Edexcel A-Level Chemistry: Amounts of Substance and Redox — Complete Revision Guide (9CH0)

Amounts of substance and redox are two topics that look like bookkeeping but are in fact the spine of every quantitative answer in chemistry. Mole calculations underlie titrations, gas volumes, atom economy and percentage yield. Redox arithmetic underlies inorganic chemistry from Group 2 to transition metals, and electrochemistry on Paper 1. Most extended numerical questions on the 9CH0 paper come back to either a moles step, a redox half-equation, or both. If your moles and redox are slow or shaky, every quantitative question takes too long and accumulates small errors. If they are automatic, you save time you can use to think about the harder parts of the paper.

This guide is a topic-by-topic walkthrough of the amounts and redox content in the 9CH0 specification. It covers the mole concept and Avogadro's number; empirical and molecular formulae; balanced equations and reacting masses; concentration, gas volume and titration calculations; atom economy and percentage yield; oxidation numbers and rules; redox half-equations and balancing; and the standard redox titrations involving manganate(VII) and iodine-thiosulfate. For each topic you will find the core skills, common pitfalls, a worked example and a link into the LearningBro Amounts of Substance and Redox course.

What the Edexcel 9CH0 Specification Covers

Edexcel A-Level Chemistry (9CH0) is examined through three written papers. Paper 1 (Inorganic and Physical) and Paper 2 (Organic and Physical) are each one hour 45 minutes for 90 marks. Paper 3 (General and Practical Principles) is two hours 30 minutes for 120 marks and tests practical and synoptic application. Amounts of substance, atom economy, redox and titration calculations appear on every paper — most heavily on Paper 1 (inorganic redox) and Paper 3 (practical titrations).

The table below maps the main sub-topics to typical paper weighting.

Sub-topic	Spec area	Typical exam weight
Mole concept and Avogadro's number	Topic 3	4-6 marks
Empirical and molecular formulae	Topic 3	3-5 marks
Balanced equations and reacting masses	Topic 3	4-6 marks
Concentration and titration calculations	Topic 3	6-10 marks
Gas volume calculations (Vm and ideal gas)	Topic 3	4-6 marks
Atom economy and percentage yield	Topic 3	3-5 marks
Oxidation numbers	Topic 5	3-5 marks
Redox half-equations and balancing	Topic 5	4-6 marks
Redox titrations (KMnO4, iodine-thiosulfate)	Topic 5	6-10 marks

These weights are estimates, modelled on the Edexcel 9CH0 paper format rather than guarantees for a single sitting. What is reliable is that a multi-step titration calculation — typically eight to ten marks — appears on essentially every Paper 3.

The Mole Concept and Avogadro's Number

A mole is the amount of substance containing as many particles as there are atoms in 12 g of carbon-12. The number of particles in one mole is Avogadro's constant, NA = 6.022 × 10^23 mol^-1. The molar mass of a substance is its relative atomic or formula mass in grams per mole. The three core formulae you must use without thinking are:

• n = m / M (moles = mass / molar mass)
• n = c × V (moles = concentration in mol dm^-3 × volume in dm^3)
• n = V / 24 (moles of gas at room temperature and pressure, V in dm^3)

The ideal gas equation, pV = nRT, is the more general version of the third (with p in Pa, V in m^3, T in K, R = 8.314 J K^-1 mol^-1). It appears on Paper 1 and Paper 3 whenever conditions are non-standard.

Worked example. Calculate the number of molecules in 4.4 g of CO2. n = 4.4 / 44 = 0.10 mol. Number of molecules = 0.10 × 6.022 × 10^23 = 6.022 × 10^22.

A common pitfall is mixing units of volume (cm^3 versus dm^3) — always convert to dm^3 before substituting into n = c × V. Another is using V/24 outside RTP conditions; if temperature or pressure is not 298 K and 100 kPa, switch to pV = nRT. See the mole concept lesson for unit drills.

Empirical and Molecular Formulae

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

The standard workflow for finding an empirical formula from percentage composition is: assume 100 g of substance, divide each element's mass by its atomic mass to get moles, divide all the moles by the smallest, multiply up if needed to get whole numbers.

Worked example. A compound is 40.0 percent C, 6.7 percent H and 53.3 percent O by mass and has Mr 60. Find its empirical and molecular formulae. Moles of C = 40.0 / 12 = 3.33; moles of H = 6.7 / 1 = 6.7; moles of O = 53.3 / 16 = 3.33. Divide through by 3.33: C = 1, H = 2, O = 1. Empirical formula is CH2O (Mr = 30). Molecular formula is twice this: C2H4O2 (Mr = 60). The compound is ethanoic acid.

A common pitfall is rounding errors that mask a 1.5 ratio (which means multiply through by 2 to clear the half) — keep at least three decimal places before dividing. See the formulae lesson for combustion-analysis problems.

Balanced Equations and Reacting Masses

Every quantitative problem starts with a balanced equation. Mass is conserved: the total mass of reactants equals the total mass of products, and the number of atoms of each element on each side must match. Balance by inspection: count each element on each side and adjust coefficients in the order metals, non-metals, oxygen, hydrogen.

Once balanced, the stoichiometry (the coefficients) tells you the mole ratio in which species react. Reacting-mass problems are then solved in three steps: (1) convert the known mass to moles using n = m/M, (2) use the mole ratio from the equation to find the moles of the unknown, (3) convert moles back to mass.

Worked example. What mass of magnesium oxide is produced when 6.0 g of magnesium burns completely in oxygen? Equation: 2Mg + O2 → 2MgO. Moles of Mg = 6.0 / 24 = 0.25 mol. Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol. Mass of MgO = 0.25 × 40 = 10 g. (The extra 4 g comes from oxygen.)

A common pitfall is forgetting the 1:1 ratio and writing 0.25 × 40 / 2 by mistake. Another is using molecular masses for ionic compounds — Mg^2+ and O^2- give MgO with Mr 40, not Mg2O2. See the reacting masses lesson for limiting-reagent practice.

Concentration and Titration Calculations

The concentration of a solution in mol dm^-3 is the number of moles of solute per cubic decimetre of solution. To convert mass concentration in g dm^-3 to mol dm^-3, divide by the molar mass: c (mol dm^-3) = c (g dm^-3) / M.

A titration uses a solution of known concentration (the standard solution) to find the concentration of an unknown solution. The four-step workflow is: (1) write the balanced equation for the reaction, (2) calculate moles of the standard from c × V, (3) use the equation's mole ratio to find moles of the unknown, (4) calculate the unknown's concentration from c = n / V.

Worked example. 25.0 cm^3 of HCl is neutralised by 28.5 cm^3 of 0.100 mol dm^-3 NaOH. Calculate the concentration of the HCl. Equation: HCl + NaOH → NaCl + H2O. Moles of NaOH = 0.100 × (28.5 / 1000) = 2.85 × 10^-3 mol. Mole ratio HCl : NaOH = 1 : 1, so moles of HCl = 2.85 × 10^-3 mol. Concentration of HCl = (2.85 × 10^-3) / (25.0 / 1000) = 0.114 mol dm^-3.

A common pitfall is forgetting to convert cm^3 to dm^3, leading to answers a thousand times too small or too large. Another is using the wrong mole ratio when the equation is not 1 : 1 — for example, when titrating sulfuric acid against sodium hydroxide (1 : 2). The titration practical procedure (rinsing burette and pipette correctly, recognising end point colour change, recording titres to the nearest 0.05 cm^3) is also examined directly. See the titration calculations lesson for full multi-step problems.

Gas Volume Calculations

For gases at room temperature and pressure (RTP), the molar volume is 24 dm^3 mol^-1 (or 24 000 cm^3 mol^-1). At any other conditions, use the ideal gas equation pV = nRT with SI units (p in pascals, V in cubic metres, T in kelvin, R = 8.314 J K^-1 mol^-1).

Worked example. Calculate the volume of CO2 produced at 25 °C and 100 kPa when 5.0 g of calcium carbonate is fully decomposed. Equation: CaCO3 → CaO + CO2. Moles of CaCO3 = 5.0 / 100 = 0.05 mol. Moles of CO2 = 0.05 mol. Volume at RTP = 0.05 × 24 = 1.2 dm^3. (Or by ideal gas: V = nRT/p = 0.05 × 8.314 × 298 / 100000 = 1.24 × 10^-3 m^3 = 1.24 dm^3, consistent within rounding.)

A common pitfall is to use 22.4 dm^3 mol^-1 (the older STP value used at 0 °C) instead of 24 dm^3 mol^-1 at RTP. Another is forgetting to convert °C to K (add 273) before using pV = nRT. See the gas volumes lesson for ideal gas drill problems.

Atom Economy and Percentage Yield

Percentage yield measures how much product you actually got compared with the theoretical maximum: percentage yield = (actual moles / theoretical moles) × 100, or equivalently in masses provided the molar masses cancel. Yields are below 100 percent for many reasons — incomplete reaction, side reactions, losses during isolation and purification, equilibrium positions short of complete conversion.

Atom economy measures how much of the reactants' mass ends up in the desired product, not in by-products: atom economy = (Mr of desired product / sum of Mr of all products) × 100, weighted by stoichiometry. A reaction with high atom economy wastes less mass and is generally more efficient and greener. Addition reactions (one product) have 100 percent atom economy; substitution and elimination reactions have lower values.

Worked example. Hydrogen for industrial use can be made from methane by either steam reforming (CH4 + H2O → CO + 3H2) or partial oxidation (2CH4 + O2 → 2CO + 4H2). Calculate the atom economy for hydrogen in steam reforming. Mass of H2 produced = 3 × 2 = 6. Total mass of products = 28 (CO) + 6 (3H2) = 34. Atom economy = 6/34 × 100 = 17.6 percent. Most of the mass leaves as CO, so steam reforming has a low atom economy for hydrogen — though CO is itself a useful product.

A common pitfall is to confuse atom economy with percentage yield — they measure different things. Yield is about how much you got versus how much you could have got; atom economy is about how much of the mass is in the right product. See the yield and atom economy lesson.

Oxidation Numbers

The oxidation number (or oxidation state) is a bookkeeping number assigned to each atom that tells you how many electrons it has gained or lost relative to its neutral state. The rules are short and exact.

Rule	Notes
Element in elemental form	0
Simple ion	Equal to the ion's charge
Group 1 in compounds	+1
Group 2 in compounds	+2
Hydrogen	+1 (except metal hydrides, -1)
Oxygen	-2 (except peroxides -1; in OF2 +2)
Fluorine	Always -1 in compounds
Sum in a neutral compound	0
Sum in a polyatomic ion	Equals the ion's charge

Worked example. Find the oxidation number of Mn in MnO4-. Sum of oxidation numbers must equal -1 (the ion's charge). Each O is -2, so 4 × (-2) = -8. Mn + (-8) = -1, so Mn = +7.

A common pitfall is forgetting that Group 1 and Group 2 metals always show their group oxidation state in compounds — this constraint helps you find unknown values quickly. See the oxidation states lesson for sample compounds and ions.

Redox Half-Equations

A redox reaction can be split into two half-equations: one oxidation (electrons on the right) and one reduction (electrons on the left). To balance a half-equation, balance atoms other than O and H first, then balance O by adding H2O, then balance H by adding H+, then balance charge by adding electrons. To combine two half-equations into a full equation, scale them so the electrons cancel and add.

Worked example. Combine the half-equations MnO4- + 8H+ + 5e- → Mn^2+ + 4H2O and Fe^2+ → Fe^3+ + e- to give the full ionic equation. The first needs five electrons; the second produces one. Multiply the second by 5 and add: MnO4- + 8H+ + 5Fe^2+ → Mn^2+ + 4H2O + 5Fe^3+. The 5e- cancel.

A common pitfall is forgetting to balance charge — many candidates balance atoms but not electrons. Another is forgetting the H2O and H+ on the reactant side when balancing oxygen-containing oxidising agents. See the half-equations lesson for permanganate and dichromate practice.

Manganate(VII) and Iodine-Thiosulfate Titrations

Acidified potassium manganate(VII) (KMnO4) is the standard oxidising agent for redox titrations on the specification. It oxidises Fe^2+ to Fe^3+, oxalate (C2O4^2-) to CO2, and ethanedioate to CO2 among other species. The end point is detected without an indicator: as long as Fe^2+ remains, MnO4- is decolourised to colourless Mn^2+; the moment all Fe^2+ has been oxidised, the next drop of MnO4- gives a permanent pale pink colour.

Worked example. 25.0 cm^3 of an acidified Fe^2+ solution requires 24.0 cm^3 of 0.0200 mol dm^-3 KMnO4 for complete oxidation. Calculate the concentration of Fe^2+. Moles of MnO4- = 0.0200 × (24.0 / 1000) = 4.80 × 10^-4 mol. Mole ratio Fe^2+ : MnO4- = 5 : 1, so moles of Fe^2+ = 5 × 4.80 × 10^-4 = 2.40 × 10^-3 mol. Concentration = 2.40 × 10^-3 / (25.0 / 1000) = 0.0960 mol dm^-3.

The iodine-thiosulfate titration is used to determine the amount of an oxidising agent that liberates iodine from KI. The half-equation 2S2O3^2- + I2 → S4O6^2- + 2I- gives the 2:1 mole ratio of thiosulfate to iodine. Starch indicator is added near the end point (when the brown iodine colour fades to pale yellow) to give a sharp blue-black to colourless transition.

A common pitfall is using the wrong mole ratio — Fe^2+ : MnO4- is 5 : 1, not 1 : 1. Another is adding starch too early in the iodine-thiosulfate titration, where it adsorbs iodine into the starch helix and gives a poor end point. See the redox titrations lesson for worked KMnO4 and iodine-thiosulfate problems.

Common Mark-Loss Patterns

• Forgetting to convert cm^3 to dm^3 in concentration calculations.
• Using V/24 outside RTP — switch to pV = nRT for non-standard conditions.
• Using the wrong mole ratio in titrations, especially Fe^2+ : MnO4- (5:1) and S2O3^2- : I2 (2:1).
• Confusing percentage yield with atom economy.
• Mis-balancing redox half-equations by neglecting charge.
• Treating oxygen as -2 in peroxides (it is -1) or in OF2 (it is +2).
• Forgetting state symbols in equations involving aqueous and gaseous species.
• Reading a burette to the nearest 0.1 cm^3 instead of 0.05 cm^3.
• Failing to discard the rough titre and average only the concordant titres.
• Misidentifying end point colours, particularly in indicator-free KMnO4 titrations.

How to Revise This Topic

• Drill mole conversions daily. Three to five short calculations a day for a fortnight will make n = m/M, n = cV and pV = nRT automatic.
• Practise titration questions in stages. Spend a session on mole ratio extraction, a session on concentration arithmetic, and a session on writing up working with units. Then combine.
• Build a redox flashcard set of the standard half-equations and full ionic equations: MnO4- / Fe^2+, Cr2O7^2- / Fe^2+, I2 / S2O3^2-, Cu^2+ / I-.
• Walk through a full practical write-up for at least one core titration: rinsing technique, recording, end-point recognition, calculation. Examiners reward this fluency.
• Always check units in the final line of working. A correct number with the wrong units loses the mark.
• Use the LearningBro practice quizzes to test under timed conditions.

Linking to Other Topics

Amounts and redox is the most heavily used skill set on the specification. Every energetics calculation begins with moles. Inorganic chemistry — Group 2, Group 7, transition metals — uses redox half-equations to write displacement and disproportionation reactions. Kinetics and equilibrium builds rate equations from concentrations measured by titration. Acids and buffers extend the titration framework with logarithmic pH calculations. Even organic synthesis questions assume you can convert mass to moles to predict yields. Time spent on this topic returns many times over.

Final Word

Amounts and redox is not glamorous, but it is the topic that earns marks on every paper. Build automaticity with the three core mole formulae, drill titration arithmetic until it is unconscious, and learn the standard redox half-equations cold. The full LearningBro Amounts of Substance and Redox course walks through every sub-topic with worked examples, practice questions and AI tutor feedback. Get this section fluent and the rest of the paper looks much shorter.