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

This study guide teaches Reactivity 2.1, “How much? The amount of chemical change”, for IB DP Chemistry. It is aligned to the IB’s DP Chemistry guide for first assessment 2025 (issued February 2023) and covers understandings Reactivity 2.1.1 to Reactivity 2.1.5. No additional higher level (AHL) material is listed for this sub-topic, so every section is for SL and HL alike, in exam sessions from 2025 onwards.

Pair it with the Reactivity 2.1 revision notes and the Reactivity 2.1 practice questions. Moles and molar concentration are recapped in our Structure 1.4 mole guide. Beyond this unit, browse every DP Chemistry topic, print the full syllabus tick-list, or sit a free diagnostic quiz.

What this unit covers

Guide statement	What you must be able to do	Level
Reactivity 2.1.1	Deduce a balanced chemical equation, with state symbols, when the reactants and products are named	SL and HL
Reactivity 2.1.2	Use the mole ratio of an equation to calculate reacting masses, gas volumes and solution concentrations	SL and HL
Reactivity 2.1.3	Identify the limiting and excess reactants; tell theoretical yield apart from experimental yield	SL and HL
Reactivity 2.1.4	Calculate percentage yield and solve problems that combine limiting reactants and yields	SL and HL
Reactivity 2.1.5	Calculate atom economy from the equation; discuss how low atom economy means more waste	SL and HL

Data used on this page: Ar values Ag 107.87, C 12.01, Cl 35.45, Fe 55.85, H 1.01, K 39.10, Mg 24.31, N 14.01, O 16.00. Molar volume of an ideal gas at STP = 22.7 dm³ mol⁻¹. The guide asks you to work with relative atomic masses to two decimal places in these calculations.

Reactivity 2.1.1 Writing balanced equations

Atoms are conserved, so every element must be counted equally on the left and the right. The coefficients then give the ratio in which particles, and so moles, react and form.

Method

1. Get every formula right first. Balancing is done with coefficients only; a subscript is never edited.
2. Start with an element that sits in just one reactant and one product. Save hydrogen, oxygen and any uncombined element for the end.
3. If an ion such as SO₄²⁻ or NO₃⁻ passes through unchanged, count it as a single block.
4. Reduce to the smallest whole-number coefficients.
5. Add state symbols, which the guide expects: (s) for a solid, (l) for a pure liquid, (g) for a gas and (aq) for a substance in aqueous solution.

Worked example 1. Solid aluminium oxide dissolves in dilute sulfuric acid, giving aqueous aluminium sulfate and water.

skeleton:  Al₂O₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
sulfate:   three SO₄ blocks on the right, so 3H₂SO₄
hydrogen:  6 H on the left, so 3H₂O
check:     Al 2 = 2; S 3 = 3; O 3 + 12 = 12 + 3 = 15; H 6 = 6
answer:    Al₂O₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂O(l)

Worked example 2. Liquid butanone, C₄H₈O, burns completely in oxygen; the products are cooled to room temperature.

Carbon gives 4CO₂ and hydrogen gives 4H₂O. The right-hand side now holds 8 + 4 = 12 oxygen atoms. The butanone supplies 1, so 11 must come from O₂, which is 5½O₂. Double everything to clear the half:

2C₄H₈O(l) + 11O₂(g) → 8CO₂(g) + 8H₂O(l)

Reactivity 2.1.2 Using the mole ratio

Every reacting-quantity problem uses three moves:

1. Turn the quantity you are given into moles (n = m/M for a mass, n = cV for a solution, n = V/Vm for a gas).
2. Multiply by the mole ratio: coefficient of the substance you want over coefficient of the substance you know.
3. Turn those moles into the unit the question asks for.

Reacting masses

Worked example 3. Magnesium nitride reacts with water to release ammonia:

Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

A 5.00 g sample of Mg₃N₂ reacts completely. Calculate the mass of magnesium hydroxide formed and the volume of ammonia released at STP.

M(Mg₃N₂) = 3(24.31) + 2(14.01) = 100.95 g mol⁻¹
n(Mg₃N₂) = 5.00 ÷ 100.95 = 0.04953 mol
n(Mg(OH)₂) = 0.04953 × 3/1 = 0.1486 mol
M(Mg(OH)₂) = 24.31 + 2(16.00 + 1.01) = 58.33 g mol⁻¹
m(Mg(OH)₂) = 0.1486 × 58.33 = 8.67 g
n(NH₃) = 0.04953 × 2/1 = 0.09906 mol
V(NH₃) = 0.09906 × 22.7 = 2.25 dm³

Gas volumes

By Avogadro’s law (Structure 1.4), gas volume measures amount when temperature and pressure are fixed.

• Mass to gas volume. Moles first, then the ratio, then multiply by Vm. Heating 5.00 g of silver oxide, 2Ag₂O(s) → 4Ag(s) + O₂(g), gives n(Ag₂O) = 5.00 ÷ 231.74 = 0.02158 mol, so n(O₂) = 0.02158 ÷ 2 = 0.01079 mol and V(O₂) = 0.245 dm³ (245 cm³) at STP.
• Gas volume to gas volume. When every substance you compare is a gas at the same temperature and pressure, the volume ratio is the mole ratio, and Vm is not needed. In a catalytic converter, 2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g). So 60 cm³ of NO removes 60 cm³ of CO and forms 60 cm³ of CO₂ and 60 × 1/2 = 30 cm³ of N₂.

Concentrations in solution

For solutions, n = cV, with V in dm³. Square brackets, as in [HNO₃], mean concentration in mol dm⁻³.

Worked example 4. A 25.00 cm³ portion of barium hydroxide solution is exactly neutralised by 18.40 cm³ of nitric acid, [HNO₃] = 0.150 mol dm⁻³.

Ba(OH)₂(aq) + 2HNO₃(aq) → Ba(NO₃)₂(aq) + 2H₂O(l)

(a) Find [Ba(OH)₂]. (b) Find the concentration of barium nitrate in the mixture at the end point.

(a) n(HNO₃) = 0.150 × 0.01840 = 0.002760 mol
    ratio Ba(OH)₂ : HNO₃ = 1 : 2, so n(Ba(OH)₂) = 0.001380 mol
    [Ba(OH)₂] = 0.001380 ÷ 0.02500 = 0.0552 mol dm⁻³
(b) n(Ba(NO₃)₂) = 0.001380 mol
    total volume = 25.00 + 18.40 = 43.40 cm³ = 0.04340 dm³
    [Ba(NO₃)₂] = 0.001380 ÷ 0.04340 = 0.0318 mol dm⁻³

The concentration of a product in a mixture uses the combined volume.

Reactivity 2.1.3 Limiting reactant and yields

The limiting reactant runs out first, so it alone decides the most product that can form. Every other reactant is in excess, and some of it remains when the reaction stops.

Method. Find the moles of each reactant. Pick one and calculate how much of the other it would need. If less of that other reactant is available than needed, it is limiting. A shortcut is to divide each amount by its coefficient; the smallest quotient marks the limiting reactant.

Worked example 5. Hot iron wool (2.80 g) is lowered into 1.50 dm³ of chlorine, measured at STP.

2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)

n(Fe) = 2.80 ÷ 55.85 = 0.05013 mol
n(Cl₂) = 1.50 ÷ 22.7 = 0.06608 mol
Cl₂ needed by all the Fe = 0.05013 × 3/2 = 0.07520 mol
0.07520 mol needed > 0.06608 mol available, so Cl₂ is limiting and Fe is in excess
n(FeCl₃) = 0.06608 × 2/3 = 0.04405 mol
M(FeCl₃) = 55.85 + 3(35.45) = 162.20 g mol⁻¹
maximum mass of FeCl₃ = 0.04405 × 162.20 = 7.15 g
Fe used = 0.04405 mol, so Fe left = (0.05013 − 0.04405) × 55.85 = 0.340 g

Notice that iron has the smaller number of moles here, yet it is chlorine that runs out. Always bring the coefficients in.

The theoretical yield is the product amount worked out from the limiting reactant, such as the 7.15 g above. The experimental yield is the amount you really obtain and weigh.

The experimental yield is usually lower because:

• the reaction does not go to completion, or reaches equilibrium
• side reactions turn some reactant into other products
• product is lost in filtering, transferring or recrystallising

It can come out higher if the product is still damp or holds impurities, such as unreacted solid.

Reactivity 2.1.4 Percentage yield

percentage yield = (experimental yield ÷ theoretical yield) × 100

Keep the units the same on top and bottom: both in grams or both in moles.

Worked example 6. A student adds excess hydrochloric acid to 3.00 g of potassium carbonate, evaporates the solution and dries the solid. The mass of potassium chloride collected is 2.91 g.

K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + H₂O(l) + CO₂(g)

M(K₂CO₃) = 2(39.10) + 12.01 + 3(16.00) = 138.21 g mol⁻¹
n(K₂CO₃) = 3.00 ÷ 138.21 = 0.02171 mol
ratio K₂CO₃ : KCl = 1 : 2, so n(KCl) = 0.04341 mol
M(KCl) = 39.10 + 35.45 = 74.55 g mol⁻¹
theoretical yield = 0.04341 × 74.55 = 3.236 g
percentage yield = 2.91 ÷ 3.236 × 100 = 89.9 %

The acid is in excess, so the potassium carbonate fixes the theoretical yield. Settle the limiting reactant before any yield calculation.

Reactivity 2.1.5 Atom economy

Atom economy compares the mass of the product you want with the total mass of everything that reacts. Green chemistry uses it to judge how efficiently a reaction turns starting materials into useful product. The guide states that the data booklet gives this equation:

atom economy = (molar mass of desired product ÷ total molar mass of all reactants) × 100

Multiply each molar mass by its coefficient from the balanced equation.

Worked example 7. Methanol, CH₃OH (M = 32.05 g mol⁻¹), can be made from carbon monoxide or from carbon dioxide.

Route 1: CO(g) + 2H₂(g) → CH₃OH(g)
  reactants = 28.01 + 2(2.02) = 32.05 g mol⁻¹
  atom economy = 32.05 ÷ 32.05 × 100 = 100 %
Route 2: CO₂(g) + 3H₂(g) → CH₃OH(g) + H₂O(g)
  reactants = 44.01 + 3(2.02) = 50.07 g mol⁻¹
  atom economy = 32.05 ÷ 50.07 × 100 = 64.0 %

When the wanted substance is the only product, atom economy is 100 %, because no atoms leave in anything else.

Atom economy and waste. The two move in opposite directions. In route 2, 36.0 % of the reactant mass ends up as water, about 0.562 t for every tonne of methanol. The lower the atom economy, the more by-product there is to separate, use or dispose of, and the more raw material each tonne of product consumes. Selling or reusing a by-product reduces real waste, but the calculated atom economy stays the same.

Atom economy is not yield. A route can have 100 % atom economy and a poor percentage yield, or the reverse. Atom economy depends on the equation alone; percentage yield depends on what happens in the reaction vessel. A full judgement of a process also weighs energy demand, rate, safety and the cost and source of the raw materials; route 2, for example, uses carbon dioxide as a feedstock.

Common errors

• Editing a subscript (H₂O₂ for H₂O) to make an equation balance.
• Treating the coefficients as a mass ratio rather than a mole ratio.
• Writing the ratio the wrong way round: wanted over known.
• Putting cm³ into n = cV without dividing by 1000.
• Calling the reactant with fewer moles limiting without dividing by the coefficients first.
• Basing a yield on the reactant in excess.
• Totalling reactant molar masses for atom economy without their coefficients.

Where to go next

Next, test your recall against the condensed notes and work through the original practice set, which has full worked answers.

Official syllabus

Diploma Programme Chemistry guide, first assessment 2025 (International Baccalaureate Organization, February 2023). This page follows the Reactivity 2.1 understandings and their guidance.