Chemical equations and stoichiometry

This page covers a single Form 4 content standard: chemical equations and stoichiometry. This is where the whole chapter comes together: a balanced equation tells you the ratio in which substances react, and the mole lets you turn that ratio into real masses and volumes. Mastering this standard means you can predict how much product a reaction gives, or how much reactant it needs, the kind of quantitative reasoning that runs through the rest of SPM Chemistry and into Form 5.

What a balanced equation tells you

A chemical equation represents a reaction using the correct chemical formulae of the reactants and products, with an arrow from reactants to products. It must be balanced: because atoms are neither created nor destroyed, the number of atoms of each element must be the same on both sides. Where required, state symbols are added, (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

Crucially, the numbers placed in front of the formulae, the coefficients, give the mole ratio (also called the stoichiometric ratio) in which the substances react and are produced. For example, in

2Mg(s) + O₂(g) → 2MgO(s)

the ratio Mg : O₂ : MgO is 2 : 1 : 2. This means 2 mol of magnesium react with 1 mol of oxygen to give 2 mol of magnesium oxide. The mole ratio is the bridge between one substance and another.

Balancing an equation

To balance, adjust the coefficients (never change a formula) until each element has equal atoms on both sides. Balance one element at a time, leave oxygen and hydrogen towards the end, and check every element when you finish. The coefficients must be the smallest whole numbers that work.

The stoichiometry method

Almost every calculation on this standard follows the same four steps:

1. Write and balance the equation.
2. Convert the given quantity to moles (mass ÷ molar mass, or gas volume ÷ molar volume, or particles ÷ the Avogadro constant).
3. Use the mole ratio from the equation to find the moles of the substance asked for.
4. Convert those moles into the quantity required (× molar mass for a mass, × molar volume for a gas volume, × the Avogadro constant for a particle count).

The mole ratio in step 3 is the only new idea; steps 2 and 4 are just the conversions you already know from standards 3.3 and 3.4.

Worked example

Question. Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed when 4.8 g of magnesium is completely burned. [Ar: Mg = 24, O = 16]

Step 1, The equation is balanced. 2Mg + O₂ → 2MgO. The mole ratio Mg : MgO is 2 : 2, i.e. 1 : 1.

Step 2, Convert the given mass to moles. Molar mass of Mg = 24 g mol⁻¹. Moles of Mg = 4.8 ÷ 24 = 0.2 mol.

Step 3, Apply the mole ratio. Since Mg : MgO = 1 : 1, moles of MgO = 0.2 mol.

Step 4, Convert moles of MgO to mass. Mr of MgO = 24 + 16 = 40, so molar mass = 40 g mol⁻¹. Mass of MgO = 0.2 × 40 = 8.0 g.

Answer. 8.0 g of magnesium oxide is formed.

Practice question

Question. Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂. Calculate the volume of carbon dioxide produced at room conditions when 10 g of calcium carbonate decomposes completely. [Ar: Ca = 40, C = 12, O = 16; molar volume at room conditions = 24 dm³ mol⁻¹ dm³ mol⁻¹]

Answer. Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol⁻¹. Moles of CaCO₃ = 10 ÷ 100 = 0.1 mol. Mole ratio CaCO₃ : CO₂ = 1 : 1, so moles of CO₂ = 0.1 mol. Volume of CO₂ = 0.1 × 24 dm³ mol⁻¹ = 2.4 dm³.

Exam tip

The order of operations is what earns marks: balance first, then work in moles. A frequent error is applying the mole ratio to masses instead of moles, the ratio only works for moles, so you must convert to moles before using it. Two more habits: include state symbols when the question asks for a full equation, and read carefully whether the final answer must be a mass (use molar mass), a gas volume (use molar volume for the stated conditions) or a number of particles (use the Avogadro constant). Writing your four steps as separate lines lets an examiner award method marks even if a single arithmetic slip creeps in.

The mole ratio is the heart of it

Everything distinctive about this standard rests on one insight: the coefficients of a balanced equation are a ratio of moles, not of masses or volumes. Two magnesium atoms react with one oxygen molecule, so two moles of magnesium react with one mole of oxygen, and because a mole is a fixed number of particles, that atom-level ratio scales up perfectly to laboratory amounts. This is exactly why the mole was invented: it lets the microscopic ratio in an equation control the macroscopic masses and volumes you measure. Once you trust the mole ratio, a stoichiometry problem becomes a short, predictable journey: into moles, across the ratio, back out to the quantity you need.

Conservation of mass in the background

A balanced equation is really a statement of the law of conservation of mass: the total mass of the reactants equals the total mass of the products, because the same atoms are simply rearranged. In the worked example, 4.8 g of magnesium combined with 3.2 g of oxygen (0.1 mol × 32) to give exactly 8.0 g of magnesium oxide, mass is conserved. Keeping this in mind gives you a quick sanity check on any stoichiometry answer, and it explains why we are allowed to balance equations by adjusting coefficients in the first place.

Where this fits

This is content standard 3.6, the culmination of the The Mole Concept, Chemical Formula and Equation chapter, drawing together relative mass, the mole, molar mass and molar volume. Practise full calculations with the worked examples and test yourself with the practice questions. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers make sure you balance confidently and never apply a mole ratio to a mass, because this standard supplies a large share of the calculation marks in the paper.

Quick recap

• A balanced equation has equal atoms of each element on both sides; add state symbols when needed.
• The coefficients give the mole ratio between substances.
• Method: balance → given amount to moles → apply mole ratio → convert to the required quantity.
• The ratio applies to moles only; balance before you calculate, and check what the answer’s units should be.