What is the significance of stoichiometry in calculating reacting masses and volumes in Chemistry?

Stoichiometry is crucial in Chemistry, especially in the Cambridge International A Levels curriculum, as it allows students to calculate the exact amounts of reactants and products involved in a chemical reaction. By using balanced chemical equations, students can determine the reacting masses and volumes of solutions and gases, ensuring that reactions are carried out with precision and efficiency.

How do you calculate the volume of a gas produced in a chemical reaction at standard temperature and pressure (STP)?

To calculate the volume of a gas produced at standard temperature and pressure (STP), you can use the ideal gas law or the molar volume concept. At STP, one mole of any gas occupies 22.4 liters. By determining the number of moles of gas produced using stoichiometry, you can multiply by 22.4 L/mol to find the volume of the gas.

What is the role of molarity in determining the volume of solutions in chemical reactions?

Molarity, defined as moles of solute per liter of solution, is essential for calculating the volume of solutions in chemical reactions. By knowing the molarity and the amount of solute needed, students can determine the volume of solution required to achieve a desired concentration, which is a key aspect of stoichiometry in the Cambridge International A Levels Chemistry syllabus.

How can you use a balanced chemical equation to find the mass of a reactant needed for a reaction?

A balanced chemical equation provides the mole ratio of reactants and products. To find the mass of a reactant needed, first calculate the moles of the desired product or reactant using stoichiometry. Then, convert moles to mass using the molar mass of the reactant. This process is fundamental in understanding reacting masses in the context of Atoms, molecules and stoichiometry in AS-Level Physical Chemistry.

Why is it important to consider the limiting reactant when calculating reacting masses and volumes?

The limiting reactant is the substance that is completely consumed first in a chemical reaction, determining the maximum amount of product that can be formed. Identifying the limiting reactant is crucial for accurately calculating reacting masses and volumes, as it ensures that calculations are based on the actual amounts of substances that will react, a key concept in the Cambridge International A Levels Chemistry curriculum.

Why is "Forgetting to convert cm³ to dm³ in n = cV." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Reading burette volumes in cm³. How to avoid it: Divide cm³ by 1000 before using n = cV. Source: 9701 Examiner Reports 2022-2024

Why is "Choosing the limiting reagent by raw moles only." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Ignoring the equation coefficients. How to avoid it: Divide each reactant's moles by its coefficient first.

Why is "Comparing the product mass to the reactant mass for % yield." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Confusing the two masses. How to avoid it: % yield compares actual product to the theoretical mass of that SAME product.

Why is "Using a 1 : 1 ratio for diprotic acids/bases." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Not reading the balanced equation. How to avoid it: H₂SO₄ : NaOH = 1 : 2; Ca(OH)₂ : HCl = 1 : 2, etc.

Why is "Using °C, kPa or dm³ directly in pV = nRT." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Not converting to SI base units. How to avoid it: Convert to Pa, m³ and K first (T/K = °C + 273; 1 dm³ = 1×10⁻³ m³). Source: 9701 Examiner Reports 2022-2024

Why is "Quoting the wrong number of significant figures or no units." a common mistake in Reacting masses and volumes (of solutions and gases)?

Why it happens: Rounding early or omitting units. How to avoid it: Match the data (usually 3 s.f.), round at the end, and always give units.

Atoms, molecules and stoichiometry(AS-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

Reacting masses and volumes (of solutions and gases)

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Reacting Masses and Volumes — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

The mole-ratio method for reacting masses, percentage yield, gas volumes (24 dm³ mol⁻¹ at r.t.p.), solution concentrations and titrations, and limiting-reagent problems — with the significant-figure discipline examiners expect.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

2.4.1(a) — Calculate reacting masses (from formulae and equations), including percentage yield.
2.4.1(b) — Calculate volumes of gases (e.g. in the burning of hydrocarbons).
2.4.1(c) — Calculate volumes and concentrations of solutions.
2.4.1(d) — Identify limiting and excess reagents.
2.4.1(e) — Deduce stoichiometric relationships from such calculations.

The mole-ratio method (the master plan)

Everything routes through moles. Convert in, use the balanced equation's ratio, convert out.

Almost every stoichiometry question follows the same three steps:

Convert the known quantity into moles (using $n=m/M$ , $n=V/24.0$ , or $n=cV$ ).
Use the mole ratio from the balanced equation to find moles of the wanted substance.
Convert moles back into mass, gas volume or concentration.

Moles sit at the centre — convert any quantity in, cross the mole ratio, and convert out.

Worked — reacting masses. What mass of CaO forms from heating $50.0\,\text{g}$ of $\text{CaCO}_3$ ? $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

Moles $\text{CaCO}_3 = 50.0/100.1 = 0.4995$ .
Ratio 1 : 1 → moles CaO $= 0.4995$ .
Mass CaO $= 0.4995 \times 56.1 = 28.0\,\text{g}$ .

In → moles → ratio → out.
Choose the right 'in' equation: $m/M$ , $V/24$ or $cV$ .
The balanced equation gives the mole ratio.

Percentage yield

Compare what you actually got to the theoretical maximum from the equation.

Real reactions rarely give 100% of the product the equation predicts. The theoretical yield is the maximum mass from the mole calculation; the actual yield is what is obtained.

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

Why yield is below 100%: the reaction may be reversible / incomplete, side reactions occur, or product is lost during purification.

Worked. A reaction predicts $8.00\,\text{g}$ of product but gives $6.40\,\text{g}$ : $\% \text{ yield} = \frac{6.40}{8.00} \times 100 = 80.0\%$

Both masses must be in the same units and refer to the same substance.

% yield = actual / theoretical × 100.
Theoretical yield comes from the mole calculation.
<100% due to incomplete/reversible reactions, side reactions, losses.

Gas volumes and combustion

At r.t.p. one mole of any gas occupies 24.0 dm³. Gas volume ratios equal mole ratios.

Molar gas volume. At room temperature and pressure (r.t.p.) one mole of any gas occupies 24.0 dm³ (24 000 cm³): $n = \frac{V}{24.0}\quad(V \text{ in dm}^3)$

Avogadro's law shortcut. For gases measured at the same temperature and pressure, volume ratio = mole ratio, so you can use the equation coefficients directly on the volumes.

Worked — combustion of a hydrocarbon. What volume of $\text{CO}_2$ (at r.t.p.) forms when $100\,\text{cm}^3$ of propane burns completely? $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

Ratio $\text{C}_3\text{H}_8 : \text{CO}_2 = 1 : 3$ .
$\text{CO}_2$ volume $= 3 \times 100 = 300\,\text{cm}^3$ .

(For a mass-to-gas-volume question, go via moles: $n=m/M$ , then $V = 24.0\,n$ .)

1 mol gas = 24.0 dm³ at r.t.p.; $n = V/24.0$ .
Same T and P → gas volume ratio = mole ratio.
Combustion: balance the equation, then apply the ratio.

Concentrations, solutions and titrations

n = c × V with V in dm³. Convert cm³ to dm³ by dividing by 1000.

Concentration is moles per unit volume: $n = c \times V \quad (c \text{ in mol dm}^{-3},\ V \text{ in dm}^3)$

Always convert cm³ → dm³ first ( $\div 1000$ ). Concentration in g dm⁻³ = concentration in mol dm⁻³ $\times M$ .

Titration method. Use the volumes and the known concentration to find moles of one reactant, apply the mole ratio, then find the unknown concentration.

Worked. $25.0\,\text{cm}^3$ of NaOH is neutralised by $20.0\,\text{cm}^3$ of $0.100\,\text{mol dm}^{-3}$ HCl. Find $[\text{NaOH}]$ . $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Moles HCl $= 0.100 \times (20.0/1000) = 2.00 \times 10^{-3}$ .
Ratio 1 : 1 → moles NaOH $= 2.00 \times 10^{-3}$ .
$[\text{NaOH}] = \dfrac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800\,\text{mol dm}^{-3}$ .

$n = cV$ , $V$ in dm³.
cm³ → dm³: ÷1000.
g dm⁻³ = mol dm⁻³ × $M$ .
Titration: known → moles → ratio → unknown concentration.

Limiting and excess reagents

Work out moles of each reactant ÷ its coefficient; the smallest controls the yield.

When the amounts of two reactants are both given, one will run out first — the limiting reagent. It determines how much product forms; the other is in excess.

Method:

Find moles of each reactant.
Divide each by its coefficient in the balanced equation.
The smallest value is the limiting reagent.
Base all product calculations on the limiting reagent.

Worked. $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , with $1.0\,\text{mol N}_2$ and $1.5\,\text{mol H}_2$ .

$\text{N}_2: 1.0/1 = 1.0$ ; $\text{H}_2: 1.5/3 = 0.50$ .
$\text{H}_2$ has the smaller value → H₂ is limiting; N₂ is in excess.
Moles $\text{NH}_3 = \frac{2}{3} \times 1.5 = 1.0\,\text{mol}$ .

Moles ÷ coefficient for each reactant.
Smallest → limiting reagent; base product on it.
The other reactant is in excess.

Significant figures and units

Match the data's significant figures and quote correct units — examiners deduct for both.

The syllabus explicitly warns that misuse of units or significant figures is penalised.

Give your answer to the same number of significant figures as the data (usually 3 s.f.).
Don't round intermediate values too early — carry an extra figure and round only at the end.
Quote the correct unit: g, dm³, cm³, mol, mol dm⁻³.
Concentrations and ratios that are dimensionless (e.g. % yield) take no unit beyond %.

Match the data's s.f. (usually 3 s.f.).
Round only at the end.
Always state the unit.

How it’s examined

This is the single most examined calculation area — present on Paper 1, the structured Paper 2, and the practical Papers 3 and 5 (titration analysis). Examiner reports repeatedly cite: forgetting to convert cm³ to dm³, using raw moles instead of moles ÷ coefficient for limiting reagent, and answers given to the wrong number of significant figures.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 2.4; 9701 Examiner Reports 2022-2024; 9701/21 & 9701/51 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Reacting masses and volumes (of solutions and gases)

Step-by-step solutions to past-paper-style questions on reacting masses and volumes (of solutions and gases), written exactly the way a tutor would explain them at the board.

1Reacting mass calculation (3 marks)
Getting started• reacting mass
Question
What mass of magnesium oxide forms when 6.0 g of magnesium burns completely in oxygen? ( $A_r$ : Mg 24.0, O 16.0)
Step-by-step solution
1. Step 1
  Equation and moles of Mg.
  $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO};\quad n(\text{Mg}) = \frac{6.0}{24.0} = 0.25$
2. Step 2
  Mole ratio Mg : MgO = 2 : 2 = 1 : 1.
  $n(\text{MgO}) = 0.25$
3. Step 3
  Mass of MgO ( $M = 40.0$ ).
  $m = 0.25 \times 40.0 = 10.0\,\text{g}$
Answer
$10.0\,\text{g}$ of MgO.
Examiner tip
Three reliable steps every time: moles of what you know → mole ratio → mass of what you want.
2Moles in a solution (2 marks)
Getting started• concentration, n=cV
Question
Calculate the amount, in moles, of HCl in 25.0 cm³ of 0.10 mol dm⁻³ hydrochloric acid.
Step-by-step solution
1. Step 1
  Convert the volume to dm³ (÷ 1000).
  $V = \frac{25.0}{1000} = 0.0250\,\text{dm}^3$
2. Step 2
  Use $n = c \times V$ .
  $n = 0.10 \times 0.0250 = 2.5 \times 10^{-3}\,\text{mol}$
Answer
$2.5 \times 10^{-3}\,\text{mol}$ .
Examiner tip
The single most common slip in solution work is forgetting to divide cm³ by 1000.
3Gas volume by Avogadro's law (3 marks)
Building confidence• gas volume, Avogadro's law
Question
25 cm³ of butane $\text{C}_4\text{H}_{10}$ is burned completely. What volume of oxygen (same conditions) is needed?
Step-by-step solution
1. Step 1
  Balance the combustion.
  $2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}$
2. Step 2
  Same T, P → volume ratio = mole ratio (Avogadro's law), $\text{C}_4\text{H}_{10} : \text{O}_2 = 2 : 13$ .
  $V(\text{O}_2) = \frac{13}{2} \times 25$
3. Step 3
  Evaluate.
  $= 162.5\,\text{cm}^3$
Answer
$162.5\,\text{cm}^3$ of O₂.
Examiner tip
For gases at the same conditions you can use volume ratios directly — no need to convert to moles.
4Percentage yield (3 marks)
Building confidence• yield
Question
16.0 g of $\text{SO}_2$ is converted to $\text{SO}_3$ ; 18.0 g of $\text{SO}_3$ is obtained. Calculate the % yield. ( $M_r$ : SO₂ 64.1, SO₃ 80.1)
Step-by-step solution
1. Step 1
  Theoretical: moles SO₂ → moles SO₃ (1 : 1).
  $n(\text{SO}_2) = \frac{16.0}{64.1} = 0.2496 = n(\text{SO}_3)$
2. Step 2
  Theoretical mass of SO₃.
  $0.2496 \times 80.1 = 20.0\,\text{g}$
3. Step 3
  % yield.
  $\frac{18.0}{20.0} \times 100 = 90.0\%$
Answer
90.0% yield.
Examiner tip
Compare actual mass to the theoretical mass of the SAME product, not to the reactant mass.
5Titration concentration, diprotic acid (4 marks)
Stretch• titration, concentration
Question
In a titration, 24.0 cm³ of $\text{H}_2\text{SO}_4$ neutralises 25.0 cm³ of 0.200 mol dm⁻³ NaOH. Find $[\text{H}_2\text{SO}_4]$ .
Step-by-step solution
1. Step 1
  Moles of NaOH.
  $n(\text{NaOH}) = 0.200 \times \frac{25.0}{1000} = 5.00 \times 10^{-3}$
2. Step 2
  Ratio $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow$ , so $\text{H}_2\text{SO}_4 : \text{NaOH} = 1 : 2$ .
  $n(\text{H}_2\text{SO}_4) = \tfrac{1}{2}(5.00 \times 10^{-3}) = 2.50 \times 10^{-3}$
3. Step 3
  Concentration.
  $c = \frac{2.50 \times 10^{-3}}{24.0/1000} = 0.104\,\text{mol dm}^{-3}$
Answer
$[\text{H}_2\text{SO}_4] = 0.104\,\text{mol dm}^{-3}$ (3 s.f.).
Examiner tip
The 1 : 2 ratio for diprotic H₂SO₄ is the marking point; convert both volumes to dm³.
6Limiting reagent and product mass (4 marks)
Stretch• limiting reagent
Question
4.0 g of $\text{H}_2$ reacts with 28.0 g of $\text{N}_2$ . Which is limiting, and what maximum mass of $\text{NH}_3$ forms? ( $A_r$ : H 1.0, N 14.0)
Step-by-step solution
1. Step 1
  Moles of each.
  $n(\text{H}_2) = \frac{4.0}{2.0} = 2.0;\quad n(\text{N}_2) = \frac{28.0}{28.0} = 1.0$
2. Step 2
  Divide by coefficients ( $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ ): N₂ 1.0/1 = 1.0; H₂ 2.0/3 = 0.67 → H₂ limiting.
3. Step 3
  Moles NH₃ from H₂ (ratio 3 : 2).
  $n(\text{NH}_3) = \tfrac{2}{3} \times 2.0 = 1.33$
4. Step 4
  Mass NH₃ ( $M = 17.0$ ).
  $m = 1.33 \times 17.0 = 22.7\,\text{g}$
Answer
H₂ is limiting; maximum 22.7 g of NH₃.
Examiner tip
Never pick the limiting reagent by raw moles — divide by the coefficient first, then work product from the limiting one.

Key Formulae — Reacting masses and volumes (of solutions and gases)

The formulae you need to memorise for reacting masses and volumes (of solutions and gases) on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Moles in solution
$n = c \times V$
$c$
concentration / mol dm⁻³
$V$
volume / dm³ (convert cm³ ÷ 1000)
When to use
Titrations and any solution calculation.
Concentration in g dm⁻³
$c\,(\text{g dm}^{-3}) = c\,(\text{mol dm}^{-3}) \times M$
$M$
molar mass / g mol⁻¹
When to use
Converting between mol dm⁻³ and g dm⁻³ concentrations.
Moles of gas at r.t.p.
$n = \frac{V}{24.0}$
$V$
gas volume / dm³ at r.t.p.
When to use
Gas volume ↔ moles at room temperature and pressure (molar gas volume = 24.0 dm³ mol⁻¹).
Ideal gas equation
$pV = nRT$
$p$
pressure / Pa
$V$
volume / m³
$n$
amount of gas / mol
$R$
gas constant, 8.31 J K⁻¹ mol⁻¹
$T$
temperature / K (= °C + 273)
When to use
Gas calculations not at r.t.p.; convert to SI units first (Pa, m³, K).
Example
0.50 mol at 100 kPa and 298 K: $V = nRT/p = (0.50 \times 8.31 \times 298)/100000 = 0.0124\,\text{m}^3 = 12.4\,\text{dm}^3$ .
Percentage yield
$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
When to use
After finding the theoretical (maximum) yield from moles.

Key Definitions and Keywords — Reacting masses and volumes (of solutions and gases)

Definitions to memorise and the exact keywords mark schemes credit for reacting masses and volumes (of solutions and gases) answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Stoichiometry
Examiner keyword
The mole ratios in which substances react, given by the balancing numbers in the equation.
Limiting reagent
Examiner keyword
The reactant that is completely used up first and so determines the maximum amount of product.
Reagent in excess
Examiner keyword
A reactant present in more than the amount required to react with the limiting reagent; some is left over.
Concentration
Examiner keyword
The amount of solute dissolved per unit volume of solution, in mol dm⁻³ (or g dm⁻³).
Theoretical yield
Examiner keyword
The maximum mass of product calculable from the limiting reagent, assuming the reaction goes to completion with no losses.
Percentage yield
Examiner keyword
Actual yield as a percentage of the theoretical yield; less than 100% due to side reactions, incomplete reaction or losses.
Avogadro's law
Examiner keyword
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules — so volume ratio = mole ratio.
Molar gas volume
The volume occupied by one mole of any gas; 24.0 dm³ mol⁻¹ at room temperature and pressure (r.t.p.).

Common Mistakes and Misconceptions — Reacting masses and volumes (of solutions and gases)

The traps other students keep falling into on reacting masses and volumes (of solutions and gases) questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to convert cm³ to dm³ in n = cV.
9701 Examiner Reports 2022-2024
Why it happens
Reading burette volumes in cm³.
How to avoid it
Divide cm³ by 1000 before using n = cV.
✕Choosing the limiting reagent by raw moles only.
Why it happens
Ignoring the equation coefficients.
How to avoid it
Divide each reactant's moles by its coefficient first.
✕Comparing the product mass to the reactant mass for % yield.
Why it happens
Confusing the two masses.
How to avoid it
% yield compares actual product to the theoretical mass of that SAME product.
✕Using a 1 : 1 ratio for diprotic acids/bases.
Why it happens
Not reading the balanced equation.
How to avoid it
H₂SO₄ : NaOH = 1 : 2; Ca(OH)₂ : HCl = 1 : 2, etc.
✕Using °C, kPa or dm³ directly in pV = nRT.
9701 Examiner Reports 2022-2024
Why it happens
Not converting to SI base units.
How to avoid it
Convert to Pa, m³ and K first (T/K = °C + 273; 1 dm³ = 1×10⁻³ m³).
✕Quoting the wrong number of significant figures or no units.
Why it happens
Rounding early or omitting units.
How to avoid it
Match the data (usually 3 s.f.), round at the end, and always give units.

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