Edexcel A-level Chemistry (9CH0) · Topic 5: Formulae, Equations and Amounts of Substance
Mini-Lesson
Formulae, Equations & Amounts of Substance
This mini-lesson covers Edexcel Topic 5: the mole and molar mass, the ideal gas equation, empirical and molecular formulae, titration calculations, and percentage yield and atom economy.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
The mole
Amount of substance
One mole is the amount of substance that contains as many particles as there are atoms in 12 g of carbon-12 — that is 6.02 × 1023 particles (the Avogadro constant, L).
n = m ÷ Mn = amount (mol) · m = mass (g) · M = molar mass (g mol⁻¹)
Molar mass is the mass of one mole. Numerically it equals the relative formula mass, but it has units g mol⁻¹.
Concentration: n = c × V, with c in mol dm⁻³ and V in dm³. Remember: V(dm³) = V(cm³) ÷ 1000.
Exam habit: write the molar mass out in full before you divide. Most lost marks in Topic 5 are arithmetic, not chemistry.
Calculate
Your turn — moles from mass
1Calculate the amount, in mol, in 4.20 g of sodium hydrogencarbonate, NaHCO3. (Ar: Na 23.0, H 1.0, C 12.0, O 16.0.) Give your answer to 3 significant figures.
mol
Hint: M(NaHCO3) = 84.0 g mol⁻¹. n = 4.20 ÷ 84.0.
Gases
The ideal gas equation
pV = nRTp in Pa · V in m³ · T in K · R = 8.31 J K⁻¹ mol⁻¹
The unit conversions are where the marks are won and lost:
kPa → Pa: × 1000 (100 kPa = 1.00 × 105 Pa)
cm³ → m³: ÷ 106 | dm³ → m³: ÷ 1000
°C → K: + 273 (25 °C = 298 K)
Worked example — volume of a gas
What volume does 0.250 mol of an ideal gas occupy at 100 kPa and 298 K?
V = nRT ÷ p = (0.250 × 8.31 × 298) ÷ (1.00 × 105)
= 619.1 ÷ 100 000 = 6.19 × 10−3 m³ = 6.19 dm³
Rearrangements you must know: n = pV/RT · V = nRT/p · M = mRT/pV (since n = m/M).
Calculate
Your turn — volume of a gas
2Calculate the volume, in dm³, occupied by 0.500 mol of an ideal gas at 100 kPa and 298 K. (R = 8.31 J K⁻¹ mol⁻¹.)
dm³
Hint: V = nRT/p = (0.500 × 8.31 × 298) ÷ (1.00 × 10⁵) m³, then multiply by 1000 to get dm³.
Calculate
Your turn — Mr of a volatile liquid
3A 0.240 g sample of a volatile liquid is vaporised and occupies 1.00 × 10−4 m³ at 1.00 × 105 Pa and 373 K. Calculate its Mr to 3 significant figures.
Hint: n = pV/RT = (1.00×10⁵ × 1.00×10⁻⁴) ÷ (8.31 × 373) = 10.0 ÷ 3099.6 = 3.23 × 10⁻³ mol. Then M = m ÷ n.
Formulae
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the real number of atoms in one molecule. The molecular formula is always a whole-number multiple of the empirical formula.
Worked example
A compound is 40.0% C, 6.7% H and 53.3% O by mass, and has Mr = 180.0.
Divide each percentage by the Ar: C 40.0/12.0 = 3.33 · H 6.7/1.0 = 6.7 · O 53.3/16.0 = 3.33
Divide by the smallest (3.33): C 1.00 : H 2.01 : O 1.00 → empirical formula CH2O (M = 30.0)
180.0 ÷ 30.0 = 6, so the molecular formula is C6H12O6 (glucose).
Watch out: if a ratio comes out as 1 : 1.5, do not round — multiply everything by 2 (so 2 : 3).
Calculate
Your turn — molecular formula
4A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and has Mr = 56.0. How many carbon atoms are in one molecule?
C atoms
Hint: C: 85.7/12.0 = 7.14; H: 14.3/1.0 = 14.3. Ratio 1 : 2, so the empirical formula is CH₂ (M = 14.0). Now do 56.0 ÷ 14.0.
Titrations
Titration calculations
A titration finds an unknown concentration by reacting it with a solution of known concentration (a standard solution). Concordant titres (within 0.10 cm³) are averaged; the rough titre is discarded.
The method is always the same three steps:
1. Moles of the substance you know everything about: n = c × V
2. Use the balanced equation to get the moles of the other substance
3. Divide by its volume (or molar mass) to get what was asked for
Worked example
25.0 cm³ of NaOH needed 22.40 cm³ of 0.100 mol dm⁻³ HCl. HCl + NaOH → NaCl + H2O
n(HCl) = 0.100 × 0.02240 = 2.240 × 10−3 mol
Ratio 1 : 1, so n(NaOH) = 2.240 × 10−3 mol
c(NaOH) = 2.240 × 10−3 ÷ 0.0250 = 0.0896 mol dm⁻³
Calculate
Your turn — a 1 : 2 titration
525.0 cm³ of 0.200 mol dm⁻³ NaOH was exactly neutralised by 18.70 cm³ of sulfuric acid. H2SO4 + 2NaOH → Na2SO4 + 2H2O Calculate the concentration of the H2SO4 to 3 significant figures.
mol dm⁻³
Hint: n(NaOH) = 0.200 × 0.0250 = 5.00 × 10⁻³ mol. The ratio is 2 NaOH : 1 H₂SO₄, so n(acid) = 2.50 × 10⁻³ mol. Divide by 0.01870 dm³.
Quick check
Which equation is correctly balanced?
6Only one of these equations balances for both atoms and charge. Which one?
Yield & economy
Percentage yield and atom economy
These two measure completely different things, and examiners love the distinction.
% yield = (actual mass ÷ theoretical mass) × 100a practical measure — how much you really got out of the reaction you ran
atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100a theoretical measure — how much of the reactant mass ends up as something useful
A reaction can have a 100% yield and a terrible atom economy (lots of by-product).
Addition reactions have an atom economy of 100% — there is only one product.
Yield is lost to: incomplete reactions, side reactions, and losses on transfer/purification.
Worked example — atom economy
CaCO3(s) → CaO(s) + CO2(g) (M: CaCO3 100.1, CaO 56.1, CO2 44.0)
Atom economy for CaO = 56.1 ÷ (56.1 + 44.0) × 100 = 56.1 ÷ 100.1 × 100 = 56.0%
Calculate
Your turn — percentage yield
712.5 g of calcium carbonate was heated and 5.60 g of calcium oxide was collected. CaCO3(s) → CaO(s) + CO2(g) (M: CaCO3 100.1, CaO 56.1) Calculate the percentage yield to 3 significant figures.
%
Hint: n(CaCO₃) = 12.5 ÷ 100.1 = 0.1249 mol. The ratio is 1 : 1, so the theoretical mass of CaO = 0.1249 × 56.1 = 7.01 g. Now do (5.60 ÷ 7.01) × 100.
Calculate
Your turn — atom economy
8Iron is extracted in the blast furnace: Fe2O3 + 3CO → 2Fe + 3CO2 Calculate the atom economy for making iron. (Ar: Fe 55.8; Mr: CO2 44.0.)
%
Hint: Mass of desired product = 2 × 55.8 = 111.6. Total mass of all products = 111.6 + (3 × 44.0) = 111.6 + 132.0 = 243.6. Now do (111.6 ÷ 243.6) × 100.
Quick check
Yield or atom economy?
9A chemist improves a process so that a much larger fraction of the reactants is converted, but the same by-products are still formed in the same balanced equation. Which quantity increases?
Quick check
Spectator ions
10Silver nitrate solution is added to sodium chloride solution and a white precipitate of silver chloride forms. Which is the correct ionic equation?
Sort it
Which route to moles?
Tap a sample, then tap the equation you would use to find the amount in moles.
🟫 n = m / M
🟩 n = c × V
🟪 n = pV / RT
Quick check
A standard solution
11A student weighs 2.65 g of anhydrous Na2CO3 (M = 106.0 g mol⁻¹) and makes it up to 250 cm³ of solution in a volumetric flask. What is the concentration?
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Recap
The big ideas to know
The mole: n = m/M · n = c × V (V in dm³) · n = pV/RT (p in Pa, V in m³, T in K, R = 8.31)
Ideal gas: pV = nRT; rearrange to M = mRT/pV to find Mr of a volatile liquid
Empirical formula: divide % (or mass) by Ar, then divide by the smallest — never round 1.5 down
Molecular formula: Mr ÷ (empirical formula mass) gives the multiplier
Titrations: moles of the known → use the balanced equation ratio → divide by volume or molar mass
% yield: (actual ÷ theoretical) × 100 — a measure of how well the reaction ran
Atom economy: (Mr of desired product ÷ total Mr of products) × 100 — fixed by the equation
You've now covered Topic 5: Formulae, Equations and Amounts of Substance of the Edexcel A-level Chemistry (9CH0) specification. Press Finish to see your score.
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