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Cambridge IGCSE β€’ Year 11 β€’ Chemistry

Stoichiometry

Formulae, equations, relative masses, moles and concentration.

Chapter 3

Verified Curriculum Topic

What is Stoichiometry?

Formulae, equations, relative masses, moles and concentration.

Stoichiometry matters because it links chemical ideas, reactions, and reasoning patterns that recur throughout the syllabus. At Year 11 level, students are often expected to define terms accurately, explain processes clearly, and connect theory to reactions, observations, or applications.

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Summary

The One Thing

Stoichiometry uses balanced chemical equations to relate the amounts of reactants and products. Relative masses, moles, concentrations and gas volumes provide the quantitative links needed to calculate these amounts.

Reactions, Processes and Experiments

What happensEquation or processWhat you observeType
Hydrogen reacts with oxygen to form water.2H2 + O2 β†’ 2H2Oβ€”Balanced chemical equation showing a mole ratio
The relative formula mass of calcium carbonate is calculated by adding the relative atomic masses of its atoms.Mr of CaCO3 = 40.1 + 12.0 + (3 Γ— 16.0) = 100.1β€”Relative formula mass calculation
An empirical formula is determined by converting each element's mass or percentage to moles, dividing by the smallest mole value and obtaining whole-number ratios.Convert masses or percentages to moles β†’ divide by the smallest value β†’ multiply to obtain whole numbers if necessary.β€”Empirical formula calculation
A molecular formula is determined from the empirical formula.molecular formula multiplier = molecular mass Γ· empirical formula massβ€”Molecular formula calculation
Percentage composition by mass is calculated for an element in a compound.Percentage composition by mass = (mass of the element in one mole of compound Γ· Mr of compound) Γ— 100%.β€”Percentage composition calculation
Percentage yield compares the actual product obtained with the theoretical maximum.Percentage yield = (actual yield Γ· theoretical yield) Γ— 100%.β€”Percentage yield calculation
Atom economy compares the desired product's formula mass with the total formula mass of the reactants.Atom economy = (Mr of desired product Γ· total Mr of reactants) Γ— 100%, using the balanced equation.β€”Atom economy calculation
A known solution reacts with a solution of unknown concentration to determine the unknown concentration.In titration calculations, use the measured titre volume in dm^3, calculate moles of the known solution, apply the balanced equation ratio, and then calculate the unknown concentration.β€”Titration
A limiting-reactant calculation compares the available amounts of reactants with the balanced-equation requirements.Compare the available mole amounts with the balanced-equation requirements; the reactant that produces the smaller amount of product is limiting.The limiting reactant is completely used up first; the excess reactant remains.Limiting-reactant calculation

Key Terms

  • Chemical formula: A symbolic representation showing which elements are in a substance and the ratio of their atoms or ions.
  • Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.
  • Molecular formula: The actual number of atoms of each element in one molecule.
  • Structural formula: A formula showing how atoms are arranged or bonded in a molecule.
  • Balanced chemical equation: An equation with the same number of atoms of each element on both sides, obeying conservation of mass.
  • State symbol: A symbol showing physical state: (s) solid, (l) liquid, (g) gas or (aq) aqueous solution.
  • Relative atomic mass (Ar): The weighted average mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.
  • Relative formula mass (Mr): The sum of the relative atomic masses of all atoms shown in a formula.
  • Mole: The amount of substance containing approximately 6.02 Γ— 10^23 particles, such as atoms, molecules or ions.
  • Avogadro constant: The number of particles in one mole: approximately 6.02 Γ— 10^23 mol^-1.
  • Molar mass: The mass of one mole of a substance, measured in g mol^-1; its numerical value equals Mr for a compound.
  • Concentration: The amount of solute dissolved per unit volume of solution.
  • Limiting reactant: The reactant that is completely used up first and therefore limits the amount of product formed.
  • Excess reactant: A reactant present in more than the amount required, so some remains after the reaction.
  • Percentage yield: The actual amount of product obtained expressed as a percentage of the theoretical maximum amount.
  • Percentage composition by mass: The percentage of a compound's total mass contributed by a particular element.
  • Titration: A method used to find the concentration of a solution by reacting it with a solution of known concentration.
  • Conservation of mass: Atoms are rearranged during a chemical reaction but are not created or destroyed.
  • Stoichiometric calculation: A calculation using a balanced equation to determine the quantitative relationship between reactants and products.
  • Mass concentration: The mass of solute dissolved per unit volume of solution, commonly expressed in g dm^-3.

Easily Confused

  • Empirical formula vs molecular formula: The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of atoms in one molecule.
  • Relative atomic mass vs relative formula mass: Relative atomic mass applies to an individual element's atoms; relative formula mass is the sum for all atoms represented in a formula.
  • Moles from mass vs mass from moles: Use n = mass Γ· molar mass to calculate moles from mass, but mass = moles Γ— molar mass to calculate mass from moles.
  • Limiting reactant vs excess reactant: The limiting reactant is completely used up first; the excess reactant remains after the reaction.
  • Concentration in mol dm^-3 vs concentration in g dm^-3: Molar concentration uses moles per dm^3; mass concentration uses grams per dm^3.
  • Volume in cm^3 vs volume in dm^3: Concentration calculations require consistent units; 1 dm^3 = 1000 cm^3, so volume in dm^3 = volume in cm^3 Γ· 1000.
  • Theoretical yield vs actual yield: The theoretical yield is the maximum calculated amount; the actual yield is the amount obtained experimentally.
  • Coefficients vs subscripts: When balancing equations, change coefficients placed before formulae; never change subscripts within formulae.
  • Gas volume vs solution volume: At room temperature and pressure, one mole of gas occupies approximately 24 dm^3; solution volumes are used in concentration equations.

What Gets Asked

  • Balancing chemical equations: Questions require the same number of atoms of each element on both sides. Changing subscripts within formulae instead of coefficients loses marks.
  • Relative mass calculations: Questions may require Mr = sum of (Ar Γ— number of each atom), including the example Mr of CaCO3 = 40.1 + 12.0 + (3 Γ— 16.0) = 100.1.
  • Mole and stoichiometric calculations: The required sequence is to balance the equation, convert the known quantity to moles, use the equation ratio, and convert to the required quantity. The equation coefficients must be used as mole ratios.
  • Concentration and titration calculations: Questions require volumes in dm^3, calculation of moles of the known solution, application of the balanced-equation ratio and calculation of the unknown concentration.
  • Limiting-reactant problems: Students must compare available mole amounts with the balanced-equation requirements and identify the reactant producing the smaller amount of product as limiting.
  • Yield, atom economy and composition: Questions may require Percentage yield = (actual yield Γ· theoretical yield) Γ— 100%, Atom economy = (Mr of desired product Γ· total Mr of reactants) Γ— 100%, or percentage composition by mass. Experimental yields may be below theoretical yields because of incomplete reactions, side reactions, product loss during separation or measurement uncertainty.

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Syllabus-verified

Learning objectives

  • 3.1Define relative atomic mass and relative molecular mass, and calculate Mr from a chemical formula.
  • 3.2Define the mole as the amount of substance containing 6.02 x 10^23 particles, and calculate moles from mass and Mr.
  • 3.3Calculate reacting masses using balanced chemical equations and molar ratios.extended
  • 3.4Define empirical formula and calculate it from given percentage composition or reacting masses data.extended
  • 3.5Calculate the concentration of a solution in g/dm3 or mol/dm3.extended
  • 3.6Calculate the volume of a gas at room temperature and pressure using the molar gas volume of 24 dm3.extended
Syllabus-verified

Practice questions

Q1. What is the relative molecular mass (Mr) of carbon dioxide, CO2? (Ar: C = 12, O = 16)1 mark Β· core
  • A. 28
  • B. 32
  • C. 44
  • D. 48

Answer: C

  • β€’ 1 mark for selecting C

Mr(CO2) = 12 + (16 x 2) = 12 + 32 = 44.

Q2. Magnesium reacts with oxygen: 2Mg + O2 β†’ 2MgO. Calculate the mass of magnesium oxide produced when 6 g of magnesium is completely burned in oxygen. (Ar: Mg = 24, O = 16)4 marks Β· extended

Answer: Moles of Mg = 6 / 24 = 0.25 mol. From the equation, moles of MgO produced = moles of Mg = 0.25 mol (1:1 ratio). Mr(MgO) = 24 + 16 = 40. Mass of MgO = 0.25 x 40 = 10 g.

  • β€’ 1 mark: moles of Mg = 6/24 = 0.25 mol
  • β€’ 1 mark: correct 1:1 mole ratio identified between Mg and MgO from the equation
  • β€’ 1 mark: Mr(MgO) = 40 calculated correctly
  • β€’ 1 mark: final mass = 0.25 x 40 = 10 g
Q3. Calculate the number of moles in 11 g of carbon dioxide, CO2. (Mr = 44)2 marks Β· core

Answer: Moles = mass / Mr = 11 / 44 = 0.25 mol.

  • β€’ 1 mark: correct method (mass Γ· Mr)
  • β€’ 1 mark: correct answer 0.25 mol
Q4. A solution contains 4.0 g of sodium hydroxide (NaOH) dissolved in 500 cm3 of solution. Calculate the concentration of the solution in g/dm3.3 marks Β· extended

Answer: Concentration = mass / volume(dm3). 500 cm3 = 0.5 dm3. Concentration = 4.0 / 0.5 = 8.0 g/dm3.

  • β€’ 1 mark: correct conversion of 500 cm3 to 0.5 dm3
  • β€’ 1 mark: correct method (mass Γ· volume in dm3)
  • β€’ 1 mark: correct final answer of 8.0 g/dm3

Key ideas to master

  • Learn the precise terms, laws, and reaction patterns associated with Stoichiometry.
  • Understand why each step or change happens instead of memorising the result only.
  • Practise writing balanced equations, comparisons, or structured explanations where relevant.
  • Revise common exceptions, observations, and applications that examiners often test.

Common exam prompts

  • Define the main idea in Stoichiometry using correct chemical terminology.
  • Write or interpret the reactions, observations, or comparisons that belong to this topic.
  • Explain why a process happens, not just what happens.
  • Summarise the high-yield facts and exceptions examiners often choose from this chapter.

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What is Stoichiometry in Cambridge IGCSE Year 11 Chemistry?

Formulae, equations, relative masses, moles and concentration.

How should I study Stoichiometry effectively?

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