Chemical formulae, equations and calculations

Q: Do I need to include state symbols in all equations I write?

Yes! The 4CH1 specification explicitly requires state symbols for all balanced equations. You will lose 1 mark per equation if you omit them, even if the equation is correctly balanced.

Q: How do I convert cm³ to dm³ for calculations?

Divide the volume in cm³ by 1000 to get the value in dm³, e.g. 250 cm³ = 0.25 dm³. Always check units before plugging values into concentration or gas volume formulae.

Q: What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole number ratio of atoms in a compound, while the molecular formula is the actual number of each atom in a molecule. The molecular formula is always a multiple of the empirical formula.

Edexcel International GCSE Chemistry· 1.25–1.36· 45 min read

1. Writing Balanced Chemical Equations with State Symbols★★☆☆☆⏱ 10 min

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Word equations list reactants on the left and products on the right. To write a balanced symbol equation, first replace each substance with its correct chemical formula, then add stoichiometric coefficients to ensure the number of atoms of each element is equal on both sides of the equation. Finally, add state symbols for every species.

📘 Definition

State Symbols

Abbreviations added after each chemical species to show its physical state: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (dissolved in water)

📐 Worked Example

Write a balanced chemical equation for the reaction of magnesium metal with hydrochloric acid to form magnesium chloride solution and hydrogen gas.

1
1. Write the unbalanced word equation: magnesium + hydrochloric acid → magnesium chloride + hydrogen
2
1. Replace with correct formulae and add initial state symbols: $Mg(s) + HCl(aq) \to MgCl_{2} (aq) + H_{2} (g)$
3
1. Balance atoms: 2 Cl atoms on the right, so add a coefficient of 2 to HCl on the left: $Mg(s) + 2 HCl(aq) \to MgCl_{2} (aq) + H_{2} (g)$
4
1. Verify counts: 1 Mg, 2 H, 2 Cl on both sides, equation is balanced with correct state symbols.

Exam tip:

You will lose 1 mark per equation if you omit state symbols, even if the stoichiometry is 100% correct.

2. Basic Mole Calculations, Mr and Percentage Yield★★★☆☆⏱ 12 min

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📘 Definition

Relative Formula Mass (Mr)

Sum of the relative atomic masses (Ar) of all atoms in a chemical formula, with no units.

M_r = \text{sum of } (A_r \times \text{number of atoms of each element})

📐 Worked Example

Calculate the Mr of sulfuric acid ( $H_{2} SO_{4}$ ) using Ar values: H=1, S=32, O=16.

1
1. Calculate total mass for each element: H = 2 × 1 = 2, S = 1 × 32 = 32, O = 4 × 16 = 64
2
1. Sum the values: 2 + 32 + 64 = 98, so $M_{r} = 98$

📘 Definition

Mole Calculation (Mass Relationship)

The amount of a substance in moles is equal to its mass in grams divided by its relative formula mass.

n = \frac{\text{mass (g)}}{M_r}

📐 Worked Example

Calculate the amount in moles of 49 g of $H_{2} SO_{4}$ (Mr = 98).

1
1. Substitute values into the formula: $n = \frac{49}{98} = 0.5$ mol

📘 Definition

Percentage Yield

The ratio of the actual experimental yield of a product to the theoretical maximum yield, expressed as a percentage.

\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100

📐 Worked Example

A reaction produces 8.2 g of copper oxide, with a theoretical yield of 10.0 g. Calculate the percentage yield.

1
1. Substitute values: $Percentage Yield = \frac{8.2}{10.0} \times 100 = 82%$

Exam tip:

You must memorise all three formulae in this section, as no formula sheet is provided in the exam.

3. Reacting Masses, Empirical & Molecular Formulae, Practical Work★★★☆☆⏱ 15 min

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Reacting mass calculations use mole ratios from balanced equations to relate the mass of one reactant or product to another. Empirical and molecular formulae are calculated from experimental mass or percentage composition data.

📐 Worked Example

What mass of magnesium oxide is produced when 4.8 g of magnesium burns in excess oxygen? Ar values: Mg=24, O=16.

1
1. Write balanced equation: $2 Mg(s) + O_{2} (g) \to 2 MgO(s)$
2
1. Calculate moles of Mg: $n = \frac{4.8}{24} = 0.2$ mol
3
1. Mole ratio Mg:MgO = 1:1, so moles of MgO = 0.2 mol
4
1. Mr of MgO = 24 + 16 = 40, mass of MgO = 0.2 × 40 = 8.0 g

📘 Definition

Empirical & Molecular Formula Relationship

Molecular formula = (Empirical formula) × n, where $n = \frac{M _{r} of compound}{Empirical formula mass}$

📐 Worked Example

A compound is 40% C, 6.7% H, 53.3% O by mass, with Mr = 60. Find its empirical and molecular formula. Ar: C=12, H=1, O=16.

1
1. Assume 100 g sample: masses = 40 g C, 6.7 g H, 53.3 g O
2
1. Convert to moles: C = 40/12 ≈ 3.33, H = 6.7/1 = 6.7, O = 53.3/16 ≈ 3.33
3
1. Divide by smallest value (3.33): C=1, H=2, O=1 → Empirical formula = $CH_{2} O$
4
1. Empirical mass = 30, n = 60/30 = 2 → Molecular formula = $C_{2} H_{4} O_{2}$

Practical to find the formula of magnesium oxide: Weigh a crucible with magnesium ribbon, heat strongly lifting the lid occasionally to let air in, reweigh until constant mass. Use mass changes to calculate the mass of magnesium and oxygen that reacted, then find the mole ratio.

Exam tip:

If you get a non-integer mole ratio (e.g. 1.5) when calculating empirical formula, multiply all values by the smallest integer to get whole numbers, do not round 1.5 to 2.

4. Higher Only: Solution Concentration Calculations★★★★☆HL only⏱ 8 min

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📘 Definition

Concentration (mol/dm³)

Amount of solute in moles dissolved per 1 cubic decimetre of solution.

c = \frac{n}{V}

Where c = concentration (mol/dm³), n = moles of solute, V = volume of solution in dm³. Convert cm³ to dm³ by dividing by 1000 before substituting values.

📐 Worked Example

Calculate the concentration of a solution formed by dissolving 0.2 mol of NaOH in 250 cm³ of water.

1
1. Convert volume to dm³: 250 / 1000 = 0.25 dm³
2
1. Substitute into formula: $c = \frac{0.2}{0.25} = 0.8$ mol/dm³

Exam tip:

Nearly all concentration questions give volume in cm³ to test you remember the unit conversion, so always check units first.

5. Higher Only: Gas Volume Calculations at RTP★★★★☆HL only⏱ 8 min

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📘 Definition

Molar Volume at RTP

1 mole of any gas occupies 24 dm³ (24000 cm³) at room temperature and pressure.

n = \frac{V}{24}

Where n = moles of gas, V = volume of gas in dm³. Convert cm³ to dm³ by dividing by 1000 if required.

📐 Worked Example

What volume of CO₂ gas is produced at rtp when 10 g of CaCO₃ decomposes? Ar: Ca=40, C=12, O=16.

1
1. Balanced equation: $CaCO_{3} (s) \to CaO(s) + CO_{2} (g)$
2
1. Mr of CaCO₃ = 100, moles of CaCO₃ = 10 / 100 = 0.1 mol
3
1. Mole ratio CaCO₃:CO₂ = 1:1, so moles of CO₂ = 0.1 mol
4
1. Volume = 0.1 × 24 = 2.4 dm³ (or 2400 cm³ if requested)

Exam tip:

You must memorise that molar volume is 24 dm³ at rtp, this value is not provided in the exam paper.

6. Common Pitfalls

Wrong move:

Omitting state symbols from balanced chemical equations

Why:

The 4CH1 specification explicitly requires state symbols, so you lose 1 mark per equation even if stoichiometry is correct

Correct move:

Add (s), (l), (g), (aq) to every species in all equations you write

Wrong move:

Forgetting to convert cm³ to dm³ for concentration/gas volume calculations

Why:

Formulae use volume in dm³, so using cm³ gives results 1000x larger/smaller than the correct value

Correct move:

Divide volume in cm³ by 1000 to get dm³ before substituting into any formula

Wrong move:

Rounding non-integer mole ratios (e.g. 1.5 to 2) for empirical formula calculations

Why:

Rounding leads to an incorrect atom ratio, and lost marks for the entire calculation

Correct move:

Multiply all mole ratios by the smallest integer to eliminate decimals (e.g. multiply by 2 for 0.5 increments, by 3 for 0.33 increments)

Wrong move:

Using Ar instead of Mr for mole calculations of compounds

Why:

This gives an incorrect mole value, leading to wrong results for all downstream calculation steps

Correct move:

Always calculate Mr for compounds first by summing Ar values of all atoms in the formula

Wrong move:

Using the wrong mole ratio from balanced equations for reacting mass calculations

Why:

The mole ratio links reactants and products, so an incorrect ratio leads to wrong values for the desired substance

Correct move:

Use the stoichiometric coefficients from the balanced equation to get the correct ratio between the two species in your calculation

7. Quick Reference Cheatsheet

Formula / Concept	Equation / Rule	Units / Notes
Balanced Equations	Equal atoms of each element on both sides	Always add state symbols (s)/(l)/(g)/(aq)
Relative Formula Mass	$M_{r} = \sum (A_{r} \times atom count)$	No units
Mole (mass)	$n = \frac{mass}{M _{r}}$	Mass in g, n in mol
Percentage Yield	$% Yield = \frac{Actual}{Theoretical} \times 100$	No units, always <100%
Empirical Formula	Simplest whole number atom ratio	Divide masses by Ar, then divide by smallest value
Molecular Formula	$Empirical \times n$ , $n = \frac{M _{r}}{Empirical Mass}$
Concentration (Higher only)	$c = \frac{n}{V}$	c in mol/dm³, V in dm³
Gas Volume (Higher only)	$n = \frac{V}{24}$ at rtp	V in dm³, 24 dm³ = 24000 cm³

8. Frequently Asked

Do I need to include state symbols in all equations I write?

Yes! The 4CH1 specification explicitly requires state symbols for all balanced equations. You will lose 1 mark per equation if you omit them, even if the equation is correctly balanced.

How do I convert cm³ to dm³ for calculations?

Divide the volume in cm³ by 1000 to get the value in dm³, e.g. 250 cm³ = 0.25 dm³. Always check units before plugging values into concentration or gas volume formulae.

What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole number ratio of atoms in a compound, while the molecular formula is the actual number of each atom in a molecule. The molecular formula is always a multiple of the empirical formula.

Going deeper

What's Next

Mastering these calculation skills is foundational for all subsequent topics in Edexcel IGCSE Chemistry, as quantitative questions appear in every exam paper across both core and higher tiers. You will apply these mole calculation principles to reactions like acid-base neutralisation, redox processes, and organic synthesis in later units. Next, practice applying these skills to past paper questions to build speed and accuracy, and make sure you memorise the required formulae as no formula sheet is provided in the exam. For higher tier students, you will use these concentration calculation skills for titration practical work in the next section of the syllabus.