The Mole & Avogadro's Constant Revision Notes | A Level AQA Chemistry

1. Why Chemists Need a Counting Unit

Picture trying to count out exactly one billion grains of sand by hand. Atoms are incomparably smaller than grains of sand — a single copper coin contains roughly $2.4 \times 10^{22}$ copper atoms. Counting them individually is not just impractical; it is physically impossible.

Chemists solved this problem the same way a shopkeeper sells eggs: not one at a time, but in fixed groups. The chemist's group is called the mole, and the fixed number in that group is Avogadro's constant. Once you see the mole as a counting unit — like a dozen, but astronomically larger — everything else in this topic clicks into place.

This topic underpins almost every quantitative calculation in A-Level chemistry: titrations, gas volumes, empirical formulae, and equilibrium constants all depend on working confidently with moles. Getting this foundation solid now pays dividends across the entire course.

2. The Mole and Avogadro's Constant

The number $6.022 \times 10^{23}$ has a name: Avogadro's constant, given the symbol $L$ (sometimes written $N_A$). Its unit is mol$^{-1}$, which means 'per mole'. AQA does not require you to recall this value in the exam — it will be provided on the data sheet — but you absolutely need to know how to use it.

The relationship between amount in moles, number of particles, and Avogadro's constant is captured in one equation. Think of it as a scaling factor: multiply moles by $L$ to scale up to the actual particle count, or divide the particle count by $L$ to scale back down to moles.

3. What Counts as a 'Particle'? Applying the Mole Broadly

Here is where students often trip up: the mole is not restricted to atoms. AQA requires you to apply the mole concept to electrons, atoms, molecules, ions, formula units, and equations. The particle type must always be specified, because the same number of moles of different particle types represents the same count but different chemical entities.

Notice that ionic solids like NaCl do not contain discrete molecules — they consist of a giant lattice of ions. Chemists refer to one 'formula unit' of NaCl, meaning one Na$^+$ ion paired with one Cl$^-$ ion. One mole of NaCl therefore contains $6.022 \times 10^{23}$ formula units, which is simultaneously $6.022 \times 10^{23}$ Na$^+$ ions and $6.022 \times 10^{23}$ Cl$^-$ ions.

Electrons are perhaps the most surprising entry on that list. In electrochemistry and half-equations, moles of electrons are transferred between species. One mole of electrons carries a charge of 96 485 C (the Faraday constant) — but that connection belongs to a later topic. For now, the key point is that the mole applies to electrons just as naturally as it applies to atoms.

4. The Mole in Chemical Equations

Chemical equations do not describe what happens to individual atoms — they describe what happens to moles of substances. The coefficients in a balanced equation are molar ratios. This is one of the most powerful ideas in quantitative chemistry.

Consider the combustion of methane: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$. The coefficients tell you that 1 mol of CH$_4$ reacts with exactly 2 mol of O$_2$ to produce 1 mol of CO$_2$ and 2 mol of H$_2$O. Scale that up: 2 mol of CH$_4$ requires 4 mol of O$_2$, and so on. The ratio is fixed by the equation.

At the particle level, those same coefficients tell you that $6.022 \times 10^{23}$ molecules of CH$_4$ react with $2 \times 6.022 \times 10^{23}$ molecules of O$_2$. The mole bridges the equation on paper with the actual particles in the flask — that is its real power.

5. Calculations Using Avogadro's Constant

The formula $N = n \times L$ is the engine of this topic. You need to be comfortable rearranging it in both directions: finding the number of particles from moles, and finding moles from a given number of particles. The worked examples below build from straightforward to multi-step, mirroring the progression you will see in AQA papers.

Worked example: Finding the Number of Particles from Moles

Calculate the number of molecules in 0.25 mol of carbon dioxide, CO$_2$(g). (Use $L = 6.022 \times 10^{23}$ mol$^{-1}$)

1. $N = n \times L$ — We want the number of particles $N$. We know the amount $n$ and Avogadro's constant $L$. [M1]
2. Particle = CO$_2$ molecule — Always state what particle you are counting. Here it is molecules of CO$_2$, not atoms.
3. $N = 0.25 \times 6.022 \times 10^{23}$ — $n = 0.25$ mol, $L = 6.022 \times 10^{23}$ mol$^{-1}$. [M1]
4. $N = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}$ — $0.25 \times 6.022 = 1.5055$. The power of ten stays at $10^{23}$. [A1]
5. $N = 1.51 \times 10^{23}$ molecules of CO$_2$ — Round to 3 significant figures to match the precision of the data given. Always state the particle type in your answer.

Answer: $1.51 \times 10^{23}$ molecules of CO$_2$

Worked example: Finding Moles from a Number of Particles

A sample of sodium chloride, NaCl(s), contains $3.613 \times 10^{24}$ formula units. Calculate the amount, in mol, of NaCl present. (Use $L = 6.022 \times 10^{23}$ mol$^{-1}$)

1. $N = n \times L \Rightarrow n = \dfrac{N}{L}$ — We know $N$ and $L$; we need $n$. Divide both sides by $L$. [M1]
2. $n = \dfrac{3.613 \times 10^{24}}{6.022 \times 10^{23}}$ — $N = 3.613 \times 10^{24}$ formula units, $L = 6.022 \times 10^{23}$ mol$^{-1}$. [M1]
3. $\dfrac{10^{24}}{10^{23}} = 10^{24-23} = 10^{1} = 10$ — Subtract exponents when dividing powers of ten. This gives a factor of 10.
4. $\dfrac{3.613}{6.022} = 0.5999... \approx 0.600$ — $3.613 \div 6.022 \approx 0.600$ (to 3 s.f.)
5. $n = 0.600 \times 10 = 6.00$ mol — Multiply the numerical result by the power-of-ten factor. [A1]

Answer: $n = 6.00$ mol of NaCl

Worked example: Multi-Step: Atoms Within Molecules

Calculate the total number of hydrogen atoms in 0.40 mol of ethanol, C$_2$H$_5$OH(l). (Use $L = 6.022 \times 10^{23}$ mol$^{-1}$)

1. C$_2$H$_5$OH has 6 H atoms per molecule (5 from C$_2$H$_5$ + 1 from OH) — Write out the full molecular formula: C$_2$H$_6$O. Count: 6 hydrogen atoms per molecule.
2. $n(\text{H atoms}) = 0.40 \times 6 = 2.40$ mol — Each mole of ethanol molecules contains 6 moles of hydrogen atoms.
3. $N = 2.40 \times 6.022 \times 10^{23}$ — Now use Avogadro's constant to convert moles of H atoms to number of H atoms. [M1]
4. $N = 2.40 \times 6.022 \times 10^{23} = 14.4528 \times 10^{23} = 1.445 \times 10^{24}$ — $2.40 \times 6.022 = 14.4528$. Rewrite in standard form: $1.445 \times 10^{24}$. [A1]
5. $N = 1.45 \times 10^{24}$ hydrogen atoms — Rounded to 3 significant figures. Always specify the particle type — here, hydrogen atoms.

Answer: $1.45 \times 10^{24}$ hydrogen atoms

6. Exam Technique and Common Mistakes

The calculations in this topic are straightforward in structure, but AQA examiners consistently award marks for precision in language and method. The mistakes below are the ones that cost students marks most frequently — not because the chemistry is hard, but because of careless phrasing or arithmetic slips with powers of ten.

AQA 'calculate' questions follow a predictable mark scheme: one method mark for writing or implying the correct formula, one method mark for correct substitution with units, and one accuracy mark for the correct final answer. Showing your working in full — even when the arithmetic feels obvious — protects your method marks if you make a slip in the final step.

7. Consolidation and Retrieval Check

You have now covered every AQA specification requirement for this topic: the definition of the mole, the role of Avogadro's constant, and the application of the mole to atoms, molecules, ions, electrons, formula units, and equations. The retrieval questions below are structured to mirror the range of question styles you will encounter in Paper 1.