Understanding Stoichiometry: IGCSE Chemistry Explained

What stoichiometry actually means

Stoichiometry is the branch of chemistry concerned with the quantities of substances involved in chemical reactions. The word itself derives from the Greek stoicheion (element) and metron (measure), and that etymology captures its purpose precisely: measuring elements and compounds in exact proportions. For IGCSE Chemistry, stoichiometry underpins nearly every calculation question on Papers 3, 4, and 6, making it one of the highest-value topics to master.

Formulae: molecular and empirical

A molecular formula states the actual number and type of atoms in one molecule of a substance. Glucose, for instance, has the molecular formula C₆H₁₂O₆. An empirical formula gives the simplest whole-number ratio of atoms. For glucose, the empirical formula is CH₂O, because 6:12:6 simplifies to 1:2:1.

Cambridge examiners regularly ask candidates to deduce empirical formulae from experimental data. The method is consistent every time:

1. Write down the mass or percentage of each element.
2. Divide each by the element's relative atomic mass (A_r).
3. Divide all results by the smallest value to obtain a ratio.
4. Round to the nearest whole numbers (multiply through if you get values like 1.5 or 2.5).

Writing and balancing equations

Chemical equations must be balanced so that the number of atoms of each element is equal on both sides. This reflects the law of conservation of mass: atoms are neither created nor destroyed in a chemical reaction. Balancing is a non-negotiable expectation on every IGCSE Chemistry paper that involves equations, and marks are routinely lost when candidates present unbalanced formulae.

The process of balancing is systematic. Begin by writing the correct formulae for all reactants and products. Then adjust coefficients (the large numbers placed before formulae) until each element appears the same number of times on both sides. Work through the elements one at a time, typically starting with metals, then non-metals, and leaving hydrogen and oxygen until last.

State symbols are required in IGCSE equations:

• (s) - solid
• (l) - liquid
• (g) - gas
• (aq) - aqueous (dissolved in water)

A balanced equation with state symbols might read:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

The coefficient 2 before HCl ensures that chlorine and hydrogen atoms balance. A common error among candidates is adjusting subscripts within a formula rather than placing coefficients in front of it. Changing HCl₂ would alter the substance itself; placing a 2 before HCl preserves the compound's identity while satisfying atom counts.

Relative atomic mass and relative molecular mass

The relative atomic mass (A_r) of an element is the weighted average mass of its naturally occurring isotopes, measured on a scale where carbon-12 has a mass of exactly 12. The relative molecular mass (M_r) of a compound is the sum of the relative atomic masses of all atoms in its molecular formula.

The mole and the Avogadro constant

A mole is the amount of substance that contains 6.02 x 10²³ particles (atoms, molecules, ions, or electrons). This number is called the Avogadro constant. One mole of any substance has a mass in grams numerically equal to its relative formula mass. One mole of CaCO₃, for example, has a mass of 100 g.

The fundamental relationship linking mass, moles, and relative formula mass is:

moles = mass (g) / M_r

This single equation, rearranged as needed, answers most quantitative questions on an IGCSE Chemistry paper.

Key formulae at a glance

Mole calculations in practice

The approach is always the same: convert the given quantity to moles, use the molar ratio from the balanced equation, then convert back to the required unit (mass, volume, or concentration).

Concentration and solution calculations

When a solute dissolves in a solvent, the resulting solution has a concentration measured in mol/dm³ (sometimes written as M). The relationship is straightforward:

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

Take care with units. Volume is frequently given in cm³ in exam questions. To convert: divide cm³ by 1000 to obtain dm³. A 250 cm³ solution is 0.250 dm³.

Gas volumes at room temperature and pressure

At room temperature and pressure (RTP, approximately 20 degrees C and 1 atm), one mole of any gas occupies 24 dm³ (or 24,000 cm³). This is the molar gas volume and applies regardless of the gas's identity. Whether the gas is hydrogen, oxygen, or carbon dioxide, one mole at RTP occupies the same volume. This principle, rooted in Avogadro's law, makes gas volume calculations remarkably straightforward.

To find the volume of gas produced or consumed:

volume (dm³) = moles x 24

Returning to the calcium carbonate example above: 0.1 mol of CO₂ would occupy 0.1 x 24 = 2.4 dm³ at RTP. In the reverse direction, if a question gives a gas volume, divide by 24 to find the number of moles. For instance, 4.8 dm³ of hydrogen at RTP corresponds to 4.8 / 24 = 0.2 mol of H₂.

Be alert to questions that give volumes in cm³ rather than dm³. If 600 cm³ of gas is collected at RTP, convert first: 600 / 1000 = 0.6 dm³, then calculate moles: 0.6 / 24 = 0.025 mol.

Percentage yield and percentage purity

In practice, reactions rarely convert 100% of reactants into products. Some product may be lost during transfer, purification, or incomplete reaction. The percentage yield compares what was actually obtained with what was theoretically possible:

% yield = (actual yield / theoretical yield) x 100

Percentage purity measures how much of a sample is the desired substance versus impurities:

% purity = (mass of pure substance / total mass of sample) x 100

IGCSE examiners frequently combine these calculations with mole problems. A question might ask for the theoretical yield first (using moles and M_r), then require percentage yield as a second step.

Common pitfalls and how to avoid them

• Forgetting to convert cm³ to dm³ - always divide by 1000 before using the concentration formula.
• Using the wrong A_r values - the periodic table provided in the exam is your reference. Do not memorise approximate values when precise ones are available.
• Ignoring molar ratios - the balanced equation gives the ratio. A 2:1 ratio means two moles of one reactant react with one mole of another, not equal masses.
• Rounding too early - carry at least three significant figures through intermediate steps and round only in the final answer.
• Confusing empirical and molecular formulae - the empirical formula is the simplest ratio; the molecular formula may be a whole-number multiple of it.

Self-check questions

1. Calculate the M_r of magnesium sulfate, MgSO₄. (A_r: Mg = 24, S = 32, O = 16)
2. A compound is 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. Determine its empirical formula.
3. What volume of hydrogen gas (at RTP) is produced when 0.5 mol of zinc reacts with excess sulfuric acid? (Zn + H₂SO₄ → ZnSO₄ + H₂)
4. A student expected to produce 7.1 g of chlorine but obtained 5.0 g. Calculate the percentage yield.

Stoichiometry rewards precision and method. Every calculation follows the same logical sequence: identify the given quantity, convert to moles, apply the equation ratio, then convert to the target unit. Candidates who practise this sequence with past IGCSE papers will find that even unfamiliar contexts yield to the same disciplined approach.