Stoichiometry and the Mole

What you'll learn

Stoichiometry forms the quantitative foundation of CIE IGCSE Chemistry, enabling precise calculations about substances in chemical reactions. This topic appears in both Paper 2 (Core) and Paper 4 (Extended), typically accounting for 15-20% of available marks. Mastery of the mole concept, reacting masses, and solution calculations is essential for success across multiple question types.

Key terms and definitions

Relative atomic mass (Ar) — the average mass of an atom of an element compared to 1/12 the mass of a carbon-12 atom, taking into account the natural abundance of isotopes.

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

Relative formula mass — the sum of the relative atomic masses in an ionic compound's formula (used instead of Mr for ionic substances).

Mole (mol) — the amount of substance containing 6.02 × 10²³ particles (the Avogadro constant), equal to the relative atomic or molecular mass expressed in grams.

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.

Stoichiometry — the quantitative relationship between reactants and products in a chemical reaction, based on balanced chemical equations.

Limiting reactant — the reactant that is completely consumed first in a chemical reaction, determining the maximum amount of product that can form.

Core concepts

Calculating relative atomic and molecular masses

Relative atomic mass values appear on the Periodic Table and must be used correctly in CIE IGCSE Chemistry calculations. To find the relative molecular mass or relative formula mass, add together all the Ar values in the formula.

Example calculations:

• Water (H₂O): (2 × 1) + 16 = 18
• Calcium carbonate (CaCO₃): 40 + 12 + (3 × 16) = 100
• Sulfuric acid (H₂SO₄): (2 × 1) + 32 + (4 × 16) = 98
• Copper(II) sulfate pentahydrate (CuSO₄·5H₂O): 64 + 32 + (4 × 16) + 5[(2 × 1) + 16] = 250

For compounds containing water of crystallisation, calculate the anhydrous compound first, then add the mass of the water molecules separately.

The mole concept and Avogadro constant

The mole provides a bridge between the atomic scale and measurable quantities. One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This value is the Avogadro constant (L or Nₐ).

The fundamental equation connecting mass, moles and molar mass:

Number of moles = mass (g) ÷ molar mass (g/mol)

or

n = m ÷ M

This equation must be rearranged depending on what the question asks:

• To find mass: m = n × M
• To find molar mass: M = m ÷ n

CIE IGCSE Chemistry questions frequently require conversion between mass and moles as a first step before proceeding to further calculations.

Reacting masses and stoichiometric calculations

Balanced equations provide the mole ratio between reactants and products. This ratio allows calculation of masses of substances involved in reactions.

Step-by-step method for reacting mass calculations:

1. Write the balanced chemical equation
2. Identify the substances whose masses you're working with
3. Calculate the molar masses of these substances
4. Convert any given mass to moles using n = m ÷ M
5. Use the mole ratio from the equation to find moles of the required substance
6. Convert moles back to mass using m = n × M

Example: Calculate the mass of calcium oxide produced when 50 g of calcium carbonate decomposes completely.

CaCO₃ → CaO + CO₂

• Mr of CaCO₃ = 100; Mr of CaO = 56
• Moles of CaCO₃ = 50 ÷ 100 = 0.5 mol
• Mole ratio shows 1:1 relationship, so moles of CaO = 0.5 mol
• Mass of CaO = 0.5 × 56 = 28 g

Empirical and molecular formulae

The empirical formula represents the simplest ratio of elements in a compound, while the molecular formula shows the actual number of atoms in one molecule.

Method for finding empirical formula from masses or percentages:

1. Write down the mass (or percentage) of each element
2. Divide each mass by the element's Ar to get moles
3. Divide all mole values by the smallest mole value
4. If necessary, multiply all values by a small whole number to obtain whole number ratios
5. Write the empirical formula using these ratios

Method for finding molecular formula:

1. Calculate the empirical formula mass
2. Divide the given Mr by the empirical formula mass
3. Multiply all subscripts in the empirical formula by this number

Example: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Mr = 180.

• C: 40 ÷ 12 = 3.33; H: 6.7 ÷ 1 = 6.7; O: 53.3 ÷ 16 = 3.33
• Divide by smallest (3.33): C: 1, H: 2, O: 1
• Empirical formula = CH₂O (mass = 30)
• Mr ÷ empirical mass = 180 ÷ 30 = 6
• Molecular formula = C₆H₁₂O₆

Solution calculations: concentration and volume

Solutions feature prominently in CIE IGCSE Chemistry practicals and calculations. Concentration can be expressed in g/dm³ or mol/dm³.

Key equations:

Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

Concentration (mol/dm³) = moles of solute (mol) ÷ volume of solution (dm³)

Rearranging:

• Mass = concentration × volume
• Moles = concentration × volume
• Volume = mass or moles ÷ concentration

Volume conversions:

• 1 dm³ = 1000 cm³
• To convert cm³ to dm³: divide by 1000
• To convert dm³ to cm³: multiply by 1000

When given volume in cm³, always convert to dm³ before using the concentration equations, or adjust the equation accordingly.

Gas volume calculations

At room temperature and pressure (r.t.p.), one mole of any gas occupies approximately 24 dm³ (or 24,000 cm³). This value is provided in CIE IGCSE Chemistry examinations when needed.

Volume of gas (dm³) = moles of gas × 24

Moles of gas = volume of gas (dm³) ÷ 24

Gas volume calculations often combine with reacting mass problems, requiring use of balanced equations to establish mole ratios.

Percentage yield and purity

Percentage yield compares actual product obtained with the theoretical maximum:

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

Percentage yield is always less than 100% in real reactions due to:

• Incomplete reactions
• Side reactions producing alternative products
• Loss of product during separation and purification
• Reversible reactions not going to completion

Percentage purity indicates the proportion of desired substance in an impure sample:

Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100%

CIE IGCSE Chemistry questions may require calculation of actual masses from percentage purity before proceeding with stoichiometric calculations.

Worked examples

Example 1: Reacting masses calculation

Question: Zinc reacts with hydrochloric acid according to the equation:

Zn + 2HCl → ZnCl₂ + H₂

Calculate the mass of zinc chloride produced when 13 g of zinc reacts completely with excess hydrochloric acid. [Ar: Zn = 65, Cl = 35.5] [3 marks]

Solution:

• Mr of ZnCl₂ = 65 + (2 × 35.5) = 136 [1]
• Moles of Zn = 13 ÷ 65 = 0.2 mol
• From equation, mole ratio Zn:ZnCl₂ = 1:1, so moles of ZnCl₂ = 0.2 mol [1]
• Mass of ZnCl₂ = 0.2 × 136 = 27.2 g [1]

Example 2: Empirical and molecular formula

Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 42. Determine the empirical formula and molecular formula of the hydrocarbon. [Ar: C = 12, H = 1] [4 marks]

Solution:

• C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3
• Divide by smallest: C: 7.14 ÷ 7.14 = 1; H: 14.3 ÷ 7.14 = 2
• Empirical formula = CH₂ [2]
• Empirical formula mass = 12 + 2 = 14
• Mr ÷ empirical mass = 42 ÷ 14 = 3 [1]
• Molecular formula = C₃H₆ [1]

Example 3: Solution concentration

Question: A student dissolves 10.6 g of sodium carbonate in water to make 500 cm³ of solution.

(a) Calculate the concentration in g/dm³. [1 mark]

(b) Calculate the concentration in mol/dm³. [Mr of Na₂CO₃ = 106] [2 marks]

Solution:

(a) Volume = 500 cm³ = 0.5 dm³

Concentration = 10.6 ÷ 0.5 = 21.2 g/dm³ [1]

(b) Moles of Na₂CO₃ = 10.6 ÷ 106 = 0.1 mol [1]

Concentration = 0.1 ÷ 0.5 = 0.2 mol/dm³ [1]

Common mistakes and how to avoid them

Forgetting to balance the equation first — always write and check the balanced equation before identifying mole ratios. An unbalanced equation gives incorrect stoichiometric relationships and wrong answers.

Not converting cm³ to dm³ — when using concentration in mol/dm³, volume must be in dm³. Divide cm³ values by 1000. A volume of 250 cm³ = 0.25 dm³, not 250 dm³.

Using Ar instead of Mr — for compounds, calculate the relative molecular or formula mass by adding all atomic masses. Using just one element's Ar gives completely incorrect mole calculations.

Rounding too early — keep at least 3-4 significant figures during intermediate steps. Only round the final answer. Early rounding accumulates errors that lose marks in CIE examinations.

Mixing up empirical and molecular formulae — the empirical formula is the simplest ratio; the molecular formula is the actual molecule. Ethene (C₂H₄) and butene (C₄H₈) share the same empirical formula (CH₂) but different molecular formulae.

Incorrect mole ratio application — carefully match the substances in the equation to those mentioned in the question. If the equation shows 2:3 ratio but you're working with the products, ensure you identify which product corresponds to which coefficient.

Exam technique for Stoichiometry and the Mole

Show all working clearly — CIE mark schemes award method marks even when the final answer is incorrect. Write out the n = m ÷ M equation, substitute values, and show each calculation step on separate lines.

Use the correct number of significant figures — unless specified otherwise, give answers to 3 significant figures. Check whether the question asks for a specific number of decimal places or significant figures.

State units in final answers — marks are lost for missing units. Mass should be in grams (g), volume in dm³ or cm³, concentration in g/dm³ or mol/dm³. Include the unit with your final numerical answer.

Command word recognition — "Calculate" requires numerical working and an answer with units (typically 2-3 marks). "Determine" means find a value through calculation or from data (2-4 marks). "State" needs only the final answer without working (1 mark).

Quick revision summary

Stoichiometry quantifies chemical reactions using the mole concept. One mole contains 6.02 × 10²³ particles and has mass equal to the Ar or Mr in grams. Use n = m ÷ M to convert between mass and moles. Balanced equations provide mole ratios for reacting mass calculations. Empirical formulae show simplest ratios from composition data; molecular formulae require Mr values. Solution concentration links moles, mass and volume. At r.t.p., one mole of gas occupies 24 dm³. Always show full working, convert units correctly, and include units in final answers for maximum marks in CIE IGCSE Chemistry examinations.