The Mole and Avogadro's Constant

This lesson introduces the concept of the mole — the chemist's counting unit — as required by the AQA GCSE Combined Science Trilogy specification (8464). You will learn what a mole is, understand Avogadro's constant, and be able to convert between mass, moles and relative formula mass.

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Why Do Chemists Need the Mole?

Atoms are incredibly small. A single carbon atom has a mass of about $2 \times 10^{-23}$2×10−23 g — far too small to weigh individually. Chemists need a way to count atoms by weighing substances, and the mole provides that bridge.

The mole connects the submicroscopic world (atoms and molecules) to the macroscopic world (grams and kilograms on a balance).

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What Is One Mole?

One mole of any substance contains exactly $6.022 \times 10^{23}$6.022×1023 particles (atoms, molecules, ions or formula units). This number is called Avogadro's constant ($N_A$NA​).

$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$NA​=6.022×1023 mol−1

Putting This Number in Perspective

• One mole of grains of sand would cover the entire surface of the Earth several metres deep.
• One mole of seconds is about 19 quadrillion years — far longer than the age of the universe.
• If you stacked one mole of pound coins, the pile would reach from Earth to the nearest star and back — millions of times.

Exam Tip (AQA 8464): You do NOT need to memorise Avogadro's constant to many decimal places. In the exam it will be given as $6.022 \times 10^{23}$6.022×1023. Focus on knowing how to use it.

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The Key Relationship: Mass, Moles and $M_r$Mr​

The most important equation in quantitative chemistry is:

$\text{moles} = \frac{\text{mass (g)}}{M_r}$moles=Mr​mass (g)​

This can be rearranged into a mole triangle — cover the quantity you want to find:

graph TD
    A["<b>mass (g)</b>"] --- B["<b>moles</b>"]
    A --- C["<b>M<sub>r</sub></b>"]
    B --- C
    style A fill:#f59e0b,color:#000,stroke:#d97706
    style B fill:#3b82f6,color:#fff,stroke:#2563eb
    style C fill:#3b82f6,color:#fff,stroke:#2563eb

To Find	Formula
Moles	$\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​
Mass	$\text{mass} = \text{moles} \times M_r$mass=moles×Mr​
$M_r$Mr​	$M_r = \dfrac{\text{mass}}{\text{moles}}$Mr​=molesmass​

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Worked Examples

Example 1 — Finding Moles from Mass

Question: How many moles are in 22 g of carbon dioxide ($\text{CO}_2$CO2​)?

$M_r$Mr​ of $\text{CO}_2 = 12 + (2 \times 16) = 44$CO2​=12+(2×16)=44

$\text{moles} = \frac{\text{mass}}{M_r} = \frac{22}{44} = 0.5 \text{ mol}$moles=Mr​mass​=4422​=0.5 mol

Example 2 — Finding Mass from Moles

Question: What is the mass of 0.25 mol of sodium hydroxide ($\text{NaOH}$NaOH)?

$M_r$Mr​ of $\text{NaOH} = 23 + 16 + 1 = 40$NaOH=23+16+1=40

$\text{mass} = \text{moles} \times M_r = 0.25 \times 40 = 10 \text{ g}$mass=moles×Mr​=0.25×40=10 g

Example 3 — Moles of a More Complex Substance

Question: How many moles are in 17.1 g of aluminium sulfate, $\text{Al}_2(\text{SO}_4)_3$Al2​(SO4​)3​?

$M_r = (2 \times 27) + (3 \times 32) + (12 \times 16) = 54 + 96 + 192 = 342$Mr​=(2×27)+(3×32)+(12×16)=54+96+192=342

$\text{moles} = \frac{17.1}{342} = 0.05 \text{ mol}$moles=34217.1​=0.05 mol

Example 4 — Working Backwards to Find $M_r$Mr​

Question: 0.2 mol of a substance has a mass of 11.2 g. What is the $M_r$Mr​?

$M_r = \frac{\text{mass}}{\text{moles}} = \frac{11.2}{0.2} = 56$Mr​=molesmass​=0.211.2​=56

This could be iron ($A_r$Ar​ = 56) or possibly CaO ($M_r$Mr​ = 56).

Example 5 — Number of Particles

Question: How many molecules are in 9 g of water ($\text{H}_2\text{O}$H2​O)?

$M_r$Mr​ of $\text{H}_2\text{O} = 18$H2​O=18

$\text{moles} = \frac{9}{18} = 0.5 \text{ mol}$moles=189​=0.5 mol

$\text{number of molecules} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$number of molecules=0.5×6.022×1023=3.011×1023

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Using the Mole Triangle

A helpful way to remember the three rearrangements is to use a formula triangle:

flowchart TD
    subgraph TRIANGLE["Mole Triangle"]
        M["mass (g)"]
        N["moles"]
        R["M_r"]
    end
    M -->|"÷ M_r"| N
    N -->|"× M_r"| M
    M -->|"÷ moles"| R
    style M fill:#ef4444,color:#fff,stroke:#dc2626
    style N fill:#3b82f6,color:#fff,stroke:#2563eb
    style R fill:#22c55e,color:#fff,stroke:#16a34a

Cover the quantity you want with your thumb:

• Cover moles → mass ÷ $M_r$Mr​
• Cover mass → moles × $M_r$Mr​
• Cover $M_r$Mr​ → mass ÷ moles

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Common Mistakes to Avoid

Mistake	Correction
Mixing up the formula — multiplying instead of dividing	Use the triangle: moles = mass ÷ $M_r$Mr​
Using $A_r$Ar​ instead of $M_r$Mr​ for a compound	$A_r$Ar​ is for single elements; $M_r$Mr​ is the total for the whole formula
Forgetting to calculate $M_r$Mr​ first	Always find $M_r$Mr​ before using the mole equation
Confusing moles with mass	Moles is a count (in units of $6.022 \times 10^{23}$6.022×1023); mass is in grams

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Summary

• The mole is the chemist's counting unit: 1 mol = $6.022 \times 10^{23}$6.022×1023 particles.
• Avogadro's constant ($N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$NA​=6.022×1023 mol−1) tells you how many particles are in one mole.
• The key equation is $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​.
• You can rearrange to find mass or $M_r$Mr​.
• Always calculate $M_r$Mr​ first, then use the mole equation.
• This topic appears on both Foundation and Higher tier AQA papers.

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Extended Worked Examples — Moles and the Mass/$M_r$Mr​ Triangle

Note on AQA Trilogy: the full mole concept ($N = n \times N_A$N=n×NA​) is Triple Science only (5.3.1.5). For Trilogy, focus on the mass–$M_r$Mr​ relationship $n = m / M_r$n=m/Mr​, which IS examinable in Trilogy as part of 5.3.2 (use of amount of substance in relation to masses of pure substances). The worked examples below keep the language consistent with Trilogy expectations.

Worked Example 6 — Sodium Carbonate

Question: How many moles are in 53 g of sodium carbonate ($\text{Na}_2\text{CO}_3$Na2​CO3​)?

Step 1 — $M_r$Mr​: $(2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106$(2×23)+12+(3×16)=46+12+48=106.

Step 2 — moles: $n = \frac{m}{M_r} = \frac{53}{106} = 0.5 \text{ mol}$n=Mr​m​=10653​=0.5 mol

Worked Example 7 — Iron Oxide Rust

Question: What mass of iron(III) oxide, $\text{Fe}_2\text{O}_3$Fe2​O3​, is 0.2 mol?

$M_r = (2 \times 56) + (3 \times 16) = 112 + 48 = 160$Mr​=(2×56)+(3×16)=112+48=160

$m = n \times M_r = 0.2 \times 160 = 32 \text{ g}$m=n×Mr​=0.2×160=32 g

Worked Example 8 — Finding $M_r$Mr​ from Experiment

Question: A chemist weighs 0.1 mol of an unknown compound and finds it has a mass of 8.5 g. What is the $M_r$Mr​?

$M_r = \frac{m}{n} = \frac{8.5}{0.1} = 85$Mr​=nm​=0.18.5​=85