Balanced Equations and Reacting Masses

Balanced chemical equations are the load-bearing skeleton of quantitative chemistry. Every prediction about how much product you can make, how much reagent to weigh out, and how much waste a process generates flows from getting the stoichiometric coefficients right and then applying mole ratios to them. The conceptual foundation is Antoine Lavoisier's law of conservation of mass (1789) — matter is neither created nor destroyed during chemical change — refined by John Dalton's atomic theory, which interprets that conservation at the particle level: atoms are simply rearranged between reactant and product formulae, never appearing or disappearing. This lesson develops four headline calculations that recur on every Paper 1 and Paper 2: (1) reacting-mass calculations using mole ratios, (2) limiting-reagent identification, (3) percentage yield, and (4) atom economy as a green-chemistry metric.

Spec mapping (AQA 7405): This lesson maps primarily to §3.1.2.5 (balanced equations, stoichiometry, percentage yield and atom economy). It builds on §3.1.2.3 (empirical and molecular formulae — lesson 5 of this course) and feeds directly into §3.1.2.4 (solutions and concentrations — lesson 7) and the Required Practical 1 titration anchored in lesson 8. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Writing balanced symbol equations from word equations, and defining percentage yield and atom economy, are AO1 recall items. Applying mole ratios in numerical calculations and identifying the limiting reagent are AO2 applications and appear on essentially every paper. Multi-step synoptic problems — for example, combining a limiting-reagent identification with a percentage-yield correction, or evaluating two synthetic routes against atom economy and yield together — and justifying the green-chemistry preference of one route over another, are AO3 reasoning items typically worth the highest single-question mark counts.

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Balancing Chemical Equations

A balanced equation has the same number of atoms of each element on both sides, and the same total charge on both sides. The first requirement is conservation of mass; the second is conservation of charge.

The Four-Step Recipe

1. Write the correct formulae for all reactants and products. Never change a formula to make the equation balance — only the coefficients.
2. Balance one element at a time, starting with the most complex species (often the one with the most atoms or the rarest element).
3. Leave hydrogen and oxygen until last, because they typically appear in water and other simple species that adjust easily.
4. Check that every element balances and that state symbols are included: (s), (l), (g), (aq).

Examples of Balanced Equations

Combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Reaction of sodium with water: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Thermal decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g)

Neutralisation of sulfuric acid by sodium hydroxide: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Reaction of iron with dilute sulfuric acid: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)

State Symbols — (l) vs (aq)

Pure liquid water at room temperature is written H₂O(l). Water that is the solvent (or any species dissolved in water) is written (aq) — aqueous. The distinction matters in ionic equations: H₂O(l) appears as the product of neutralisation, while NaCl(aq) signals dissociated Na⁺(aq) and Cl⁻(aq) ions. Examiners often allocate a mark for correct use of state symbols, particularly (l) vs (aq) in neutralisation and precipitation equations.

Balancing Ionic Equations

Many reactions in solution are clearer when written as ionic equations, where spectator ions are cancelled from both sides. From the full molecular equation, dissociate every (aq) species into its constituent ions, then cancel anything that appears unchanged on both sides.

Example — neutralisation: Full: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Dissociated: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l) Cancel Na⁺ and Cl⁻ (spectators): H⁺(aq) + OH⁻(aq) → H₂O(l)

Example — precipitation of silver chloride: Full: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) Ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Ionic equations show what is chemically happening and reveal that any two compounds containing the relevant ions will produce the same reaction.

Balancing Redox Half-Equations

Redox reactions are best balanced by splitting them into oxidation and reduction half-equations, balanced separately for mass and charge, then recombined so that the electrons cancel. The procedure differs slightly for acidic and alkaline media — this is developed in depth in the redox and electrochemistry course; here we set out the bare scaffolding.

Acidic medium — five-step procedure:

1. Balance all atoms other than O and H.
2. Balance O by adding H₂O.
3. Balance H by adding H⁺.
4. Balance charge by adding electrons (e⁻).
5. Multiply the two half-equations so that electrons cancel, then add.

Worked half-equation — reduction of permanganate in acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Alkaline medium: balance as if acidic, then add the same number of OH⁻ to both sides as there are H⁺, converting H⁺ + OH⁻ → H₂O and cancelling spare water. This sequence is fully developed in the later redox and electrochemistry course.

Exam Tip: State symbols are essential in many exam mark schemes. Always include them. (aq) means dissolved in water; (l) means pure liquid. Marks are routinely deducted when (aq) is used for water itself or when (l) is written for a dissolved species.

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Mole Ratios and Reacting Mass Calculations

The coefficients in a balanced equation give the mole ratio of reactants and products. This is the single most important fact in stoichiometry: the equation tells you nothing about masses directly — only about moles. To use it for masses, every quantity must first be converted to moles (using n = m/M, n = V/24.0 at RTP, n = c × V for solutions, or n = pV/RT for non-RTP gases).

Method for Reacting Mass Calculations

1. Write the balanced equation with state symbols.
2. Calculate the moles of the substance you know (using n = m/M, or one of the other mole equations).
3. Use the mole ratio from the equation to find the moles of the substance you want.
4. Convert back to mass (m = n × M) or volume as required.
5. Quote the answer to an appropriate number of significant figures (usually 3 s.f.).

Worked Example 1: Thermal Decomposition (1:1 Ratio)

Calculate the mass of carbon dioxide produced when 5.00 g of calcium carbonate is decomposed completely.

CaCO₃(s) → CaO(s) + CO₂(g)

Step 1: M(CaCO₃) = 40.1 + 12.0 + 3 × 16.0 = 100.1 g mol⁻¹. n(CaCO₃) = 5.00 / 100.1 = 0.04995 mol.

Step 2: From the equation, 1 mol CaCO₃ produces 1 mol CO₂ (ratio 1:1). So n(CO₂) = 0.04995 mol.

Step 3: M(CO₂) = 44.0 g mol⁻¹. m(CO₂) = 0.04995 × 44.0 = 2.20 g (3 s.f.).

Notice how the mole ratio carries the whole calculation. If the ratio had been 2 CaCO₃ : 1 CO₂ (it isn't, but hypothetically), the answer would have halved.

Worked Example 2: Combustion (1:3 Ratio)

Calculate the mass of water produced when 4.60 g of ethanol (C₂H₅OH) undergoes complete combustion.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

M(C₂H₅OH) = 2 × 12.0 + 6 × 1.0 + 16.0 = 46.0 g mol⁻¹. n(C₂H₅OH) = 4.60 / 46.0 = 0.100 mol.

From the equation: 1 mol C₂H₅OH produces 3 mol H₂O.

n(H₂O) = 0.100 × 3 = 0.300 mol.

m(H₂O) = 0.300 × 18.0 = 5.40 g.

Worked Example 3: Neutralisation (Acid-Base, 1:2 Ratio)

Calculate the mass of sodium sulfate formed when 4.90 g of sulfuric acid reacts completely with sodium hydroxide solution.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

M(H₂SO₄) = 2 × 1.0 + 32.1 + 4 × 16.0 = 98.1 g mol⁻¹. n(H₂SO₄) = 4.90 / 98.1 = 0.04995 mol.

From the equation: 1 mol H₂SO₄ produces 1 mol Na₂SO₄.

n(Na₂SO₄) = 0.04995 mol.

M(Na₂SO₄) = 2 × 23.0 + 32.1 + 4 × 16.0 = 142.1 g mol⁻¹.

m(Na₂SO₄) = 0.04995 × 142.1 = 7.10 g (3 s.f.).

Worked Example 4: Displacement Reaction Producing a Gas

Calculate the volume of hydrogen gas (at RTP) produced when 0.460 g of sodium reacts with excess water.

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

n(Na) = 0.460 / 23.0 = 0.0200 mol.

From the equation: 2 mol Na produces 1 mol H₂.

n(H₂) = 0.0200 / 2 = 0.0100 mol.

V(H₂) = 0.0100 × 24.0 = 0.240 dm³ (or 240 cm³ at RTP).

The 2:1 ratio is a classic exam trap. The mass of sodium gives 0.0200 mol of Na, but the question asks for hydrogen, so you must halve not double. Reading the ratio in the wrong direction is the most common mark-loss pattern in 2:1 / 1:2 stoichiometry.

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Limiting Reagent

In most real reactions, one reagent is used up before the others — this is the limiting reagent, and it caps the maximum amount of product. The other reagent(s) are said to be in excess. Identifying the limiting reagent correctly is the difference between getting an exam question right and writing a beautiful calculation based on the wrong reagent.

Method 1: n ÷ Coefficient

1. Calculate the moles of each reactant.
2. Divide each by its coefficient in the balanced equation.
3. The reactant with the smallest value is the limiting reagent.

Worked Example 5: Identifying the Limiting Reagent

5.40 g of aluminium reacts with 16.0 g of iron(III) oxide. Which is the limiting reagent, and what mass of iron is produced?

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

n(Al) = 5.40 / 27.0 = 0.200 mol → 0.200 / 2 = 0.100 n(Fe₂O₃) = 16.0 / 159.7 = 0.1002 mol → 0.1002 / 1 = 0.1002

Al gives the smaller value (0.100), so aluminium is the limiting reagent.

The maximum moles of Fe produced equal n(Al) directly (the Al : Fe ratio is 2 : 2 = 1 : 1): n(Fe) = 0.200 mol; m(Fe) = 0.200 × 55.8 = 11.2 g.

The Fe₂O₃ is in slight excess: 0.1002 − 0.100 = 0.0002 mol of Fe₂O₃ is left unreacted.

Method 2: The ICE-Style Table

A more transparent — and exam-friendly — method is to build a "moles before / change / moles after" table, sometimes called an ICE table (Initial, Change, Equilibrium/End). This is the same structure used later in equilibrium calculations, so practising it here pays dividends in the A2 equilibrium course.

For the same reaction with 0.200 mol Al and 0.1002 mol Fe₂O₃:

	2Al	Fe₂O₃	Al₂O₃	2Fe
Initial	0.200	0.1002	0	0
Change	−0.200	−0.100	+0.100	+0.200
End	0	0.0002	0.100	0.200

Al goes to zero first → Al is limiting. Direct read-off: n(Fe) = 0.200 mol; m(Fe) = 11.2 g. The table also reveals that 0.0002 mol of Fe₂O₃ remains.

Worked Example 6: Limiting Reagent with a Solution

3.00 g of magnesium reacts with 100 cm³ of 1.00 mol dm⁻³ hydrochloric acid. Calculate the volume of hydrogen produced at RTP and identify the limiting reagent.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

n(Mg) = 3.00 / 24.3 = 0.1235 mol → 0.1235 / 1 = 0.1235 n(HCl) = 1.00 × 100/1000 = 0.100 mol → 0.100 / 2 = 0.0500

HCl gives the smaller value, so HCl is the limiting reagent.

n(H₂) = 0.100 / 2 = 0.0500 mol (using the ratio 2 HCl : 1 H₂). V(H₂) = 0.0500 × 24.0 = 1.20 dm³ at RTP.

Worked Example 7: Synoptic — Limiting Reagent Paired with Percentage Yield

15.0 g of methanol is reacted with 30.0 g of ethanoic acid in an esterification: CH₃COOH + CH₃OH → CH₃COOCH₃ + H₂O The reaction proceeds to an actual yield of 18.5 g of methyl ethanoate. Calculate the percentage yield.

Stage 1 — limiting reagent. M(CH₃OH) = 32.0; n(CH₃OH) = 15.0 / 32.0 = 0.469 mol → / 1 = 0.469. M(CH₃COOH) = 60.0; n(CH₃COOH) = 30.0 / 60.0 = 0.500 mol → / 1 = 0.500. Methanol is limiting (0.469 < 0.500).

Stage 2 — theoretical yield. n(methyl ethanoate) = 0.469 mol (1:1 ratio with methanol). M(CH₃COOCH₃) = 74.0; theoretical mass = 0.469 × 74.0 = 34.7 g.

Stage 3 — percentage yield. % yield = (18.5 / 34.7) × 100 = 53.3%.

A low percentage yield is typical for esterifications because the reaction is reversible and reaches equilibrium well before all reagents are consumed (Le Chatelier's principle predicts this; lesson 7 onwards explores it).

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Percentage Yield

The percentage yield compares the actual (experimental) yield to the theoretical (calculated) maximum yield based on the limiting reagent.

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

Percentage yield is always less than or equal to 100%. The four standard reasons examiners credit for yield < 100% are:

• The reaction may be reversible and reach equilibrium before reactants are consumed (this is the dominant reason for esterifications, the Haber process, and the Contact process).
• Side reactions may produce unwanted by-products (e.g. over-oxidation of a primary alcohol to a carboxylic acid when an aldehyde is wanted).
• Product is lost during purification or transfer — filtration losses, residue on glassware, droplets adhering during recrystallisation, evaporation during drying.
• The reaction may be incomplete because insufficient time, heat, or catalyst was supplied — equivalent to stopping the reaction before it reaches its endpoint.

Worked Example 8: Straightforward Percentage Yield

In a thermite reaction, 14.0 g of iron was obtained. The theoretical yield from the masses of Al and Fe₂O₃ used was 16.8 g. Calculate the percentage yield.

% yield = (14.0 / 16.8) × 100 = 83.3%.

Worked Example 9: Rearrangement — Theoretical Yield from Percentage Yield

The reaction of ethanol with ethanoic acid produces ethyl ethanoate. If 0.500 mol of ethanol reacts (ethanoic acid in excess) and the percentage yield is 67.0%, calculate the mass of ethyl ethanoate produced.

CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Theoretical n(ethyl ethanoate) = 0.500 mol (1:1 with ethanol). M(CH₃COOC₂H₅) = 88.0 g mol⁻¹. Theoretical mass = 0.500 × 88.0 = 44.0 g. Actual mass = 44.0 × 67.0 / 100 = 29.5 g (3 s.f.).

Worked Example 10: Reverse — Finding Required Reactant Mass from Yield Target

A chemist needs 5.00 g of aspirin (C₉H₈O₄, M = 180.0). The percentage yield of the salicylic-acid acetylation is typically 70.0%. What mass of salicylic acid (C₇H₆O₃, M = 138.0) is needed? C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

Required actual n(aspirin) = 5.00 / 180.0 = 0.02778 mol. Theoretical n(aspirin) needed = 0.02778 / 0.700 = 0.03968 mol. n(salicylic acid) = 0.03968 mol (1:1 with aspirin). m(salicylic acid) = 0.03968 × 138.0 = 5.48 g.

The principle: scaling up for yield happens at the limiting-reagent stage — divide the required actual product moles by the fractional yield to obtain the theoretical moles, then back-translate to reagent mass.

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Atom Economy

Atom economy measures how efficiently a reaction incorporates reactant atoms into the desired product. It is a pure stoichiometric property of the balanced equation — completely independent of yield, conditions, or scale.

Atom economy = (Mᵣ of desired product / Σ Mᵣ of all products) × 100%

Equivalently (since total reactant mass equals total product mass in a balanced equation):

Atom economy = (Mᵣ of desired product / Σ Mᵣ of reactants) × 100%

Green Chemistry Context

Atom economy was elevated as a design principle by Paul Anastas and John Warner in their 1998 framework, The 12 Principles of Green Chemistry. The framework names atom economy as Principle 2: synthetic methods should be designed to maximise the incorporation of all reagent atoms into the final product. The other principles — including prevention of waste, use of safer solvents, energy efficiency, and design for degradation — collectively reframe synthetic chemistry as a sustainability-aware discipline. Atom economy is the metric most easily calculated at A-Level from a balanced equation, which is why it features in exam questions.

Why Yield and Atom Economy Are Both Needed

Yield tells you what fraction of theoretical product you actually obtained. Atom economy tells you what fraction of reactant atoms ended up in the desired product even if the yield were 100%. A reaction with 100% atom economy but 20% yield wastes most of its reactants. A reaction with 95% yield but 30% atom economy generates a lot of by-product waste even when "successful". Industrial-scale process selection requires both metrics.

Worked Example 11: Atom Economy of the Thermite Reaction

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

Desired product = 2Fe; Mᵣ contribution = 2 × 55.8 = 111.6. Total Mᵣ of all products = Mᵣ(Al₂O₃) + Mᵣ(2Fe) = 102.0 + 111.6 = 213.6.

Atom economy = (111.6 / 213.6) × 100 = 52.2%.

About half of the reactant atom-mass ends up in the desired product (Fe); the rest is "locked up" in the by-product Al₂O₃.

Worked Example 12: Comparing Two Routes to Ethanol

(a) Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (b) Hydration of ethene: C₂H₄ + H₂O → C₂H₅OH

(a) Fermentation: Desired product = 2 × C₂H₅OH, Mᵣ = 2 × 46.0 = 92.0. Total products Mᵣ = 92.0 + 2 × 44.0 = 180.0. Atom economy = (92.0 / 180.0) × 100 = 51.1%.

(b) Hydration of ethene: Desired product = C₂H₅OH, Mᵣ = 46.0. Total products Mᵣ = 46.0 (single product). Atom economy = (46.0 / 46.0) × 100 = 100%.

The hydration route has perfect atom economy because addition reactions, with only one product, are by definition 100% atom-economical. The fermentation route loses about half its mass to CO₂ as a by-product.

Addition vs Substitution vs Condensation vs Elimination