Chemical formulae equations and calculations

Writing Word Equations

A word equation names the reactants on the left and the products on the right, separated by an arrow. The arrow (→) means "produces" or "forms." Reactants are the substances that react together; products are the substances formed.

Word equations do not include formulae, numbers, or state symbols — only the names of substances.

MisconceptionStudents sometimes write "+" on the product side even when there is only one product. Only use "+" to separate two or more distinct substances on the same side.

Exam TipRead the question carefully to identify every product before writing the equation.

Balanced Chemical Equations with State Symbols

A balanced chemical equation uses chemical formulae and ensures the same number of each type of atom appears on both sides. This reflects the law of conservation of mass — atoms are neither created nor destroyed in a chemical reaction.

State symbols indicate the physical state of each substance:

To balance an equation, adjust only the large numbers (coefficients) placed before formulae. Never change the small subscript numbers inside a formula, because that changes the substance itself.

Balancing strategy: count atoms of each element on both sides, then adjust coefficients one element at a time. Start with the most complex formula. Leave single-element substances (such as O₂) until last.

Examples of balanced equations for reactions in this specification:

For unfamiliar reactions, the exam provides formulae. The task is to determine the correct coefficients so that atoms balance on each side and to add state symbols.

Examiner InsightExaminers award separate marks for the correct formulae, correct balancing, and correct state symbols. A common lost mark is omitting state symbols when the question says "include state symbols."

Exam TipAlways re-count every element after balancing to confirm totals match.

MisconceptionChanging subscripts (e.g. writing H₃O instead of H₂O) to balance an equation is incorrect. Only coefficients in front of formulae may be altered.

Exam TipIf atoms do not balance, adjust the number before the formula, never the small number within it.

Relative Formula Mass

The relative formula mass (Mᵣ) of a compound equals the sum of the relative atomic masses (Aᵣ) of all atoms shown in its formula. For molecules, Mᵣ may also be called the relative molecular mass — the calculation is identical.

Aᵣ values are found on the periodic table. Each Aᵣ is the mass of one atom relative to ¹⁄₁₂ the mass of a carbon-12 atom, so Aᵣ and Mᵣ have no units.

Worked example — Mᵣ of calcium hydroxide, Ca(OH)₂:

Representation note: In a formula such as Ca(OH)₂, the subscript 2 outside the bracket multiplies everything inside the bracket. So (OH)₂ means 2 oxygen atoms and 2 hydrogen atoms.

Identify the atoms present and their counts:

$${M}_{r}=40+(2\times 16)+(2\times 1)$$

$${M}_{r}=40+32+2=74$$

MisconceptionStudents sometimes forget to multiply by the subscript outside brackets. In Ca(OH)₂, there are 2 oxygen atoms and 2 hydrogen atoms, not 1 of each.

Exam TipExpand brackets in the formula before summing Aᵣ values.

The Mole

The mole (mol) is the unit for the amount of a substance. One mole of any substance contains the same number of particles (atoms, molecules, ions, or formula units). This number is approximately 6.02 × 10²³, but the Edexcel IGCSE syllabus requires only the understanding that the mole is a counting unit that links mass to formula mass.

The mass of one mole of a substance in grams is numerically equal to its Aᵣ (for elements) or Mᵣ (for compounds). For example, one mole of carbon (Aᵣ = 12) has a mass of 12 g; one mole of water (Mᵣ = 18) has a mass of 18 g.

Examiner InsightQuestions often state "amount" rather than "number of moles." In chemistry, "amount of substance" is the quantity measured in moles.

Exam TipWhen a question asks for "amount," give your answer in mol.

Calculations Involving Amount, Aᵣ, and Mᵣ

Key Equations

Amount of substance: n = m ÷ Mᵣ

n = amount of substance (mol)

m = mass (g)

Mᵣ = relative formula mass (no unit)

SI unit of n: mol

Rearrangements of this equation give:

$$m=n\times {M}_{r}$$

$${M}_{r}=\frac{m}{n}$$

Calculations involving amount of substance use the relationship between mass, moles, and relative formula mass. The triangle below summarises the three rearrangements: m sits on top, n and Mᵣ on the bottom. Cover the quantity needed to see the required operation.

Worked example — Calculate the amount, in moles, of 11 g of carbon dioxide (CO₂). Aᵣ: C = 12, O = 16.

Carbon dioxide CO₂

First, calculate Mᵣ:

$${M}_{r}=12+(2\times 16)=44$$

State the equation:

$$n=\frac{m}{{M}_{r}}$$

Substitute:

$$n=\frac{11}{44}$$

$$n=0.25\text{ mol}$$

Worked example — Calculate the mass of 0.40 mol of sodium hydroxide (NaOH). Aᵣ: Na = 23, O = 16, H = 1.

Sodium hydroxide NaOH

$${M}_{r}=23+16+1=40$$

$$m=n\times {M}_{r}$$

$$m=0.40\times 40$$

$$m=16\text{ g}$$

Reacting Masses

Reacting masses can be calculated from a balanced equation because the coefficients give the mole ratio of reactants and products. This ratio links the mass of one substance to the mass of another.

The general method:

1. Write the balanced equation.

2. Calculate the amount (in mol) of the substance whose mass is given.

3. Use the mole ratio from the equation to find the amount of the required substance.

4. Convert that amount back to mass using m = n × Mᵣ.

Worked example — What mass of magnesium oxide forms when 4.8 g of magnesium burns in excess oxygen? Aᵣ: Mg = 24, O = 16.

Balanced equation:

2Mg(s) + O₂(g) → 2MgO(s)

Magnesium Mg

$${M}_{r}=24$$

$$n=\frac{m}{{M}_{r}}$$

$$n=\frac{4.8}{24}$$

$$n=0.20\text{ mol}$$

Use the mole ratio from the equation: 2 mol Mg produces 2 mol MgO, so the ratio is 1 : 1.

Magnesium oxide MgO

$$n=0.20\text{ mol}$$

$${M}_{r}=24+16=40$$

$$m=n\times {M}_{r}$$

$$m=0.20\times 40$$

$$m=8.0\text{ g}$$

Examiner InsightWhen the question says "excess," it signals that the other reagent is the limiting reactant. Base your calculation on the substance that is not in excess.

Exam TipIdentify the limiting reactant first, then use its moles to find the answer.

Percentage Yield

Key Equations

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

actual yield = mass of product actually obtained (g)

theoretical yield = maximum mass of product calculated from the equation (g)

SI unit: % (dimensionless)

The percentage yield compares the mass of product actually obtained in an experiment with the maximum mass predicted by the balanced equation. In practice, yield is almost always less than 100 % because of losses during transfer, incomplete reactions, or side reactions.

$$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$$

Worked example — A student calculated that 8.0 g of MgO should form. The experiment produced 6.8 g. Calculate the percentage yield.

$$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100$$

$$\text{percentage yield}=\frac{6.8}{8.0}\times 100$$

$$\text{percentage yield}=85\%$$

MisconceptionPercentage yield can never exceed 100 %. If a calculated yield exceeds 100 %, the product is impure or the actual mass was measured incorrectly.

Exam TipIf your answer is over 100 %, recheck your working.

Empirical Formula and Molecular Formula

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule.

For example, glucose has an empirical formula of CH₂O and a molecular formula of C₆H₁₂O₆. The molecular formula is always a whole-number multiple of the empirical formula.

To calculate an empirical formula from experimental data (masses or percentages):

1. Write each element and its mass (or percentage).

2. Divide each mass by the element's Aᵣ to find moles.

3. Divide every mole value by the smallest mole value.

4. If the resulting ratios are not whole numbers, multiply all by the smallest factor that converts them to whole numbers.

Worked example — A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula. Aᵣ: C = 12, H = 1.

Empirical formula = CH₄

Finding the molecular formula from the empirical formula:

Divide the Mᵣ of the molecular formula by the Mᵣ of the empirical formula to find the multiplier n.

$$n=\frac{{M}_{r}\text{ (molecular)}}{{M}_{r}\text{ (empirical)}}$$

Then multiply each subscript in the empirical formula by n.

Worked example — The empirical formula of a compound is CH₂O and its Mᵣ is 60. Find the molecular formula.

$${M}_{r}{\text{ of CH}}_{2}\text{O}=12+2+16=30$$

$$n=\frac{60}{30}=2$$

Molecular formula = C₂H₄O₂

Examiner InsightExam questions may give data as percentages instead of masses. Use the percentages directly as if they were masses (in grams) for a 100 g sample.

Exam TipIf given percentages, treat each percentage as a mass in grams.

MisconceptionAn empirical formula is not always different from the molecular formula. If the molecular formula is already in its simplest ratio (e.g. H₂O), then the empirical and molecular formulae are the same.

Exam TipAfter calculating the empirical formula, check whether the question also asks for the molecular formula.

Obtaining Formulae Experimentally

The formulae of simple compounds can be obtained experimentally by measuring the masses of elements that combine. This applies to metal oxides, water, and salts containing water of crystallisation.

For a metal oxide, the mass of the metal and the mass of oxygen that combine are measured. The mass of oxygen equals the increase in mass when the metal burns, or the decrease in mass when the oxide is reduced. These masses are used to calculate the empirical formula.

For water (H₂O), electrolysis produces hydrogen and oxygen in a 2 : 1 volume ratio, confirming the formula.

For salts containing water of crystallisation [water molecules bonded within the crystal structure], heating drives off the water. The mass before and after heating gives the mass of water lost. The formula of the anhydrous salt is known, so the number of moles of water per mole of salt can be calculated. The result is written as, for example, CuSO₄·5H₂O.

Worked example — 2.00 g of hydrated copper(II) sulfate (CuSO₄·xH₂O) is heated until all water is removed. The residue has a mass of 1.28 g. Find x. Mᵣ: CuSO₄ = 160, H₂O = 18.

Mass of water lost:

$${m}_{\text{water}}=2.00-1.28=0.72\text{ g}$$

Anhydrous copper(II) sulfate CuSO₄

$$n=\frac{1.28}{160}=0.0080\text{ mol}$$

Water H₂O

$$n=\frac{0.72}{18}=0.040\text{ mol}$$

Find the ratio:

$$\frac{0.040}{0.0080}=5$$

Therefore x = 5, and the formula is CuSO₄·5H₂O.

Calculations Involving Concentration

Key Equations

Concentration equation: n = c × V

n = amount of substance (mol)

c = concentration (mol dm⁻³)

V = volume of solution (dm³)

SI unit of c: mol dm⁻³

Concentration measures how much solute is dissolved per unit volume of solution. In the equation n = c × V, the volume must be in dm³. To convert cm³ to dm³, divide by 1000.

Representation note: dm³ means cubic decimetres. 1 dm³ = 1000 cm³ = 1 litre.

$$V{\text{ (dm}}^{3}\text{)}=\frac{V{\text{ (cm}}^{3}\text{)}}{1000}$$

Rearrangements:

$$c=\frac{n}{V}$$

$$V=\frac{n}{c}$$

Worked example — Calculate the amount, in moles, of sodium hydroxide in 25.0 cm³ of 0.10 mol dm⁻³ NaOH solution.

Convert volume to dm³:

$$V=\frac{25.0}{1000}=0.0250{\text{ dm}}^{3}$$

$$n=c\times V$$

$$n=0.10\times 0.0250$$

$$n=0.0025\text{ mol}$$

Worked example — 0.050 mol of HCl is dissolved in 500 cm³ of solution. Calculate the concentration.

Convert volume:

$$V=\frac{500}{1000}=0.500{\text{ dm}}^{3}$$

$$c=\frac{n}{V}$$

$$c=\frac{0.050}{0.500}$$

$$c=0.10{\text{ mol dm}}^{-3}$$

MisconceptionForgetting to convert cm³ to dm³ causes answers to be 1000 times too large or too small. The equation only works when V is in dm³.

Exam TipIf the volume is given in cm³, divide by 1000 as the very first step.

Calculations Involving Gas Volumes and Molar Volume

Key Equations

Gas volume equation: V = n × Vₘ

V = volume of gas (dm³ or cm³)

n = amount of substance (mol)

Vₘ = molar volume (24 dm³ or 24 000 cm³ at rtp)

SI unit of V: dm³ (or cm³)

At room temperature and pressure (rtp), one mole of any gas occupies 24 dm³ (24 000 cm³). This value is the molar volume and applies to all gases regardless of their identity.

$$n=\frac{V}{{V}_{m}}$$

$$V=n\times {V}_{m}$$

Representation note: rtp refers to approximately 20 °C and 1 atm. At rtp the molar volume Vₘ = 24 dm³ mol⁻¹ = 24 000 cm³ mol⁻¹.

Match units consistently: if V is in dm³, use 24 dm³; if V is in cm³, use 24 000 cm³.

Worked example — Calculate the volume of 0.30 mol of hydrogen gas at rtp.

$$V=n\times {V}_{m}$$

$$V=0.30\times 24$$

$$V=7.2{\text{ dm}}^{3}$$

Worked example — Calculate the amount, in moles, of CO₂ in 480 cm³ at rtp.

$$n=\frac{V}{{V}_{m}}$$

$$n=\frac{480}{24 000}$$

$$n=0.020\text{ mol}$$

Examiner InsightQuestions may combine gas volume calculations with reacting mass calculations. Calculate moles of the solid reactant first, use the mole ratio, then convert to gas volume.

Exam TipIdentify whether the question gives or asks for a gas volume — this signals the use of Vₘ = 24 dm³.

Determining the Formula of a Metal Oxide

PRACTICAL: Determining the Formula of a Metal Oxide

Aim: Determine the empirical formula of a metal oxide by measuring the masses of the metal and oxygen that combine, using either combustion (e.g. magnesium oxide) or reduction (e.g. copper(II) oxide).

Method (combustion — magnesium oxide):

1. Weigh a clean, dry crucible with its lid. Record the mass.

2. Add a known mass of magnesium ribbon to the crucible and record the total mass.

3. Heat the crucible strongly with the lid slightly raised to allow air (oxygen) in but prevent loss of the white magnesium oxide smoke.

4. Lift the lid periodically to admit more air, then replace it.

5. Continue heating until there is no further change in appearance (all magnesium has reacted).

6. Allow the crucible to cool, then reweigh.

7. Repeat heating and reweighing until a constant mass is obtained.

Method (reduction — copper(II) oxide):

1. Weigh a dry combustion boat/tube containing a known mass of copper(II) oxide.

2. Pass hydrogen gas (or natural gas) over the heated copper(II) oxide in a glass tube.

3. Continue heating until the black powder has completely changed to a pink-brown colour (copper).

4. Allow to cool in the stream of hydrogen to prevent re-oxidation.

5. Reweigh the boat to find the mass of copper remaining.

Variables

Independent variable (IV): the type of metal used (e.g. magnesium or copper)

Dependent variable (DV): mass of the product (measured by reweighing the crucible/boat using a balance, in grams)

Control variables (CV): form of metal (ribbon/powder to ensure consistent surface area); heating to constant mass; same balance used throughout

Expected results

For magnesium, the mass increases because oxygen from air combines with the metal. The ratio of moles of Mg to moles of O is approximately 1 : 1, giving an empirical formula of MgO. For copper(II) oxide reduction, the mass decreases because oxygen is removed. The ratio of moles of Cu to moles of O is approximately 1 : 1, giving CuO.

Precaution

If the crucible lid is lifted too high during combustion, magnesium oxide smoke escapes, decreasing the final mass and giving an inaccurate Mg : O ratio. Heating to constant mass minimises the error of incomplete reaction.

SafetyMagnesium burns with an intense white light that can damage eyesight. Avoid looking directly at the burning magnesium; view through the crucible lid gap only.

Worked example — In a combustion experiment, 0.48 g of magnesium produced 0.80 g of magnesium oxide. Calculate the empirical formula. Aᵣ: Mg = 24, O = 16.

Mass of oxygen that combined:

$${m}_{\text{O}}=0.80-0.48=0.32\text{ g}$$

Magnesium Mg

$$n=\frac{0.48}{24}=0.020\text{ mol}$$

Oxygen O

$$n=\frac{0.32}{16}=0.020\text{ mol}$$

Ratio: Mg : O = 0.020 : 0.020 = 1 : 1

Empirical formula = MgO

Examiner InsightPractical questions often ask why the lid is lifted then replaced. The examiner expects two points: lifting admits air/oxygen for the reaction; replacing prevents loss of magnesium oxide smoke.

Exam TipAlways state both reasons when explaining the lid technique.