What is the mole in Chemistry and why is it important in stoichiometry?

In Chemistry, a mole is a fundamental unit used to measure the amount of substance. It is important in stoichiometry because it allows chemists to count particles like atoms, molecules, or ions in a given sample by relating mass to the number of particles. This is crucial for balancing chemical equations and calculating reactant and product quantities in chemical reactions, which is a key aspect of stoichiometry.

How is the Avogadro Constant defined and what is its significance?

The Avogadro Constant, approximately 6.022 x 10^23 mol^-1, defines the number of particles in one mole of a substance. It is significant because it provides a bridge between the macroscopic scale of grams and the microscopic scale of atoms and molecules, allowing chemists to convert between mass and number of particles, which is essential for stoichiometric calculations.

How do you calculate the number of moles from a given mass?

To calculate the number of moles from a given mass, you use the formula: moles = mass (g) / molar mass (g/mol). The molar mass is the mass of one mole of a substance and can be found on the periodic table as the atomic or molecular weight. This calculation is a fundamental step in stoichiometry for determining how much of a substance is involved in a chemical reaction.

What is the relationship between the mole, molar mass, and Avogadro's number in stoichiometry?

The relationship between the mole, molar mass, and Avogadro's number is central to stoichiometry. One mole of any substance contains Avogadro's number of particles and has a mass equal to its molar mass. This relationship allows chemists to convert between mass, moles, and number of particles, facilitating the quantitative analysis of chemical reactions.

Why is understanding the concept of the mole crucial for Cambridge IGCSE Chemistry students?

Understanding the concept of the mole is crucial for Cambridge IGCSE Chemistry students because it is foundational for mastering stoichiometry, which involves calculating reactants and products in chemical reactions. The mole concept helps students grasp how chemical equations are balanced and how to predict the outcomes of reactions, which are essential skills for success in the IGCSE Chemistry curriculum.

Why is "Using $\text{cm}^{3}$ directly in $c = n/V$" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Questions usually quote burette/pipette volumes in cm^3. How to avoid it: Convert first: V(dm^3) = V(cm^3) ÷ 1000. (e.g. 25.0\,cm^3 = 0.0250\,dm^3.) Source: 0620/42 — recurring across series

Why is "Applying equation ratios to masses instead of moles" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Forgetting that the coefficients in a balanced equation are ratios of MOLES, not grams. How to avoid it: Always convert mass → moles first, apply the mole ratio, then convert back to mass if a mass is required.

Why is "Writing the Avogadro constant as $6.02 \times 10^{22}$" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Misremembering the exponent. How to avoid it: Learn it as 6.02 × 10^23\,mol^-1 — the exponent is 23.

Why is "Rounding a $1:1.5$ empirical-formula ratio to $1:2$" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Wanting whole numbers, so a .5 value gets rounded off. How to avoid it: Multiply the WHOLE ratio (here by 2) to clear the decimal — 1:1.5 becomes 2:3 (e.g. Fe_2O_3).

Cambridge IGCSE Chemistry (0620)

The Mole and the Avagadro Constant

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Short Notes - The Mole and the Avagadro Constant

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Detailed Study Notes

Detailed notes on Stoichiometry for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

The Mole and the Avogadro Constant — Cambridge IGCSE 0620 Chemistry Extended (2026)

$n = m/M_r$ , $n_{\text{gas}} = V/24\,\text{dm}^{3}$ at r.t.p., concentration $c = n/V$ . The mole is the bridge between equation coefficients and lab quantities.

At a glance

Mole = $6.02 \times 10^{23}$ particles (Avogadro's constant).
$n = \dfrac{m}{M_r}$ : moles from mass.
$n_{\text{gas}} = \dfrac{V}{24\,\text{dm}^{3}}$ at room temp and pressure.
Concentration $c = \dfrac{n}{V}$ . Units: $\text{mol/dm}^{3}$ .
Equation coefficients give MOLE RATIOS — e.g. $2\text{H}_{2} + \text{O}_{2} \to 2\text{H}_{2}\text{O}$ means 2 mol H₂ : 1 mol O₂ : 2 mol H₂O.
Limiting reagent: the reactant that runs out first.

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

3.3 — Define the mole and Avogadro's constant.
3.3 — Convert between mass, moles, $M_r$ , gas volume, and concentration.
3.3 — Use mole ratios in balanced equations.
3.3 — Identify the limiting reagent in a reaction (Extended).

What is a mole?

$6.02 \times 10^{23}$ particles. Defined as the amount in $12\,\text{g}$ of $^{12}$ C.

Mole. The amount of substance containing the same number of particles as there are atoms in $12\,\text{g}$ of $^{12}\text{C}$ . That number is Avogadro's constant: $N_{A} = 6.02 \times 10^{23}\,\text{mol}^{-1}.$

Why $12\,\text{g}$ of $^{12}\text{C}$ ? It's a definition. The point is: 1 mole of any substance has $6.02 \times 10^{23}$ particles.

Worked. How many atoms in $0.5\,\text{mol}$ of iron?

$0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ atoms.

Mole and mass. The mass of 1 mole of an element (in grams) is numerically equal to its $A_r$ . So 1 mole of carbon = $12\,\text{g}$ ; 1 mole of iron = $56\,\text{g}$ .

For compounds, 1 mole has a mass equal to $M_r$ in grams. 1 mol of $\text{H}_{2}\text{O}$ = $18\,\text{g}$ .

Conversion: mass ↔ moles. $n = \dfrac{m}{M_r}, \quad m = n \times M_r.$

Always convert TO moles first — the mole is the hub that links mass, gas volume, particles and concentration.

Worked. Convert $36\,\text{g}$ of water to moles.

$n = 36/18 = 2\,\text{mol}$ .

Worked. Convert $0.25\,\text{mol}$ of CO₂ to mass.

$m = 0.25 \times 44 = 11\,\text{g}$ .

$1\,\text{mol} = 6.02 \times 10^{23}$ particles.
Mass of 1 mol = $A_r$ or $M_r$ in grams.
$n = m/M_r$ .
Conversion is the most-tested skill.

Gas volume and concentration

1 mol of gas at r.t.p. = $24\,\text{dm}^{3}$ . Concentration = mol/dm³.

Gas volume at r.t.p. Room temperature and pressure (about $20\degree\text{C}$ , $1\,\text{atm}$ ): 1 mole of any gas occupies $24\,\text{dm}^{3}$ ( $24{,}000\,\text{cm}^{3}$ ). $n_{\text{gas}} = \dfrac{V}{24\,\text{dm}^{3}/\text{mol}}.$

Worked. What volume does $0.5\,\text{mol}$ of CO₂ occupy at r.t.p.?

$V = 0.5 \times 24 = 12\,\text{dm}^{3}$ .

Concentration of solutions. $c = \dfrac{n}{V}, \quad n = c \times V,$ where $V$ is in $\text{dm}^{3}$ ( $1\,\text{dm}^{3} = 1000\,\text{cm}^{3} = 1\,\text{L}$ ). Units: $\text{mol/dm}^{3}$ (sometimes written $M$ ).

Worked. $250\,\text{cm}^{3}$ of $0.4\,\text{mol/dm}^{3}$ HCl. Find moles of HCl.

$V = 250/1000 = 0.25\,\text{dm}^{3}$ .
$n = 0.4 \times 0.25 = 0.1\,\text{mol}$ .

Cambridge tip. Watch the volume unit — convert $\text{cm}^{3}$ to $\text{dm}^{3}$ by dividing by $1000$ .

1 mol gas at r.t.p. = $24\,\text{dm}^{3}$ .
$n = V/24$ (with $V$ in dm³).
Concentration: $c = n/V$ , units mol/dm³.
Convert cm³ → dm³ by dividing by 1000.

Using mole ratios from balanced equations

Coefficients in a balanced equation are mole ratios. Use them to find quantities of products from quantities of reactants.

Steps for stoichiometric calculations.

Write the BALANCED equation.
Convert known quantity to moles.
Apply the mole ratio from the equation.
Convert the answer back to the required units (mass or volume).

Never skip the mole ratio — moving straight from one mass to another loses marks.

Worked. What mass of magnesium oxide forms from $4.8\,\text{g}$ of magnesium?

Equation: $2\text{Mg} + \text{O}_{2} \to 2\text{MgO}$ .
Moles of Mg: $4.8/24 = 0.2\,\text{mol}$ .
Mole ratio Mg : MgO = $2 : 2 = 1 : 1$ .
Moles of MgO: $0.2\,\text{mol}$ .
Mass of MgO: $0.2 \times 40 = 8\,\text{g}$ .

Worked. Volume of CO₂ from burning $0.5\,\text{mol}$ of methane?

Equation: $\text{CH}_{4} + 2\text{O}_{2} \to \text{CO}_{2} + 2\text{H}_{2}\text{O}$ .
Mole ratio CH₄ : CO₂ = $1 : 1$ .
Moles of CO₂: $0.5\,\text{mol}$ .
Volume at r.t.p.: $0.5 \times 24 = 12\,\text{dm}^{3}$ .

Limiting reagent (Extended). When two reactants are mixed in non-stoichiometric amounts, one runs out first — the limiting reagent. The product yield is determined by the limiting reagent.

Worked. $0.4\,\text{mol}$ Mg reacted with $0.1\,\text{mol}$ O₂. Equation: $2\text{Mg} + \text{O}_{2} \to 2\text{MgO}$ .

Mg available: $0.4\,\text{mol}$ . To use it all, you'd need $0.2\,\text{mol}$ O₂.
O₂ available: only $0.1\,\text{mol}$ .
O₂ is LIMITING.
Product: $0.1 \times 2 = 0.2\,\text{mol}$ MgO.
$0.2\,\text{mol}$ Mg leftover (unreacted).

Coefficients = mole ratios.
Find moles → use ratio → convert to required units.
Limiting reagent = runs out first.
Product yield set by limiting reagent.

Quick recap

Mole = $6.02 \times 10^{23}$ particles.
$n = m/M_r$ .
$V_{\text{gas}} = n \times 24$ at r.t.p.
$c = n/V$ (V in dm³).
Coefficients = mole ratios.
Limiting reagent runs out first.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Mole — amount of substance containing $6.02 \times 10^{23}$ particles.
Avogadro's constant — $6.02 \times 10^{23}$ particles per mole.
Concentration — moles of solute per dm³ of solution.
Limiting reagent — reactant that determines the maximum product yield.
Percentage yield — actual yield as a percentage of the theoretical maximum.

How it’s examined

Mole calculations appear every Paper 4 (8-12 marks) and most Paper 2s (3-5 marks). Cambridge stacks $n = m/M_r$ , gas volume, concentration, and stoichiometry into multi-step problems. Examiner reports flag dm³/cm³ unit confusion and forgetting the mole ratio (just using mass directly).

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (3.3); 0620/42 Oct/Nov 2024 — Q8 (mole + limiting reagent); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — The Mole and the Avagadro Constant

Step-by-step solutions to past-paper-style questions on the mole and the avagadro constant, written exactly the way a tutor would explain them at the board.

Question type:

1Convert a mass into moles
Getting startedDirect calculation• n=m/Mr
Question
Calculate the amount, in moles, in $8.0\,\text{g}$ of sodium hydroxide, $\text{NaOH}$ . ( $M_{r}$ of $\text{NaOH} = 40$ .)
Step-by-step solution
1. Step 1
  Use $n = m / M_{r}$ , with $m = 8.0\,\text{g}$ and $M_{r} = 40$ .
  $n = \dfrac{m}{M_{r}} = \dfrac{8.0}{40}$
2. Step 2
  Divide to get the amount in moles.
  $n = 0.20\,\text{mol}$
Answer
$0.20\,\text{mol}$
Examiner tip
Always quote the unit 'mol'. $M_{r}$ for $\text{NaOH} = 23 + 16 + 1 = 40$ — build it from the relative atomic masses if it is not given.
2Number of particles using the Avogadro constant
Getting startedDirect calculation• Avogadro constant
Question
How many molecules are present in $0.25\,\text{mol}$ of carbon dioxide, $\text{CO}_{2}$ ? ( $N_{A} = 6.02 \times 10^{23}\,\text{per mole}$ .)
Step-by-step solution
1. Step 1
  Number of particles $=$ moles $\times N_{A}$ .
  $N = n \times N_{A} = 0.25 \times 6.02 \times 10^{23}$
2. Step 2
  Multiply out.
  $N = 1.505 \times 10^{23} = 1.51 \times 10^{23}\,\text{molecules}$
Answer
$1.51 \times 10^{23}$ molecules of $\text{CO}_{2}$
Examiner tip
One mole of ANY substance contains $6.02 \times 10^{23}$ particles. The 'particles' may be atoms, molecules or ions — read the question.
3Reacting mass from a balanced equation
Building confidenceMulti-step problem• Adapted from 0620/42 May/Jun 2024 Q5• reacting mass, mole ratio
Question
What mass of magnesium oxide is formed when $6.0\,\text{g}$ of magnesium burns completely in oxygen? $2\text{Mg} + \text{O}_{2} \to 2\text{MgO}$ . ( $A_{r}$ : $\text{Mg} = 24$ ; $M_{r}$ : $\text{MgO} = 40$ .)
Step-by-step solution
1. Step 1
  Find the moles of magnesium.
  $n(\text{Mg}) = \dfrac{6.0}{24} = 0.25\,\text{mol}$
2. Step 2
  The equation shows a $2\!:\!2$ (i.e. $1\!:\!1$ ) mole ratio of $\text{Mg}$ to $\text{MgO}$ .
  $n(\text{MgO}) = 0.25\,\text{mol}$
3. Step 3
  Convert moles of $\text{MgO}$ back to a mass with $m = n \times M_{r}$ .
  $m(\text{MgO}) = 0.25 \times 40 = 10\,\text{g}$
Answer
$10\,\text{g}$ of $\text{MgO}$
Examiner tip
Mole ratios come from the BIG numbers (coefficients) in the balanced equation, never from grams. Route: mass → moles → ratio → moles → mass.
4Concentration in mol/dm³ and g/dm³
Building confidenceMulti-step problem• concentration, c=n/V
Question
$0.10\,\text{mol}$ of sodium hydroxide is dissolved to make $250\,\text{cm}^{3}$ of solution. Find the concentration in (a) $\text{mol/dm}^{3}$ and (b) $\text{g/dm}^{3}$ . ( $M_{r}$ of $\text{NaOH} = 40$ .)
Step-by-step solution
1. Step 1
  Convert the volume from $\text{cm}^{3}$ to $\text{dm}^{3}$ (divide by 1000).
  $V = \dfrac{250}{1000} = 0.250\,\text{dm}^{3}$
2. Step 2
  (a) Apply $c = n / V$ .
  $c = \dfrac{0.10}{0.250} = 0.40\,\text{mol/dm}^{3}$
3. Step 3
  (b) Convert to $\text{g/dm}^{3}$ by multiplying by $M_{r}$ .
  $c = 0.40 \times 40 = 16\,\text{g/dm}^{3}$
Answer
(a) $0.40\,\text{mol/dm}^{3}$ ; (b) $16\,\text{g/dm}^{3}$
Examiner tip
The commonest error is forgetting to convert $\text{cm}^{3}$ to $\text{dm}^{3}$ . $\text{concentration in g/dm}^{3} = \text{concentration in mol/dm}^{3} \times M_{r}$ .
5Empirical formula from percentage composition
StretchMulti-step problem• Adapted from 0620/42 Oct/Nov 2023 Q4• empirical formula
Question
An oxide of iron contains $70\%$ iron and $30\%$ oxygen by mass. Determine its empirical formula. ( $A_{r}$ : $\text{Fe} = 56$ , $\text{O} = 16$ .)
Step-by-step solution
1. Step 1
  Assume $100\,\text{g}$ , so the masses are $70\,\text{g}$ of $\text{Fe}$ and $30\,\text{g}$ of $\text{O}$ . Find moles of each.
  $n(\text{Fe}) = \dfrac{70}{56} = 1.25; \quad n(\text{O}) = \dfrac{30}{16} = 1.875$
2. Step 2
  Divide both by the smaller value ( $1.25$ ) to get the simplest ratio.
  $\text{Fe} : \text{O} = \dfrac{1.25}{1.25} : \dfrac{1.875}{1.25} = 1 : 1.5$
3. Step 3
  Multiply by $2$ to remove the decimal and obtain whole numbers.
  $\text{Fe} : \text{O} = 2 : 3$
Answer
Empirical formula $= \text{Fe}_{2}\text{O}_{3}$
Examiner tip
Never round a $.5$ ratio to a whole number — multiply the WHOLE ratio (here by 2) until every figure is a whole number.
6Limiting reactant, gas volume and percentage yield
StretchMulti-step problem• limiting reactant, gas volume, percentage yield
Question
$5.85\,\text{g}$ of calcium carbonate is heated and decomposes: $\text{CaCO}_{3} \to \text{CaO} + \text{CO}_{2}$ . (a) Calculate the maximum volume of $\text{CO}_{2}$ produced at rtp. (b) Only $1.20\,\text{dm}^{3}$ of gas is actually collected — find the percentage yield. ( $M_{r}$ of $\text{CaCO}_{3} = 100$ ; molar gas volume $= 24\,\text{dm}^{3}/\text{mol}$ .)
Step-by-step solution
1. Step 1
  Find the moles of $\text{CaCO}_{3}$ (the only reactant — it is the limiting reactant).
  $n(\text{CaCO}_{3}) = \dfrac{5.85}{100} = 0.0585\,\text{mol}$
2. Step 2
  The $1\!:\!1$ ratio gives the same number of moles of $\text{CO}_{2}$ .
  $n(\text{CO}_{2}) = 0.0585\,\text{mol}$
3. Step 3
  (a) Convert moles of gas to a volume at rtp with $V = n \times 24$ .
  $V = 0.0585 \times 24 = 1.404\,\text{dm}^{3}$
4. Step 4
  (b) Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100$ .
  $\% \text{ yield} = \dfrac{1.20}{1.404} \times 100 = 85.5\%$
Answer
(a) $1.40\,\text{dm}^{3}$ of $\text{CO}_{2}$ ; (b) $85.5\%$ yield
Examiner tip
Percentage yield compares like with like — here actual and theoretical are both gas VOLUMES, so no further conversion is needed. Yield can never exceed $100\%$ .

Model Answers — The Mole and the Avagadro Constant

High-scoring sample answers for the mole and the avagadro constant on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 4 short-answer style1 mark
State the value of the Avogadro constant. (1 mark)
Model answer
$6.02 \times 10^{23}$ (per mole) — the number of particles in one mole of any substance.
Why this scores
One mark for $6.02 \times 10^{23}$ . Watch the exponent: it is $10^{23}$ , not $10^{22}$ . The unit is 'per mole' ( $\text{mol}^{-1}$ ).
Question 2
Paper 4 structured style2 marks
Calculate the amount, in moles, in $11\,\text{g}$ of carbon dioxide, $\text{CO}_{2}$ . ( $M_{r}$ of $\text{CO}_{2} = 44$ .) (2 marks)
Model answer
$n = \dfrac{m}{M_{r}} = \dfrac{11}{44} = 0.25\,\text{mol}$ .
Why this scores
One mark for the correct method ( $n = m/M_{r}$ ) and one for the answer with the unit 'mol'. Quoting $0.25$ without 'mol' loses a mark.
Question 3
Paper 4 structured style3 marks
Calculate the volume of hydrogen, in $\text{dm}^{3}$ at rtp, produced when $0.30\,\text{mol}$ of zinc reacts completely with excess hydrochloric acid. $\text{Zn} + 2\text{HCl} \to \text{ZnCl}_{2} + \text{H}_{2}$ . (Molar gas volume $= 24\,\text{dm}^{3}/\text{mol}$ .) (3 marks)
Model answer
The mole ratio of $\text{Zn}$ to $\text{H}_{2}$ is $1\!:\!1$ , so $0.30\,\text{mol}$ of $\text{Zn}$ gives $0.30\,\text{mol}$ of $\text{H}_{2}$ . Volume $= n \times 24 = 0.30 \times 24 = 7.2\,\text{dm}^{3}$ .
Why this scores
Three marks: correct $1\!:\!1$ ratio giving $0.30\,\text{mol}$ of $\text{H}_{2}$ ; correct use of $V = n \times 24$ ; final answer $7.2\,\text{dm}^{3}$ with unit.
Question 4
Paper 4 (Extended) structured style4 marks
$25.0\,\text{cm}^{3}$ of sodium hydroxide solution is exactly neutralised by $20.0\,\text{cm}^{3}$ of $0.100\,\text{mol/dm}^{3}$ hydrochloric acid. $\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_{2}\text{O}$ . Calculate the concentration of the sodium hydroxide in $\text{mol/dm}^{3}$ . (4 marks)
Model answer
Moles of $\text{HCl} = c \times V = 0.100 \times \dfrac{20.0}{1000} = 0.00200\,\text{mol}$ . The equation has a $1\!:\!1$ ratio, so moles of $\text{NaOH} = 0.00200\,\text{mol}$ . Concentration of $\text{NaOH} = \dfrac{n}{V} = \dfrac{0.00200}{25.0/1000} = \dfrac{0.00200}{0.0250} = 0.0800\,\text{mol/dm}^{3}$ .
Why this scores
Four marks: moles of $\text{HCl}$ (with $\text{cm}^{3} \to \text{dm}^{3}$ conversion); $1\!:\!1$ ratio for moles of $\text{NaOH}$ ; rearranged $c = n/V$ ; final answer $0.0800\,\text{mol/dm}^{3}$ . The most-penalised slip is leaving the volume in $\text{cm}^{3}$ .
Question 5
Paper 4 (Extended) structured style5 marks
A compound has the empirical formula $\text{CH}_{2}\text{O}$ and a relative molecular mass of $180$ . Determine its molecular formula. ( $A_{r}$ : $\text{C} = 12$ , $\text{H} = 1$ , $\text{O} = 16$ .) (5 marks)
Model answer
Mass of one empirical-formula unit $\text{CH}_{2}\text{O} = 12 + (2 \times 1) + 16 = 30$ . The number of empirical units in one molecule $= \dfrac{M_{r}}{\text{empirical mass}} = \dfrac{180}{30} = 6$ . Multiplying the empirical formula by $6$ gives the molecular formula $\text{C}_{6}\text{H}_{12}\text{O}_{6}$ (glucose).
Why this scores
Five marks: empirical mass $=30$ ; divide $M_{r}$ by empirical mass to get the multiplier $6$ ; multiply each subscript by $6$ ; correct molecular formula $\text{C}_{6}\text{H}_{12}\text{O}_{6}$ ; reasoning shown. The molecular formula is always a whole-number multiple of the empirical formula.
Question 6
Paper 4 (Extended) structured style6 marks
Iron is extracted by reducing iron(III) oxide with carbon monoxide: $\text{Fe}_{2}\text{O}_{3} + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_{2}$ . $160\,\text{g}$ of $\text{Fe}_{2}\text{O}_{3}$ is reduced and $90\,\text{g}$ of iron is obtained. Calculate the percentage yield of iron. ( $M_{r}$ of $\text{Fe}_{2}\text{O}_{3} = 160$ ; $A_{r}$ of $\text{Fe} = 56$ .) (6 marks)
Model answer
Moles of $\text{Fe}_{2}\text{O}_{3} = \dfrac{160}{160} = 1.0\,\text{mol}$ . From the equation, $1\,\text{mol}$ of $\text{Fe}_{2}\text{O}_{3}$ gives $2\,\text{mol}$ of $\text{Fe}$ , so the theoretical amount of iron $= 2.0\,\text{mol}$ . Theoretical mass of iron $= n \times A_{r} = 2.0 \times 56 = 112\,\text{g}$ . Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100 = \dfrac{90}{112} \times 100 = 80.4\%$ (to 3 s.f.).
Why this scores
Six marks: moles of $\text{Fe}_{2}\text{O}_{3}$ ; use of the $1\!:\!2$ mole ratio; moles of $\text{Fe} = 2.0$ ; theoretical mass $112\,\text{g}$ ; correct yield formula; final answer $\approx 80\%$ . A yield above $100\%$ signals an arithmetic error.

Key Formulae — The Mole and the Avagadro Constant

The formulae you need to memorise for the mole and the avagadro constant on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.

Moles from mass
$n = \dfrac{m}{M_{r}}$
$n$
amount of substance in mol
$m$
mass in g
$M_{r}$
relative formula mass
When to use
Convert between mass and moles (rearrange to $m = n M_{r}$ ).
Example
$8.0\,\text{g}$ of $\text{NaOH}$ ( $M_{r}=40$ ) $= 8.0/40 = 0.20\,\text{mol}$ .
Number of particles
$N = n \times N_{A}, \quad N_{A} = 6.02 \times 10^{23}\,\text{mol}^{-1}$
$N$
number of particles (atoms, molecules or ions)
$n$
amount in mol
$N_{A}$
Avogadro constant
When to use
Find how many particles are in a given amount of substance.
Concentration (mol/dm³)
$c = \dfrac{n}{V}$
$c$
concentration in mol/dm³
$n$
amount of solute in mol
$V$
volume of solution in dm³
When to use
Solutions and titrations. Convert $\text{cm}^{3}$ to $\text{dm}^{3}$ first ( $\div 1000$ ).
Concentration (g/dm³)
$\text{conc (g/dm}^{3}) = \text{conc (mol/dm}^{3}) \times M_{r}$
$M_{r}$
relative formula mass of the solute
When to use
Convert a concentration between $\text{mol/dm}^{3}$ and $\text{g/dm}^{3}$ .
Molar gas volume (rtp)
$V = n \times 24\,\text{dm}^{3}/\text{mol}$
$V$
volume of gas in dm³ at rtp
$n$
amount of gas in mol
When to use
Find a gas volume at room temperature and pressure (rtp).
Example
$0.5\,\text{mol}$ of gas $= 0.5 \times 24 = 12\,\text{dm}^{3}$ at rtp.
Percentage yield & percentage purity
$\% \text{ yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100; \quad \% \text{ purity} = \dfrac{\text{mass of pure substance}}{\text{mass of sample}} \times 100$
When to use
Compare an experimental amount with the maximum possible amount.

Key Definitions and Keywords — The Mole and the Avagadro Constant

Definitions to memorise and the exact keywords mark schemes credit for the mole and the avagadro constant answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Mole
Examiner keyword
The unit of amount of substance. One mole contains $6.02 \times 10^{23}$ particles (the Avogadro number) of the substance.
Avogadro constant ( $N_{A}$ )
Examiner keyword
The number of particles in one mole of a substance: $6.02 \times 10^{23}\,\text{mol}^{-1}$ .
Molar gas volume
Examiner keyword
The volume occupied by one mole of any gas at room temperature and pressure (rtp) $= 24\,\text{dm}^{3}$ .
Empirical formula
Examiner keyword
The simplest whole-number ratio of the atoms of each element in a compound (e.g. $\text{CH}_{2}\text{O}$ for glucose).
Limiting reactant
Examiner keyword
The reactant that is completely used up first; it determines the maximum amount of product that can form.

Common Mistakes and Misconceptions — The Mole and the Avagadro Constant

The traps other students keep falling into on the mole and the avagadro constant questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Using $\text{cm}^{3}$ directly in $c = n/V$
0620/42 — recurring across series
Why it happens
Questions usually quote burette/pipette volumes in $\text{cm}^{3}$ .
How to avoid it
Convert first: $V(\text{dm}^{3}) = V(\text{cm}^{3}) \div 1000$ . (e.g. $25.0\,\text{cm}^{3} = 0.0250\,\text{dm}^{3}$ .)
✕Applying equation ratios to masses instead of moles
Why it happens
Forgetting that the coefficients in a balanced equation are ratios of MOLES, not grams.
How to avoid it
Always convert mass → moles first, apply the mole ratio, then convert back to mass if a mass is required.
✕Writing the Avogadro constant as $6.02 \times 10^{22}$
Why it happens
Misremembering the exponent.
How to avoid it
Learn it as $6.02 \times 10^{23}\,\text{mol}^{-1}$ — the exponent is $23$ .
✕Rounding a $1:1.5$ empirical-formula ratio to $1:2$
Why it happens
Wanting whole numbers, so a $.5$ value gets rounded off.
How to avoid it
Multiply the WHOLE ratio (here by 2) to clear the decimal — $1:1.5$ becomes $2:3$ (e.g. $\text{Fe}_{2}\text{O}_{3}$ ).