# Chemical formulae, equations and calculations

> Edexcel International GCSE Chemistry · 4CH1 (2017)
> Source: https://www.owlsprep.com/study/edexcel-igcse-chemistry-s1-chemical-formulae-equations-and-calculations/

This guide covers all core calculation and equation skills for Edexcel IGCSE Chemistry 4CH1 section 1(e), including balanced equations, mole calculations, empirical formulae, percentage yield, and Higher-only solution and gas volume problems.

**Prerequisites:** [Atomic structure and relative atomic mass](https://www.owlsprep.com/study/edexcel-igcse-chemistry-s1-atomic-structure/); [Common ion charges and simple compound formulae](https://www.owlsprep.com/study/edexcel-igcse-chemistry-s1-elements-compounds-mixtures/)

## Learning objectives

- Write balanced chemical equations with state symbols for familiar and unfamiliar reactions
- Calculate relative formula mass (Mr) from given Ar values
- Use the mole relationship n = mass / Mr for basic calculations
- Calculate reacting masses from balanced chemical equations
- Calculate percentage yield of chemical reactions
- Determine empirical and molecular formulae from experimental data
- Perform solution concentration calculations (Higher tier only)
- Perform gas volume calculations at rtp (Higher tier only)
- Describe the practical to determine the formula of a metal oxide

## Writing Balanced Chemical Equations with State Symbols

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.

**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. 2. Replace with correct formulae and add initial state symbols: $\text{Mg(s)} + \text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$
3. 3. Balance atoms: 2 Cl atoms on the right, so add a coefficient of 2 to HCl on the left: $\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$
4. 4. 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.

*Calculator:* allowed

## Basic Mole Calculations, Mr and Percentage Yield

**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 ($\text{H}_2\text{SO}_4$) 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. 2. Sum the values: 2 + 32 + 64 = 98, so $M_r = 98$

**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 $\text{H}_2\text{SO}_4$ (Mr = 98).

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

**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: $\text{Percentage Yield} = \frac{8.2}{10.0} \times 100 = 82\%$

> **Exam tip:** You must memorise all three formulae in this section, as no formula sheet is provided in the exam.

*Calculator:* allowed

## Reacting Masses, Empirical & Molecular Formulae, Practical Work

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: $2\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)}$
2. 2. Calculate moles of Mg: $n = \frac{4.8}{24} = 0.2$ mol
3. 3. Mole ratio Mg:MgO = 1:1, so moles of MgO = 0.2 mol
4. 4. Mr of MgO = 24 + 16 = 40, mass of MgO = 0.2 × 40 = 8.0 g

**Empirical & Molecular Formula Relationship** — Molecular formula = (Empirical formula) × n, where $n = \frac{M_r \text{ of compound}}{\text{Empirical formula mass}}$

**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. 2. Convert to moles: C = 40/12 ≈ 3.33, H = 6.7/1 = 6.7, O = 53.3/16 ≈ 3.33
3. 3. Divide by smallest value (3.33): C=1, H=2, O=1 → Empirical formula = $\text{CH}_2\text{O}$
4. 4. Empirical mass = 30, n = 60/30 = 2 → Molecular formula = $\text{C}_2\text{H}_4\text{O}_2$

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.

*Calculator:* allowed

## Higher Only: Solution Concentration Calculations

**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. 2. 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.

*Calculator:* allowed

## Higher Only: Gas Volume Calculations at RTP

**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: $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$
2. 2. Mr of CaCO₃ = 100, moles of CaCO₃ = 10 / 100 = 0.1 mol
3. 3. Mole ratio CaCO₃:CO₂ = 1:1, so moles of CO₂ = 0.1 mol
4. 4. 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Omitting state symbols from balanced chemical equations
  - Why it fails: The 4CH1 specification explicitly requires state symbols, so you lose 1 mark per equation even if stoichiometry is correct
  - Correct: Add (s), (l), (g), (aq) to every species in all equations you write
- **Wrong:** Forgetting to convert cm³ to dm³ for concentration/gas volume calculations
  - Why it fails: Formulae use volume in dm³, so using cm³ gives results 1000x larger/smaller than the correct value
  - Correct: Divide volume in cm³ by 1000 to get dm³ before substituting into any formula
- **Wrong:** Rounding non-integer mole ratios (e.g. 1.5 to 2) for empirical formula calculations
  - Why it fails: Rounding leads to an incorrect atom ratio, and lost marks for the entire calculation
  - Correct: 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:** Using Ar instead of Mr for mole calculations of compounds
  - Why it fails: This gives an incorrect mole value, leading to wrong results for all downstream calculation steps
  - Correct: Always calculate Mr for compounds first by summing Ar values of all atoms in the formula
- **Wrong:** Using the wrong mole ratio from balanced equations for reacting mass calculations
  - Why it fails: The mole ratio links reactants and products, so an incorrect ratio leads to wrong values for the desired substance
  - Correct: Use the stoichiometric coefficients from the balanced equation to get the correct ratio between the two species in your calculation

## 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 | $M_r = \sum(A_r \times \text{atom count})$ | No units |
| Mole (mass) | $n = \frac{\text{mass}}{M_r}$ | Mass in g, n in mol |
| Percentage Yield | $\% \text{Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100$ | No units, always <100% |
| Empirical Formula | Simplest whole number atom ratio | Divide masses by Ar, then divide by smallest value |
| Molecular Formula | $\text{Empirical} \times n$, $n = \frac{M_r}{\text{Empirical Mass}}$ |  |
| 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 |

## 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.

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