Quantitative Chemistry (Moles, Titrations, Yield)

Key Definitions

Relative formula mass (M_r): 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

Atoms are neither created nor destroyed in a chemical reaction: only rearranged into different substances.

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.

Relative Formula Mass (M_r)

The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula. A_r values are found on the periodic table.

Calculate the M_r of calcium carbonate (CaCO₃)

Ca = 40, C = 12, O = 16

M_r = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100

Calculate the M_r of magnesium hydroxide (Mg(OH)₂)

Mg = 24, O = 16, H = 1

M_r = 24 + 2 × (16 + 1) = 24 + 34 = 58

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 M_r of aluminium sulfate, Al₂(SO₄)₃

Al = 27, S = 32, O = 16

There are: 2 Al, 3 S, 12 O (3 × 4 = 12)

M_r = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342

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).

moles = mass ÷ M_r

The Moles Triangle

mass = mol × M_r
mol = mass ÷ M_r
M_r = mass ÷ mol

How many moles in 11 g of CO₂?

M_r of CO₂ = 12 + (2 × 16) = 44

Moles = 11 ÷ 44 = 0.25 mol

What mass is 3 moles of water (H₂O)?

M_r of H₂O = (2 × 1) + 16 = 18

Mass = mol × M_r = 3 × 18 = 54 g

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: M_r 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

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

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).

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.

Dilute vs concentrated solutions at the same total volume.

concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)

The Concentration Triangle

mass = conc × vol
conc = mass ÷ vol
vol = mass ÷ conc

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

concentration (mol/dm³) = moles of solute ÷ volume (dm³)

4 g of NaOH is dissolved in 250 cm³ of water. Find the concentration in mol/dm³.

Step 1: M_r 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

If a question gives you mass in grams and asks for mol/dm³, you must first convert mass to moles using mol = mass ÷ M_r, 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).

% yield = (actual yield ÷ theoretical yield) × 100

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.

atom economy = (M_r of desired product ÷ total M_r of all products) × 100

High atom economy means fewer waste products and maximum efficiency.

High atom economy is desirable. It means less waste, lower costs, and a more sustainable process.

Yield vs Atom Economy

Feature	% Yield	Atom Economy
Measures	Lab performance	Pathway efficiency
Needs experiment?	Yes	No (equation only)
Type of waste	Spills, losses	By-products
Ideal value	100%	100%

Calculate the atom economy for producing hydrogen from the reaction:

Zn + H₂SO₄ → ZnSO₄ + H₂

Step 1: M_r of desired product (H₂) = 2

Step 2: M_r 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).

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 burette delivers a measured volume of acid into the alkali below.

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).

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

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).

volume (dm³) = moles × 24

moles = volume (dm³) ÷ 24

Gas Volume Triangle

vol = mol × 24
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: M_r 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)

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.

Students Also Studied

Previous Topic

Topic 2: Bonding, Structure & Properties

Ionic, covalent and metallic bonding. Giant and simple structures.

Next Topic

Topic 4: Chemical Changes

Reactivity series, extraction of metals, acids, bases and electrolysis.

Blog Post

How to Calculate Moles: 3 Formulas

The three mole formulas with five worked examples and common mistakes to avoid.