Topic Core Terms
- Relative formula mass (Mr)
- The sum of the relative atomic masses of all atoms in a formula.
- Mole
- The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).
- Concentration
- The amount of solute dissolved in a given volume of solution (g/dm³ or mol/dm³).
- Limiting reactant
- The reactant that is completely used up first and determines the amount of product formed.
- Percentage yield
- The actual yield as a percentage of the theoretical (maximum) yield.
- Atom economy
- The percentage of reactant atoms that form the desired (useful) product.
Conservation of Mass
The law of conservation of mass states that no atoms are lost or made during a chemical reaction. The total mass of the products equals the total mass of reactants.
This is because the same atoms are present before and after the reaction: they have just been rearranged. This is why we must balance chemical equations.
The same atoms exist before and after: mass is always conserved
Key Fact
Atoms are neither created nor destroyed in a chemical reaction: only rearranged into different substances.
Conservation of Mass and Relative Formula Mass (Mᵣ)
Because mass is conserved, in any balanced chemical equation, the sum of the Mᵣ of all reactants equals the sum of the Mᵣ of all products (taking stoichiometry numbers into account).
Example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
- Reactants Mᵣ sum: Mᵣ(CH₄) + 2 × Mᵣ(O₂) = 16 + (2 × 32) = 80
- Products Mᵣ sum: Mᵣ(CO₂) + 2 × Mᵣ(H₂O) = 44 + (2 × 18) = 80
Deducing Balancing Coefficients from Reacting Masses [HT Only]
8.1 g of zinc (Aᵣ = 65) reacts with 3.6 g of oxygen gas (Mᵣ = 32) to form 11.7 g of zinc oxide (Mᵣ = 81). Deduce the balanced equation.
Step 1 (Moles): Calculate moles of each substance:
Moles of Zn = 8.1 / 65 = 0.125 mol
Moles of O₂ = 3.6 / 32 = 0.1125 mol → 0.125 / 0.1125 ... (simplified: 0.1246 : 0.0623 : 0.1246 = 2 : 1 : 2)
Step 2 (Simplest Whole Number Ratio): Divide all moles by the smallest mole value (0.0625):
Zn: 0.125 / 0.0625 = 2 | O₂: 0.0625 / 0.0625 = 1 | ZnO: 0.125 / 0.0625 = 2
Balanced Equation: 2Zn(s) + O₂(g) → 2ZnO(s)
Apparent Changes in Mass
Sometimes, the mass of a reaction vessel appears to change:
- If a gas escapes (e.g., CO₂ from a thermal decomposition), mass appears to decrease.
- If a gas is gained from the air (e.g., oxygen during oxidation of metals), mass appears to increase.
In reality, if the system were sealed, the total mass would remain unchanged.
Chemical Measurements & Uncertainty
Whenever a measurement is made, there is always some uncertainty about the result obtained. No measurement is perfectly accurate; all instruments have limitations and all humans introduce small errors when reading scales.
Improving the Quality of Measurements
To increase confidence in results, scientists:
- Repeat measurements and calculate a mean (average).
- Identify and exclude anomalous results (outliers): values that don't fit the pattern. These are usually caused by human error or equipment malfunction.
- Use the range of the measurements about the mean as a measure of uncertainty.
Key Fact
Uncertainty = range ÷ 2. The range is the difference between the highest and lowest values in a set of repeat measurements (after removing anomalies).
A student records three titre values: 24.50, 24.60, and 24.40 cm³. Calculate the uncertainty.
Step 1: Mean = (24.50 + 24.60 + 24.40) ÷ 3 = 24.50 cm³
Step 2: Range = 24.60 - 24.40 = 0.20 cm³
Step 3: Uncertainty = 0.20 ÷ 2 = ± 0.10 cm³
Result: 24.50 ± 0.10 cm³
Examiner Trap
If a question asks you to "estimate the uncertainty," calculate half the range. A smaller uncertainty means the results are more precise (close together). Don't confuse precision with accuracy: accurate means close to the true value.
Relative Formula Mass (Mr)
The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula. Ar values are found on the periodic table.
Calculate the Mr of calcium carbonate (CaCO₃)
Ca = 40, C = 12, O = 16
Mr = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
Calculate the Mr of magnesium hydroxide (Mg(OH)₂)
Mg = 24, O = 16, H = 1
Mr = 24 + 2 × (16 + 1) = 24 + 34 = 58
Examiner Trap
Watch out for brackets! In Mg(OH)₂, the subscript 2 applies to both the O and the H inside the brackets, so there are 2 oxygen atoms and 2 hydrogen atoms.
Calculate the Mr of aluminium sulfate, Al₂(SO₄)₃
Al = 27, S = 32, O = 16
There are: 2 Al, 3 S, 12 O (3 × 4 = 12)
Mr = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342
Exam Technique
For formulae like Al₂(SO₄)₃, multiply everything inside the brackets by the subscript outside. Here the 3 applies to one S and four O atoms, giving 3 S and 12 O.
Moles & Avogadro's Constant
A mole is simply a number: 6.02 × 10²³ (Avogadro's constant). One mole of any substance contains exactly this number of particles (atoms, molecules, or ions).
The Moles Triangle
mol = mass ÷ Mr
Mr = mass ÷ mol
How many moles in 11 g of CO₂?
Mr of CO₂ = 12 + (2 × 16) = 44
Moles = 11 ÷ 44 = 0.25 mol
What mass is 3 moles of water (H₂O)?
Mr of H₂O = (2 × 1) + 16 = 18
Mass = mol × Mr = 3 × 18 = 54 g
Exam Technique
This triangle is your best friend for calculations. Cover what you want to find, and the remaining two values show you what to do. If they are side by side, multiply. If one is on top, divide.
Check your mole calculations instantly with our Moles Calculator - converts between mass, moles, and molar mass.
Reacting Masses
Use a balanced equation plus the moles formula to predict masses used or produced in reactions. Follow these three steps every time:
What mass of magnesium oxide is produced from 6 g of magnesium?
Equation: 2Mg + O₂ → 2MgO
Step 1: Moles of Mg = 6 ÷ 24 = 0.25 mol
Step 2: Ratio: 2Mg : 2MgO = 1:1, so moles of MgO = 0.25 mol
Step 3: Mass of MgO = 0.25 × 40 = 10 g
What mass of iron is produced when 80 g of iron(III) oxide is reduced?
Equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂
Step 1: Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 160.
Moles = 80 ÷ 160 = 0.5 mol
Step 2: Ratio: 1 Fe₂O₃ : 2 Fe, so moles of Fe = 0.5 × 2 = 1.0 mol
Step 3: Mass of Fe = 1.0 × 56 = 56 g
Exam Technique
Always check the mole ratio carefully - it is NOT always 1:1! Read the balanced equation coefficients to find the correct ratio.
Limiting Reactants
In many reactions, one reactant is used up before the others. This reactant is called the limiting reactant: it limits the amount of product that can be formed.
The other reactant(s) are said to be in excess.
Limiting
Used up first
Excess
Left over
Key Fact
The amount of product formed is always determined by the limiting reactant, not the one in excess.
4.8 g of Mg reacts with 14.6 g of HCl. Which is the limiting reactant?
Equation: Mg + 2HCl → MgCl₂ + H₂
Step 1: Moles of Mg = 4.8 ÷ 24 = 0.2 mol
Step 2: Moles of HCl = 14.6 ÷ 36.5 = 0.4 mol
Step 3: The ratio is 1 Mg : 2 HCl. So 0.2 mol Mg needs 0.2 × 2 = 0.4 mol HCl.
Result: We have exactly 0.4 mol HCl - both reactants run out at the same time. Neither is in excess.
If we only had 0.3 mol HCl, then HCl would be limiting (not enough to react with all the Mg).
Exam Technique
To identify the limiting reactant: (1) calculate moles of each reactant, (2) use the balanced equation ratios to compare, (3) the one that runs out first is the limiting reactant.
Concentration
Concentration tells you how much solute is dissolved in a given volume of solution.
The Concentration Triangle
conc = mass ÷ vol
vol = mass ÷ conc
Key Fact
Remember: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.
2.5 g of NaOH dissolved in 500 cm³. Find concentration in g/dm³.
Volume = 500 ÷ 1000 = 0.5 dm³
Concentration = 2.5 ÷ 0.5 = 5 g/dm³
Concentration in mol/dm³
4 g of NaOH is dissolved in 250 cm³ of water. Find the concentration in mol/dm³.
Step 1: Mr of NaOH = 23 + 16 + 1 = 40
Step 2: Moles = 4 ÷ 40 = 0.1 mol
Step 3: Volume = 250 ÷ 1000 = 0.25 dm³
Step 4: Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³
Exam Technique
If a question gives you mass in grams and asks for mol/dm³, you must first convert mass to moles using mol = mass ÷ Mr, then divide by the volume in dm³. Always convert cm³ to dm³ first!
Need to work through a titration calculation? Use our Titration Calculator for step-by-step working with automatic unit conversion.
Percentage Yield Chemistry Only
The percentage yield compares the actual amount of product obtained to the theoretical maximum (calculated from stoichiometry).
Three main reasons why yields are always less than 100%:
- The reaction is reversible and does not go to completion.
- Some product is lost during transfer (e.g., filtration, evaporation).
- Side reactions produce unwanted by-products.
A student calculates they should make 10 g of copper sulfate. They actually collect 7.5 g. What is the percentage yield?
% yield = (7.5 ÷ 10) × 100 = 75%
Atom Economy Chemistry Only
Atom economy measures the proportion of reactant atoms that become useful product. It evaluates the efficiency of the reaction pathway itself.
High atom economy is desirable. It means less waste, lower costs, and a more sustainable process.
Yield vs Atom Economy
Calculate the atom economy for producing hydrogen from the reaction:
Zn + H₂SO₄ → ZnSO₄ + H₂
Step 1: Mr of desired product (H₂) = 2
Step 2: Mr of all products = ZnSO₄ (161) + H₂ (2) = 163
Step 3: Atom economy = (2 ÷ 163) × 100 = 1.2%
This is very low - most of the atoms end up in the by-product (ZnSO₄), not in the desired product (H₂).
Exam Technique
You need both high percentage yield and high atom economy for a truly efficient, sustainable reaction. AQA loves comparing these two metrics.
Titrations Chemistry Only
A titration is a technique used to find the concentration of an unknown acid or alkali by reacting it with one of known concentration.
The Method
- Use a pipette to measure a fixed volume of alkali into a conical flask.
- Add a few drops of indicator (e.g., phenolphthalein or methyl orange).
- Fill a burette with acid of known concentration.
- Add acid gradually, swirling, until the indicator permanently changes colour: the end point.
- Record the titre (volume of acid added). Repeat until concordant results are achieved (within 0.10 cm³).
Exam Technique
When describing the titration method, always mention: using a pipette for the fixed volume, a burette for the variable volume, and an indicator for the end point. Repeat to get concordant results.
Titration Calculations
25 cm³ of NaOH (unknown concentration) is neutralised by 20 cm³ of 0.5 mol/dm³ HCl. Find the concentration of NaOH.
Equation: NaOH + HCl → NaCl + H₂O
Step 1: Moles of HCl = conc × vol = 0.5 × (20 ÷ 1000)
= 0.5 × 0.02 = 0.01 mol
Step 2: Ratio NaOH : HCl = 1:1, so moles of NaOH = 0.01 mol
Step 3: Volume of NaOH = 25 ÷ 1000 = 0.025 dm³
Step 4: Concentration of NaOH = 0.01 ÷ 0.025 = 0.4 mol/dm³
Exam Technique
Titration calculations always follow the same pattern: find moles of the known substance → use the mole ratio → calculate the unknown concentration. Be careful to convert cm³ to dm³!
Molar Volume of Gases Chemistry Only Higher Tier
At room temperature and pressure (RTP: 20°C, 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).
Gas Volume Triangle
mol = vol ÷ 24
(at RTP)
What volume does 0.5 mol of oxygen gas occupy at RTP?
Volume = 0.5 × 24 = 12 dm³
What volume of CO₂ is produced when 5 g of CaCO₃ is decomposed? (at RTP)
Equation: CaCO₃ → CaO + CO₂
Step 1: Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100
Step 2: Moles of CaCO₃ = 5 ÷ 100 = 0.05 mol
Step 3: Ratio 1:1, so moles of CO₂ = 0.05 mol
Step 4: Volume = 0.05 × 24 = 1.2 dm³ (or 1200 cm³)
Exam Technique
The molar volume (24 dm³) only applies at RTP. If a question states different conditions, you may need the ideal gas equation instead (not required at GCSE). Questions often combine mass→moles→volume in multi-step problems.
Explore gas behaviour with our Gas Law Calculator - see how pressure, volume, and temperature are connected.
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Reactivity series, extraction of metals, acids, bases and electrolysis.
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Try These Tools
Moles Calculator
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Titration Calculator
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Topic 3 Flashcards
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