The Mole Concept

The mole is the chemist's counting unit. It links the microscopic world of atoms and molecules to the macroscopic quantities we measure in the laboratory. This lesson covers the Avogadro constant, molar mass, and calculations involving moles of solids, liquids, solutions, and gases.

Spec mapping (AQA 7405): This lesson maps to §3.1.2.1 (relative atomic mass and relative molecular mass), §3.1.2.2 (the Avogadro constant), §3.1.2.3 (the mole and molar mass), and §3.1.2.5 (the ideal gas equation). Solutions and concentrations (§3.1.2.4) are developed in lesson 7 of this course; titration calculations (the anchor for Required Practical 1) in lesson 8. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Definition of the mole and the Avogadro constant are AO1 recall items. Calculations using n = m/M, gas volumes at RTP, and the ideal gas equation pV = nRT are AO2 and feature on every Paper 1 and Paper 2. Multi-step problems (e.g. finding the molar mass of an unknown gas from p, V, T and mass data) test AO3 reasoning and synthesis.

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The Avogadro Constant

One mole of any substance contains exactly 6.022 × 10²³ particles (atoms, molecules, ions, electrons, etc.). This number is called the Avogadro constant (Nₐ) or Avogadro's number.

Nₐ = 6.022 × 10²³ mol⁻¹

One mole of carbon-12 has a mass of exactly 12.00 g. One mole of any element has a mass in grams equal to its relative atomic mass.

Key Point: Always specify the type of particle when using moles. For example, "1 mole of water" means 6.022 × 10²³ water molecules, but it also contains 2 × 6.022 × 10²³ hydrogen atoms and 1 × 6.022 × 10²³ oxygen atoms.

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Molar Mass

The molar mass (M) of a substance is the mass of one mole of that substance, measured in g mol⁻¹. Numerically, it equals the relative formula mass (Mᵣ).

Substance	Formula	Mᵣ	Molar mass (g mol⁻¹)
Water	H₂O	18.0	18.0
Carbon dioxide	CO₂	44.0	44.0
Sodium chloride	NaCl	58.5	58.5
Sulfuric acid	H₂SO₄	98.1	98.1
Calcium carbonate	CaCO₃	100.1	100.1
Ethanol	C₂H₅OH	46.1	46.1

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Calculating Moles from Mass

The fundamental equation linking moles and mass is:

n = m / M

where:

• n = number of moles (mol)
• m = mass (g)
• M = molar mass (g mol⁻¹)

This can be rearranged to:

• m = n × M (to find mass)
• M = m / n (to find molar mass)

Worked Example 1: Moles from Mass

Calculate the number of moles in 11.0 g of carbon dioxide (CO₂).

M(CO₂) = 12.0 + (2 × 16.0) = 44.0 g mol⁻¹

n = m / M = 11.0 / 44.0 = 0.250 mol

Worked Example 2: Mass from Moles

Calculate the mass of 0.350 mol of sodium hydroxide (NaOH).

M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹

m = n × M = 0.350 × 40.0 = 14.0 g

Worked Example 3: Number of Particles

How many molecules are in 9.00 g of water?

M(H₂O) = 18.0 g mol⁻¹

n = 9.00 / 18.0 = 0.500 mol

Number of molecules = n × Nₐ = 0.500 × 6.022 × 10²³ = 3.01 × 10²³ molecules

How many atoms in total? Each water molecule has 3 atoms (2H + 1O).

Total atoms = 3.01 × 10²³ × 3 = 9.03 × 10²³ atoms

Exam Tip: If a question asks for the number of atoms rather than molecules, multiply by the number of atoms per molecule. This is a common source of error.

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Moles of Gases

Molar Volume at Room Temperature and Pressure (RTP)

At room temperature and pressure (approximately 298 K and 101.3 kPa), one mole of any gas occupies approximately 24.0 dm³ (or 24 000 cm³).

n = V / 24.0 (where V is in dm³)

n = V / 24 000 (where V is in cm³)

Worked Example 4: Volume of Gas at RTP

Calculate the volume of 0.150 mol of oxygen gas at RTP.

V = n × 24.0 = 0.150 × 24.0 = 3.60 dm³

Worked Example 5: Moles from Volume of Gas at RTP

Calculate the number of moles in 480 cm³ of ammonia gas at RTP.

n = V / 24 000 = 480 / 24 000 = 0.0200 mol

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The Ideal Gas Equation

For conditions other than RTP, we use the ideal gas equation:

pV = nRT

where:

• p = pressure in Pa (pascals)
• V = volume in m³
• n = number of moles (mol)
• R = gas constant = 8.314 J K⁻¹ mol⁻¹
• T = temperature in K (kelvin)

Unit Conversions

Conversion	Relationship
kPa → Pa	× 1000
atm → Pa	× 101 325
dm³ → m³	÷ 1000
cm³ → m³	÷ 1 000 000
°C → K	+ 273

Common Misconception: Students frequently forget to convert units. The ideal gas equation ONLY works when p is in Pa, V is in m³, and T is in K. Getting units wrong is the most common source of error.

Worked Example 6: Using pV = nRT to Find Volume

Calculate the volume occupied by 0.200 mol of nitrogen gas at 300 K and 100 kPa.

p = 100 kPa = 100 000 Pa n = 0.200 mol T = 300 K R = 8.314 J K⁻¹ mol⁻¹

pV = nRT

V = nRT / p = (0.200 × 8.314 × 300) / 100 000

V = 498.84 / 100 000

V = 4.99 × 10⁻³ m³ = 4.99 dm³

Worked Example 7: Using pV = nRT to Find Moles

A gas occupies 2.50 dm³ at 25 °C and 110 kPa. Calculate the number of moles.

p = 110 000 Pa V = 2.50 / 1000 = 2.50 × 10⁻³ m³ T = 25 + 273 = 298 K R = 8.314 J K⁻¹ mol⁻¹

n = pV / RT = (110 000 × 2.50 × 10⁻³) / (8.314 × 298)

n = 275.0 / 2477.6

n = 0.111 mol (to 3 significant figures)

Worked Example 8: Using pV = nRT to Find Molar Mass

A gas has a mass of 1.28 g and occupies 500 cm³ at 100 kPa and 20 °C. Calculate its molar mass.

p = 100 000 Pa V = 500 / 1 000 000 = 5.00 × 10⁻⁴ m³ T = 20 + 273 = 293 K R = 8.314 J K⁻¹ mol⁻¹

n = pV / RT = (100 000 × 5.00 × 10⁻⁴) / (8.314 × 293)

n = 50.0 / 2436.0

n = 0.02053 mol

M = m / n = 1.28 / 0.02053 = 62.3 g mol⁻¹

Exam Tip: When using pV = nRT, always write out the unit conversion steps explicitly. Marks are often awarded for correct conversions even if the final answer is wrong.

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Summary of Key Equations

Calculation	Equation
Moles from mass	n = m / M
Moles of gas at RTP	n = V(dm³) / 24.0
Ideal gas equation	pV = nRT
Number of particles	N = n × Nₐ

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Practice Problems

1. Calculate the number of moles in 4.00 g of NaOH. [M = 40.0 g mol⁻¹] Answer: n = 4.00 / 40.0 = 0.100 mol

2. What mass of CaCO₃ contains 0.0500 mol? [M = 100.1 g mol⁻¹] Answer: m = 0.0500 × 100.1 = 5.01 g

3. Calculate the volume of 3.20 g of oxygen (O₂) at RTP. [M = 32.0 g mol⁻¹] Answer: n = 3.20 / 32.0 = 0.100 mol. V = 0.100 × 24.0 = 2.40 dm³

4. A gas occupies 1.00 dm³ at 298 K and 100 kPa. How many moles? Answer: n = (100 000 × 1.00 × 10⁻³) / (8.314 × 298) = 100 / 2477.6 = 0.0404 mol

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Synoptic Links

The mole concept underpins essentially every quantitative calculation in A-Level Chemistry:

• empirical-molecular-formulae (next lesson, §3.1.2.3 continued): empirical and molecular formulae are derived from mole ratios. The skills practised here are directly transferred.
• balanced-equations-calculations (§3.1.2.5 continued): stoichiometry — using mole ratios in balanced equations to predict masses or volumes of products — is impossible without fluency in the mole concept.
• solutions-concentrations and titration-calculations (§3.1.2.4, §3.1.2.5): n = c × V (the solution-concentration equation) is the third fundamental mole equation alongside n = m/M and pV = nRT. Anchors Required Practical 1.
• energetics (§3.1.4, A2): enthalpy changes are reported per mole of reaction; the mole concept is therefore essential for calorimetry calculations.
• kinetics and equilibrium (§3.1.5–§3.1.6, A2): rate constants and equilibrium constants both have units derived from concentrations (mol dm⁻³) — the mole appears explicitly in every kinetic and equilibrium expression.

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Specimen Exam Question — modelled on the AQA A-Level Chemistry paper format

Question 1. [11 marks total]

(a) State the meaning of the Avogadro constant. [1 mark]

(b) Calculate the number of molecules in 4.40 g of carbon dioxide (Mᵣ = 44.0). [3 marks]

(c) A sample of an unknown gas X has a mass of 2.42 g and occupies 1.20 dm³ at 100 kPa and 295 K. Calculate the molar mass of X to three significant figures and suggest its identity. [4 marks]

(d) Explain why 1.00 mol of nitrogen gas (N₂) and 1.00 mol of carbon dioxide gas (CO₂) occupy the same volume at the same temperature and pressure, even though their molar masses differ. [3 marks]

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Mark Scheme — Specimen Question 1 (modelled, not a real paper)

(a) Meaning of Avogadro constant [1 mark, AO1]

• 1 mark: the number of particles (atoms, molecules, ions, etc.) in one mole of a substance, equal to 6.022 × 10²³ mol⁻¹.

Accept any equivalent precise wording. Do not award for "the number of atoms in 12 g of carbon-12" alone — that's an out-of-date definition (post-2019 SI redefinition the constant is defined directly).

(b) Molecules in 4.40 g CO₂ [3 marks, AO2]

• 1 mark: n = m/M = 4.40/44.0 = 0.100 mol.
• 1 mark: number of molecules = n × Nₐ = 0.100 × 6.022 × 10²³.
• 1 mark: final answer 6.022 × 10²² (or 6.02 × 10²², accepting 3 s.f.).

A common error is to give 6.022 × 10²³ (forgetting to multiply by 0.100). Read the question — it asks for molecules in 4.40 g, not in 1 mole.

(c) Molar mass of X [4 marks, AO2 + AO3]