IB DP Chemistry Reactivity 2.1: How much? The amount of chemical change – Revision Notes

These are condensed notes. The Reactivity 2.1 study guide has the full explanations and longer worked examples.

The notes cover Reactivity 2.1, “How much? The amount of chemical change”, as laid out in the IB’s DP Chemistry guide for first assessment 2025 (issued February 2023): understandings Reactivity 2.1.1 to Reactivity 2.1.5. Reactivity 2.1 carries no AHL content, so SL and HL students learn exactly the same material here, for exam sessions from 2025 onwards.

Afterwards, test yourself with the Reactivity 2.1 practice questions. For the mole itself, see the Structure 1.4 summary. For other topics, use the DP Chemistry topic list, the syllabus tick-list plus free diagnostic quizzes.

Key definitions

• Balanced equation: correct formulas for all reactants and products, with coefficients chosen so that no element gains or loses atoms between the two sides. The coefficients give the mole ratio.
• State symbols: (s) solid, (l) pure liquid, (g) gas, (aq) in aqueous solution. The guide expects them in your equations.
• Limiting reactant: the reactant that is used up first. It caps the amount of product.
• Excess reactant: a reactant present beyond what is needed; some is left at the end.
• Theoretical yield: the amount of product calculated from the limiting reactant, assuming complete reaction and no losses.
• Experimental yield: the amount of product actually collected.
• Percentage yield: experimental yield as a percentage of theoretical yield.
• Atom economy: molar mass of the desired product as a percentage of the total molar mass of all reactants in the balanced equation; a green chemistry measure of efficiency.

Formulas

Quantity	Relationship	Notes
Amount from a mass	n = m/M	mass in grams; molar mass in g mol⁻¹
Amount in solution	n = cV	c in mol dm⁻³; V in dm³ (cm³ ÷ 1000)
Amount of gas	n = V/Vm	Vm = 22.7 dm³ mol⁻¹ at STP (given as data on this page)
Gas to gas	volume ratio = mole ratio	only for gases at the same temperature and pressure
Percentage yield	(experimental ÷ theoretical) × 100	same units top and bottom
Atom economy	(M of desired product ÷ total M of reactants) × 100	multiply each M by its coefficient

Two points come straight from the guide: work with Ar values to two decimal places, and the atom economy equation is supplied in the data booklet.

Method in steps

Balancing (Reactivity 2.1.1)

1. Correct formulas first. Change coefficients only, never subscripts.
2. Balance an element found in one substance on each side first; leave H, O or an uncombined element until last.
3. Count an unchanged polyatomic ion as one unit.
4. Lowest whole numbers, then state symbols.

Reacting quantities (Reactivity 2.1.2)

1. Given quantity → moles.
2. Moles wanted = moles known × (wanted coefficient ÷ known coefficient).
3. Moles wanted → mass, volume or concentration.

Limiting reactant (Reactivity 2.1.3)

1. Moles of each reactant.
2. Divide each by its coefficient. The smallest result is the limiting reactant.
3. Base every product calculation on the limiting reactant.
4. Excess left over = moles present − moles used.

Percentage yield (Reactivity 2.1.4)

1. Identify the limiting reactant.
2. Theoretical yield from its moles and the ratio.
3. Experimental ÷ theoretical, × 100.

Atom economy (Reactivity 2.1.5)

1. Balanced equation for the route in question.
2. Add up the molar masses of all reactants, each multiplied by its coefficient.
3. Divide the molar mass of the wanted product (times its coefficient) by that total, then × 100.

Worked reminders

Limiting check. 4Li(s) + O₂(g) → 2Li₂O(s), with 0.50 mol Li and 0.15 mol O₂. Li: 0.50 ÷ 4 = 0.125. O₂: 0.15 ÷ 1 = 0.15. Lithium gives the smaller value, so it is limiting. Li₂O formed = 0.50 × 2/4 = 0.25 mol, and 0.15 − 0.125 = 0.025 mol O₂ is left.

Gas volumes without Vm. 2C₃H₆(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(l), all gases at the same temperature and pressure. 10 cm³ of propene needs 10 × 9/2 = 45 cm³ of oxygen and makes 10 × 6/2 = 30 cm³ of carbon dioxide. Liquid water takes up no gas volume.

Atom economy in one line. If the wanted substance is the only product, atom economy is 100 %. Each by-product pulls it down.

Must-know distinctions

• Mole ratio vs mass ratio. A coefficient of 2 means twice the moles, not twice the grams.
• Limiting vs excess. Fewer grams, or even fewer moles, does not by itself make a reactant limiting. Divide by the coefficients before you decide.
• Theoretical vs experimental yield. One is calculated; the other is weighed.
• Below 100 % vs above 100 %. Losses, side reactions and incomplete reaction lower the yield. A damp or impure product can push the recorded yield over 100 %.
• Percentage yield vs atom economy. Yield depends on how the reaction is carried out. Atom economy is fixed by the equation, so better technique cannot change it.
• Atom economy vs waste. Inversely related: lower atom economy means a bigger share of reactant mass leaves as by-products.
• (aq) vs (l). A substance dissolved in water is (aq); a pure liquid, such as water itself or ethanol, is (l).

Quick self-test

Use these Ar values: Cu 63.55, H 1.01, O 16.00, Zn 65.38. Vm = 22.7 dm³ mol⁻¹ at STP.

1. Methylbenzene, C₇H₈, is a liquid. Write its equation for complete combustion, including state symbols, with the products at room temperature.
2. Aqueous calcium nitrate and aqueous sodium carbonate are mixed; solid calcium carbonate forms. Give the equation with state symbols.
3. CuO(s) + H₂(g) → Cu(s) + H₂O(g). Calculate the mass of copper formed from 7.95 g of CuO.
4. Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). Calculate the volume of hydrogen at STP, in cm³, when 0.654 g of zinc reacts with excess acid.
5. 2NO(g) + O₂(g) → 2NO₂(g). 40 cm³ of NO reacts with excess oxygen, all volumes at the same conditions. What volume of oxygen is used, and what volume of NO₂ forms?
6. 22.50 cm³ of KOH(aq), concentration 0.120 mol dm⁻³, neutralises 15.00 cm³ of sulfuric acid: 2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l). Find [H₂SO₄].
7. N₂(g) + 3H₂(g) → 2NH₃(g). 0.50 mol N₂ is mixed with 1.20 mol H₂. Identify the limiting reactant, the maximum amount of NH₃ and the amount of excess reactant left.
8. State the difference between theoretical yield and experimental yield.
9. A preparation has a theoretical yield of 12.4 g. The student collects 9.30 g. Calculate the percentage yield.
10. Water can be electrolysed to make hydrogen: 2H₂O(l) → 2H₂(g) + O₂(g). Calculate the atom economy for hydrogen.
11. Explain why adding HBr to propene has an atom economy of 100 %.

Answers

1. C₇H₈(l) + 9O₂(g) → 7CO₂(g) + 4H₂O(l)
2. Ca(NO₃)₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaNO₃(aq)
3. M(CuO) = 79.55 g mol⁻¹; n = 7.95 ÷ 79.55 = 0.09994 mol; 1 : 1 ratio; m(Cu) = 0.09994 × 63.55 = 6.35 g
4. n(Zn) = 0.654 ÷ 65.38 = 0.01000 mol = n(H₂); V = 0.01000 × 22.7 = 0.227 dm³ = 227 cm³
5. 20 cm³ of O₂ is used and 40 cm³ of NO₂ forms (ratio 2 : 1 : 2)
6. n(KOH) = 0.120 × 0.02250 = 0.002700 mol; n(H₂SO₄) = 0.001350 mol; [H₂SO₄] = 0.001350 ÷ 0.01500 = 0.0900 mol dm⁻³
7. N₂: 0.50 ÷ 1 = 0.50; H₂: 1.20 ÷ 3 = 0.40. H₂ is limiting; n(NH₃) = 1.20 × 2/3 = 0.80 mol; N₂ left = 0.50 − 0.40 = 0.10 mol
8. Theoretical yield is calculated from the limiting reactant, assuming complete reaction and no losses; experimental yield is the amount actually obtained.
9. 9.30 ÷ 12.4 × 100 = 75.0 %
10. Wanted: 2 × 2.02 = 4.04; reactants: 2 × 18.02 = 36.04; atom economy = 4.04 ÷ 36.04 × 100 = 11.2 %
11. Propene and HBr add together: C₃H₆ + HBr → C₃H₇Br. The equation has only one product, so every reactant atom ends up in it and none leaves in a by-product.

Where marks are usually lost

• Altering a formula, such as writing KOH₂, to force an equation to balance.
• Missing or wrong state symbols: (aq) on a precipitate, or (g) on water at room temperature.
• Writing the mole ratio upside down; the wanted coefficient goes on top.
• Using cm³ directly in n = cV.
• Converting gas volumes to moles and back when the volume ratio alone was enough, then rounding twice.
• Choosing the limiting reactant from masses instead of moles divided by coefficients.
• Calculating a yield from the reactant in excess.
• Mixing grams and moles in the percentage yield fraction.
• Leaving coefficients out when totalling reactant molar masses for atom economy.
• Claiming that selling a by-product raises the atom economy; the value comes only from the equation.

Official syllabus

Diploma Programme Chemistry guide, first assessment 2025 (International Baccalaureate Organization, February 2023). These notes follow its Reactivity 2.1 understandings.