Formulae, Equations & Calculations

For many students, chemistry calculations are the most feared part of the course. But here is the good news: almost every calculation question uses the same small set of ideas, applied in the same logical order. Once you can write formulae, balance equations and use the mole, you can answer questions on reacting masses, yields, gas volumes and concentrations with confidence. This guide takes you from the very beginning to exam level, with clear worked examples at every step.

Part 1: Chemical Formulae

A chemical formula uses element symbols and numbers to show the composition of a substance.

• H₂O contains 2 hydrogen atoms and 1 oxygen atom. The small number after a symbol, called a subscript, applies only to the element directly before it.

• CO₂ contains 1 carbon atom and 2 oxygen atoms.

• In Ca(OH)₂, the 2 after the bracket multiplies everything inside it, so there are 1 calcium, 2 oxygen and 2 hydrogen atoms.

• A large number in front, called a coefficient, multiplies the whole formula: 3H₂O means three water molecules, containing 6 hydrogen and 3 oxygen atoms in total.

Formulae of elements to remember

Most elements are written simply as their symbol, such as Fe, Mg and C. But seven elements exist as diatomic molecules and must be written with a 2: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂. A useful memory aid is "Have No Fear Of Ice Cold Beer".

Writing formulae of ionic compounds

Ionic compounds are neutral, so the total positive charge must equal the total negative charge. You need to know the charges on common ions:

The fastest method is to swap and drop the charge numbers:

• Aluminium oxide: Al³⁺ and O²⁻. Swap the numbers to get Al₂O₃.

• Calcium chloride: Ca²⁺ and Cl⁻. Swap to get CaCl₂.

• Magnesium oxide: Mg²⁺ and O²⁻. Swapping gives Mg₂O₂, which simplifies to MgO.

• Calcium hydroxide: Ca²⁺ and OH⁻. You need two hydroxide ions, so use brackets: Ca(OH)₂.

• Ammonium sulfate: NH₄⁺ and SO₄²⁻ gives (NH₄)₂SO₄.

Brackets are needed only when there is more than one of a compound ion.

Part 2: Chemical Equations

Word equations

A word equation names the reactants and products:

magnesium + oxygen → magnesium oxide

Reactants are on the left of the arrow and products on the right. The arrow means "reacts to form".

Symbol equations and the law of conservation of mass

In a chemical reaction, atoms are not created or destroyed; they are simply rearranged. This is the law of conservation of mass: the total mass of the products equals the total mass of the reactants. A balanced symbol equation must therefore have the same number of each type of atom on both sides.

How to balance an equation step by step

Example 1: magnesium burning in oxygen

• Write the correct formulae: Mg + O₂ → MgO

• Count the atoms. Left: 1 Mg, 2 O. Right: 1 Mg, 1 O. The oxygen does not balance.

• Put a 2 in front of MgO: Mg + O₂ → 2MgO. Now there are 2 Mg on the right.

• Put a 2 in front of Mg: 2Mg + O₂ → 2MgO. Balanced.

Example 2: methane burning

• CH₄ + O₂ → CO₂ + H₂O

• Carbon balances. Hydrogen: 4 on the left, 2 on the right, so write 2H₂O.

• Oxygen: now 2 + 2 = 4 on the right, so write 2O₂ on the left.

• CH₄ + 2O₂ → CO₂ + 2H₂O. Balanced.

Golden rule: never change the small numbers inside a formula. Changing H₂O to H₂O₂ turns water into hydrogen peroxide, a completely different substance. Only change the large numbers in front.

Tips for balancing tricky equations

• Balance elements that appear in only one compound on each side first.

• Leave elements that appear on their own, such as O₂ or Fe, until last.

• Treat compound ions such as SO₄ as a single unit if they appear unchanged on both sides.

• Check every element at the end.

State symbols

State symbols show the physical state of each substance:

• (s) solid

• (l) liquid

• (g) gas

• (aq) aqueous, meaning dissolved in water

For example: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Ionic equations

An ionic equation shows only the particles that actually change. For every neutralisation between an acid and an alkali:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The ions that do not change, such as Na⁺ and Cl⁻, are spectator ions and are left out.

Part 3: Relative Formula Mass (M_r)

The relative atomic mass (A_r) of each element is found in the Periodic Table. The relative formula mass (M_r) of a compound is the sum of the relative atomic masses of all the atoms in its formula.

Worked example: find the M_r of calcium carbonate, CaCO₃ (A_r: Ca = 40, C = 12, O = 16).

M_r = 40 + 12 + (3 × 16) = 100

Worked example: find the M_r of Ca(OH)₂ (H = 1).

M_r = 40 + 2 × (16 + 1) = 74

M_r has no units, because it is a relative value.

Percentage by mass of an element

% of element = (A_r × number of atoms of that element ÷ M_r) × 100

Example: percentage of nitrogen in ammonium nitrate, NH₄NO₃ (M_r = 80).

% N = (14 × 2 ÷ 80) × 100 = 35%

Farmers compare fertilisers using calculations like this.

Part 4: The Mole

Atoms are far too small to count individually, so chemists count them in huge batches called moles. One mole of any substance contains 6.02 × 10²³ particles, a number called the Avogadro constant.

The key idea: one mole of a substance has a mass in grams equal to its M_r. One mole of carbon has a mass of 12 g; one mole of water has a mass of 18 g; one mole of calcium carbonate has a mass of 100 g.

The most important formula in chemistry

moles = mass (g) ÷ M_r

Rearranged: mass = moles × M_r, and M_r = mass ÷ moles.

Example: how many moles are in 36 g of water (M_r = 18)? Moles = 36 ÷ 18 = 2 mol.

Example: what is the mass of 0.25 mol of sodium hydroxide, NaOH (M_r = 40)? Mass = 0.25 × 40 = 10 g.

Part 5: Reacting Mass Calculations

Balanced equations tell you the ratio of moles that react. You can use this to calculate the mass of product formed or reactant needed.

The three-step method

• Step 1: Calculate the moles of the substance you know, using moles = mass ÷ M_r.

• Step 2: Use the balanced equation to find the moles of the substance you want (the mole ratio).

• Step 3: Convert moles back to mass, using mass = moles × M_r.

Worked example: What mass of calcium oxide is produced when 50 g of calcium carbonate is heated?

CaCO₃ → CaO + CO₂

• Moles of CaCO₃ = 50 ÷ 100 = 0.5 mol.

• The ratio of CaCO₃ to CaO is 1 : 1, so 0.5 mol of CaO forms.

• Mass of CaO = 0.5 × 56 = 28 g.

Worked example with a different ratio: What mass of hydrogen is produced when 4.8 g of magnesium reacts with excess hydrochloric acid?

Mg + 2HCl → MgCl₂ + H₂

• Moles of Mg = 4.8 ÷ 24 = 0.2 mol.

• The ratio of Mg to H₂ is 1 : 1, so 0.2 mol of H₂ forms.

• Mass of H₂ = 0.2 × 2 = 0.4 g.

Limiting reactants

When two reactants are mixed, the one that is completely used up is the limiting reactant; it controls how much product forms. The other is in excess. To find the limiting reactant, calculate the moles of each and compare them with the ratio in the equation.

Part 6: Empirical and Molecular Formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one molecule. For example, glucose has the molecular formula C₆H₁₂O₆ and the empirical formula CH₂O.

Worked example: a compound contains 2.4 g of carbon and 0.6 g of hydrogen. Find its empirical formula.

The empirical formula is CH₃. If the M_r of the compound is 30, the empirical formula mass is 15, so the molecular formula is twice that: C₂H₆ (ethane).

Part 7: Percentage Yield

The theoretical yield is the maximum mass of product you could get, calculated from the equation. The actual yield is what you really obtain. In practice it is always lower because some product is lost during filtering and transferring, the reaction may be reversible, or side reactions may occur.

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

Example: the theoretical yield of calcium oxide is 28 g, but only 21 g is collected. Percentage yield = (21 ÷ 28) × 100 = 75%.

Part 8: Gas Volumes

At room temperature and pressure (rtp), one mole of any gas occupies 24 dm³ (24,000 cm³).

volume of gas (dm³) = moles × 24

Example: what volume of carbon dioxide is produced when 0.5 mol of calcium carbonate decomposes? The ratio is 1 : 1, so 0.5 mol of CO₂ forms. Volume = 0.5 × 24 = 12 dm³.

Part 9: Concentration of Solutions

Concentration tells you how much solute is dissolved in a volume of solution.

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

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

Remember to convert cm³ to dm³ by dividing by 1000.

Example: 0.1 mol of sodium hydroxide is dissolved to make 250 cm³ of solution. Volume = 250 ÷ 1000 = 0.25 dm³. Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³.

These formulae are used in titrations to find the unknown concentration of an acid or alkali. For example, if 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid, the moles of acid are 0.100 × 0.0200 = 0.00200 mol. The equation NaOH + HCl → NaCl + H₂O has a 1 : 1 ratio, so the concentration of the alkali is 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.

Common Calculation Mistakes

• Changing subscripts instead of coefficients when balancing.

• Forgetting that the bracket number multiplies everything inside.

• Using the mass of an element instead of the M_r of the whole compound.

• Ignoring the mole ratio in the balanced equation.

• Forgetting to convert cm³ to dm³.

• Rounding too early. Keep full numbers until the final answer.

Key Takeaways

• Formulae show the elements and number of atoms; ionic formulae balance the charges.

• Balanced equations have equal numbers of each atom on both sides; only change coefficients.

• M_r is the sum of the A_r values of all atoms in a formula.

• Moles = mass ÷ M_r is the key to every calculation.

• Use the mole ratio from the equation for reacting masses, gas volumes and yields.

Practice is the secret to mastering calculations. Try our Chemistry Calculations Practice Book for hundreds of exam-style questions with full worked answers.