glucose).

Empirical = simplest whole-number ratio.
Steps: % → ÷ Ar → ÷ smallest → whole-number ratio.
Molecular = empirical × ( $M_r$ (compound) / $M_r$ (empirical)).
Paper 2 only.

See the full worked example for chemical formulae, equations and calculations →

Percentage yield and atom economy (spec 1.42P, 1.45P — Paper 2)

Yield = how well it WORKED; atom economy = how clean the EQUATION is.

Both are Paper 2 calculations. Conceptually distinct, often confused.

Percentage yield — how much of the theoretical maximum mass was actually obtained?

$\% \text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Theoretical yield = calculated mass of product from the balanced equation, assuming 100% conversion.
Actual yield = mass actually obtained at the bench.
Yield is always ≤ 100% in practice.

Why yield is less than 100%.

The reaction may not go to completion (reversible or slow).
Side reactions form unwanted by-products.
Practical losses during transfers, filtering, drying, etc.

Worked example. 6.0 g Mg burns in excess O₂ to give 8.0 g of MgO. Theoretical yield (from earlier) = 10 g. Actual yield = 8.0 g.

$\% \text{ yield} = \dfrac{8.0}{10} \times 100 = 80 \%$

Atom economy — what fraction of the reactant mass ends up in the DESIRED product (rather than wasted as by-products)?

$\% \text{ atom economy} = \dfrac{M_r(\text{desired product})}{M_r(\text{all products})} \times 100$

(Remember to include the STOICHIOMETRIC COEFFICIENTS in the Mr sums.)

Worked example — blast furnace. $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$

$M_r$ (desired, Fe) including coefficient = 2 × 56 = 112.
$M_r$ (all products) = 2(56) + 3(44) = 112 + 132 = 244.
Atom economy = 112 / 244 × 100 = 45.9 %.

The other 54.1% of the reactant mass becomes CO₂ — a waste product.

Why does atom economy matter? High atom economy = sustainable / green process. A process with high yield but low atom economy still wastes raw materials; the goal in modern chemistry is to design routes that maximise BOTH.

% yield = actual / theoretical × 100 (depends on the experiment).
% atom economy = $M_r$ (desired) / $M_r$ (all products) × 100 (depends only on the equation).
Both Paper 2 only.
Higher atom economy = more sustainable.

See the full worked example for chemical formulae, equations and calculations →

How it’s examined

Chemical formulae / equations / calculations is the largest topic in 4CH1 — it appears on every paper, often as the highest-value cluster of marks. Paper 1 (4CH1/1C): balancing equations (2-3 marks), $n = m/M_r$ (3 marks), reacting masses (5-6 marks), concentration calculations (4-5 marks), gas volumes at rtp (3-4 marks). Paper 2 (4CH1/2C): all of the above PLUS empirical-formula questions (5 marks), molecular-formula deduction, percentage yield (3-4 marks), atom economy (3-4 marks). Worth 20-30 marks across both papers in a typical session.

Sources: Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4CH1 Examiner Reports 2022-2024; 4CH1/1C May/Jun 2024 question paper and mark scheme; 4CH1/1C January 2024 question paper and mark scheme; 4CH1/2C May/Jun 2024 question paper and mark scheme; 4CH1/2C January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Chemical Formulae, Equations and Calculations

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1Balancing a chemical equation (3 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q3• balancing, spec-1.22
Question
Balance the following equation, then add state symbols (the reaction is the combustion of propane in oxygen, all reactants and products are gaseous except liquid water): $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ (3 marks)
Step-by-step solution
1. Step 1
  Balance C first (1 M). Propane has 3 C atoms → need 3 CO₂.
  $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}$
2. Step 2
  Balance H next. 8 H atoms → need 4 H₂O.
  $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
3. Step 3
  Balance O last (1 A). Right-hand side: 3 × 2 + 4 × 1 = 10 O atoms → need 5 O₂.
  $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
4. Step 4
  Add state symbols (1 B). Propane (g) + oxygen (g) → carbon dioxide (g) + water (l).
  $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$
Answer
$\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$
Examiner tip
Edexcel will only credit the balanced equation if EVERY coefficient is in the smallest whole-number ratio. NEVER alter the subscripts inside a formula (e.g. don't change CO₂ to CO₂.₅). State symbols are worth their own mark on most papers — practise (s), (l), (g), (aq).
2Mass to moles using $n = m / M_r$ (3 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q5• moles, spec-1.30, Mr
Question
Calculate the number of moles of sodium hydroxide (NaOH) in a 24.0 g sample. ( $A_r$ : Na = 23, O = 16, H = 1) (3 marks)
Step-by-step solution
1. Step 1
  Calculate $M_r$ of NaOH (1 M). $M_r$ = 23 + 16 + 1.
  $M_r(\text{NaOH}) = 40$
2. Step 2
  Use $n = m / M_r$ (1 M).
  $n = \dfrac{m}{M_r} = \dfrac{24.0}{40}$
3. Step 3
  Evaluate (1 A).
  $n = 0.60\ \text{mol}$
Answer
Number of moles of NaOH = 0.60 mol.
Examiner tip
Edexcel marks: 1 for $M_r$ , 1 for $n = m/M_r$ , 1 for the final answer. Always quote the unit 'mol'. The 'amount of substance' is technically the correct phrase; 'number of moles' is also credited.
3Reacting mass (stoichiometry) (5 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q7• stoichiometry, spec-1.31
Question
Calculate the maximum mass of magnesium oxide (MgO) that can be produced when 6.0 g of magnesium burns in excess oxygen. ( $A_r$ : Mg = 24, O = 16). The balanced equation is $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ . (5 marks)
Step-by-step solution
1. Step 1
  Moles of Mg (1 M).
  $n(\text{Mg}) = \dfrac{6.0}{24} = 0.25\ \text{mol}$
2. Step 2
  Use mole ratio from balanced equation (1 M). 2 Mg : 2 MgO is a 1 : 1 ratio.
  $n(\text{MgO}) = 0.25\ \text{mol}$
3. Step 3
  $M_r$ of MgO (1 A).
  $M_r(\text{MgO}) = 24 + 16 = 40$
4. Step 4
  Mass of MgO = $n \times M_r$ (1 B).
  $m = n \times M_r = 0.25 \times 40$
5. Step 5
  Final answer with unit (1 B).
  $m(\text{MgO}) = 10\ \text{g}$
Answer
Maximum mass of MgO = 10 g.
Examiner tip
Five-mark reacting-mass questions are a near-guaranteed Paper 1 item. Mark schemes split as {moles of reactant} / {use ratio} / {Mr of product} / {mass = n × Mr} / {answer + unit}. ECF applies through each step — show the working line by line so partial credit is awarded.
4Gas volume at rtp from moles (3 marks, Paper 1)
Core• Adapted from 4CH1/1CR Jan 2024 Q6• gas volume, rtp, spec-1.36
Question
Calculate the volume of carbon dioxide produced at room temperature and pressure (rtp) when 5.0 g of calcium carbonate is heated and fully decomposed. Molar volume of any gas at rtp = 24 dm³/mol. ( $A_r$ : Ca = 40, C = 12, O = 16). Balanced equation: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ . (3 marks)
Step-by-step solution
1. Step 1
  $M_r$ of CaCO₃ and moles (1 M).
  $M_r(\text{CaCO}_3) = 40 + 12 + 48 = 100; \quad n = \dfrac{5.0}{100} = 0.050\ \text{mol}$
2. Step 2
  Mole ratio 1 : 1 (1 M). So moles of CO₂ = 0.050 mol.
3. Step 3
  Volume at rtp = $n \times 24$ dm³ (1 A).
  $V = 0.050 \times 24 = 1.2\ \text{dm}^3$
Answer
Volume of CO₂ at rtp = 1.2 dm³ (or 1200 cm³).
Examiner tip
Edexcel mark schemes accept either dm³ or cm³ — but you must write the correct unit alongside your value. Molar volume 24 dm³ = 24,000 cm³. Don't forget to use the mole ratio from the balanced equation.
5Concentration of a solution (4 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q8• concentration, spec-1.38
Question
4.00 g of sodium hydroxide (NaOH) is dissolved in water to make 250 cm³ of solution. Calculate the concentration of the solution in mol/dm³. ( $M_r$ : NaOH = 40) (4 marks)
Step-by-step solution
1. Step 1
  Moles of NaOH (1 M).
  $n = \dfrac{4.00}{40} = 0.100\ \text{mol}$
2. Step 2
  Convert volume to dm³ (1 M). 250 cm³ ÷ 1000 = 0.250 dm³.
  $V = 0.250\ \text{dm}^3$
3. Step 3
  Apply $c = n / V$ (1 A).
  $c = \dfrac{n}{V} = \dfrac{0.100}{0.250}$
4. Step 4
  Final answer with unit (1 B).
  $c = 0.400\ \text{mol/dm}^3$
Answer
Concentration = 0.400 mol/dm³.
Examiner tip
The volume MUST be in dm³ for concentration in mol/dm³. cm³ → dm³ is ÷ 1000. Examiner reports highlight unit conversion as the most-common-lost mark on this question. State the unit conversion line explicitly.
6Empirical formula from percentage composition (5 marks, Paper 2)
Extended• Adapted from 4CH1/2C May/Jun 2024 Q4• empirical formula, spec-1.39P, spec-1.40P
Question
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate the empirical formula. The relative molecular mass of the compound is 180. State its molecular formula. ( $A_r$ : C = 12, H = 1, O = 16) (5 marks)
Step-by-step solution
1. Step 1
  Convert % to moles (per 100 g) — divide each % by $A_r$ (1 M).
  $n(C) = \dfrac{40.0}{12} = 3.33;\ n(H) = \dfrac{6.7}{1} = 6.7;\ n(O) = \dfrac{53.3}{16} = 3.33$
2. Step 2
  Divide each by the smallest (1 M). Smallest = 3.33.
  $C : H : O = \dfrac{3.33}{3.33} : \dfrac{6.7}{3.33} : \dfrac{3.33}{3.33} = 1 : 2.01 : 1$
3. Step 3
  Round to whole-number ratio (1 A). Empirical formula CH₂O.
4. Step 4
  Compare empirical Mr to molecular Mr (1 B). $M_r$ (CH₂O) = 12 + 2 + 16 = 30. Multiplier = 180 / 30 = 6.
5. Step 5
  Multiply subscripts by 6 (1 B). Molecular formula = C₆H₁₂O₆ (glucose).
Answer
Empirical formula = CH₂O. Molecular formula = C₆H₁₂O₆.
Examiner tip
Five-mark mark scheme: 1 each for (1) divide % by Ar, (2) divide by smallest, (3) ratio → empirical formula, (4) Mr(empirical) vs Mr(molecular) ratio, (5) molecular formula. Spec 1.39P-1.40P appear ONLY on Paper 2. ECF flows through every step — show all working.
7Percentage yield and atom economy (5 marks, Paper 2)
Extended• Adapted from 4CH1/2C Jan 2024 Q7• yield, atom economy, spec-1.42P, spec-1.45P
Question
Iron is extracted from iron(III) oxide using carbon monoxide in a blast furnace: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ An industrial process uses 8.0 tonnes of iron(III) oxide and produces 4.2 tonnes of iron. ( $M_r$ : Fe₂O₃ = 160, Fe = 56) (a) Calculate the percentage yield. (b) Calculate the atom economy of the desired product (iron). (5 marks)
Step-by-step solution
1. Step 1
  (a) Theoretical mass of Fe (1 M). 160 tonnes Fe₂O₃ would give 2 × 56 = 112 tonnes Fe. So 8.0 tonnes Fe₂O₃ → (8.0/160) × 112 = 5.6 tonnes Fe (theoretical).
2. Step 2
  Percentage yield formula (1 M).
  $\% \text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$
3. Step 3
  Substitute and evaluate (1 A).
  $\% \text{ yield} = \dfrac{4.2}{5.6} \times 100 = 75 \%$
4. Step 4
  (b) Atom economy formula (1 B).
  $\% \text{ atom economy} = \dfrac{M_r(\text{desired product})}{M_r(\text{all products})} \times 100$
5. Step 5
  Substitute and evaluate (1 B). $M_r$ of Fe (desired) = 2 × 56 = 112. $M_r$ of all products = 2 × 56 + 3 × 44 = 112 + 132 = 244.
  $\% \text{ atom economy} = \dfrac{112}{244} \times 100 = 45.9\%$
Answer
(a) Percentage yield = 75 %. (b) Atom economy for Fe = 45.9 % (the rest is CO₂, a waste product).
Examiner tip
Edexcel mark schemes: yield = actual / theoretical × 100; atom economy = Mr(desired) / Mr(all products) × 100 — and stoichiometric coefficients MUST be included in the Mr sum. Atom economy doesn't depend on the actual yield — only on the equation. Higher atom economy → more sustainable process.

Key Formulae — Chemical Formulae, Equations and Calculations

The formulae you need to memorise for chemical formulae, equations and calculations on the Pearson Edexcel IGCSE 4CH1 paper, with every variable defined in plain English and a note on when to use it.

Moles from mass (spec 1.30)
$n = \dfrac{m}{M_r}$
When to use
Use to convert mass (g) to moles (mol). $M_r$ = relative formula mass (sum of $A_r$ values of all atoms in the formula). Rearrange to $m = n \times M_r$ to find mass when moles are known.
Avogadro's constant (spec 1.29)
$L = 6.02 \times 10^{23}\ \text{mol}^{-1}$
When to use
Number of particles in 1 mole of substance. Number of particles = moles × $L$ . For example, 0.5 mol of water = $0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ water molecules.
Gas volume at room temperature and pressure (spec 1.36)
$V_{\text{gas (dm}^3\text{)}} = n \times 24$
When to use
One mole of ANY gas occupies 24 dm³ (= 24,000 cm³) at room temperature and pressure (rtp). Volume in cm³: use $V = n \times 24000$ . Convert: dm³ × 1000 = cm³.
Concentration of a solution (spec 1.38)
$c = \dfrac{n}{V_{\text{dm}^3}}$
When to use
Units: c in mol/dm³, n in mol, V in dm³. The volume MUST be in dm³ (cm³ ÷ 1000). Rearrange to $n = c \times V$ for titration calculations.
Percentage composition by mass
$\% \text{ of X} = \dfrac{\text{number of X atoms} \times A_r(X)}{M_r(\text{compound})} \times 100\%$
When to use
Used to find the mass-percentage of an element in a compound. For example, % nitrogen in NH₃: (1 × 14) / (14 + 3) × 100 = 14/17 × 100 = 82.4 %.
Percentage yield (spec 1.42P, Paper 2)
$\% \text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$
When to use
Actual yield = mass of product actually obtained. Theoretical yield = mass predicted by stoichiometry assuming 100% conversion. Yield is always ≤ 100% in practice. Used on Paper 2.
Atom economy (spec 1.45P, Paper 2)
$\% \text{ atom economy} = \dfrac{M_r(\text{desired product})}{M_r(\text{all products})} \times 100$
When to use
Use balanced equation Mr values, INCLUDING stoichiometric coefficients. Atom economy is a property of the EQUATION, not of the actual yield — a process with high atom economy is more sustainable / less wasteful.

Model Answers — Chemical Formulae, Equations and Calculations

High-scoring sample answers for chemical formulae, equations and calculations on the Pearson Edexcel IGCSE 4CH1 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4CH1/1C-style — balancing equations5 marks
Explain why chemical equations must be balanced and describe the rules for balancing using the combustion of methane ( $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ ) as an example. (5 marks)
Model answer
Why balance. Chemical equations must be balanced because atoms are neither created nor destroyed in a chemical reaction — this is the law of conservation of mass. The total number of atoms of each element must be the same on both sides of the equation, and so must the total mass.

Rules.

NEVER change the subscripts within a chemical formula (this would change the compound itself).

Only adjust the stoichiometric coefficients (the numbers in front of formulae) to balance.

Use the smallest whole-number ratio.

Add state symbols at the end: (s), (l), (g), (aq).

Worked example — methane combustion.

Step 1. Start with the unbalanced equation:

$\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$

Step 2. Balance C (already balanced, 1 = 1).

Step 3. Balance H. 4 H on the left → 2 H₂O on the right:

$\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Step 4. Balance O. Right-hand side: 2 + 2 = 4 O atoms. Need 2 O₂ on the left:

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Step 5. Add state symbols:

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

The equation is now balanced — 1 C, 4 H and 4 O on each side.
Why this scores
Edexcel 5-mark mark scheme: 1 for 'law of conservation of mass' / 'atoms neither created nor destroyed'; 1 for the rule about subscripts; 1 for showing a balanced equation; 1 for explicit working (counting atoms); 1 for state symbols. The phrase 'law of conservation of mass' is a credited keyword.
Question 2
4CH1/2C-style — moles and Avogadro4 marks
Explain what is meant by 'one mole' of a substance, including a reference to Avogadro's constant. Use 1.0 mol of water (H₂O) as an example. (4 marks)
Model answer
The mole. One mole of any substance contains the same number of particles as there are atoms in exactly 12 g of carbon-12. This number is called Avogadro's constant, $L$ (or $N_A$ ), and has the value $6.02 \times 10^{23}$ per mole (1 M + 1 A).

For water (H₂O). $M_r$ (H₂O) = 2 + 16 = 18. So 1 mole of water has a mass of 18 g (1 B) and contains $6.02 \times 10^{23}$ water molecules (1 B).

Each water molecule contains 2 H atoms and 1 O atom, so 1 mole of water contains:

$2 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$ hydrogen atoms.

$1 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}$ oxygen atoms.

Why useful. The mole is a way of "counting" particles by weighing — it's impractical to count $10^{23}$ atoms one by one, but the mole lets us scale up reaction equations (which talk about atoms / molecules) to laboratory-scale masses (which we can weigh on a balance).
Why this scores
4-mark mark scheme: 1 for the C-12 reference / Avogadro's number; 1 for the value $6.02 \times 10^{23}$ ; 1 for Mr(H₂O) = 18; 1 for '1 mole H₂O = 18 g'. The Edexcel spec uses the symbol $L$ for Avogadro's constant — $N_A$ is also accepted.
Question 3
4CH1/2C-style — titration / concentration5 marks
In a titration, 25.0 cm³ of dilute hydrochloric acid is exactly neutralised by 28.5 cm³ of 0.100 mol/dm³ sodium hydroxide solution. The equation for the reaction is $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ . Calculate the concentration of the hydrochloric acid in mol/dm³, giving your answer to 3 significant figures. (5 marks)
Model answer
Step 1 — Moles of NaOH. Convert volume to dm³: 28.5 cm³ ÷ 1000 = 0.0285 dm³.

$n(\text{NaOH}) = c \times V = 0.100 \times 0.0285 = 2.85 \times 10^{-3}\ \text{mol}$

Step 2 — Mole ratio. From the balanced equation, HCl : NaOH = 1 : 1. So:

$n(\text{HCl}) = n(\text{NaOH}) = 2.85 \times 10^{-3}\ \text{mol}$

Step 3 — Concentration of HCl. Convert acid volume to dm³: 25.0 cm³ ÷ 1000 = 0.0250 dm³.

$c(\text{HCl}) = \dfrac{n}{V} = \dfrac{2.85 \times 10^{-3}}{0.0250} = 0.114\ \text{mol/dm}^3$

Final answer. $c(\text{HCl}) = 0.114$ mol/dm³ (3 sig figs).
Why this scores
Mark scheme: 1 for n(NaOH); 1 for converting volume to dm³; 1 for mole ratio (1 : 1); 1 for the c = n/V calculation; 1 for the final 3 sig fig answer with units. ECF applies. Tip: always do the unit conversion FIRST (cm³ → dm³ = ÷ 1000) before substituting.
Question 4
4CH1/2C-style — yield reasons3 marks
A student carries out a reaction in the laboratory but obtains a percentage yield of only 65%. Explain THREE possible reasons why the actual yield is less than the theoretical yield, even when the chemicals are pure and the equation is correct. (3 marks)
Model answer
1. The reaction may not go to completion. Many reactions are reversible ( $\rightleftharpoons$ ) — the product partly converts back to reactants, so less product is obtained than predicted. Other reactions are slow and incomplete in the time allowed.

2. Side reactions may occur. Reactants can take alternative reaction paths and form unwanted by-products instead of (or as well as) the desired product. The reactants used for the side reaction can no longer make the main product.

3. Practical losses during handling. Small amounts of product are inevitably lost when transferring between containers, filtering, washing, drying or crystallising. For example a few mg may stick to the filter paper, the side of the beaker or the spatula.

(Any one further valid reason — e.g. impure reactants giving lower-than-expected mass, evaporation losses, incomplete drying — would also be credited.)
Why this scores
Edexcel 3-mark mark scheme: 1 mark for each of three distinct reasons, chosen from (1) reversible / incomplete reaction; (2) side reactions / by-products; (3) practical losses in handling. Don't repeat the same reason in different words — examiners only credit DIFFERENT reasons.

Key Definitions and Keywords — Chemical Formulae, Equations and Calculations

Definitions to memorise and the exact keywords mark schemes credit for chemical formulae, equations and calculations answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4CH1 sitting.

Relative formula mass ( $M_r$ )
Examiner keyword
The sum of the $A_r$ values of all atoms in a chemical formula. For H₂O: $M_r$ = 2(1) + 16 = 18. No units.
Mole
Examiner keyword
The SI unit for amount of substance. One mole contains $6.02 \times 10^{23}$ particles (Avogadro's constant). 1 mole of any substance has a mass equal to its $M_r$ in grams.
Avogadro's constant
Examiner keyword
The number of particles in one mole: $L = 6.02 \times 10^{23}$ per mole. The same number applies to atoms, molecules, ions, or formula units.
Room temperature and pressure (rtp)
Examiner keyword
Approximately 25 °C and 1 atm (1 atmospheric pressure). At rtp, 1 mole of any gas occupies 24 dm³ (= 24,000 cm³).
Empirical formula (Paper 2)
Examiner keyword
The simplest whole-number ratio of atoms of each element in a compound. e.g. C₆H₁₂O₆ has empirical formula CH₂O.
Molecular formula (Paper 2)
Examiner keyword
The actual number of atoms of each element in one molecule of the compound. Always a whole-number multiple of the empirical formula (e.g. glucose C₆H₁₂O₆ = 6 × CH₂O).
Percentage yield (Paper 2)
Examiner keyword
The actual mass of product obtained as a percentage of the theoretical maximum (calculated from the equation). A measure of the efficiency of an experiment.
Atom economy (Paper 2)
Examiner keyword
The percentage of the reactant mass that ends up in the DESIRED product. High atom economy = sustainable process with little waste.
State symbols
Examiner keyword
Letters in brackets after a chemical formula indicating physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water). Required in balanced equations.

Common Mistakes and Misconceptions — Chemical Formulae, Equations and Calculations

The traps other students keep falling into on chemical formulae, equations and calculations questions — taken from recent Pearson Edexcel IGCSE 4CH1 examiner reports and mark schemes — and how to avoid them.

✕Changing subscripts inside a formula when balancing
4CH1 Examiner Reports 2022-2024
Why it happens
Trying to balance H by changing H₂O to H₃O.
How to avoid it
NEVER change subscripts — they define the compound's identity. Only adjust the coefficients (the numbers IN FRONT). H₃O is a different species from H₂O!
✕Using cm³ in $c = n/V$ when concentration is in mol/dm³
4CH1 Examiner Reports 2022-2024
Why it happens
Forgetting to convert.
How to avoid it
ALWAYS convert volume from cm³ to dm³ first: divide by 1000. 25 cm³ = 0.025 dm³. Then apply $c = n/V$ .
✕Using a 1:1 mole ratio when the balanced equation says otherwise
Why it happens
Rushing past the balanced equation.
How to avoid it
Always WRITE the balanced equation first and identify the mole ratio explicitly. For $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ , the ratio Mg : O₂ is 2 : 1.
✕Confusing atom economy with percentage yield
4CH1 Examiner Reports 2022-2024
Why it happens
Both are percentages relating to product.
How to avoid it
Yield = actual / theoretical (depends on the actual experiment). Atom economy = Mr(desired product) / Mr(all products) (depends ONLY on the equation — same value every time). A reaction with atom economy 50% can still have yield 95%.
✕Rounding empirical-formula ratios too aggressively (e.g. 2.5 → 3)
Why it happens
Assuming small deviations from whole numbers are experimental error.
How to avoid it
If your ratio is 1 : 2.5, multiply ALL values by 2 to get 2 : 5 — a whole-number ratio. 2.5 should not be rounded to 3.
✕Forgetting state symbols
Why it happens
Treating them as optional decoration.
How to avoid it
Most Edexcel marks for 'balance and add state symbols' questions split the state-symbols mark separately. Always add (s), (l), (g) or (aq) — it's a free mark.

Practice questions

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

Past paper style quiz

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

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Chemical Formulae, Equations and Calculations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Chemical Formulae, Equations and Calculations

Short Study Notes in the form of Slides

Short Study Notes — Chemical Formulae, Equations and Calculations

Detailed Notes

What you’ll learn

How it’s examined

1Balancing a chemical equation (3 marks, Paper 1)

2Mass to moles using n=m/Mrn = m / M_rn=m/Mr​ (3 marks, Paper 1)

3Reacting mass (stoichiometry) (5 marks, Paper 1)

4Gas volume at rtp from moles (3 marks, Paper 1)

5Concentration of a solution (4 marks, Paper 1)

6Empirical formula from percentage composition (5 marks, Paper 2)

7Percentage yield and atom economy (5 marks, Paper 2)

Moles from mass (spec 1.30)

Avogadro's constant (spec 1.29)

Gas volume at room temperature and pressure (spec 1.36)

Concentration of a solution (spec 1.38)

Percentage composition by mass

Percentage yield (spec 1.42P, Paper 2)

Atom economy (spec 1.45P, Paper 2)

Relative formula mass (MrM_rMr​)

Mole

Avogadro's constant

Room temperature and pressure (rtp)

Empirical formula (Paper 2)

Molecular formula (Paper 2)

Percentage yield (Paper 2)

Atom economy (Paper 2)

State symbols

✕Changing subscripts inside a formula when balancing

✕Using cm³ in c=n/Vc = n/Vc=n/V when concentration is in mol/dm³

✕Using a 1:1 mole ratio when the balanced equation says otherwise

✕Confusing atom economy with percentage yield

✕Rounding empirical-formula ratios too aggressively (e.g. 2.5 → 3)

✕Forgetting state symbols

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Chemical Formulae, Equations and Calculations — frequently asked questions

2Mass to moles using $n = m / M_r$ (3 marks, Paper 1)

Relative formula mass ( $M_r$ )

✕Using cm³ in $c = n/V$ when concentration is in mol/dm³