What is the 'amount of substance' in chemistry, and why is it important in the Edexcel International A Levels curriculum?

The 'amount of substance' is a fundamental concept in chemistry that refers to the quantity of entities, such as atoms, molecules, or ions, in a given sample. It is measured in moles, which is a standard unit in the International System of Units (SI). Understanding the amount of substance is crucial for calculating chemical reactions and stoichiometry, which are key components of the Edexcel International A Levels Chemistry curriculum.

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

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 relative atomic or molecular mass. This calculation is essential for solving problems related to the amount of substance in the Edexcel International A Levels Chemistry exams.

What is Avogadro's constant, and how is it used in calculating the amount of substance?

Avogadro's constant, approximately 6.022 x 10^23 mol^-1, is the number of atoms, molecules, or particles in one mole of a substance. It is used to convert between the number of particles and the amount of substance in moles. This constant is fundamental in understanding the mole concept and is frequently used in calculations involving the amount of substance in the Edexcel International A Levels Chemistry syllabus.

How is the concept of molar volume applied in the context of gases in chemistry?

Molar volume is the volume occupied by one mole of a gas at a given temperature and pressure, typically 24 dm^3 at room temperature and pressure (RTP). This concept is used to relate the volume of a gas to the amount of substance in moles, which is particularly useful in gas stoichiometry problems. Understanding molar volume is important for the Edexcel International A Levels Chemistry curriculum, especially in topics involving gas laws and reactions.

Why is the empirical formula important, and how is it determined using the amount of substance?

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is determined by calculating the moles of each element in a sample and then dividing by the smallest number of moles to find the simplest ratio. This process involves the concept of the amount of substance and is crucial for understanding chemical composition and reactions, as emphasized in the Edexcel International A Levels Chemistry syllabus.

Why is "Not converting cm³ to dm³ in concentration calculations" a common mistake in Amount of Substance?

Why it happens: Titration volumes are read in cm³ from the burette, so students plug the raw value straight into n = c × V. How to avoid it: Always **divide cm³ by 1000** to get dm³ before using n = c × V or c = n/V. A quick check: sensible lab concentrations are ~0.01–2 mol dm⁻³; an answer of 800 or 0.0001 signals a factor-of-1000 slip. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Why is "Using kPa, cm³ or °C in the ideal gas equation pV = nRT" a common mistake in Amount of Substance?

Why it happens: Data is often given in convenient units and students forget R = 8.31 requires strict SI units. How to avoid it: Convert **before** substituting: pressure to **Pa** (×1000 from kPa), volume to **m³** (cm³ ×10⁻⁶, dm³ ×10⁻³), temperature to **K** (°C + 273). The raw answer for V is in m³ — convert back to dm³ (×1000) or cm³ (×10⁶) as required. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Why is "Confusing percentage yield with atom economy" a common mistake in Amount of Substance?

Why it happens: Both are percentages about 'efficiency', so students use the wrong formula or treat them as the same thing. How to avoid it: **Percentage yield** = actual ÷ theoretical product (depends on the experiment). **Atom economy** = Mr of desired product ÷ Mr of all products (fixed by the balanced equation). A reaction can be 100% yield yet low atom economy. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Why is "Ignoring the mole ratio from the balanced equation (assuming 1:1)" a common mistake in Amount of Substance?

Why it happens: Students rush from moles of reactant to moles of product without checking the stoichiometric coefficients. How to avoid it: Read the **balancing numbers** from the equation. For H₂SO₄ + 2NaOH the ratio is 1:2; for 2Mg + O₂ it is 2:1. Multiply/divide the moles by the correct ratio before finding mass or volume. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Why is "Basing the product amount on the reactant in excess instead of the limiting reagent" a common mistake in Amount of Substance?

Why it happens: Students calculate moles of both reactants but pick the larger number, or fail to test against the mole ratio. How to avoid it: Compare the **ratio required** by the equation with the **ratio available**. The reagent that runs out first (needs more than is present) is limiting — always base the product on it. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Why is "Rounding intermediate values too early and losing significant figures" a common mistake in Amount of Substance?

Why it happens: Students round each step to 2 figures, so accumulated errors shift the final answer. How to avoid it: Keep **full accuracy in the calculator** through the whole calculation and round only the **final answer**, matching the significant figures of the data (usually 3 s.f.). Always state the correct **unit**. Source: Edexcel IAL Chemistry Unit 1 examiner reports

Edexcel International A Levels Chemistry (XCH11-YCH11)

Amount of Substance

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Short Notes - Amount of Substance

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Detailed Study Notes

Detailed notes on Structure, Bonding and Introduction to Organic Chemistry for Edexcel International A Levels Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

Amount of Substance (the Mole) — Pearson Edexcel International A Level Chemistry (XCH11/YCH11) Study Notes (Unit 1)

The mole and Avogadro constant, moles from mass and concentration, gas volumes (molar gas volume and the ideal gas equation pV = nRT), reacting masses and limiting reagents, percentage yield and atom economy, and Core Practical 1 (molar volume of a gas). Every relationship reduces to converting between mass, moles, particles, gas volume and solution volume.

At a glance

The mole is the amount containing $6.02 \times 10^{23}$ particles (the Avogadro constant, $N_A$ ).
Moles from mass: $n = m/M$ . Number of particles: $N = n \times N_A$ .
Concentration: $n = c \times V$ with V in dm³ — always convert cm³ → dm³ by ÷1000.
Gases at RTP: volume $= n \times 24$ dm³ mol⁻¹. Any conditions: $pV = nRT$ in strict SI units (Pa, m³, K, R = 8.31).
Reacting masses / limiting reagent: convert to moles → use the balanced-equation ratio → convert back.
Percentage yield = (actual ÷ theoretical) × 100 — measures practical efficiency.
Atom economy = (Mr of desired product ÷ Mr of all products) × 100 — measures how little waste is made.
CP1 determines the molar volume of a gas by reacting a known mass and measuring the gas collected.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XCH11-YCH11 syllabus (2018-onwards).

1.1 — Understand the terms relative atomic mass and relative molecular/formula mass, based on the ¹²C scale.
1.2–1.3 — Use the Avogadro constant and the mole; calculate amounts of substance from masses, gas volumes and solution volumes ( $n = m/M$ , $n = cV$ , $n = V/24$ ).
1.4 — Use the ideal gas equation $pV = nRT$ with quantities in SI units.
1.5 — Carry out calculations involving reacting masses, volumes of gases and solutions, and use balanced equations to identify limiting reagents.
1.6 — Calculate percentage yield and atom economy, and understand their importance for sustainable ('green') chemistry.
CP1 — Measure the molar volume of a gas, for example by reacting a metal or a carbonate and collecting the gas produced.

The mole, the Avogadro constant and n = m/M

The mole is chemistry's counting unit; n = m/M links mass to number of particles.

Atoms and molecules are far too small and numerous to count individually, so chemists count them in moles. One mole is the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12 — the Avogadro constant, $N_A = 6.02 \times 10^{23}$ particles per mole.

Two master relationships flow from this:

Moles from mass: $n = \frac{m}{M}$ where $m$ is the mass in grams and $M$ is the molar mass in g mol⁻¹ (numerically equal to the relative formula mass).
Number of particles: $N = n \times N_A$

Always be clear what the particles are. "One mole of oxygen" is ambiguous: 1 mol of O₂ molecules contains $6.02 \times 10^{23}$ molecules but $1.20 \times 10^{24}$ atoms.

The diagram below is the "conversion map" that underlies almost every Unit 1 calculation. Moles sits at the centre; each arrow is one formula. Get a quantity into moles, travel across the middle, and come out as whatever the question asks for.

The mole conversion map. Every Unit 1 amount-of-substance calculation is a journey to moles and out again: mass (÷M), particles (÷N_A), gas volume at RTP (÷24 dm³ mol⁻¹) and solution (n = cV).

Mole = 6.02 × 10²³ particles (the Avogadro constant, N_A).
n = m/M links mass and moles; N = n × N_A links moles and particle count.
State the particle type — moles of O₂ ≠ moles of O atoms.

See the full worked example for amount of substance →

Concentration of solutions and titrations

n = c × V, with V in dm³ — the source of the most common mark loss.

The concentration of a solution is the amount of solute per unit volume, usually in mol dm⁻³:

$n = c \times V \qquad\text{(V in dm}^3\text{)} \qquad\text{so}\qquad c = \frac{n}{V}$

The single most important habit: volumes measured in cm³ (burettes, pipettes, measuring cylinders) must be converted to dm³ by dividing by 1000 before using these equations.

$1 \text{ dm}^3 = 1000 \text{ cm}^3 \qquad\Rightarrow\qquad V(\text{dm}^3) = \frac{V(\text{cm}^3)}{1000}$

A standard solution is one of accurately known concentration, prepared by dissolving a weighed mass of solute and making it up to a precise volume in a volumetric flask.

Titration calculations follow a fixed four-step routine:

Write the balanced equation and read off the mole ratio of the two reacting solutions.
Find the moles of the substance you know completely (concentration and volume) using $n = cV$ .
Use the mole ratio to find the moles of the other substance.
Divide by its volume (in dm³) to get its concentration, or multiply by its molar mass to get a mass.

The ratio is decisive. For a strong acid–base reaction it is not always 1:1:

Reaction	Ratio (acid : base)
HCl + NaOH → NaCl + H₂O	1 : 1
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O	1 : 2
2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂	2 : 1

n = c × V; c = n/V; V must be in dm³ (cm³ ÷ 1000).
Titration routine: equation → moles known → ratio → concentration of unknown.
Read the mole ratio from the balanced equation — it is often not 1:1.

See the full worked example for amount of substance →

Gas volumes — molar gas volume and pV = nRT

24 dm³ per mole at RTP; pV = nRT everywhere else — in strict SI units.

At RTP (room temperature and pressure, ~20–25 °C and 100 kPa), one mole of any gas occupies 24 dm³ (24 000 cm³). So:

$\text{volume of gas} = n \times 24 \text{ dm}^3 \qquad\Rightarrow\qquad n = \frac{V(\text{dm}^3)}{24}$

This shortcut only holds at RTP. For any other temperature or pressure, use the ideal gas equation:

$pV = nRT$

The equation only balances if every quantity is in the correct SI unit — this is where the marks are won or lost. Keep this conversion table in your head:

Symbol	Quantity	SI unit needed	Common trap
$p$	pressure	pascal, Pa	kPa given → ×1000
$V$	volume	cubic metre, m³	cm³ → ×10⁻⁶; dm³ → ×10⁻³
$n$	amount	mole, mol	—
$R$	gas constant	8.31 J K⁻¹ mol⁻¹	fixed
$T$	temperature	kelvin, K	°C → + 273

Before touching pV = nRT, convert pressure to pascals, volume to cubic metres and temperature to kelvin. The volume you calculate is in m³ — convert it back to dm³ or cm³ for the final answer.

At RTP: V = n × 24 dm³ (= 24 000 cm³); n = V/24.
Otherwise pV = nRT with p in Pa, V in m³, T in K, R = 8.31.
Answer for V from pV = nRT is in m³ — convert back to dm³/cm³.

See the full worked example for amount of substance →

Reacting masses and limiting reagents

Convert to moles → apply the balanced-equation ratio → convert back.

Almost every quantitative question follows the same three-step spine:

Convert what you are given into moles (using $n = m/M$ , $n = cV$ , $n = V/24$ , or $pV = nRT$ ).
Use the balanced equation's mole ratio to find the moles of what you want.
Convert the moles back into mass, volume or concentration.

Limiting reagents. When the amounts of two reactants are both fixed, one usually runs out first — the limiting reagent. It controls how much product forms; the other reactant is in excess and some is left over. To find it:

Work out the moles of each reactant.
Compare the ratio you have with the ratio the equation requires.
The reactant that would need more than is available runs out first — it is limiting.

Always base the product amount on the limiting reagent, never on the reagent in excess. For example, in Mg + 2HCl → MgCl₂ + H₂, if you have 0.10 mol Mg and only 0.15 mol HCl, the Mg would need 0.20 mol HCl — but only 0.15 mol is present, so HCl is limiting and the product is calculated from the HCl.

Empirical and molecular formulae use the same moles-first logic in reverse: divide each element's mass (or %) by its A_r to get a mole ratio, simplify to the smallest whole numbers for the empirical formula, then scale up using the relative molecular mass to get the molecular formula.

Three-step spine: to moles → ratio → back to mass/volume.
Limiting reagent = the one that runs out first; base products on it.
Test by comparing the available ratio with the equation's required ratio.
Empirical formula: ÷ A_r → simplest whole-number ratio; scale up for molecular formula.

See the full worked example for amount of substance →

Percentage yield and atom economy

Two different 'efficiency' measures — do not confuse them.

Real reactions rarely give the maximum possible product, and often make unwanted by-products. Two separate measures capture this.

Percentage yield compares what you actually got with the theoretical maximum from the limiting reagent:

$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

It is a practical measure — reduced by incomplete reactions, reversible reactions, competing side-reactions, and product lost during filtering, transferring or purifying. It can only be found by doing the experiment.

Atom economy measures how much of the reactants' mass ends up in the desired product, using only the balanced equation:

$\% \text{ atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$

It is a theoretical, fixed property of the reaction. A reaction with only one product has 100% atom economy; addition reactions tend to be high, whereas substitution and elimination reactions that make by-products are lower.

Why they differ: a reaction can go essentially to completion (high yield) yet still make a lot of by-product (low atom economy), or vice versa. Both matter for sustainable ('green') chemistry — high atom economy means fewer raw materials wasted and less waste to dispose of, while high yield means less reactant is thrown away unreacted. Industry aims to maximise both, and to sell or recycle any by-products (raising the effective atom economy).

% yield = actual ÷ theoretical × 100 (practical, from experiment).
Atom economy = Mr(desired) ÷ Mr(all products) × 100 (theoretical, from equation).
One product → 100% atom economy; by-products lower it.
High yield with low atom economy is possible — they are independent.

See the full worked example for amount of substance →

Core Practical 1 — the molar volume of a gas

React a known mass, collect the gas, and work back to 24 dm³ mol⁻¹.

Core Practical 1 determines the molar volume of a gas experimentally. A common version reacts a weighed mass of a metal (e.g. magnesium) or a carbonate with excess acid and collects the gas in a gas syringe (or over water in an inverted measuring cylinder).

Method outline:

Weigh an accurate mass of the solid (e.g. 0.12 g of magnesium).
Add excess dilute acid so the solid is the limiting reagent and reacts completely.
Collect the gas and record its volume once the reaction stops and the apparatus has returned to room temperature.

Working back to the molar volume:

Calculate moles of solid ( $n = m/M$ ) and use the equation's ratio to get moles of gas.
$\text{molar volume} = \dfrac{\text{volume of gas collected}}{\text{moles of gas}}$ — this should come out close to 24 000 cm³ mol⁻¹ (24 dm³ mol⁻¹) at RTP.

Sources of error / why results fall short:

Gas escaping before the bung/syringe is connected, or leaks in the apparatus.
The magnesium having a surface layer of oxide (MgO), so less reactive metal than weighed.
Measuring the volume before the gas has cooled to room temperature (warm gas reads high), or some gas dissolving in the solution.

This practical ties the whole topic together: a mass measurement and a gas-volume measurement, connected by moles and the balanced equation.

React a weighed solid with excess acid; collect the gas in a syringe.
moles of gas from n = m/M and the equation ratio.
molar volume = volume collected ÷ moles of gas ≈ 24 000 cm³ mol⁻¹ at RTP.
Errors: gas escaping/leaks, oxide layer on metal, measuring before cooling.

See the full worked example for amount of substance →

Quick recap

The mole = 6.02 × 10²³ particles; n = m/M and N = n × N_A convert between mass, moles and particle count.
Concentration: n = c × V with V in dm³ — convert cm³ → dm³ by ÷1000.
At RTP a gas occupies 24 dm³ mol⁻¹; otherwise use pV = nRT in SI units (Pa, m³, K, R = 8.31).
Reacting-mass problems: to moles → balanced-equation ratio → back to mass/volume.
The limiting reagent runs out first and controls the product; base all product amounts on it.
Percentage yield = actual ÷ theoretical × 100; atom economy = Mr(desired) ÷ Mr(all products) × 100 — they are different.
CP1: react a known mass, collect and measure the gas, and work back to the molar volume (~24 dm³ mol⁻¹).

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

A mole contains 6.02 × 10²³ particles (the Avogadro constant, N_A).
n = m/M ; N = n × N_A ; n = c × V (V in dm³) ; n = V/24 at RTP.
pV = nRT with p in Pa, V in m³, n in mol, R = 8.31 J K⁻¹ mol⁻¹, T in K.
1 dm³ = 1000 cm³ ; convert cm³ → dm³ by dividing by 1000.
Percentage yield = (actual yield ÷ theoretical yield) × 100.
Atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100.
The limiting reagent is used up first and determines the amount of product.

How it’s examined

Amount of substance is examined on Unit 1 (Structure, Bonding and Introduction to Organic Chemistry) and threads through every later calculation paper. Expect: short 'state/define' questions on the mole and Avogadro constant (1–2 marks); single-formula calculations (n = m/M, n = cV, volume at RTP; 2–3 marks); multi-step calculations combining limiting reagent, reacting masses and percentage yield (4–6 marks); ideal-gas questions requiring unit conversion (4–6 marks); and a 'compare/explain' item on percentage yield versus atom economy (green chemistry). Method marks are awarded for correct working even if a numerical slip occurs, so always show formula → substitution with units → answer, and quote the correct significant figures and units to secure the accuracy mark. CP1 provides the practical context for molar-volume and error-analysis questions.

Sources: Pearson Edexcel International Advanced Level Chemistry (YCH11/XCH11) specification (Issue 1, 2018 onwards); Edexcel IAL Chemistry Unit 1 examiner reports; Edexcel IAL Chemistry Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-07-07.

Take this whole topic with you

Step-by-step worked examples — Amount of Substance

Step-by-step solutions to past-paper-style questions on amount of substance, written exactly the way a tutor would explain them at the board.

Question type:

1Moles from mass, n = m/M (2 marks)
Getting startedDirect calculation• mole, n=m/M, AO2, Unit 1
Question
Calculate the amount, in moles, of carbon dioxide molecules in 8.8 g of CO₂. (A_r: C = 12.0, O = 16.0) (2 marks)
Step-by-step solution
1. Step 1
  Work out the molar mass of CO₂ by adding the relative atomic masses.
  $M(CO_2) = 12.0 + (2 \times 16.0) = 44.0 \text{ g mol}^{-1}$
2. Step 2
  Substitute the mass and molar mass into $n = m/M$ .
  $n = \frac{m}{M} = \frac{8.8}{44.0} = 0.20 \text{ mol}$
Answer
n(CO₂) = 0.20 mol
Examiner tip
Mark scheme: (1) M(CO₂) = 44.0 g mol⁻¹; (2) n = 8.8/44.0 = 0.20 mol. A* discriminator: quote the unit (mol) and the correct significant figures. The commonest error is dividing 8.8 by 12 (mass of C only) — you must use the full molar mass of the compound.
2Number of particles using the Avogadro constant (2 marks)
Getting startedDirect calculation• Avogadro, particles, AO2, Unit 1
Question
Calculate the number of molecules present in 0.20 mol of carbon dioxide. (Avogadro constant N_A = 6.02 × 10²³ mol⁻¹) (2 marks)
Step-by-step solution
1. Step 1
  Number of particles = amount in moles × Avogadro constant.
  $N = n \times N_A$
2. Step 2
  Substitute the values.
  $N = 0.20 \times (6.02 \times 10^{23}) = 1.20 \times 10^{23} \text{ molecules}$
Answer
1.20 × 10²³ molecules of CO₂
Examiner tip
Mark scheme: (1) correct method N = n × N_A; (2) 1.20 × 10²³. A* discriminator: the question asks for molecules — if it had asked for the number of oxygen atoms you would multiply by a further 2 (there are 2 O atoms per CO₂ molecule), giving 2.41 × 10²³. Read whether it wants molecules, atoms or ions.
3Concentration of a solution (3 marks)
Getting startedMulti-step problem• concentration, solutions, AO2, Unit 1
Question
5.85 g of sodium chloride is dissolved in water and the solution made up to 250 cm³. Calculate the concentration of the solution in mol dm⁻³. (A_r: Na = 23.0, Cl = 35.5) (3 marks)
Step-by-step solution
1. Step 1
  Find the amount of NaCl in moles using $n = m/M$ .
  $M(NaCl) = 23.0 + 35.5 = 58.5;\quad n = \frac{5.85}{58.5} = 0.100 \text{ mol}$
2. Step 2
  Convert the volume from cm³ to dm³ by dividing by 1000.
  $V = \frac{250}{1000} = 0.250 \text{ dm}^3$
3. Step 3
  Apply $c = n/V$ .
  $c = \frac{n}{V} = \frac{0.100}{0.250} = 0.400 \text{ mol dm}^{-3}$
Answer
0.400 mol dm⁻³
Examiner tip
Mark scheme: (1) n(NaCl) = 0.100 mol; (2) volume converted to 0.250 dm³; (3) c = 0.400 mol dm⁻³. A* discriminator: the single biggest mark loss is forgetting to convert cm³ → dm³ (÷1000). Using 250 instead of 0.250 gives an answer 1000× too small.
4Volume of gas at RTP (3 marks)
Building confidenceMulti-step problem• molar gas volume, RTP, AO2, Unit 1
Question
Calculate the volume of carbon dioxide, measured at room temperature and pressure (RTP), released when 0.20 mol of calcium carbonate is completely decomposed. Molar gas volume at RTP = 24 dm³ mol⁻¹. (3 marks)
Step-by-step solution
1. Step 1
  Write the balanced equation and read off the mole ratio.
  $CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \quad (\text{1 : 1})$
2. Step 2
  The ratio CaCO₃ : CO₂ is 1 : 1, so 0.20 mol of CaCO₃ gives 0.20 mol of CO₂.
  $n(CO_2) = 0.20 \text{ mol}$
3. Step 3
  Volume of gas = amount in moles × molar gas volume.
  $V = n \times 24 = 0.20 \times 24 = 4.8 \text{ dm}^3$
Answer
4.8 dm³ of CO₂ at RTP
Examiner tip
Mark scheme: (1) correct 1:1 mole ratio from a balanced equation; (2) n(CO₂) = 0.20 mol; (3) V = 4.8 dm³. A* discriminator: 24 dm³ mol⁻¹ is the volume of ONE mole at RTP — multiply by the moles of gas, not by the moles of solid reactant unless the ratio is 1:1. Watch the units: 24 dm³ (not 24 000 cm³ unless the answer is asked for in cm³).
5Finding a concentration by titration (5 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Chemistry Unit 1• titration, concentration, AO2, Unit 1
Question
In a titration, 25.0 cm³ of sodium hydroxide solution was exactly neutralised by 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. (5 marks)
Step-by-step solution
1. Step 1
  Write the balanced neutralisation equation.
  $NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) \quad (\text{1 : 1})$
2. Step 2
  Calculate the moles of HCl using $n = c \times V$ (volume in dm³).
  $n(HCl) = 0.100 \times \frac{20.0}{1000} = 2.00 \times 10^{-3} \text{ mol}$
3. Step 3
  The ratio NaOH : HCl is 1 : 1, so the moles of NaOH equal the moles of HCl.
  $n(NaOH) = 2.00 \times 10^{-3} \text{ mol}$
4. Step 4
  Rearrange $c = n/V$ using the NaOH volume in dm³.
  $c(NaOH) = \frac{2.00 \times 10^{-3}}{25.0/1000} = \frac{2.00 \times 10^{-3}}{0.0250}$
5. Step 5
  Evaluate to the correct significant figures.
  $c(NaOH) = 0.0800 \text{ mol dm}^{-3}$
Answer
0.0800 mol dm⁻³
Examiner tip
Mark scheme: (1) balanced equation / 1:1 ratio; (2) n(HCl) = 2.00 × 10⁻³ mol; (3) n(NaOH) equal by ratio; (4) correct rearrangement; (5) 0.0800 mol dm⁻³ (3 s.f.). A* discriminator: both volumes must be converted to dm³, and the mole ratio comes from the balanced equation — for a reaction like H₂SO₄ + 2NaOH the ratio is 1:2, not 1:1.
6Limiting reagent then percentage yield (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Chemistry Unit 1• limiting reagent, percentage yield, AO2, Unit 1
Question
2.4 g of magnesium was added to 100 cm³ of 1.5 mol dm⁻³ hydrochloric acid. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). (a) Show, by calculation, which reactant is the limiting reagent. (b) Calculate the maximum mass of MgCl₂ that could form. (c) The actual mass of MgCl₂ obtained was 6.0 g. Calculate the percentage yield. (A_r: Mg = 24.0, Cl = 35.5) (6 marks)
Step-by-step solution
1. Step 1
  Find the moles of each reactant.
  $n(Mg) = \frac{2.4}{24.0} = 0.100 \text{ mol};\quad n(HCl) = 1.5 \times \frac{100}{1000} = 0.150 \text{ mol}$
2. Step 2
  Use the 1 : 2 ratio to test. 0.100 mol Mg would need 0.200 mol HCl, but only 0.150 mol HCl is available — so HCl runs out first.
  $\text{HCl needed} = 2 \times 0.100 = 0.200 > 0.150 \Rightarrow \textbf{HCl is limiting}$
3. Step 3
  Base the product on the limiting reagent. From 2HCl : 1MgCl₂, moles of MgCl₂ = ½ × moles of HCl.
  $n(MgCl_2) = \frac{0.150}{2} = 0.0750 \text{ mol}$
4. Step 4
  Find the molar mass of MgCl₂ and the theoretical mass.
  $M(MgCl_2) = 24.0 + (2 \times 35.5) = 95.0;\quad m = 0.0750 \times 95.0 = 7.13 \text{ g}$
5. Step 5
  Percentage yield = (actual ÷ theoretical) × 100.
  $\% \text{ yield} = \frac{6.0}{7.13} \times 100 = 84.2\%$
Answer
(a) HCl is the limiting reagent. (b) Theoretical mass of MgCl₂ = 7.13 g. (c) Percentage yield = 84.2%.
Examiner tip
Mark scheme: (1) both amounts n(Mg) = 0.100, n(HCl) = 0.150; (2) HCl identified as limiting with justification via the 1:2 ratio; (3) n(MgCl₂) = 0.0750 mol; (4) theoretical mass = 7.13 g; (5) % yield = 84.2%. A* discriminator: candidates who forget the 2 in front of HCl treat the ratio as 1:1 and wrongly conclude Mg is limiting. Always base the product amount on the LIMITING reagent, never the reagent in excess.
7Using the ideal gas equation pV = nRT (5 marks)
StretchMulti-step problem• ideal gas, pV=nRT, AO2, Unit 1
Question
A sealed flask contains 0.25 mol of nitrogen gas at 27 °C and a pressure of 150 kPa. Calculate the volume of the flask, giving your answer in dm³. (R = 8.31 J K⁻¹ mol⁻¹) (5 marks)
Step-by-step solution
1. Step 1
  Convert every quantity to SI units required by pV = nRT: pressure in pascals, temperature in kelvin.
  $p = 150 \text{ kPa} = 150\,000 \text{ Pa};\quad T = 27 + 273 = 300 \text{ K}$
2. Step 2
  Rearrange the ideal gas equation to make V the subject.
  $pV = nRT \Rightarrow V = \frac{nRT}{p}$
3. Step 3
  Substitute the values in SI units.
  $V = \frac{0.25 \times 8.31 \times 300}{150\,000}$
4. Step 4
  Evaluate — the answer comes out in cubic metres (m³).
  $V = \frac{623.25}{150\,000} = 4.16 \times 10^{-3} \text{ m}^3$
5. Step 5
  Convert m³ to dm³ by multiplying by 1000 (1 m³ = 1000 dm³).
  $V = 4.16 \times 10^{-3} \times 1000 = 4.16 \text{ dm}^3$
Answer
V = 4.16 × 10⁻³ m³ = 4.16 dm³
Examiner tip
Mark scheme: (1) p converted to Pa AND T converted to K; (2) correct rearrangement V = nRT/p; (3)(4) substitution and 4.16 × 10⁻³ m³; (5) converted to 4.16 dm³. A* discriminator: pV = nRT only works in SI units when R = 8.31 J K⁻¹ mol⁻¹ — kPa, cm³ or °C give answers out by factors of 1000 or nonsense. The raw answer is in m³; converting to dm³ (×1000) or cm³ (×10⁶) is where most marks are dropped.
8Atom economy of an extraction (5 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Chemistry Unit 1• atom economy, green chemistry, AO2, Unit 1
Question
Titanium is extracted by reducing titanium(IV) chloride with magnesium: TiCl₄ + 2Mg → Ti + 2MgCl₂. Calculate the atom economy of this process with respect to titanium, and comment on the value. (A_r: Ti = 47.9, Mg = 24.3, Cl = 35.5) (5 marks)
Step-by-step solution
1. Step 1
  Write the atom-economy formula: mass of the DESIRED product over the total mass of ALL products.
  $\% \text{ atom economy} = \frac{M_r(\text{desired product})}{\sum M_r(\text{all products})} \times 100$
2. Step 2
  Find the Mr of the desired product, titanium.
  $M_r(Ti) = 47.9$
3. Step 3
  Find the total Mr of all products (Ti + 2MgCl₂).
  $M_r(MgCl_2) = 24.3 + (2 \times 35.5) = 95.3;\quad \text{total} = 47.9 + (2 \times 95.3) = 238.5$
4. Step 4
  Substitute into the formula.
  $\% \text{ atom economy} = \frac{47.9}{238.5} \times 100 = 20.1\%$
5. Step 5
  Comment: only about a fifth of the reactant mass ends up as the wanted product; the rest (MgCl₂) is by-product. A low atom economy is wasteful of raw materials and produces large amounts of waste to treat or recycle.
Answer
Atom economy = 20.1%. This is low, so most of the starting material becomes waste (MgCl₂) — economically and environmentally inefficient.
Examiner tip
Mark scheme: (1) correct formula; (2) Mr(Ti) = 47.9; (3) total product Mr = 238.5; (4) 20.1%; (5) sensible comment linking the low value to waste/inefficiency. A* discriminator: atom economy uses Mr values from the balanced equation only — it is fixed by the equation and is NOT the same as percentage yield (which depends on how much product you actually isolate). A reaction can have 100% yield yet low atom economy.

Model Answers — Amount of Substance

High-scoring sample answers for amount of substance on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by the term mole and give the value of the Avogadro constant.
Model answer
A mole is the amount of substance that contains the same number of particles (atoms, molecules, ions or electrons) as there are atoms in exactly 12 g of carbon-12. This number is the Avogadro constant, $N_A = 6.02 \times 10^{23}$ particles per mole (mol⁻¹).

When you state a number of moles you must also state what the particles are — for example "0.5 mol of oxygen molecules" contains twice as many oxygen atoms.
Why this scores
Why this scores 2/2. (1) mole = amount containing 6.02 × 10²³ particles (or 'as many particles as atoms in 12 g of carbon-12'); (2) N_A = 6.02 × 10²³ mol⁻¹. A* answers specify the type of particle, since 'a mole of oxygen' is ambiguous between O and O₂.
Question 2
2 marks
Q (2 marks). Calculate the amount, in moles, of sodium hydroxide in 20.0 g of NaOH. (A_r: Na = 23.0, O = 16.0, H = 1.0)
Model answer
First find the molar mass:

$M(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0 \text{ g mol}^{-1}$

Then apply $n = m/M$ :

$n = \dfrac{20.0}{40.0} = 0.500 \text{ mol}$
Why this scores
Why this scores 2/2. (1) M(NaOH) = 40.0 g mol⁻¹; (2) n = 0.500 mol with the unit shown. Dropping the unit or giving 0.5 without significant-figure care is where casual answers lose the A mark on harder questions.
Question 3
3 marks
Q (3 marks). Calculate the mass of potassium hydroxide, KOH, needed to make 500 cm³ of a 0.200 mol dm⁻³ solution. (A_r: K = 39.1, O = 16.0, H = 1.0)
Model answer
Convert the volume to dm³ and find the moles of KOH required:

$V = \dfrac{500}{1000} = 0.500 \text{ dm}^3$

$n = c \times V = 0.200 \times 0.500 = 0.100 \text{ mol}$

Find the molar mass and then the mass:

$M(\text{KOH}) = 39.1 + 16.0 + 1.0 = 56.1 \text{ g mol}^{-1}$

$m = n \times M = 0.100 \times 56.1 = 5.61 \text{ g}$
Why this scores
Why this scores 3/3. (1) volume converted and n = 0.100 mol; (2) M(KOH) = 56.1 g mol⁻¹; (3) m = 5.61 g. A* discriminator: the cm³ → dm³ conversion (÷1000) is the classic dropped mark, and the final mass must carry the correct significant figures and unit.
Question 4
4 marks
Q (4 marks). Magnesium burns in oxygen: 2Mg(s) + O₂(g) → 2MgO(s). Calculate the mass of magnesium oxide formed when 6.0 g of magnesium is burned completely in excess oxygen. (A_r: Mg = 24.0, O = 16.0)
Model answer
Find the moles of magnesium:

$n(\text{Mg}) = \dfrac{6.0}{24.0} = 0.25 \text{ mol}$

From the balanced equation the ratio Mg : MgO is 2 : 2, i.e. 1 : 1, so:

$n(\text{MgO}) = 0.25 \text{ mol}$

Find the molar mass of MgO and the mass:

$M(\text{MgO}) = 24.0 + 16.0 = 40.0 \text{ g mol}^{-1}$

$m(\text{MgO}) = n \times M = 0.25 \times 40.0 = 10.0 \text{ g}$
Why this scores
Why this scores 4/4. (1) n(Mg) = 0.25 mol; (2) correct 1:1 mole ratio from the balanced equation; (3) n(MgO) = 0.25 mol; (4) mass = 10.0 g. A* discriminator: the ratio must be read from the balanced equation (the 2:1:2 stoichiometry simplifies to 1:1 for Mg:MgO here) — do not assume every ratio is 1:1.
Question 5
5 marks
Q (5 marks). Explain the difference between percentage yield and atom economy, and state why a reaction with a high percentage yield can still have a low atom economy.
Model answer
Percentage yield compares how much product you actually obtain with the maximum you could theoretically obtain from the limiting reagent:

$\% \text{ yield} = \frac{\text{actual amount (or mass) of product}}{\text{theoretical (maximum) amount}} \times 100$

It measures practical efficiency — losses from incomplete reactions, side-reactions, or product left behind during purification lower the yield.

Atom economy measures how much of the mass of the reactants ends up in the desired product, based only on the balanced equation:

$\% \text{ atom economy} = \frac{M_r \text{ of desired product}}{\sum M_r \text{ of all products}} \times 100$

It measures how efficiently the reactants' atoms are used, i.e. how little by-product is made.

A reaction can have a high percentage yield but low atom economy if it goes essentially to completion (so almost all the limiting reagent becomes product), yet the balanced equation produces large amounts of unwanted by-product alongside the desired product. In that case nearly all reactant is converted (high yield), but much of the reactant mass ends up as by-product rather than as the wanted product (low atom economy). Reactions with high atom economy are preferred in industry because they generate less waste and use raw materials more sustainably.
Why this scores
Why this scores 5/5. (1) correct percentage-yield definition/equation (actual vs theoretical); (2) yield reflects practical losses; (3) correct atom-economy definition/equation (desired product Mr ÷ total product Mr); (4) atom economy reflects proportion of atoms in the wanted product; (5) clear reason the two differ — a single-product reaction is 100% atom economy, but a multi-product reaction can be 100% yield yet low atom economy. A* answers include both equations and note the green-chemistry/waste benefit.
Question 6
4 marks
Q (4 marks). 25.0 cm³ of 0.0500 mol dm⁻³ sulfuric acid was exactly neutralised by 20.0 cm³ of sodium hydroxide solution. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the sodium hydroxide.
Model answer
Moles of sulfuric acid:

$n(\text{H}_2\text{SO}_4) = c \times V = 0.0500 \times \dfrac{25.0}{1000} = 1.25 \times 10^{-3} \text{ mol}$

The ratio H₂SO₄ : NaOH is 1 : 2, so:

$n(\text{NaOH}) = 2 \times 1.25 \times 10^{-3} = 2.50 \times 10^{-3} \text{ mol}$

Concentration of NaOH (volume in dm³):

$c(\text{NaOH}) = \dfrac{2.50 \times 10^{-3}}{20.0/1000} = \dfrac{2.50 \times 10^{-3}}{0.0200} = 0.125 \text{ mol dm}^{-3}$
Why this scores
Why this scores 4/4. (1) n(H₂SO₄) = 1.25 × 10⁻³ mol; (2) correct use of the 1:2 ratio; (3) n(NaOH) = 2.50 × 10⁻³ mol; (4) c = 0.125 mol dm⁻³. A* discriminator: the 1:2 stoichiometry is essential — treating it as 1:1 halves the answer. This is the most common titration slip in examiner reports.
Question 7
6 marks
Q (6 marks). 0.72 g of a gaseous hydrocarbon occupies a volume of 500 cm³ at a temperature of 90 °C and a pressure of 100 kPa. (a) Calculate the amount, in moles, of the gas. (b) Hence calculate the molar mass of the gas. (c) Suggest a molecular formula for the hydrocarbon. (R = 8.31 J K⁻¹ mol⁻¹, A_r: C = 12.0, H = 1.0)
Model answer
(a) Convert to SI units: $p = 100\,000$ Pa, $V = 500 \text{ cm}^3 = 500 \times 10^{-6} = 5.00 \times 10^{-4} \text{ m}^3$ , $T = 90 + 273 = 363 \text{ K}$ .

Rearrange $pV = nRT$ :

$n = \frac{pV}{RT} = \frac{100\,000 \times 5.00 \times 10^{-4}}{8.31 \times 363} = \frac{50.0}{3016.5} = 1.66 \times 10^{-2} \text{ mol}$

(b) Molar mass:

$M = \frac{m}{n} = \frac{0.72}{1.66 \times 10^{-2}} = 43.4 \text{ g mol}^{-1}$

(c) A hydrocarbon with $M \approx 44$ g mol⁻¹ is propane, C₃H₈ ( $M = (3 \times 12.0) + (8 \times 1.0) = 44.0$ g mol⁻¹), which matches the calculated value.
Why this scores
Why this scores 6/6. (a) 3 marks: all units converted (Pa, m³, K) AND n = 1.66 × 10⁻² mol; (b) 2 marks: M = m/n rearranged and ≈ 43–44 g mol⁻¹; (c) 1 mark: C₃H₈ identified as consistent with M ≈ 44. A* discriminator: the cm³ → m³ conversion (×10⁻⁶) and °C → K conversion are the mark-critical steps; a molar mass near 44 must be matched to a real hydrocarbon formula, not just left as a number.
Question 8
Adapted from Edexcel IAL Chemistry Unit 1 Core Practical 16 marks
Q (6 marks). (Based on Core Practical 1.) A student added an excess of dilute hydrochloric acid to 0.120 g of magnesium and collected the hydrogen gas produced in a gas syringe. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). (a) Calculate the amount, in moles, of hydrogen expected. (b) Calculate the volume of hydrogen that should be collected at RTP (molar gas volume = 24 dm³ mol⁻¹ = 24 000 cm³ mol⁻¹). (c) The student collected 108 cm³. Suggest two reasons why the measured volume was less than expected. (A_r: Mg = 24.0)
Model answer
(a) Moles of magnesium (limiting, since acid is in excess):

$n(\text{Mg}) = \dfrac{0.120}{24.0} = 5.00 \times 10^{-3} \text{ mol}$

From Mg : H₂ = 1 : 1:

$n(\text{H}_2) = 5.00 \times 10^{-3} \text{ mol}$

(b) Volume at RTP:

$V = n \times 24\,000 = 5.00 \times 10^{-3} \times 24\,000 = 120 \text{ cm}^3$

(c) The measured 108 cm³ is less than the expected 120 cm³. Possible reasons include:

Some hydrogen escaped before the syringe was connected / gas leaked from the apparatus.

The magnesium was partly oxidised (a layer of MgO on the surface), so less than 0.120 g was actually reactive metal.

The reaction was not complete / some gas dissolved, or gas was still being measured while cooling.

(Any two valid reasons score the marks.)
Why this scores
Why this scores 6/6. (a) 2 marks: n(Mg) = 5.00 × 10⁻³ mol and 1:1 ratio to H₂; (b) 2 marks: V = n × 24 000 = 120 cm³ (correct RTP value in cm³); (c) 2 marks: any two sensible sources of loss (leaks/escape before sealing, oxide layer on Mg, incomplete reaction). A* discriminator: recognising the acid is in excess so Mg is limiting, and using 24 000 cm³ mol⁻¹ (not 24) because the answer is wanted in cm³.
Question 9
Adapted from Edexcel IAL Chemistry Unit 17 marks
Q (7 marks). Ammonia is oxidised industrially: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). In a trial, 68 g of ammonia was reacted with 200 g of oxygen. (a) Determine which reactant is limiting. (b) Calculate the maximum mass of nitrogen monoxide, NO, that could be produced. (c) 90 g of NO was actually collected — calculate the percentage yield. (A_r: N = 14.0, H = 1.0, O = 16.0)
Model answer
(a) Moles of each reactant:

$M(\text{NH}_3) = 14.0 + (3 \times 1.0) = 17.0 \Rightarrow n(\text{NH}_3) = \dfrac{68}{17.0} = 4.0 \text{ mol}$

$M(\text{O}_2) = 32.0 \Rightarrow n(\text{O}_2) = \dfrac{200}{32.0} = 6.25 \text{ mol}$

The ratio required is NH₃ : O₂ = 4 : 5. For 4.0 mol NH₃ we would need $\frac{5}{4} \times 4.0 = 5.0$ mol O₂. There is 6.25 mol O₂ available (more than 5.0), so NH₃ is the limiting reagent (O₂ is in excess).

(b) From NH₃ : NO = 4 : 4 = 1 : 1:

$n(\text{NO}) = 4.0 \text{ mol}$

$M(\text{NO}) = 14.0 + 16.0 = 30.0 \Rightarrow m(\text{NO}) = 4.0 \times 30.0 = 120 \text{ g}$

(c) Percentage yield:

$\% \text{ yield} = \frac{90}{120} \times 100 = 75\%$
Why this scores
Why this scores 7/7. (a) 3 marks: n(NH₃) = 4.0 mol, n(O₂) = 6.25 mol, and NH₃ identified as limiting via the 4:5 ratio comparison; (b) 3 marks: n(NO) = 4.0 mol from the 1:1 ratio, M(NO) = 30.0, theoretical mass = 120 g; (c) 1 mark: 75%. A* discriminator: the limiting-reagent test must compare the required ratio (4:5) against what is available, not just compare raw moles. Basing the NO amount on O₂ (the excess reagent) is the classic error.

Key Definitions and Keywords — Amount of Substance

Definitions to memorise and the exact keywords mark schemes credit for amount of substance answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Mole (mol)
Examiner keyword
The amount of substance that contains as many elementary particles (atoms, molecules, ions or electrons) as there are atoms in exactly 12 g of carbon-12 — that is, 6.02 × 10²³ particles.
Example
1 mol of CO₂ contains 6.02 × 10²³ molecules and has a mass of 44.0 g.
Avogadro constant (N_A)
Examiner keyword
The number of particles in one mole of a substance, 6.02 × 10²³ mol⁻¹. Number of particles = amount in moles × N_A.
Molar mass (M)
Examiner keyword
The mass of one mole of a substance, in g mol⁻¹. Numerically equal to the relative atomic, molecular or formula mass. Linked to moles by n = m/M.
Relative atomic mass (A_r)
Examiner keyword
The weighted mean mass of an atom of an element relative to 1/12 the mass of a carbon-12 atom. It has no units.
Relative molecular / formula mass (M_r)
Examiner keyword
The sum of the relative atomic masses of all the atoms shown in the molecular formula (molecular mass) or formula unit (formula mass) of a substance. No units.
Concentration
Examiner keyword
The amount of solute dissolved per unit volume of solution, usually in mol dm⁻³ (or g dm⁻³). Related to moles by n = c × V, with V in dm³ (cm³ ÷ 1000).
Standard solution
Examiner keyword
A solution of accurately known concentration, made by dissolving a weighed mass of solute and making it up to a known volume in a volumetric flask. Used in titrations.
Molar gas volume
Examiner keyword
The volume occupied by one mole of any gas under stated conditions — 24 dm³ mol⁻¹ (24 000 cm³ mol⁻¹) at RTP. Volume of gas = amount in moles × molar gas volume.
Ideal gas equation (pV = nRT)
Examiner keyword
Relates pressure p (Pa), volume V (m³), amount n (mol), gas constant R = 8.31 J K⁻¹ mol⁻¹ and temperature T (K) for an ideal gas. All quantities must be in these SI units.
RTP (room temperature and pressure)
Approximately 20–25 °C and 100 kPa (1 atm). At RTP one mole of gas occupies 24 dm³.
Limiting reagent
Examiner keyword
The reactant that is completely used up first in a reaction; it determines the maximum amount of product that can form. The other reactant(s) are in excess.
Theoretical yield
Examiner keyword
The maximum mass (or amount) of product that could be obtained from the limiting reagent, assuming the reaction goes fully to completion with no losses.
Percentage yield
Examiner keyword
(actual yield ÷ theoretical yield) × 100. Measures how much of the maximum possible product was actually obtained; reduced by incomplete reactions, side-reactions and losses on purification.
Atom economy
Examiner keyword
(M_r of desired product ÷ sum of M_r of all products) × 100. Measures the proportion of reactant atoms that end up in the useful product; a high atom economy means less waste.
Empirical formula
Examiner keyword
The simplest whole-number ratio of atoms of each element in a compound (e.g. CH₂ for many alkenes). Found from masses or percentage composition by dividing by A_r and simplifying the ratio.
Molecular formula
Examiner keyword
The actual number of atoms of each element in a molecule (e.g. C₄H₈). It is a whole-number multiple of the empirical formula, found using the relative molecular mass.

Common Mistakes and Misconceptions — Amount of Substance

The traps other students keep falling into on amount of substance questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Not converting cm³ to dm³ in concentration calculations
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Titration volumes are read in cm³ from the burette, so students plug the raw value straight into n = c × V.
How to avoid it
Always divide cm³ by 1000 to get dm³ before using n = c × V or c = n/V. A quick check: sensible lab concentrations are ~0.01–2 mol dm⁻³; an answer of 800 or 0.0001 signals a factor-of-1000 slip.
✕Using kPa, cm³ or °C in the ideal gas equation pV = nRT
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Data is often given in convenient units and students forget R = 8.31 requires strict SI units.
How to avoid it
Convert before substituting: pressure to Pa (×1000 from kPa), volume to m³ (cm³ ×10⁻⁶, dm³ ×10⁻³), temperature to K (°C + 273). The raw answer for V is in m³ — convert back to dm³ (×1000) or cm³ (×10⁶) as required.
✕Confusing percentage yield with atom economy
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Both are percentages about 'efficiency', so students use the wrong formula or treat them as the same thing.
How to avoid it
Percentage yield = actual ÷ theoretical product (depends on the experiment). Atom economy = Mr of desired product ÷ Mr of all products (fixed by the balanced equation). A reaction can be 100% yield yet low atom economy.
✕Ignoring the mole ratio from the balanced equation (assuming 1:1)
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Students rush from moles of reactant to moles of product without checking the stoichiometric coefficients.
How to avoid it
Read the balancing numbers from the equation. For H₂SO₄ + 2NaOH the ratio is 1:2; for 2Mg + O₂ it is 2:1. Multiply/divide the moles by the correct ratio before finding mass or volume.
✕Basing the product amount on the reactant in excess instead of the limiting reagent
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Students calculate moles of both reactants but pick the larger number, or fail to test against the mole ratio.
How to avoid it
Compare the ratio required by the equation with the ratio available. The reagent that runs out first (needs more than is present) is limiting — always base the product on it.
✕Rounding intermediate values too early and losing significant figures
Edexcel IAL Chemistry Unit 1 examiner reports
Why it happens
Students round each step to 2 figures, so accumulated errors shift the final answer.
How to avoid it
Keep full accuracy in the calculator through the whole calculation and round only the final answer, matching the significant figures of the data (usually 3 s.f.). Always state the correct unit.

Amount of Substance — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Amount of Substance

Short Notes - Amount of Substance

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Moles from mass, n = m/M (2 marks)

2Number of particles using the Avogadro constant (2 marks)

3Concentration of a solution (3 marks)

4Volume of gas at RTP (3 marks)

5Finding a concentration by titration (5 marks)

6Limiting reagent then percentage yield (6 marks)

7Using the ideal gas equation pV = nRT (5 marks)

8Atom economy of an extraction (5 marks)

Mole (mol)

Avogadro constant (N_A)

Molar mass (M)

Relative atomic mass (A_r)

Relative molecular / formula mass (M_r)

Concentration

Standard solution

Molar gas volume

Ideal gas equation (pV = nRT)

RTP (room temperature and pressure)

Limiting reagent

Theoretical yield

Percentage yield

Atom economy

Empirical formula

Molecular formula

✕Not converting cm³ to dm³ in concentration calculations

✕Using kPa, cm³ or °C in the ideal gas equation pV = nRT

✕Confusing percentage yield with atom economy

✕Ignoring the mole ratio from the balanced equation (assuming 1:1)

✕Basing the product amount on the reactant in excess instead of the limiting reagent

✕Rounding intermediate values too early and losing significant figures

Amount of Substance — frequently asked questions