Formulae and calculations (part 2)
Formulae and calculations (part 2) — the chapter map
SB p.64Part 2 of the formulae-and-calculations chapter extends the mole toolkit you built in Chapter 5 from masses to gases and solutions. This material stands squarely on part 1: everything here is the same "calculate the moles, use the equation, convert to what's asked" three-stage method, applied to two new ways of quoting an amount. By the end you will learn to calculate what volume of gas a reaction produces, and how to work with solution concentrations quoted in mol/dm³.
The two objectives of record for this lesson are both Chemistry-only specification statements — they are assessed in Paper 2 only and are not shared with Science Double Award:
- calculations involving amount of substance, volume and concentration (in mol/dm³) of solution (spec 1.34C);
- calculations involving gas volumes and the molar volume of a gas (24 dm³ and 24 000 cm³ at room temperature and pressure, rtp) (spec 1.35C).
The chapter falls into two clean halves: gas volumes first (Avogadro's law and the molar volume), then solutions (concentration units, the four worked examples, and titration calculations). Master both halves and every gas-and-solution calculation on the course becomes routine.
You need the part-1 toolkit before starting — the chapter is built on it. Concretely, from Chapter 5 (u1:c05): the mole as the unit for amount of substance (spec 1.27); moles from mass via mass ÷ mass of 1 mol (spec 1.28); calculating Mᵣ from Aᵣ values (spec 1.26); the big-numbers-in-front-of-formulae mole ratios from a balanced equation; and the three-stage method for reacting-mass problems (spec 1.29). If any of those feel shaky, revisit the relevant sections of Lesson u1:c05 before going further — every calculation here leans on them.
Both of this chapter's spec points carry the C suffix (1.34C, 1.35C), which in Edexcel IGCSE Chemistry means: examined in Paper 2 (2C) only, and never in Science Double Award. Practically: gas-volume and concentration calculations are Paper-2 bankable marks — the standard titration calculation in particular appears nearly every series, and it is the same three-stage method every time.
"Part 2 is a new topic, so it needs new methods." It doesn't — it is the same mole ladder with two new rungs. Mass, gas volume and concentration are three different currencies for "amount of substance"; the exchange rates are Mᵣ (for mass), the molar volume 24 dm³ (for gases at rtp) and 1 dm³ of solution (for concentration). Learners who treat gases and solutions as separate topics memorise twice; learners who see them as conversions solve unfamiliar problems.
Question-bank join: spec 1.34C and 1.35C items — the two-objective map, the Paper-2-only route badge, and the three-stage method carried over from part 1.
Moles part 2: the same ladder now reaches into solutions and gases. One mole of any gas occupies 24 dm3 at room temperature and pressure; in solution, concentration counts moles per dm3 - moles = concentration x volume. Excess and limiting decide which reactant runs out first, and the amount of product follows the limiting reactant, never the excess.
Examiner note - excess is not the answer
Excess means more than enough of one reactant, so the other reactant runs out first and caps the product. The examiner phrase is 'in excess' - never write 'it all reacts'. Calculations that hand you both masses are testing exactly this: find the limiting reactant by moles, then compute the product from that one.
Step throughblock 3
Dissolve 20.0 g of NaOH (Mr = 40) into water and make the solution up to 250 cm3. What is the concentration?
1•Moles of NaOH dissolved (n = m / Mr)
(mol)2•The same volume expressed in dm3 (1 dm3 = 1000 cm3)
3•Concentration (c = n / V)
Page-true figure region in the book pane — Worked example: reading a concentration calculation off the page (EXAMPLE 1).
Simulationblock 5
Molarity
Dissolve solute, watch the concentration read-out, and see how moles and solution volume trade off against c = n / V.
Interactive simulation — loads from PhET (University of Colorado Boulder) when you start it.
Fill the tableblock 6
Titration vocabulary - one short phrase per cell.
| Piece of apparatus or term | What it is / carries |
|---|---|
| Burette | |
| Titre | |
| End point |
Check yourselfblock 7
Watchblock 8
Concentration Calculations (grams per dm³) (2027/28 exams)
Concentration calculations — moles per dm³, the mole ladder's top rung.
Watch on YouTubeVideo © Cognito (cognito.org) — embedded via YouTube
Watchblock 9
GCSE Chemistry - Gas Calculations - Volume & Moles | Mass, Moles & Mr (2027/28 exams)
Gas volume calculations — the 24 dm³ molar volume shortcut done properly.
Watch on YouTubeVideo © Cognito (cognito.org) — embedded via YouTube
Calculations involving gas volumes
SB p.64The gas-volume half of the chapter opens with one of the most striking ideas in the course: picture three identical flasks holding different gases at the same temperature and pressure — all three contain equal numbers of molecules. Not similar numbers: equal. The identity of the gas and the size of its molecules simply do not matter.
That single figure is the whole foundation for this half-chapter. If equal volumes hold equal numbers of molecules, then volumes of gases tell you directly about numbers of molecules — and since moles are just counted molecules (the Avogadro constant of them per mole), volumes of gases tell you about moles with no Mᵣ and no weighing step at all. The next section names this statement properly (Avogadro's law) and starts using it on real equations.
When a question gives you volumes of gases and a balanced equation, the volumes are already "in moles" at the same ratio — do not convert them to masses first. The next section's methane example walks the full reasoning: 100 cm³ of methane needs exactly 200 cm³ of oxygen and yields exactly 100 cm³ of carbon dioxide, because the big numbers say 1 : 2 : 1.
"Flaskfuls of chlorine must contain more molecules than flaskfuls of hydrogen — chlorine molecules are bigger and heavier." Size and mass are properties of each molecule, not of the space between molecules. In a gas the molecules occupy a negligible fraction of the container; the volume is almost entirely empty space, and at the same temperature and pressure every gas spaces its molecules identically. Equal volumes → equal molecule counts, every time.
Question-bank join: spec 1.35C items — equal volumes at the same temperature and pressure containing equal numbers of molecules, and reading reacting volumes straight from a balanced equation.
Avogadro's law
SB p.64The law fits in one line, and the line is worth memorising verbatim: avogadros-law — equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So 100 cm³ of hydrogen holds exactly as many molecules as 100 cm³ of Cl₂ or any other gas under the same conditions, irrespective of the size of the molecules. One refinement: for helium and the other noble gases, say "atoms" rather than "molecules", because they exist as single atoms.
The law becomes a calculating tool the moment you apply it to a balanced equation. Methane burns:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
The big numbers say 1 methane molecule reacts with 2 oxygen molecules to make 1 carbon dioxide molecule (and 2 water molecules — which, as steam, would also count as a gas volume; as written here water is the liquid). So gas volumes react and form in the same ratio as the big numbers: react 100 cm³ of methane and you need exactly 200 cm³ of oxygen, and exactly 100 cm³ of carbon dioxide gas forms. No moles arithmetic needed at this stage — the volume ratio is the mole ratio.
Two flags: the vocabulary and the single most important consequence. The cubic decimetre is the preferred unit (not the litre) when you mean 1000 cm³. And from the law it follows that 1 mole of any gas occupies 24 dm³ (24 000 cm³) at rtp — rtp meaning room temperature and pressure, usually taken as 20–25 °C and 1 atmosphere. That consequence is the molar volume, and the next-but-one section turns it into a calculating tool.
Compressed: for helium and the other noble gases we say atoms instead of molecules; the cubic decimetre (dm³) is the preferred unit for 1000 cm³ — not the litre; and the most important consequence of Avogadro's law is that 1 mole of any gas occupies 24 dm³ (24 000 cm³) at rtp.
"State the ratio in which the gases react" — read it straight off the big numbers and quote the volumes in the same ratio, as in the methane example: 100 cm³ of CH₄ needs 200 cm³ of O₂ and forms 100 cm³ of CO₂. Always name the condition — the ratio holds at the same temperature and pressure — and count only the substances that are actually gases in that equation (the state symbols tell you which: here H₂O is (l), so it contributes no gas volume).
"100 cm³ of methane plus 200 cm³ of oxygen must give 300 cm³ of products." Mass is conserved, but gas volume is not — volume depends on how many molecules of gas you end up with, not on total mass. In the methane reaction the oxygen's volume "disappears" because the water product is liquid: only 100 cm³ of CO₂(g) remains. Count gas-phase molecules from the equation; never add volumes like masses.
Question-bank join: spec 1.35C items — stating Avogadro's law, predicting reacting/formed gas volumes from big numbers at the same temperature and pressure, and the 24 dm³ per mole consequence.
- Avogadro's law
- — Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, irrespective of the size of the molecules.
- room temperature and pressure
- — The conditions at which gas-volume calculations are quoted in this course: about 20-25 °C and 1 atmosphere pressure, abbreviated rtp.
Units of volume — cm³ and dm³
SB p.65Volumes of gases and of solutions are measured in cubic centimetres (cm³) or cubic decimetres (dm³) — you may also meet the litre, but the last section already told you the house preference: 1 litre = 1 dm³ = 1000 cm³, and you should quote cubic decimetres rather than litres. Everything in this chapter runs on that one conversion, so it is worth being automatic about it.
Visualise the conversion as a ladder. Because 1 dm³ is equivalent to 1000 cm³, converting dm³ to cm³ means going from 1 to 1000, so you multiply by 1000; converting cm³ to dm³ means going from 1000 down to 1, so you divide by 1000. The most important thing you need to be able to do here is exactly this conversion — the molar volume comes in two sizes (24 dm³ and 24 000 cm³) and the concentration unit is built on the dm³, so every calculation in this chapter will ask you to pick the right one.
1 dm³ = 1000 cm³ = 1 litre. To convert dm³ → cm³, multiply by 1000; to convert cm³ → dm³, divide by 1000.
The lost-mark scan in every gas-volume or concentration answer: which unit is my volume in, and which unit does my constant want? Molar volume 24 dm³ pairs with volumes in dm³; 24 000 cm³ pairs with volumes in cm³. Mixing them throws answers out by a factor of 1000 — look carefully: is the volume given in cm³ or dm³? That error is the most common one in the paper.
"dm³ means cubic metres." It doesn't — the d stands for deci-, a tenth: a decimetre is one tenth of a metre, so a cubic decimetre is (0.1 m)³ = one thousandth of a cubic metre. The cubic metre (m³) is 1000 dm³ — a thousand times bigger again. Confusing dm³ with m³ is a factor-of-1000 error in the opposite direction from the cm³ slip.
The conversion is a unit-prefix ladder, the same skill as mm → cm → m in physics: each step of 10³. Write the ladder once (cm³ —×1000→ dm³ —×1000→ m³) and read off the direction for any problem. Exam-style: convert 350 cm³ to dm³ (350 ÷ 1000 = 0.350 dm³); convert 2.4 dm³ to cm³ (2.4 × 1000 = 2400 cm³). Quote the unit in the answer — an unlabelled number is a lost mark in every calculation question of this chapter.
Question-bank join: spec 1.34C and 1.35C items — the cm³/dm³ conversions that underpin molar-volume and concentration calculations.
- cubic decimetre
- — The volume unit 1 dm³ = 1000 cm³ = 1 litre. Gas molar volumes and solution concentrations are both quoted per dm³; convert cm³ to dm³ by dividing by 1000.
The molar volume — one mole of any gas occupies 24 dm³
SB p.65Avogadro's law has one consequence that is the most important in the whole chapter: since 1 mole of any gas contains the same number of molecules as 1 mole of any other gas, it must occupy the same volume. At room temperature and pressure (rtp — usually 20–25 °C and 1 atmosphere) that volume is approximately 24 dm³, or 24 000 cm³. This is the molar-volume. It belongs to the conditions, not to the gas: 1 mole of hydrogen, 1 mole of chlorine, 1 mole of carbon dioxide — 24 dm³ each at rtp. (Extension: strictly the molar volume has units of dm³/mol — a unit you are unlikely to meet at International GCSE.)
Two triangles convert between volume and moles the same way the Chapter-5 mass triangle does:
- volume = number of moles × molar volume
- number of moles = volume of gas ÷ molar volume
Worked straight through: the volume of 0.20 mol of CO₂ at rtp is 0.20 × 24 = 4.8 dm³ (using 24 because the answer is wanted in dm³). Going the other way, 0.01 g of hydrogen (Aᵣ: H = 1) is 0.01 ÷ 2 = 0.005 mol, which occupies 0.005 × 24 000 = 120 cm³ — using 24 000 because the question asks for cm³. And 120 cm³ of carbon dioxide is 120 ÷ 24 000 = 0.005 mol. Same triangle every time; the only decision is which size of the molar volume to substitute.
Now run the full three-stage method — calculate the moles of what you can, use the equation to deduce the moles of what you want, convert to the quantity asked — on a gas-volume version of the Chapter-5 reacting-mass problem: what volume of CO₂ forms when excess dilute hydrochloric acid is added to 1.00 g of calcium carbonate? (Aᵣ: C = 12, O = 16, Ca = 40; molar volume = 24 dm³ at rtp.) The equation is CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). The acid is in excess — more than enough to react with all the calcium carbonate — so the CaCO₃ sets the yield. Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 100, so 1.00 g is 1.00 ÷ 100 = 0.0100 mol. The equation says 1 mol CaCO₃ produces 1 mol CO₂, so 0.0100 mol of CO₂ forms; converting with the molar volume in dm³ gives 0.0100 × 24 = 0.24 dm³. The closing warning is the exam-hint below.
1 mole of any gas occupies 24 dm³ (24 000 cm³) at rtp — the same for every gas. The unit law: volume given in cm³ → use 24 000 cm³; volume given in dm³ → use 24 dm³.
"It is really important to include the units here because the examiner will not know whether you mean cm³ or dm³ unless you write it down" — the whole point for the 0.0100 × 24 = 0.24 dm³ answer. Quote the unit with every intermediate and final volume, and match the constant to the unit before you multiply, not after.
"24 dm³ is a property of the gas — hydrogen occupies a different molar volume from chlorine." At the same temperature and pressure, no: the molar volume is fixed by Avogadro's law, not by the gas's identity. What does change with conditions is the value itself — 24 dm³ is an rtp number; at STP (0 °C, 1 atm) the molar volume is different. Read the conditions in the question and use the value the question pins (every 4CH1 question at rtp gives or implies 24 dm³ / 24 000 cm³).
Question-bank join: spec 1.35C items — moles ↔ volume conversions via 24 dm³ / 24 000 cm³, and three-stage gas-volume problems with an excess reagent.
- molar volume
- — The volume occupied by 1 mole of any gas: 24 dm³ (24 000 cm³) at rtp. The same for every gas, because equal moles hold equal numbers of molecules.
- room temperature and pressure
- — The conditions at which gas-volume calculations are quoted in this course: about 20-25 °C and 1 atmosphere pressure, abbreviated rtp.
- excess
- — More than enough of a reactant to react with all of the other reactant; some is left over. 'Added to excess' means the other substance reacts completely.
- molarity
- — The symbol M for concentration in mol/dm³: '2 M' hydrochloric acid means 2 mol/dm³, read as '2 molar'.
Worked problem — making hydrogen from aluminium
SB p.67The chapter's first full gas-volume problem runs the three stages in the direction that feels backwards: from a volume of gas wanted to a mass of solid needed. Aluminium reacts with dilute hydrochloric acid:
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
What mass of aluminium would you need to add to an excess of dilute hydrochloric acid so that you produced 100 cm³ of hydrogen at room temperature and pressure? (Aᵣ of Al = 27; molar volume = 24 000 cm³ at rtp.)
Stage 1 — moles of what we can. We can only calculate the moles of the hydrogen; the mass of aluminium is the unknown. The volume is in cm³, so use the molar volume as 24 000 cm³: number of moles of H₂ = 100 ÷ 24 000 = 0.00417 mol, written in standard form as 4.17 × 10⁻³ mol.
Stage 2 — equation ratio. The big numbers say 3 mol of H₂ are produced from 2 mol of Al. So the moles of Al are two-thirds of the moles of H₂: 0.00417 × 2/3 = 0.00278 mol of aluminium.
Stage 3 — convert to the asked quantity. mass = number of moles × mass of 1 mol = 0.00278 × 27 = 0.075 g of Al.
Two boxes below earn their place as examiner advice, and they target the two classic slips: mangling the 3-to-2 ratio, and then double-counting the 2 in the final mass step.
The ratio step: to go from 3 (H₂) to 2 (Al), divide the 3 by 3 to get 1 and multiply by 2 to get 2 — in other words, multiply the moles of H₂ by 2/3. The same divide-then-multiply routine works for any ratio the equation hands you, forwards or backwards.
"Moles of Al = 0.00417 × 54, because the equation has 2Al and Aᵣ of Al is 27." That double-counts the 2 — the correction: multiply by 27, not 54. The 2 in front of the Al was already used in the ratio step (2 mol Al per 3 mol H₂). The mass of 0.00278 mol of Al is fixed at 0.00278 × 27: the mass of a given number of moles cannot depend on what the aluminium is reacting with.
Maths — standard form and ratios. The intermediate answer 0.00417 mol is written in standard form, 4.17 × 10⁻³ mol: small mole counts in gas problems regularly demand confident standard form, and the "divide by 3, multiply by 2" ratio move is the same proportion skill as map scales and recipe scaling. Both are Paper-2 quantitative skills shared with the maths qualification.
Question-bank join: spec 1.35C items — mass-to-volume and volume-to-mass problems through an equation ratio at rtp.
- excess
- — More than enough of a reactant to react with all of the other reactant; some is left over. 'Added to excess' means the other substance reacts completely.
Working with solution concentrations
SB p.67The solutions half of the chapter begins with the two concentration units. Concentrations can be measured in g/dm³ or in mol/dm³ — also written g dm⁻³ and mol dm⁻³, and read as "grams per cubic decimetre" and "moles per cubic decimetre". Both units ride on the cubic-decimetre from the last unit section: remember that 1 dm³ is the same as 1 litre. (You may also meet the symbol molarity — a concentration quoted as "2 M" means 2 mol/dm³.)
The core skill is converting between g/dm³ and mol/dm³, and the point is reassuring: this is no different from converting moles into grams and vice versa — multiply or divide by the Mᵣ. It makes no difference that the substance is dissolved in 1 dm³ of solution; the arithmetic is the Chapter-5 mass↔moles ladder wearing a per-dm³ hat.
The worked case: sea water with a sodium chloride concentration of 35.1 g/dm³ — find the concentration in mol/dm³ (Aᵣ: Na = 23, Cl = 35.5). 1 mol of NaCl has a mass of 58.5 g. Every dm³ of solution holds 35.1 g, so the moles in every dm³ are 35.1 ÷ 58.5 = 0.600 mol. The concentration is therefore 0.600 mol/dm³ — sea water holds about 0.6 moles of NaCl per cubic decimetre.
Two concentration units, one bridge: g/dm³ → mol/dm³ is ÷ Mᵣ; mol/dm³ → g/dm³ is × Mᵣ (per dm³ of solution, exactly as mass ↔ moles in Chapter 5).
Quote concentration answers with the unit the question asked for — a bare "0.600" scores nothing when the question said mol/dm³. And when a question hands you a concentration in one unit and the equation mole-ratio arithmetic needs moles, convert to mol/dm³ first: mole ratios only ever apply to moles, never to grams.
"Concentration in g/dm³ tells you how many moles you have." Not until you divide by Mᵣ — 35.1 g/dm³ of NaCl and 35.1 g/dm³ of NaHCO₃ are different mole counts, because the formula units have different masses. The g/dm³ number is a mass in every dm³; only ÷ Mᵣ turns it into an amount of substance.
The two-unit ladder is a proportional-reasoning chain: g/dm³ —(÷ Mᵣ)→ mol/dm³, and dm³ of solution —(× concentration)→ moles. Exam-style with those numbers: 35.1 g/dm³ of NaCl (Mᵣ 58.5) = 0.600 mol/dm³; then 2.0 dm³ of that sea water holds 2.0 × 0.600 = 1.2 mol of NaCl. Three conversions, one skill: keep the units attached to every number and let them cancel like fractions.
Question-bank join: spec 1.34C items — the two concentration units, the Mᵣ bridge between them, and concentration → moles for a stated volume.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
- molarity
- — The symbol M for concentration in mol/dm³: '2 M' hydrochloric acid means 2 mol/dm³, read as '2 molar'.
- cubic decimetre
- — The volume unit 1 dm³ = 1000 cm³ = 1 litre. Gas molar volumes and solution concentrations are both quoted per dm³; convert cm³ to dm³ by dividing by 1000.
Worked example 1 — concentration from mol/dm³ to g/dm³
SB p.68Example 1 runs the × Mᵣ direction of the bridge. What is the concentration of a 0.050 mol/dm³ solution of sodium carbonate, Na₂CO₃, in g/dm³? (Aᵣ: C = 12, O = 16, Na = 23.)
Reason per dm³ of solution — the cleanest way to see the units do the work:
- 1 dm³ of solution contains 0.050 mol of Na₂CO₃.
- Mᵣ of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 106, so 1 mol weighs 106 g.
- 0.050 mol weighs 0.050 × 106 = 5.3 g.
- 1 dm³ of solution contains 5.3 g — so the concentration is 5.3 g/dm³.
Notice the structure: the question was per-dm³ in, per-dm³ out, and the only arithmetic was the moles → grams conversion you have used since Chapter 5. Every "convert a mol/dm³ concentration into g/dm³" question in the paper is this four-line shape with different numbers.
Mark-scheme-friendly layout: write the Mᵣ line explicitly ((2 × 23) + 12 + (3 × 16) = 106) — the Mᵣ itself usually carries a mark — then the multiply line with units. The × Mᵣ direction check: mol/dm³ → g/dm³ must make the number bigger (you are counting formula-unit masses); if your g/dm³ answer comes out smaller than your mol/dm³ number, you divided instead.
"0.050 mol/dm³ is a small number, so 0.050 g/dm³ is the answer." The two units are not interchangeable decimals — the Mᵣ bridges them and it is never
- Here 0.050 mol/dm³ becomes 5.3 g/dm³, a hundredfold bigger, because each formula unit has a mass of 106 g. Always × or ÷ by the Mᵣ; never equate the two numbers.
Question-bank join: spec 1.34C items — mol/dm³ → g/dm³ conversions with an explicit Mᵣ step.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
Worked example 2 — concentration in mol/dm³ from grams
SB p.68Example 2 runs the ÷ Mᵣ direction, with one extra twist: the solution volume is not 1 dm³. What is the concentration in mol/dm³ of a solution containing 2.1 g of sodium hydrogencarbonate, NaHCO₃, in 250 cm³ of solution? (Aᵣ: H = 1, C = 12, O = 16, Na = 23.)
- Mᵣ of NaHCO₃ = 23 + 1 + 12 + (3 × 16) = 84, so 1 mol has a mass of 84 g.
- 2.1 g is 2.1 ÷ 84 = 0.025 mol.
- That 0.025 mol sits in 250 cm³ — but concentration wants per 1000 cm³. There are four lots of 250 cm³ in 1000 cm³, so scale up: 4 × 0.025 = 0.10 mol in 1 dm³.
- The concentration is 0.10 mol/dm³.
The solution triangle is the shortcut for steps 3 and 4 — the same triangle the next section turns into the single most-used equation of the topic.
That hint is aimed at the questions you will actually meet: "when doing calculations involving solutions you will most often be asked to calculate a number of moles given the volume of the solution and its concentration", via:
number of moles = volume of solution (dm³) × concentration (mol/dm³)
— and because concentrations come in mol/dm³ while volumes usually come in cm³, you usually have to convert the volume to dm³ by dividing by 1000 first. The HINT's reassurance about unfamiliar formulae is worth keeping too: meet an equation containing, say, CH₃COOH (ethanoic acid) in a later question, and the concentration arithmetic does not care — it only needs the volume, the concentration and the big numbers.
The scaling shortcut generalises: whatever the volume, dividing by it and multiplying by 1000 is the same as "per dm³". 250 cm³ → ×4; 200 cm³ → ×5; 50 cm³ → ×20. But in the exam the safest habit is the triangle's algebra: moles = (volume ÷ 1000) × concentration — write the ÷1000 line visibly, because that is the mark the marker looks for.
"0.025 mol in 250 cm³ means a concentration of 0.025 mol/dm³." That ignores the per-dm³ scale-up: the concentration must state what is in 1000 cm³. The tell is the unit sanity check — a solution made from 2.1 g in a quarter of a dm³ reading "0.025 mol/dm³" would hold just 0.025 mol in a full dm³, i.e. less in more, which contradicts dissolving the same solid. Scale to the dm³ before quoting the unit.
Question-bank join: spec 1.34C items — grams → mol/dm³ with a non-dm³ solution volume, and the moles = volume × concentration law.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
Worked example 3 — moles from volume and concentration
SB p.69Example 3 is that most-used equation in action — the single most reusable line of the solutions half. Calculate the number of moles of NaOH in 50 cm³ of 0.10 mol/dm³ solution.
- Convert the volume: 50 ÷ 1000 = 0.050 dm³.
- Apply the equation: number of moles = volume of solution (dm³) × concentration (mol/dm³).
- Number of moles = 0.050 × 0.10 = 0.0050 mol.
That three-line shape — ÷1000, then × concentration — is the standard opener of every solution calculation from here to the end of the course: titrations (s12–s14), the limescale problem (s11), and later units' acid-base work all start by converting "V cm³ of c mol/dm³" into a mole count this way. Note the trailing zeros (0.050, 0.0050): they are significant figures, and preserving them keeps the working honest about precision.
The two-step dance appears in every mark scheme as two separate marks: M1 for converting the volume (÷1000) or for the correct substitution into moles = V × c, and A1 for the evaluated mole count. Writing 50/1000 × 0.10 in one visible line earns the first mark even if the arithmetic later slips — which is why the layout converts first, then multiplies, as separate displayed steps.
"50 cm³ of 0.10 mol/dm³ contains 5 mol — 50 × 0.10." The volume must be in dm³, and 50 cm³ is 0.050 dm³, not 50 dm³. The wrong answer is a thousandfold too big. Quick sense-check that catches this every time: 1 dm³ of 0.10 mol/dm³ holds 0.10 mol, so 50 cm³ — a twentieth of a dm³ — must hold a twentieth of that, 0.0050 mol.
Question-bank join: spec 1.34C items — moles from volume × concentration with a cm³ → dm³ conversion.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
Calculations with equations involving solutions
SB p.69Now the two halves of the chapter merge: solution arithmetic feeding the three-stage method through a balanced equation. The context is domestic chemistry — limescale (calcium carbonate) removed from kettles by a dilute acid, such as the ethanoic acid in vinegar:
CaCO₃(s) + 2CH₃COOH(aq) → (CH₃COO)₂Ca(aq) + CO₂(g) + H₂O(l)
What mass of calcium carbonate can be removed by 50 cm³ of a solution of ethanoic acid that has a concentration of 2 mol/dm³? (Aᵣ: C = 12, O = 16, Ca = 40.)
Stage 1 — moles of what we can. We know the volume and concentration of the acid: convert the volume to dm³ by dividing by 1000, then number of moles of ethanoic acid = (50 ÷ 1000) × 2 = 0.1 mol.
Stage 2 — equation ratio. The equation has a 2 in front of the CH₃COOH but no number (meaning 1) in front of the CaCO₃, so 2 mol of acid react with 1 mol of CaCO₃ — the moles of CaCO₃ are half the moles of acid: 0.1 ÷ 2 = 0.05 mol of CaCO₃.
Stage 3 — convert to the asked quantity. Mᵣ of CaCO₃ = 100, so mass = 0.05 × 100 = 5 g of calcium carbonate removed.
This is the same skeleton as the hydrogen problem (s06) and the CaCO₃ problem (s05) — only the stage-1 conversion changes, from molar volume to volume × concentration. "Again, we will use our three stages" is deliberate: by now the method, not the chemistry, is the transferable skill.
The ratio extraction deserves its own written line: "2 mol CH₃COOH : 1 mol CaCO₃, so moles CaCO₃ = moles acid ÷ 2". Mark schemes award the ratio line even when the arithmetic fails, and the half-it habit (÷2 when your substance has the bigger coefficient) is safer than cross-multiplying in your head. State the equation's big numbers before using them.
"The acid's coefficient 2 means the acid is twice as concentrated, so use 2 mol/dm³ differently." No — the 2 is already spent the moment the concentration is multiplied in: 0.050 dm³ × 2 mol/dm³ is 0.1 mol of acid, full stop. The coefficient re-enters only in stage 2 as the reacting ratio between the substances. Never fold equation coefficients into the concentration; keep stage 1 (how many moles of this solution) and stage 2 (how the substances relate) cleanly apart.
Question-bank join: spec 1.34C items — three-stage problems where a solution's volume × concentration opens the chain.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
Calculations from titrations
SB p.70Everything so far has been calculation technique; the titration is the experiment that feeds it. A titration finds out how much of one solution reacts with a certain volume of another solution of known concentration. The apparatus: a solution of alkali is measured into a conical flask with a pipette (one accurate, fixed volume), and the acid is run in from a burette (a variable, read-off volume), swirling the flask constantly. Towards the end the acid goes in a drop at a time until the indicator just changes colour — the endpoint. Acid–alkali titration technique itself is developed in Chapters 16 and 17; here it is the numbers that matter.
The logic of the calculation: if you know the concentration of either the acid or the alkali, the titration hands you the other piece of the pair — a volume of solution of known concentration reacting exactly with a volume of the unknown. That is precisely one volume × concentration term and one unknown concentration: the standard three stages with the burette readings as stage-1 data.
Two hints on titration practice are examiner-neutral and worth pinning now, before the standard calculation uses them.
The two hints, in substance: (1) the acid and alkali can be added the other way round, with the alkali in the burette — the concentration calculation does not care which reagent sat where, only which volumes reacted; (2) remember to divide the volume by 1000 to convert to dm³ — burette and pipette readings come in cm³, and every concentration in this family is per dm³.
"The endpoint tells you the acid and alkali are equal." The endpoint tells you the reaction is just complete — the reacting amounts are in the equation's mole ratio, which is 1 : 1 for NaOH + HCl but 2 : 1 for 2NaOH + H₂SO₄. Equal volumes of the two solutions are a coincidence of concentrations, not a law; the mole ratio, never the volume equality, drives stage 2.
Question-bank join: spec 1.34C items — titration apparatus roles and the volume-data logic of the titration calculation.
- titration
- — A technique that measures exactly what volume of one solution reacts with a volume of another of known concentration, so an unknown concentration can be calculated.
- burette
- — The long graduated tube with a tap that runs one titration solution into the other, delivering a variable measured volume (read to 0.05 cm³).
- pipette
- — The calibrated tube that measures one accurate fixed volume of solution (e.g. 25.0 cm³ of alkali) into the conical flask before titrating.
- endpoint
- — The point in a titration at which the indicator just changes colour — the reaction is just complete. Approached by running the burette solution in drop by drop.
The standard titration calculation
SB p.70Here is the shape of a standard titration problem — learn this layout and nearly every titration question in the paper is a re-skin:
25.00 cm³ of 0.100 mol/dm³ sodium hydroxide solution required 23.50 cm³ of dilute hydrochloric acid for neutralisation. Calculate the concentration of the hydrochloric acid.
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
Stage 1 — moles of what we can. We know both the volume and the concentration of the sodium hydroxide: number of moles of NaOH = (25.00 ÷ 1000) × 0.100 = 0.00250 mol.
Stage 2 — equation ratio. The equation says 1 mol of NaOH reacts with 1 mol of HCl, so 0.00250 mol of NaOH reacts with 0.00250 mol of HCl.
Stage 3 — convert to the asked quantity. That 0.00250 mol of HCl is what was in the 23.50 cm³ the titration measured out — so the acid's concentration = moles ÷ volume in dm³ = 0.00250 ÷ (23.50 ÷ 1000) = 0.106 mol/dm³.
The closing line for this example is the checkpoint: we have the volume and number of moles of HCl and can work out the concentration. Stage 3 in a titration is always the concentration rearrangement — moles ÷ (volume ÷ 1000) — rather than a mass conversion, which is the only structural difference from s11's limescale problem.
The titration mark scheme, in this order every time: M1 moles of the known solution (volume ÷ 1000 × concentration); M1 the equation ratio quoted and applied; M1 concentration = moles ÷ volume with the other volume converted to dm³; A1 the value with mol/dm³. Carry at least 3 significant figures through the intermediate steps (0.00250, not 0.003) — rounding before the last line is the classic titration mark leak. Note the data style too: 25.00 cm³ and 23.50 cm³ — burette readings to 2 decimal places, exactly what Chapter 16's technique will demand.
"The concentration of the acid is 0.00250 ÷ 25.00..." — mixing up which volume pairs with which moles. The 0.00250 mol of HCl sits in the acid's 23.50 cm³, not in the alkali's 25.00 cm³; the alkali's data was already used in stage 1. Anchor each intermediate to its substance ("0.00250 mol of HCl in 23.50 cm³ of acid") and the rearrangement cannot pick the wrong volume.
Question-bank join: spec 1.34C items — the standard 1 : 1 titration concentration calculation with cm³ → dm³ conversions.
- titration
- — A technique that measures exactly what volume of one solution reacts with a volume of another of known concentration, so an unknown concentration can be calculated.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
A very slightly harder titration — the 2 : 1 ratio
SB p.71Same apparatus, same stages, one twist: the equation's ratio is not 1 : 1. 25.0 cm³ of sodium hydroxide solution of unknown concentration was titrated with dilute sulfuric acid of concentration 0.050 mol/dm³. 20.0 cm³ of the acid was required to neutralise the alkali. Find the concentration of the sodium hydroxide solution in mol/dm³.
2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
Stage 1 — moles of what we can. This time everything is known about the sulfuric acid: moles of H₂SO₄ = (20.0 ÷ 1000) × 0.050 = 0.0010 mol.
Stage 2 — the ratio is the whole difficulty. The equation proportions aren't 1 : 1 this time — that's what makes the calculation slightly different from the last one. The equation shows 1 mol of sulfuric acid reacts with 2 mol of sodium hydroxide, so moles of NaOH = 2 × 0.0010 = 0.0020 mol.
Stage 3 — concentration. That 0.0020 mol must have been in the 25.0 cm³ (25.0 ÷ 1000 = 0.025 dm³) of sodium hydroxide: concentration = 0.0020 ÷ 0.025 = 0.080 mol/dm³.
The reverse family exists too — instead of calculating a concentration from titration results, you may be asked to calculate what volume of a solution is needed to neutralise something else — and Example 4 in the next section runs that shape.
When the ratio is 2 : 1, say which way it doubles in words before multiplying: "1 mol acid reacts with 2 mol alkali, so moles NaOH = 2 × moles H₂SO₄". Mark schemes routinely place the ratio mark before the arithmetic — and note the trap direction: here the unknown (NaOH) gets the 2, so the answer 0.080 mol/dm³ is bigger than the acid's 0.050 mol/dm³. A quick proportion sanity check (more concentrated alkali for the same neutralising power in less volume... here slightly more volume, so mildly more concentrated) catches inverted ratios.
"Balance the equation to make the ratio 1 : 1, then divide evenly." You cannot re-arrange the chemistry to suit the arithmetic — 2NaOH + H₂SO₄ is the reaction; the 2 : 1 mole ratio is a fact of the substances, not a nuisance. Doubling the NaOH data or halving the acid data to "simplify" the ratio fabricates a different experiment. Apply the ratio; don't remove it.
Question-bank join: spec 1.34C items — titration calculations with a 2 : 1 (or 1 : 2) equation ratio in either direction.
- titration
- — A technique that measures exactly what volume of one solution reacts with a volume of another of known concentration, so an unknown concentration can be calculated.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
Worked example 4 — reversing the calculation, and the chapter close
SB p.71The chapter's last worked example runs the stages backwards: instead of a concentration, the unknown is a volume of solution. Calculate the volume of 0.100 mol/dm³ sodium hydrogencarbonate (NaHCO₃) solution needed to neutralise 20.0 cm³ of 0.125 mol/dm³ hydrochloric acid (HCl).
NaHCO₃(aq) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l)
Stage 1. Only the acid is fully known: moles of HCl = (20.0 ÷ 1000) × 0.125 = 0.00250 mol — we do not have enough information to calculate the number of moles of anything else, so must start here.
Stage 2. The equation is 1 : 1, so the same 0.00250 mol of NaHCO₃ is needed.
Stage 3 — volume instead of mass or concentration. Rearrange the concentration equation: volume = moles ÷ concentration = 0.00250 ÷ 0.100 = 0.0250 dm³ (the volume comes out in dm³ because the concentration is in mol/dm³), then × 1000 for the answer the question wants: 25.0 cm³ of sodium hydrogencarbonate solution.
The chapter closes with a self-check of everything it built — gas volumes at rtp (including potassium nitrate and chlorine preparation problems), g/dm³ ↔ mol/dm³ conversions, mass-from-concentration problems, and a full titration set. This lesson's practice is wired to the question bank below.
In one page: gases — Avogadro's law says equal volumes (same temperature and pressure) hold equal molecules, so 1 mol of any gas occupies 24 dm³ (24 000 cm³) at rtp; moles = volume ÷ molar volume (matching cm³ to 24 000 and dm³ to 24), and gas volumes react in the equation's big-number ratio. Solutions — concentration is per dm³: g/dm³ ÷ Mᵣ = mol/dm³ (and × Mᵣ back); moles = volume (÷1000) × concentration. Everywhere — the three-stage method: moles of what you can, equation ratio, convert to what is asked. Titration = the experiment that measures the reacting volumes.
- State Avogadro's law, and say why 1 mole of any gas occupies 24 dm³ at rtp? (1.35C)
- Convert 250 cm³ to dm³, and 2.4 dm³ to cm³, without hesitation? (1.35C)
- Calculate the volume of a stated number of moles (or mass) of gas at rtp — like the 0.20 mol CO₂ and 0.01 g H₂ examples? (1.35C)
- Work a full gas-volume problem through an equation ratio, like the aluminium/hydrogen problem? (1.35C)
- Convert a concentration between g/dm³ and mol/dm³ using Mᵣ, both directions? (1.34C)
- Calculate moles from a volume and concentration — with the ÷1000 — like Example 3? (1.34C)
- Run the standard titration calculation, including a 2 : 1 ratio, and a reversed volume question like Example 4? (1.34C)
The Paper-2 pattern across this chapter: one gas-volume calculation (molar volume at rtp, usually via an equation) and one solution concentration calculation (usually a titration) are near-permanent fixtures. Both are three-stage questions — full marks come from the layout, not luck: convert first (÷1000 or ÷ molar volume), quote the ratio from the equation, rearrange last, units on every line.
"Example 4 needs a new method because the answer is a volume of solution." It is the same rearrangement as the molar-volume triangle, pointed at the solution triangle: moles ÷ concentration = volume in dm³. The only genuinely new content in the whole chapter was two conversion constants — the molar volume and the mol/dm³ definition — and both now sit inside the same three-stage method you brought from part 1.
Question-bank join: spec 1.34C and 1.35C items across the whole chapter. Past-paper anchor: the standard titration calculation and the rtp molar-volume calculation are the two recurring Paper-2 question archetypes of this chapter.
- concentration
- — The amount of substance dissolved per dm³ of solution, in mol/dm³; or the mass per dm³, in g/dm³. Convert between them with Mᵣ: divide g/dm³ by Mᵣ, multiply mol/dm³ by Mᵣ.
- titration
- — A technique that measures exactly what volume of one solution reacts with a volume of another of known concentration, so an unknown concentration can be calculated.