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*Revision Materials* 1 Atomic Structure 2 Atoms, molecules and stoichiometry 3 Chemical Bonding 4 States of matter 5 Chemical energetics 6 Electrochemistry 7 Equilibria 8 Reaction kinetics 9 The Periodic Table, chemical periodicity 10 Group 2 11 Group 17 12 Nitrogen and sulfur 13 Organic 14 Hydrocarbons 15 Halogen compounds 16 Hydroxy compounds 17 Carbonyl compounds 18 Carboxylic acids and derivatives 19 Nitrogen compounds 20 Polymerisation 21 Organic synthesis 22 Analytical techniques 23 Chemical energetics 24 Electrochemistry 25 Equilibria 26 Reaction kinetics 27 Group 2 28 Chemistry of transition elements 29 Organic 30 Hydrocarbons 31 Halogen compounds 32 Hydroxy compounds 33 Carboxylic acids and derivatives 34 Nitrogen compounds 35 Polymerisation 36 Organic synthesis 37 Analytical techniques

2 Atoms, molecules and stoichiometry

2.1 Relative masses of atoms and molecules 2.2 The mole and the Avogadro constant 2.3 Formulas 2.4 Reacting masses and volumes (of solutions and gases)

Reacting Masses and Volumes (of Solutions and Gases)

Specification Reference Physical Chemistry: Atoms, molecules and stoichiometry 2.4

Quick Notes

  • We can use the mole concept to calculate:
    • Masses of reactants/products
    • Volumes of gases (at room temperature and pressure)
    • Volumes and concentrations of solutions
  • Key formulas:
    • n = m ÷ Mr
    • CIE A-Level Chemistry formula triangle showing n = m ÷ Mr for solid substances.
    • n = V ÷ 24 (gas, dm³) or n = V ÷ 24000 (gas, cm³)
    CIE A-Level Chemistry formula triangle showing n = V ÷ 24 for gases in dm³.
    • n = c × V (solution, in dm³)
    CIE A-Level Chemistry formula showing n = c × V for solutions in dm³.
  • The limiting reagent is used up first and determines how much product forms.
  • Percentage yield = (actual ÷ theoretical) × 100
    • CIE A-Level Chemistry percentage yield formula diagram showing actual over theoretical × 100.
  • Always match significant figures to data in the question.
  • We can use mole ratios from balanced equations for stoichiometric relationships.

Full Notes

Reacting Masses

We can use the mole formula to link mass and amount:

CIE A-Level Chemistry formula triangle showing n = m ÷ Mr for solid substances.

n = m ÷ Mr

Worked Example

How much CO2 is formed when 12 g of carbon is burned?
C + O2 → CO2
C + O2 → CO2
moles of C = 12 ÷ 12 = 1 mol
From the equation, 1 mol C gives 1 mol CO2
→ Mass of CO2 = 1 × 44 = 44 g

Volumes of Gases

At room temperature and pressure (r.t.p.), 1 mol of gas = 24 dm³. This is the molar volume of a gas.

CIE A-Level Chemistry formula triangle showing n = V ÷ 24 for gases in dm³.

24 dm³ is the same as 24000 cm³, so we can also use:

CIE A-Level Chemistry formula triangle showing n = V ÷ 24000 for gases in cm³.
Photo of Matt
Matt’s exam tip

The molar volume of a gas changes with temperature and pressure. It is only 24 dm³ at room temperature and pressure. Make sure a question gives those exact conditions, otherwise don’t assume the molar volume is 24.


Worked Example

What volume of O2 is needed to react with 2 mol of H2?
2H2 + O2 → 2H2O
2 mol H2 reacts with 1 mol O2 → 1 mol O2 = 24 dm³
Answer: 24 dm³

Volumes and Concentrations of Solutions

CIE A-Level Chemistry formula triangle showing n = c × V for solutions in dm³.

Formula: n = c × V

To convert cm³ to dm³: divide by 1000.

Worked Example

What volume of 2.0 mol dm−3 HCl is needed to react with 0.05 mol of NaOH?

NaOH + HCl → NaCl + H2O
1:1 ratio → 0.05 mol HCl needed
V = n ÷ c = 0.05 ÷ 2.0 = 0.025 dm³ or 25 cm³

Limiting and Excess Reagents

In a reaction with two reactants, the limiting reagent is the one that gets used up first — it determines how much product is formed.

Steps to identify the limiting reagent:

Example 2 mol H2 reacts with 1 mol O2.
If you mix 2 mol H2 and 2 mol O2:
From the ratio, H2 is the limiting reagent because 2 mol H2 needs only 1 mol O2. When the 2 mol H2 react, there will still be 1 mol O2 left (excess).

Percentage Yield

Percentage yield compares the actual amount of product obtained to the maximum possible amount (theoretical yield).

CIE A-Level Chemistry percentage yield formula diagram showing actual over theoretical × 100.

Formula: Percentage yield = (actual yield ÷ theoretical yield) × 100

Worked Example Simple Percentage Yield Calculation

In the reaction: Mg + 2HCl → MgCl2 + H2, if 2.40 g of Mg is reacted with excess HCl, and only 5.80g of MgCl2 is obtained, what is the percentage yield?

CIE A-Level Chemistry worked example showing percentage yield calculation for Mg + 2HCl reaction giving MgCl2 product.

Photo of Matt
Matt’s exam tip

Percentage yields should always be less than 100%. If a value is over 100%, the product won't be pure (e.g. crystals may be wet). Common reasons for low yields include incomplete or reversible reactions, side reactions, or product lost during filtration or transfer.

Stoichiometric Relationships

Stoichiometry is the ratio of amounts in a balanced equation. We can use it to:

Significant Figures

Your final answer in any calculation should be rounded to the same number of significant figures as the least accurate data value in the question.

For Example: If a mass is given as 12 g (2 sig figs) and an Mr value as 12.0 (3 sig figs), the answer should be to 2 sig figs.

Summary