Chemical Reactions and the Mole

What you'll learn

This revision guide covers the quantitative aspects of chemical reactions at WJEC GCSE level. You'll master how to use the mole as the fundamental unit for measuring amounts of substances, calculate reacting masses, work out empirical and molecular formulae, and determine the efficiency of chemical reactions through percentage yield and atom economy calculations.

Key terms and definitions

Mole (mol) — the unit for amount of substance; one mole contains 6.02 × 10²³ particles (Avogadro's constant)

Relative formula mass (Mr) — the sum of the relative atomic masses of all atoms in a formula

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 of a compound

Limiting reactant — the reactant that is completely used up first in a chemical reaction, determining the maximum amount of product formed

Percentage yield — the actual mass of product obtained expressed as a percentage of the theoretical maximum mass

Atom economy — a measure of how many atoms in the reactants form the desired product, expressed as a percentage

Concentration — the amount of solute dissolved in a given volume of solution, typically measured in g/dm³ or mol/dm³

Core concepts

The mole and Avogadro's constant

The mole is the chemist's counting unit, linking the microscopic world of atoms to measurable quantities in the laboratory. One mole of any substance contains exactly 6.02 × 10²³ particles (atoms, molecules, ions or electrons) — this is Avogadro's constant.

Key relationships:

• Mass in grams = moles × relative formula mass (Mr)
• Moles = mass in grams ÷ Mr
• Number of particles = moles × Avogadro's constant

For example, one mole of carbon atoms (C) has a mass of 12 g because carbon's relative atomic mass is 12. One mole of water molecules (H₂O) has a mass of 18 g because Mr(H₂O) = (2 × 1) + 16 = 18.

The mole allows us to count atoms by weighing. If you have 24 g of carbon, you have 24 ÷ 12 = 2 moles of carbon atoms, which equals 2 × 6.02 × 10²³ = 1.204 × 10²⁴ atoms.

Calculating reacting masses

Chemical equations show the ratio in which substances react. The balanced equation provides the mole ratio of reactants and products.

Steps for reacting mass calculations:

1. Write the balanced equation
2. Identify the known and unknown substances
3. Calculate moles of the known substance (moles = mass ÷ Mr)
4. Use the equation to find the mole ratio
5. Calculate moles of the unknown substance
6. Convert to mass (mass = moles × Mr)

Example: What mass of magnesium oxide forms when 6 g of magnesium burns completely?

2Mg + O₂ → 2MgO

• Moles of Mg = 6 ÷ 24 = 0.25 mol
• From the equation: 2 mol Mg produces 2 mol MgO (1:1 ratio)
• Moles of MgO = 0.25 mol
• Mass of MgO = 0.25 × 40 = 10 g

When dealing with reactions involving solutions, remember:

• Volume in dm³ = volume in cm³ ÷ 1000
• Moles = concentration (mol/dm³) × volume (dm³)

Empirical and molecular formulae

The empirical formula represents the simplest ratio of elements in a compound. The molecular formula shows the actual number of atoms in one molecule and is always a whole number multiple of the empirical formula.

Method for finding empirical formula from percentage composition or mass data:

1. List the elements and their masses (or percentages)
2. Divide each mass by the relative atomic mass to get moles
3. Divide all mole values by the smallest mole value
4. If necessary, multiply all numbers by the same factor to get whole numbers
5. Write the empirical formula

Example: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find the empirical formula.

Element	Mass	÷ Ar	÷ smallest	Ratio
C	40	40 ÷ 12 = 3.33	3.33 ÷ 3.33 = 1	1
H	6.7	6.7 ÷ 1 = 6.7	6.7 ÷ 3.33 = 2	2
O	53.3	53.3 ÷ 16 = 3.33	3.33 ÷ 3.33 = 1	1

Empirical formula = CH₂O

To find the molecular formula, you need the relative molecular mass (Mr) of the compound. Divide the Mr by the mass of the empirical formula, then multiply each subscript in the empirical formula by this number.

If the compound above has Mr = 180:

• Mass of empirical formula CH₂O = 12 + 2 + 16 = 30
• 180 ÷ 30 = 6
• Molecular formula = C₆H₁₂O₆

Limiting reactants

In many reactions, one reactant is present in excess. The limiting reactant is completely used up and determines the maximum amount of product that can form.

To identify the limiting reactant:

1. Calculate the moles of each reactant
2. Use the balanced equation to determine which reactant produces less product
3. The reactant that produces less product is the limiting reactant

Example: 5.4 g of aluminium reacts with 16 g of oxygen. Which is the limiting reactant?

4Al + 3O₂ → 2Al₂O₃

• Moles of Al = 5.4 ÷ 27 = 0.2 mol
• Moles of O₂ = 16 ÷ 32 = 0.5 mol
• From equation: 4 mol Al requires 3 mol O₂
• 0.2 mol Al requires (3/4) × 0.2 = 0.15 mol O₂
• We have 0.5 mol O₂, which is more than needed
• Therefore, aluminium is the limiting reactant (oxygen is in excess)

Percentage yield

In industrial and laboratory processes, the actual yield (mass of product obtained) is usually less than the theoretical yield (maximum possible mass). Percentage yield measures this efficiency:

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

Reasons for yields less than 100%:

• Incomplete reactions
• Side reactions producing unwanted products
• Loss of product during separation and purification
• Reversible reactions that don't go to completion

Example: The theoretical yield of a reaction is 12.5 g but only 9.4 g of product is obtained.

Percentage yield = (9.4 ÷ 12.5) × 100% = 75.2%

Higher percentage yields are economically desirable in industry as they reduce waste and cost. However, achieving very high yields may require expensive conditions that make the process uneconomical.

Atom economy

Atom economy measures the proportion of reactant atoms that become useful products. It's a measure of sustainability — reactions with high atom economy produce less waste.

Atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100%

For reactions with multiple products:

Atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100%

Example: Ethanol (C₂H₅OH) can be produced by fermentation or by hydration of ethene:

C₂H₄ + H₂O → C₂H₅OH

• Mr of ethanol = 46
• Mr of reactants = 28 + 18 = 46
• Atom economy = (46 ÷ 46) × 100% = 100%

Compare with producing ethanol from glucose fermentation, which also produces carbon dioxide as a waste product — the atom economy is much lower.

High atom economy is important for:

• Reducing waste (environmental benefit)
• Reducing costs of raw materials
• Reducing costs of waste disposal
• Sustainable chemistry

Addition reactions typically have 100% atom economy because all reactant atoms end up in the single product. Reactions producing multiple products usually have lower atom economy unless by-products can be sold.

Concentration calculations

Concentration expresses the amount of solute in a given volume of solution.

Key equations:

• Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
• Concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³)
• Moles = concentration (mol/dm³) × volume (dm³)

Converting between units:

• 1 dm³ = 1000 cm³
• To convert cm³ to dm³: divide by 1000
• To convert mol/dm³ to g/dm³: multiply by Mr

Example: What is the concentration in mol/dm³ of a solution containing 20 g of sodium hydroxide (NaOH) in 500 cm³?

• Mr(NaOH) = 23 + 16 + 1 = 40
• Moles = 20 ÷ 40 = 0.5 mol
• Volume = 500 ÷ 1000 = 0.5 dm³
• Concentration = 0.5 ÷ 0.5 = 1 mol/dm³

Worked examples

Example 1: Reacting mass calculation (3 marks)

Question: Calcium carbonate decomposes when heated: CaCO₃ → CaO + CO₂

Calculate the mass of calcium oxide produced when 25 g of calcium carbonate decomposes completely. (Relative atomic masses: Ca = 40, C = 12, O = 16)

Solution:

• Mr(CaCO₃) = 40 + 12 + (3 × 16) = 100
• Moles of CaCO₃ = 25 ÷ 100 = 0.25 mol ✓
• From equation: 1 mol CaCO₃ produces 1 mol CaO
• Moles of CaO = 0.25 mol ✓
• Mr(CaO) = 40 + 16 = 56
• Mass of CaO = 0.25 × 56 = 14 g ✓

Example 2: Empirical formula (4 marks)

Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Calculate its empirical formula. (Relative atomic masses: C = 12, H = 1)

Solution:

Element	Mass	÷ Ar	÷ smallest
C	85.7	85.7 ÷ 12 = 7.14 ✓	7.14 ÷ 7.14 = 1
H	14.3	14.3 ÷ 1 = 14.3 ✓	14.3 ÷ 7.14 = 2

Ratio is C:H = 1:2 ✓ Empirical formula = CH₂ ✓

Example 3: Percentage yield and atom economy (5 marks)

Question: Copper sulfate can be made by reacting copper oxide with sulfuric acid: CuO + H₂SO₄ → CuSO₄ + H₂O

A student reacted 8.0 g of copper oxide with excess sulfuric acid. The theoretical yield is 16.0 g but the student obtained 12.8 g of copper sulfate crystals.

(a) Calculate the percentage yield. (2 marks) (b) Calculate the atom economy for producing copper sulfate. (3 marks) (Mr: CuO = 80, H₂SO₄ = 98, CuSO₄ = 160, H₂O = 18)

Solution: (a) Percentage yield = (actual yield ÷ theoretical yield) × 100% ✓ = (12.8 ÷ 16.0) × 100% = 80% ✓

(b) Sum of Mr of all reactants = 80 + 98 = 178 ✓ Atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100% ✓ = (160 ÷ 178) × 100% = 89.9% (or 90%) ✓

Common mistakes and how to avoid them

• Forgetting to balance equations before calculations — Always check the equation is balanced. The coefficients give you the mole ratios needed for calculations.

• Confusing Mr with Ar — Relative atomic mass (Ar) is for single elements; relative formula mass (Mr) is the sum of all Ar values in a compound's formula.

• Not converting cm³ to dm³ — When using concentration in mol/dm³, volume must be in dm³. Always divide cm³ by 1000.

• Using percentages directly in empirical formula calculations — Treat percentage values as if they were masses in grams, then divide by Ar values.

• Calculating atom economy incorrectly for reactions with multiple products — Use the sum of Mr of all products in the denominator, not just reactants. Alternatively, use sum of Mr of all reactants.

• Rounding too early in multi-step calculations — Keep full calculator values until the final answer, then round appropriately (usually to 2 or 3 significant figures).

Exam technique for "Chemical Reactions and the Mole"

• Show all working clearly — In calculations, examiners award method marks even if your final answer is incorrect. Write out each step: moles calculated, mole ratio identified, final answer with units.

• Use the correct units — Mass in grams (g), moles in mol, concentration in g/dm³ or mol/dm³, volume in dm³ or cm³. Many students lose marks by omitting units.

• Command words matter — "Calculate" requires a numerical answer with working. "Determine" or "find" are similar. "Explain" requires reasoning, not just calculations.

• Check mark allocations — A 3-mark calculation typically requires: identifying moles of known substance (1 mark), using mole ratio (1 mark), calculating final answer (1 mark). Plan your answer accordingly.

Quick revision summary

The mole (6.02 × 10²³ particles) connects atomic-scale chemistry to measurable quantities. Use moles = mass ÷ Mr for conversions. Balanced equations give mole ratios for reacting mass calculations. Empirical formulae show simplest atom ratios; molecular formulae show actual numbers. The limiting reactant determines maximum product. Percentage yield = (actual ÷ theoretical) × 100%. Atom economy = (Mr of desired product ÷ sum of Mr of reactants) × 100% measures sustainability. For solutions: moles = concentration (mol/dm³) × volume (dm³).