# Formulae, Equations and Amount of Substance

> Edexcel International A-Level Chemistry · IAS Unit 1 (WCH11)
> Source: https://www.owlsprep.com/study/edexcel-ial-chemistry-u1-formulae-equations-and-amount-of/

This guide covers all Edexcel IAL Chemistry Unit 1 Topic 1 content: balanced equations, mole calculations, gas laws, percentage yield, atom economy, and Core Practical 1 for molar volume of gases.

**Prerequisites:** [Basic understanding of atomic structure](https://www.owlsprep.com/study/edexcel-ial-chemistry-u1-atomic-structure/); Familiarity with SI unit conversion rules

## Learning objectives

- Define key terms including empirical/molecular formula, mole, and Avogadro constant
- Write balanced full and ionic equations with required state symbols
- Calculate relative masses, molar mass, concentration, and ppm values
- Derive empirical and molecular formulae from experimental data
- Perform reacting mass, gas volume, pV=nRT, % yield, and atom economy calculations
- Understand Core Practical 1 methodology for measuring molar volume of a gas

## Key Terms and Formula Fundamentals

**Empirical vs Molecular Formula** — Empirical formula = simplest whole number ratio of atoms of each element in a compound. Molecular formula = actual number of atoms of each element in one molecule of the compound, which is a whole-number multiple of the empirical formula.

*Example:* Ethane: empirical formula = CH₃, molecular formula = C₂H₆

Foundational terms for this topic include: **atom** (smallest unit of an element), **element** (substance made of only one type of atom), **ion** (charged particle from gain/loss of electrons), **molecule** (neutral particle of covalently bonded atoms), **compound** (substance made of two or more elements chemically bonded). The mole is the standard unit for amount of substance, with 1 mole containing 6.02×10²³ particles (Avogadro constant).

**Worked example:** A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate its empirical formula.

1. Assume a 100g sample, so masses: C = 40.0g, H = 6.7g, O = 53.3g
2. Calculate moles of each element: $n(C) = 40.0/12.0 ≈ 3.33$, $n(H) = 6.7/1.0 = 6.7$, $n(O) = 53.3/16.0 ≈ 3.33$
3. Divide all mole values by the smallest value (3.33): C = 1, H = 2, O = 1
4. Final empirical formula: CH₂O

> **Exam tip:** Always round mole ratios to the nearest whole number, and confirm no common factors remain in your final empirical formula.

## Balanced Full and Ionic Equations

**Ionic Equations** — Equations that show only reacting ions in a reaction, omitting spectator ions that do not take part. They must be balanced for both mass and total charge, and include state symbols for all species (s/l/g/aq).

To write an ionic equation: first write a balanced full equation with state symbols, split all aqueous ionic compounds into their constituent ions, cancel spectator ions that appear on both sides, then check balance of mass and charge.

**Worked example:** Write the ionic equation for the reaction between aqueous silver nitrate and aqueous sodium chloride, forming solid silver chloride and aqueous sodium nitrate.

1. Full balanced equation: $AgNO_3(aq) + NaCl(aq) → AgCl(s) + NaNO_3(aq)$
2. Split aqueous species into ions: $Ag^+(aq) + NO_3^-(aq) + Na^+(aq) + Cl^-(aq) → AgCl(s) + Na^+(aq) + NO_3^-(aq)$
3. Cancel spectator ions (Na⁺ and NO₃⁻) from both sides
4. Final balanced ionic equation: $Ag^+(aq) + Cl^-(aq) → AgCl(s)$

> **Exam tip:** Never split solids, pure liquids, or gases into ions, only aqueous ionic compounds. Always check total charge on both sides of ionic equations matches.

## Mole Calculations: Mass, Concentration and ppm

**Molar Quantities** — Molar mass ($M$) = mass of 1 mole of substance, units g mol⁻¹, equal to the Ar/Mr of the substance. Concentration can be measured in mol dm⁻³ (moles of solute per dm³ of solution) or g dm⁻³ (mass of solute per dm³ of solution). ppm (parts per million) is used for trace concentrations.

- $n = \frac{m}{M}$ where $n$ = moles, $m$ = mass (g), $M$ = molar mass (g mol⁻¹)
- Concentration (mol dm⁻³) = $\frac{n}{V}$ where $V$ = volume (dm³, divide cm³ by 1000 to convert)
- ppm = $\frac{\text{mass of component}}{\text{total mass of mixture}} \times 10^6$

**Worked example:** Calculate the concentration in g dm⁻³ of a 0.200 mol dm⁻³ solution of NaOH.

1. Calculate Mr of NaOH: $23.0 + 16.0 + 1.0 = 40.0$ g mol⁻¹
2. Multiply molar concentration by Mr: $0.200 \times 40.0 = 8.00$ g dm⁻³

> **Exam tip:** Round final calculation answers to the same number of significant figures as the least precise data given in the question.

## Gas Volume Calculations and pV=nRT

**Ideal Gas Equation** — At room temperature and pressure (RTP = 298 K, 101 kPa), 1 mole of any gas occupies 24.0 dm³. For non-RTP conditions, use the ideal gas equation $pV = nRT$, where $p$ = pressure (Pa), $V$ = volume (m³), $n$ = moles, $R$ = 8.31 J K⁻¹ mol⁻¹, $T$ = temperature (K).

**Worked example:** Calculate the volume occupied by 0.0200 mol of carbon dioxide gas at RTP.

1. Use the molar volume relationship: $V = n \times V_m$
2. Substitute values: $V = 0.0200 \times 24.0 = 0.480$ dm³ (or 480 cm³)

**Worked example:** Calculate the number of moles of gas in a 500 cm³ container at 27°C and 100 kPa pressure.

1. Convert units: $V = 500 \times 10^{-6} = 5 \times 10^{-4}$ m³, $T = 27 + 273 = 300$ K, $p = 100 \times 10^3 = 1 \times 10^5$ Pa
2. Rearrange $pV = nRT$ to $n = \frac{pV}{RT}$
3. Calculate: $n = \frac{1 \times 10^5 \times 5 \times 10^{-4}}{8.31 \times 300} ≈ 0.0201$ mol

> **Exam tip:** Always convert all units to SI before using pV=nRT to avoid order of magnitude errors.

## % Yield, Atom Economy and Core Practical 1

**Yield and Atom Economy** — % yield measures the efficiency of product formation: $\% yield = \frac{\text{actual yield}}{\text{theoretical maximum yield}} \times 100$. % atom economy measures the sustainability of a reaction: $\% atom economy = \frac{Mr \text{ of desired product}}{\text{sum of Mr of all reactants}} \times 100$.

**Worked example:** Calculate the atom economy for the production of ethanol from ethene and water: $C_2H_4 + H_2O → C_2H_5OH$.

1. Calculate Mr of desired product (ethanol): $(2 \times 12) + (6 \times 1) + 16 = 46$
2. Sum of Mr of reactants: $C_2H_4 = 28$, $H_2O = 18$, total = 46
3. Atom economy = $\frac{46}{46} \times 100 = 100\%$ (no waste products for this addition reaction)

**Core Practical 1: Molar volume of gas** method: React a known mass of magnesium ribbon with excess hydrochloric acid, collect the hydrogen gas produced over water, measure its volume at RTP. Calculate moles of Mg (and thus moles of H₂ from the 1:1 reaction ratio), then $V_m = \frac{\text{volume of H₂}}{\text{moles of H₂}}$. Expected result = ~24 dm³ mol⁻¹ at RTP.

> **Exam tip:** Common sources of error for Core Practical 1 include gas escaping during collection, measuring gas volume before it cools to room temperature, and inaccurate mass measurement of magnesium.

## Common pitfalls

- **Wrong:** Forgetting to add state symbols to equations
  - Why it fails: Exam questions explicitly require state symbols for full marks, you will lose 1 mark per equation missing them
  - Correct: Always add (s)/(l)/(g)/(aq) to every species in all equations you write
- **Wrong:** Using cm³ directly in pV=nRT without converting to m³
  - Why it fails: pV=nRT requires SI units for volume, using cm³ will give an answer 1 million times too large
  - Correct: Convert cm³ to m³ by multiplying by 10⁻⁶, or dm³ to m³ by multiplying by 10⁻³
- **Wrong:** Balancing ionic equations only for mass, not charge
  - Why it fails: Ionic equations require equal total charge on both sides to be chemically correct, unbalanced charge will lose marks
  - Correct: After balancing mass, check total charge on left equals total charge on right, adjust coefficients if needed
- **Wrong:** Using total Mr of all products for the atom economy denominator
  - Why it fails: Atom economy uses total mass of reactants, not products, so using products gives an incorrect value
  - Correct: Use the sum of Mr of all reactants in the denominator of the atom economy formula
- **Wrong:** Leaving empirical formula ratios not fully simplified
  - Why it fails: Empirical formula is defined as the simplest whole number ratio, non-simplified ratios are invalid
  - Correct: Divide all mole ratios by the smallest mole value, then confirm no common factors remain between all numbers

## Cheatsheet

| Formula | Units/Notes | Use Case |
| --- | --- | --- |
| $n = \frac{m}{M}$ | n: mol, m: g, M: g mol⁻¹ | Calculate moles from mass of substance |
| $Conc (mol dm^{-3}) = \frac{n}{V}$ | V: dm³ (divide cm³ by 1000) | Calculate concentration or moles in solution |
| ppm = $\frac{\text{component mass}}{\text{total mass}} \times 10^6$ | No units | Calculate trace concentration values |
| $V = n \times 24.0$ dm³ | At RTP (298 K, 101 kPa) | Calculate gas volume at room temperature |
| $pV = nRT$ | p: Pa, V: m³, T: K, R=8.31 J K⁻¹ mol⁻¹ | Calculate gas quantities at non-RTP conditions |
| % Yield = $\frac{\text{actual}}{\text{theoretical}} \times 100$ | % value | Calculate efficiency of product formation |
| % Atom Economy = $\frac{\text{Mr(desired)}}{\text{sum Mr(reactants)}} \times 100$ | % value | Calculate sustainability of a reaction |

## What's next

This foundational stoichiometry content is the building block for all quantitative chemistry in both AS and A2 Edexcel IAL Chemistry, so practice plenty of calculation questions to build confidence. You will use these mole calculation skills extensively in future topics including energetics, equilibrium, kinetics, and organic chemistry. Mastery of this topic will help you avoid common calculation errors across all units, and make more advanced content much easier to grasp.

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