The Mole

This lesson introduces the concept of the mole — the chemist's counting unit — as required by the Edexcel GCSE Combined Science specification (1SC0). 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.

Exam Tip: 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 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 — Moles from Mass

Question: How many moles are in 44 g of carbon dioxide ($\text{CO}_2$CO2​)?

Step 1: Calculate $M_r$Mr​ of $\text{CO}_2$CO2​: $M_r = 12 + (2 \times 16) = 12 + 32 = 44$Mr​=12+(2×16)=12+32=44

Step 2: Use the formula: $\text{moles} = \frac{\text{mass}}{M_r} = \frac{44}{44} = 1 \text{ mol}$moles=Mr​mass​=4444​=1 mol

Example 2 — Mass from Moles

Question: What is the mass of 0.5 mol of sodium hydroxide ($\text{NaOH}$NaOH)?

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

Step 2: Use the formula: $\text{mass} = \text{moles} \times M_r = 0.5 \times 40 = 20 \text{ g}$mass=moles×Mr​=0.5×40=20 g

Example 3 — Moles from Mass (Non-Whole Number)

Question: Calculate the number of moles in 7.1 g of chlorine gas ($\text{Cl}_2$Cl2​).

Step 1: Calculate $M_r$Mr​ of $\text{Cl}_2$Cl2​: $M_r = 2 \times 35.5 = 71$Mr​=2×35.5=71

Step 2: Use the formula: $\text{moles} = \frac{7.1}{71} = 0.1 \text{ mol}$moles=717.1​=0.1 mol

Example 4 — Finding Mass of a Given Number of Moles

Question: What is the mass of 2.5 mol of calcium carbonate ($\text{CaCO}_3$CaCO3​)?

Step 1: Calculate $M_r$Mr​ of $\text{CaCO}_3$CaCO3​: $M_r = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100$Mr​=40+12+(3×16)=40+12+48=100

Step 2: Use the formula: $\text{mass} = 2.5 \times 100 = 250 \text{ g}$mass=2.5×100=250 g

Example 5 — Finding the Number of Particles

Question: How many molecules are in 0.25 mol of water?

$\text{number of particles} = \text{moles} \times N_A = 0.25 \times 6.022 \times 10^{23} = 1.506 \times 10^{23}$number of particles=moles×NA​=0.25×6.022×1023=1.506×1023

Exam Tip: Always write the equation first, then substitute, then calculate. This three-step approach earns full marks and avoids errors.

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One Mole of Different Substances

One mole of any substance has a mass (in grams) equal to its $M_r$Mr​:

Substance	Formula	$M_r$Mr​	Mass of 1 Mole
Carbon	C	12	12 g
Water	$\text{H}_2\text{O}$H2​O	18	18 g
Carbon dioxide	$\text{CO}_2$CO2​	44	44 g
Sodium chloride	NaCl	58.5	58.5 g
Calcium carbonate	$\text{CaCO}_3$CaCO3​	100	100 g

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Common Mistakes to Avoid

Mistake	Correction
Using $A_r$Ar​ of one atom instead of $M_r$Mr​ of the whole formula	For $\text{H}_2\text{O}$H2​O, use $M_r$Mr​ = 18, not $A_r$Ar​ of O = 16
Forgetting to calculate $M_r$Mr​ for diatomic elements	Chlorine gas is $\text{Cl}_2$Cl2​ ($M_r$Mr​ = 71), not Cl ($A_r$Ar​ = 35.5)
Mixing up the rearrangements	Use the triangle: cover what you want to find
Leaving out units	Moles are in mol, mass is in g

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Summary

• The mole is the unit for amount of substance. 1 mol = $6.022 \times 10^{23}$6.022×1023 particles (Avogadro's constant).
• The key formula is: $\text{moles} = \dfrac{\text{mass (g)}}{M_r}$moles=Mr​mass (g)​.
• One mole of any substance has a mass in grams numerically equal to its $M_r$Mr​.
• Always calculate $M_r$Mr​ first, then use the formula.
• Show all working: equation → substitution → answer with unit.

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Extended Worked Examples

These examples push beyond the basic triangle. Each one is written in the layout an examiner wants to see: equation first, then substitution, then answer with units. Read through once, then cover the solution and redo the arithmetic yourself.

Example 6 — Moles of Carbon in Diamond

Question: A diamond ring contains a 0.30 g diamond made of pure carbon. How many moles and how many carbon atoms does it contain?

Step 1 — write the equation for moles.

$n = \frac{m}{M_r}$n=Mr​m​

Step 2 — substitute ($M_r$Mr​ of C = 12).

$n = \frac{0.30}{12} = 0.025 \text{ mol}$n=120.30​=0.025 mol

Step 3 — use Avogadro's constant for number of atoms.

$\text{atoms} = n \times N_A = 0.025 \times 6.022 \times 10^{23} = 1.51 \times 10^{22}\,\text{atoms (3 s.f.)}$atoms=n×NA​=0.025×6.022×1023=1.51×1022atoms (3 s.f.)

Example 7 — Moles of a Mixed-Element Compound

Question: How many moles of atoms are in 6.0 g of water, $\text{H}_2\text{O}$H2​O?

Step 1 — moles of water molecules. $M_r$Mr​ of $\text{H}_2\text{O}$H2​O = 18.

$n(\text{H}_2\text{O}) = \frac{6.0}{18} = 0.333... \text{ mol}$n(H2​O)=186.0​=0.333... mol

Step 2 — each water molecule contains 3 atoms (2 H + 1 O).

$n(\text{atoms}) = 3 \times 0.333... = 1.00\,\text{mol of atoms}$n(atoms)=3×0.333...=1.00mol of atoms

Or $6.022 \times 10^{23}$6.022×1023 atoms in 6.0 g of water — a surprising amount.

Example 8 — Mass for an Exact Number of Moles