R2.1 How much? The amount of chemical changeIB Chemistry HL: Revision notes
Section 1
Balanced equations and mole ratios
A balanced equation shows the mole ratio in which substances react and form. Balance atoms, never change formulas, and add state symbols. When reactants and products are named, write the formulas first, then balance: e.g. glucose → ethanol + carbon dioxide gives C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.
Section 2
Reacting masses and gas volumes
Three-step method: (1) convert what you know to moles; (2) use the mole ratio; (3) convert to the quantity asked for.
- n = m ÷ M
- gases: n = V ÷ V_m, with V_m = 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa)
Example: 65.0 g NaN₃ (M = 65.02) is 1.00 mol; 2NaN₃ → 2Na + 3N₂ gives 1.50 mol N₂ = 34.0 dm³ at STP.
Keep unrounded values in your calculator and round only the final answer, to the number of significant figures in the data.
Section 3
Solutions and titrations
Concentration: c = n ÷ V, with V in dm³ (divide cm³ by 1000).
In a titration, find n of the reagent of known concentration, use the mole ratio, then find the unknown concentration. Remember any scale-up: a 25.0 cm³ portion of a 250.0 cm³ solution contains one-tenth of the total. This lets you find the water of crystallisation in a hydrated salt: M(hydrate) = mass ÷ n, then x = (M − M(anhydrous)) ÷ 18.02.
Forgetting the scale-up from the titrated portion to the whole volumetric flask.
Section 4
Limiting and excess reactants
The limiting reactant is completely used up and fixes the maximum amount of product; any other reactant is in excess. To identify it, convert each reactant to moles and divide by its coefficient: the smallest value is limiting. Adding more of an excess reactant does not increase the theoretical yield.
Section 5
Percentage yield
Percentage yield = experimental yield ÷ theoretical yield × 100
Yields are below 100% because of incomplete reactions, reversible reactions reaching equilibrium, side reactions and losses during transfer, filtration and purification (e.g. recrystallisation). A yield above 100% usually means the product is impure or still wet.
Section 6
Atom economy
Atom economy = M(desired product) ÷ M(all reactants) × 100, using the coefficients in the balanced equation. It measures how much of the reactant mass ends up in the product, so it is a key idea in green chemistry. Addition reactions (C₂H₄ + H₂O → C₂H₅OH) have 100% atom economy; fermentation of glucose to ethanol has 51.2% because CO₂ is also formed.
Atom economy comes from the equation; percentage yield comes from the experiment. A process can be high in one and low in the other.
Must know
- Moles → ratio → answer.
- n = m/M; n = V/22.7 at STP; c = n/V (V in dm³).
- Smallest moles ÷ coefficient = limiting reactant.
- % yield = experimental ÷ theoretical × 100.
- Atom economy = M(desired product) ÷ M(all reactants) × 100.
That's the notes covered.
Carry on to the next subtopic.