1. Balancing Equations
A full equation shows every reactant and product. When it is balanced, there is the same number of atoms of each element on both sides. In an ionic equation, the total charge is also the same on both sides.
Balancing an unfamiliar equation
- Write the correct formula for every reactant and product. Never change a formula to make it balance.
- Balance one element at a time by changing the numbers in front of the formulae. Start with an element that is in only one reactant and one product.
- Balance elements that appear on their own (such as O₂ or H₂) last.
- Check every element again. Add state symbols if the question asks for them.
Step 1 (C): 3 C on the left, so 3CO₂.
Step 2 (H): 8 H on the left, so 4H₂O.
Step 3 (O): the right has 3 × 2 + 4 = 10 O atoms, so 5O₂.
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
Step 1 (O): O₂ has 2 O and Al₂O₃ has 3 O. The lowest common multiple is 6, so 3O₂ and 2Al₂O₃.
Step 2 (Al): 2Al₂O₃ contains 4 Al, so 4Al.
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
2. Net Ionic Equations with State Symbols
In aqueous reactions, soluble ionic substances dissociate into individual hydrated ions. Net ionic equations omit non-reacting spectator ions to focus on the actual chemical transformation:
Step-by-Step Writing Rule
- Write full balanced stoichiometric equation with state symbols.
- Split only strong electrolytes in aqueous solution (\((aq)\)) into ions. Leave solids \((s)\), liquids \((l)\), and gases \((g)\) intact.
- Cancel identical spectator ions appearing on both sides.
- Ensure charges and atoms balance.
Full: \(\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
Ionic: \(2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow 2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
Net Ionic: \[\mathbf{\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g)}\]
3. Reacting Masses and Gas Volumes
Every reacting-quantity calculation uses the same three steps. The mole formulas are on the Amount of Substance page.
Three-step method
- Find the moles of the substance you know (n = m/M, n = cV or n = pV/RT).
- Use the ratio in the balanced equation to find the moles of the substance you want.
- Convert to the quantity asked for: mass, concentration or gas volume.
If amounts of two reactants are given, first find the limiting reagent. It is used up first, so it decides how much product forms.
Step 1: n(CaCO₃) = 5.00 / 100.1 = 0.04995 mol
Step 2: the ratio is 1 : 1, so n(CO₂) = 0.04995 mol
Step 3: V = nRT / p = 0.04995 × 8.31 × 298 / 100 000 = 1.237 × 10⁻³ m³ = 1240 cm³ (3 significant figures)
4. Percentage Yield and Atom Economy
Two numbers describe how efficient a reaction is:
Percentage Yield
\[ \% \text{ Yield} = \frac{\text{Actual mass of product}}{\text{Theoretical mass calculated}} \times 100 \]
Measures practical efficiency. Why is yield almost always less than 100%?
- Incomplete reaction reaching dynamic equilibrium.
- Unwanted side reactions producing by-products.
- Mechanical losses during filtration, transfer, or recrystallisation.
Atom Economy
\[ \% \text{ atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all reactants}} \times 100 \]
It shows how much of the mass of the reactants ends up in the desired product. Use the amounts in the balanced equation: multiply each Mr by its number in the equation.
Addition reactions always have 100% atom economy because all the reactant atoms end up in a single product.
Step 1: theoretical mass of CaO = 5.00 / 100.1 × 56.1 = 2.802 g
Step 2: percentage yield = 2.52 / 2.802 × 100 = 89.9%
Do not round the theoretical mass before dividing. Using 2.80 g gives 90.0%, which may lose the mark: keep the unrounded value.
Mass of desired product = 3 × 2.0 = 6.0
Sum of molecular masses of all reactants = 16.0 + 18.0 = 34.0
% atom economy = 6.0 / (16.0 + 18.0) × 100 = 17.6%
Why a high atom economy matters
- Economic: less of the starting material is wasted, so raw materials cost less. There are fewer by-products to separate and dispose of.
- Environmental: less waste is produced, so there is less pollution and less waste to treat.
- Ethical: finite raw materials, such as crude oil, are used more sustainably. This benefits society and future generations.
A process can have a high percentage yield but a low atom economy. Industry looks at both.
5. Titration Calculations
In a titration, a solution of known concentration reacts with a measured volume of another solution. The volumes and the balanced equation give the unknown concentration. The practical method is RP1: volumetric solution and acid-base titration.
Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Step 1: n(H₂SO₄) = 0.100 × (22.50 / 1000) = 2.25 × 10⁻³ mol
Step 2: the ratio is 1 : 2, so n(NaOH) = 2 × 2.25 × 10⁻³ = 4.50 × 10⁻³ mol
Step 3: c(NaOH) = 4.50 × 10⁻³ / (25.0 / 1000) = 0.180 mol dm⁻³
6. Preparing a Standard Volumetric Solution
A solution of accurately known concentration prepared using a primary chemical standard.
Standard Laboratory Protocol (RP1 Alignment)
This is part of RP1.
- Weigh by Difference: Weigh a weighing boat containing the primary standard (e.g. anhydrous Na₂CO₃ or H₂C₂O₄) on a balance reading to at least 2 decimal places. Transfer the solid to a clean 250 cm³ glass beaker. Reweigh the empty boat to account for any residual solid: \[ m(\text{transferred}) = m(\text{boat + solid}) - m(\text{empty boat}) \]
- Dissolution: Add approximately 100 cm³ of deionised water to the beaker. Stir thoroughly with a glass rod until all crystals dissolve completely.
- Quantitative Transfer: Pour the solution through a clean filter funnel into a 250.0 cm³ volumetric flask. Rinse the beaker, glass rod, and funnel with deionised water, collecting all washings in the volumetric flask.
- Make up to Mark: Add deionised water until the water level approaches the graduation line. Use a dropping pipette to add the final drops until the bottom of the curved meniscus rests exactly on the horizontal graduation line at eye level.
- Homogenisation: Stopper the flask tightly and invert 10 to 15 times to ensure thorough mixing and uniform concentration throughout.
7. Volumetric Titration Protocols
Titrations accurately determine the concentration of an unknown solution by reacting it with a standard solution of known concentration.
Apparatus Preparation Rules
- Pipette: Rinse with deionised water, then rinse with the solution to be pipetted. Deliver exactly 25.0 cm³ into a conical flask.
- Burette: Rinse with deionised water, then rinse with the titrant solution. Ensure the space below the tap is completely filled with solution and free of air bubbles before recording initial volume.
- Conical Flask: Rinse with deionised water ONLY. A conical flask is used rather than a beaker because its sloping walls prevent splashing during swirling. Placed on a white tile to observe sharp indicator colour changes.
Concordant Titres
Perform a preliminary rough titration, followed by accurate titrations. Concordant titres are results that agree within 0.10 cm³ of each other. Only concordant titres are averaged when calculating the mean titre volume.
8. Back Titrations
A back titration is employed when an analyte is insoluble in water (such as CaCO₃), reacts too slowly with the titrant, or has no direct visual endpoint.
Step 1: Calculate moles of NaOH used in the titration:
\[ n(\text{NaOH}) = c \times V = 0.100 \times (22.40 / 1000) = 0.002240\text{ mol} \]
Step 2: Find moles of unreacted HCl in the 25.0 cm³ aliquot:
Reaction: HCl + NaOH → NaCl + H₂O (1:1 ratio)
\[ n(\text{HCl in 25 cm}^3) = 0.002240\text{ mol} \]
Step 3: Scale up to find total unreacted HCl in the 250.0 cm³ volumetric flask:
\[ n(\text{HCl unreacted total}) = 0.002240 \times \left(\frac{250.0}{25.0}\right) = 0.02240\text{ mol} \]
Step 4: Calculate initial moles of HCl added:
\[ n(\text{HCl initial}) = 1.000 \times (50.0 / 1000) = 0.05000\text{ mol} \]
Step 5: Calculate moles of HCl that reacted with CaCO₃:
\[ n(\text{HCl reacted}) = 0.05000 - 0.02240 = 0.02760\text{ mol} \]
Step 6: Calculate moles and mass of pure CaCO₃:
Reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
\[ n(\text{CaCO}_3) = n(\text{HCl reacted}) / 2 = 0.02760 / 2 = 0.01380\text{ mol} \]
\[ m(\text{pure CaCO}_3) = 0.01380 \times 100.1 = 1.38138\text{ g} \]
Step 7: Calculate percentage purity of the sample:
\[ \% \text{ Purity} = \frac{\text{Mass of pure }\text{CaCO}_3}{\text{Mass of impure sample}} \times 100 = \frac{1.38138\text{ g}}{1.500\text{ g}} \times 100 = \mathbf{92.1\%} \]
Reported to 3 significant figures: 92.1%.
9. Vertical Quantitative Integration
The Quantitative Spine of A-Level Chemistry
Stoichiometry is not an isolated topic. The principles mastered in Unit 1 directly feed into every subsequent modular paper:
- Unit 1 (Energetics): Calculating molar enthalpy change \(\Delta H = -q / (n \times 1000)\) relies on accurate mole determination from solution mass or limiting reactants.
- Unit 2 (Equilibria): Setting up ICE tables (Initial, Change, Equilibrium) for Kc requires strict mole conversion before evaluating equilibrium concentrations.
- Unit 4 (Kp): Partial pressures are calculated from mole fractions, so you must find the equilibrium moles of each gas first.
- Unit 3 (Acids and Bases): pH calculations start from the moles of acid and base that react.
- Unit 5 & Required Practicals (RP1 & RP7): Unit 5 practical evaluation demands flawless titration arithmetic, back-titration deductions, and percentage uncertainty error budgets.
10. Strict Examiner Rounding Rule
Examiners routinely penalise candidates who round numbers during intermediate calculation steps. Intermediate numbers must remain stored in your calculator registers.
Mathematical Demonstration: Why Early Rounding Fails
Consider calculating \(X = (10.0 / 3.0) \times 6.0\):
- Correct unrounded method: 10.0/3.0 = 3.333333.... Multiply by 6.0 gives exactly 20.0.
- Erroneous premature rounding: If rounded to 2 significant figures mid-calculation as 3.3: 3.3 × 6.0 = 19.8.
The premature rounding creates an error of 1.0%. An error this size may take the final answer outside the range the mark scheme accepts, so keep unrounded values.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
11. Exam-Style Practice Questions
Show Answer & Mark Scheme
Equation: \(\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\)
Step 1: Calculate moles of reactants:
- \(n(\text{HCl}) = c \times V = 0.200 \times (50.0 / 1000) = 0.0100\text{ mol}\)
- \(n(\text{Mg}) = m / A_r = 0.150 / 24.3 = 0.0061728\text{ mol}\)
Step 2: Determine limiting reagent:
0.0061728 mol of Mg requires 2 × 0.0061728 = 0.01235 mol of HCl. We only have 0.0100 mol of HCl. Therefore, HCl is the limiting reactant.
Step 3: Calculate moles of H₂ gas:
From stoichiometry, \(n(\text{H}_2) = n(\text{HCl}) / 2 = 0.0100 / 2 = 0.00500\text{ mol}\).
Step 4: Calculate gas volume via pV = nRT:
\[ V = \frac{nRT}{p} = \frac{0.00500 \times 8.314 \times 295}{100000} = \frac{12.26315}{100000} = 1.2263 \times 10^{-4}\text{ m}^3 \]
Convert to cm³: 1.2263 × 10⁻⁴ × 10⁶ = 123 cm³ (to 3 significant figures).
Show Answer & Mark Scheme
Solution:
\[ \mathbf{\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)} \]
(1 mark for correct species and balance; 1 mark for correct state symbols).
(a) Calculate the percentage atom economy for iron.
(b) 100 g of Fe₂O₃ gives 63.0 g of iron. Calculate the percentage yield.
Show Answer & Mark Scheme
(a) mass of desired product = 2 × 55.8 = 111.6; sum of reactants = 159.6 + 3 × 28.0 = 243.6 (1 mark)
% atom economy = 111.6 / 243.6 × 100 = 45.8% (1 mark)
(b) theoretical mass of Fe = 100 / 159.6 × 2 × 55.8 = 69.92 g (1 mark)
% yield = 63.0 / 69.92 × 100 = 90.1% (1 mark)
Show Answer & Mark Scheme
- n(Na₂CO₃) = 0.0500 × (25.0 / 1000) = 1.25 × 10⁻³ mol (1 mark)
- n(HCl) = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol (1 mark)
- c(HCl) = 2.50 × 10⁻³ / (23.80 / 1000) = 0.105 mol dm⁻³ (1 mark)