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Reactivity 2.1 · How much? The amount of chemical change

1Chemical equations and state symbols 2.1.1 SL + HL

The syllabus statement: Chemical equations show the ratio of reactants and products in a reaction. The skill: deduce chemical equations when reactants and products are specified, including state symbols.

A car travelling at speed stops suddenly. Within about thirty milliseconds a sensor fires an electrical igniter inside the steering wheel, a small charge of solid sodium azide decomposes, and a nylon bag inflates with nitrogen before the driver moves forward. The designer of that airbag had to answer one question with precision: how much sodium azide produces exactly enough gas to fill the bag, and no more? Too little and the bag is soft; too much and it bursts. The answer comes from a balanced equation:

Reactivity 2.1 is about reading such an equation quantitatively. Structure 1.4 introduced the mole as the chemist's counting unit; this sub-topic uses it to convert between the particles in an equation and the grams, cubic decimetres and moles per cubic decimetre that are actually measured in a laboratory or a factory.

What an equation tells you

A chemical equation is a statement about particles. The formulas identify the substances; the coefficients (the large numbers in front of the formulas) give the ratio in which the particles react and form. Read at the particle level, the airbag equation says that two formula units of sodium azide give two atoms of sodium and three molecules of nitrogen. Because a mole contains the same number of particles of every substance (6.02 × 10²³), the same ratio applies to amounts: 2 mol of NaN₃ gives 2 mol of Na and 3 mol of N₂. This ratio of amounts is the mole ratio (or stoichiometric ratio), and it is the bridge used in every calculation in this chapter.

What an equation does not show is equally important. It says nothing about how fast the reaction happens (Reactivity 2.2), whether it goes to completion (Reactivity 2.3), the mechanism by which the particles actually meet (Reactivity 3), or the conditions needed, unless these are written over the arrow. An equation also says nothing directly about masses: 2 mol of NaN₃ has a mass of 130.04 g, but 3 mol of N₂ has a mass of 84.06 g. Mass is conserved in the reaction as a whole (130.04 g = 45.98 g + 84.06 g), but the masses of individual substances are never in the ratio of the coefficients.

Balancing: conservation of atoms

Atoms are neither created nor destroyed in a chemical reaction; they are rearranged. An equation is therefore balanced when every element has the same number of atoms on both sides and, for ionic equations, the total charge is also the same on both sides. Balancing is done by changing coefficients only. Changing a subscript changes the substance: writing H₂O₂ instead of 2H₂O “balances” the oxygen but describes hydrogen peroxide, not water.

State symbols

State symbols record the physical state of each substance under the conditions of the reaction: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water. They are not decoration. They distinguish H₂O(l) from H₂O(g), which have different enthalpies (Reactivity 1.2); they show which substances contribute to a pressure change or to an equilibrium expression (Reactivity 2.3); and they turn an equation into a prediction of what will be seen — a (g) product means bubbles or fizzing, an (s) product formed from two (aq) reactants means a precipitate.

Deducing equations from a description

The skill in the guide is to deduce an equation when the reactants and products are named, which may happen in an unfamiliar context. The chemistry needed is almost always one of a small number of patterns, summarized below; the new context only changes the formulas.

An ionic equation keeps only the species that change. Spectator ions, present unchanged on both sides (Na⁺ and NO₃⁻ in the precipitation above), are removed, and the equation must then balance in charge as well as atoms. The reaction of any strong acid with any strong alkali has the same ionic equation, H⁺(aq) + OH⁻(aq) → H₂O(l), which is why their enthalpies of neutralization are almost identical (Reactivity 1.1).

Exam focus · what the published papers show

2Reacting masses, gas volumes and concentrations 2.1.2 SL + HL

The syllabus statement: The mole ratio of an equation can be used to determine the masses and/or volumes of reactants and products; the concentrations of reactants and products for reactions occurring in solution. The Ar values in the data booklet, to two decimal places, are used.

Every quantitative question in this sub-topic has the same three-step structure. Convert the quantity you are given into an amount in moles; use the mole ratio from the balanced equation to find the amount of the substance you want; convert that amount into the quantity asked for. The only thing that varies is which conversion is used at each end.

Reacting masses

The mole ratio compares amounts, so masses must be turned into amounts before the ratio is used, and turned back afterwards. Comparing grams directly is the most common conceptual error in stoichiometry: 1 g of iron(III) oxide does not give 2 g of iron simply because the equation has “2Fe”.

The check in this example is worth noticing. When a single element passes unchanged from one compound into the product, the mass can be found from the percentage by mass without the equation at all. The equation is essential when two or more substances share an element, or when a gas or solution is involved.

Gas volumes

Avogadro's law (Structure 1.4) states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. Two consequences follow. First, for gases measured at the same temperature and pressure, the ratio of volumes equals the mole ratio, so gas-volume problems can be solved without converting to moles at all. Secondly, at a fixed temperature and pressure every gas has the same molar volume, 22.7 dm³ mol⁻¹ at STP, which allows volumes and amounts to be interconverted.

The airbag can now be designed. Suppose the bag must be filled with 60.0 dm³ of nitrogen, and — as a simplifying assumption — the gas is at STP. Then n(N₂) = 60.0 ÷ 22.7 = 2.643 mol; the ratio NaN₃ : N₂ = 2 : 3 gives n(NaN₃) = 1.762 mol; and with M(NaN₃) = 22.99 + 3(14.01) = 65.02 g mol⁻¹, the mass needed is 115 g. In a real bag the gas is hot, so fewer moles fill the same volume; the calculation would use pV = nRT with the actual temperature. The sodium metal formed is dangerously reactive, and real airbags contain other substances that convert it into harmless silicates.

Concentrations and titrations

For a reaction in solution, the amount of each dissolved reactant is c × V. A titration uses exactly this: a solution of known concentration (the standard solution) is added from a burette until the reaction is just complete, shown by an indicator changing colour at the end point. The volume used gives the amount of the known reactant; the mole ratio gives the amount of the unknown; dividing by the volume of the unknown gives its concentration.

Exam focus · what the published papers show

Past-paper practice · Practice set R2A · Chemical equations and reacting quantities

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R2A.1IB · May 2019 · SL Paper 1 · TZ2 · Q2 · [1]

R2A.2IB · November 2023 · SL Paper 1 · TZ1 · Q4 · [1]

R2A.3IB · November 2018 · SL Paper 1 · Q1 · [1]

R2A.4IB · May 2019 · SL Paper 1 · TZ2 · Q1 · [1]

R2A.5IB · November 2022 · SL Paper 1 · Q3 · [1]

R2A.6IB · May 2021 · SL Paper 1 · TZ1 · Q3 · [1]

R2A.7IB · May 2023 · SL Paper 1 · TZ2 · Q3 · [1]

R2A.8IB · May 2018 · SL Paper 2 · TZ1 · Q1(c), (f) · [4]

Urea, (H₂N)₂CO, is excreted by mammals and can be used as a fertilizer.

(c) Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl: KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq). Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution. [2]

(f) The combustion of urea produces water, carbon dioxide and nitrogen. Formulate a balanced equation for the reaction. [2]

Solutions and mark-scheme guidance · Set R2A

R2A.1 D

Balanced: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O; 2 + 9 + 6 + 6 = 23. The report records that 88 % balanced the equation and added the coefficients correctly.

R2A.2 B

4NH₃ + 5O₂ → 4NO + 6H₂O. Balance N and H first (4NH₃ gives 4NO and 6H₂O), then O: 4 + 6 = 10 O atoms, so 5O₂. Ratio 4 : 5.

R2A.3 A

n(SO₂) = 32 ÷ 64 = 0.50 mol; FeS₂ : SO₂ = 4 : 8 = 1 : 2, so n(FeS₂) = 0.25 mol. B assumes a 1 : 1 ratio.

R2A.4 C

Mg(OH)₂ : NH₃ = 3 : 2, so 0.50 × 3/2 = 0.75 mol. B inverts the ratio.

R2A.5 B

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O. n(C₂H₄) = 7.0 ÷ 28 = 0.25 mol; 2 mol CO₂ per mol, so 0.50 mol.

R2A.6 B

n(butane) = 5.8 ÷ (4 × 12.01 + 10 × 1.01); × 13/2 for O₂; × 22.7 dm³ mol⁻¹. The expression is 5.8 × 13 × 22.7 ÷ [(4 × 12.01 + 10 × 1.01) × 2]. D divides by 1000 unnecessarily: the molar volume is already in dm³.

R2A.7 D

Gas volumes are in the mole ratio. 10 cm³ of A reacted with 20 cm³ of B to give 10 cm³ of product: A + 2B → AB₂, so x = 1, y = 2.

R2A.8 [4]

(c) n(KNCO) = 0.0500 dm³ × 0.100 mol dm⁻³ = 5.00 × 10⁻³ mol ✔; ratio 1 : 1, M(urea) = 60.07 g mol⁻¹, mass = 0.300 g ✔. Award [2] for the correct final answer.

(f) 2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g): correct coefficients on the left ✔, on the right ✔. The scheme accepts any correct ratio, including the equation with 3/2 O₂. N₂, not N, is the product.

3The limiting reactant and theoretical yield 2.1.3 SL + HL

The syllabus statement: The limiting reactant determines the theoretical yield. The skill: identify the limiting and excess reactants from given data and distinguish between the theoretical yield and the experimental yield.

Reactants are rarely mixed in exactly the ratio of the equation. A sandwich shop with 20 slices of bread and 6 slices of cheese can make only 6 cheese sandwiches, however much bread is left over: the cheese limits the output. In the same way, when one reactant is used up the reaction stops, and the remaining reactants are left over. The reactant that is completely consumed is the limiting reactant; any reactant left at the end is in excess.

Identifying the limiting reactant

The limiting reactant is not necessarily the one present in the smaller mass, nor the one present in the smaller number of moles. What matters is the amount relative to the coefficient. Divide the amount of each reactant by its coefficient in the equation: the smallest result identifies the limiting reactant. An equivalent method is to take one reactant, calculate how much of the other it needs, and compare with what is available.

The calculation of the amount left over deserves attention because it is where marks are lost. One report records that candidates accurately deduced the limiting reagent and backed this up with an appropriate calculation, but that calculating the amount of reagent in excess proved much more challenging. The excess is found in two steps — how much reacts, then how much is left — and both are needed.

How product depends on the amount of one reactant

If one reactant is kept fixed and increasing amounts of the other are added, the amount of product first rises in proportion to the added reactant, which is limiting. Once enough has been added to react with all of the fixed reactant, the fixed reactant becomes limiting and the amount of product stays constant, however much more is added. The graph is therefore a straight line from the origin followed by a horizontal plateau. The same reasoning explains why the final volume of gas in a rate experiment depends only on the amount of limiting reactant (Reactivity 2.2).

Theoretical and experimental yield

The theoretical yield is a calculation; the experimental yield is a measurement. They differ because real experiments are not ideal. The guide asks what errors might make the experimental yield lower or higher than the theoretical yield, and the answer is found by thinking about what happens to the product and what else ends up in the weighing container.

A percentage yield above 100 % is therefore not impossible to record, but it is always evidence of an error, never of extra product created. The explanation must say which extra substance was weighed.

4Percentage yield 2.1.4 SL + HL

The syllabus statement: The percentage yield is calculated from the ratio of experimental yield to theoretical yield. The skill: solve problems involving reacting quantities, limiting and excess reactants, theoretical, experimental and percentage yields.

Both yields must be in the same unit, either mass or amount. Because the theoretical yield is the maximum, the percentage yield of a correctly performed experiment lies between 0 and 100 %. It is a measure of how much of the possible product was actually obtained; it says nothing about how much waste the chosen reaction produces, which is the job of atom economy (2.1.5).

The check line records the most expensive error in this outcome. On one structured item a report describes many being awarded no marks because they just calculated the final mass as a percentage of the initial. Masses of different substances cannot be compared; the theoretical yield must be found through the mole ratio first. A percentage yield can also be calculated from amounts: 0.01027 mol of aspirin obtained out of 0.01448 mol possible gives the same 70.9 %.

5Atom economy 2.1.5 SL + HL

The syllabus statement: The atom economy is a measure of efficiency in green chemistry. The skill: calculate the atom economy from the stoichiometry of a reaction, including the inverse relationship between atom economy and wastage in industrial processes.

A reaction can give a 100 % yield and still be wasteful. Ethanol is made industrially in two ways. Fermentation of glucose gives ethanol and carbon dioxide; the hydration of ethene gives ethanol only. Even if both went perfectly, the fermentation route would send almost half of the mass of the starting material out of the reactor as carbon dioxide. Atom economy measures this: it is the percentage of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

The equation is given in the data booklet. The molar masses are multiplied by the coefficients in the equation, so the calculation uses the stoichiometric masses of reactants and of the wanted product. Because mass is conserved, the total mass of reactants equals the total mass of all products, so atom economy can equally be written as the mass of desired product divided by the mass of all products.

Addition reactions, in which two molecules combine into one, always have an atom economy of 100 %; this includes addition polymerization, and a report notes that most recognized that an addition polymerization occurs and hence the atom economy of the first step is 100%. Substitution and elimination reactions always produce a second product and have lower atom economies. Where a process gives several useful products, the atom economy depends on which products are counted as desired.

The inverse relationship with waste

Whatever mass of reactants does not end up in the desired product ends up as by-products. A low atom economy therefore means a high proportion of waste, per kilogram of product, even when the yield is perfect: an atom economy of 51 % means that 49 % of the reactant mass becomes something else. The by-products must be separated, treated, stored or sold, and they consume raw materials and energy that do not become product. Green chemistry aims to design reactions with high atom economy so that waste is prevented rather than treated.

Yield and atom economy measure different things

A full assessment of the efficiency of a process, which the guide invites, also considers the energy used and the temperature and pressure required, the rate of reaction, whether the catalyst can be recovered, the hazards of reactants, solvents and by-products, whether the feedstock is renewable, and the cost of separating the product. Atom economy and yield are two numbers in that judgement, not the whole of it.

Past-paper practice · Practice set R2B · Limiting reactant, yield and atom economy

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R2B.1IB · May 2021 · SL Paper 1 · TZ2 · Q1 · [1]

R2B.2IB · November 2016 · SL Paper 1 · Q4 · [1]

R2B.3IB · May 2016 · SL Paper 1 · Q3 · [1]

R2B.4IB · May 2018 · SL Paper 1 · TZ1 · Q4 · [1]

R2B.5IB · November 2023 · SL Paper 1 · TZ1 · Q29 · [1]

R2B.6IB · November 2018 · SL Paper 2 · Q1(a) · [4]

3.26 g of iron powder are added to 80.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s).

(i) Determine the limiting reactant showing your working. [2]

(ii) The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper. [2]

R2B.7IB · May 2019 · SL Paper 2 · TZ1 · Q3(b) · [2]

Sodium peroxide, Na₂O₂, is formed by the reaction of sodium oxide with oxygen: 2Na₂O(s) + O₂(g) → 2Na₂O₂(s). Calculate the percentage yield of sodium peroxide if 5.00 g of sodium oxide produces 5.50 g of sodium peroxide. [2]

R2B.8 HL paperIB · November 2023 · HL Paper 2 · TZ1 · Q3(b) · [2]

Methanoic acid can be converted into methyl methanoate, HCOOCH₃. 1.72 g of methyl methanoate is produced from 2.83 g of methanoic acid and excess of the other reagent. Determine the percentage yield. [2]

R2B.9 PracticePractice · written for these notes · [3]

Chloroethane can be made in two ways: (1) C₂H₄(g) + HCl(g) → C₂H₅Cl(g); (2) C₂H₆(g) + Cl₂(g) → C₂H₅Cl(g) + HCl(g). Calculate the atom economy of each route and state, with a reason, which produces less waste. [3]

R2B.10 PracticePractice · written for these notes · [3]

Titanium is extracted by the reaction TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(l). (a) Calculate the atom economy for titanium. (b) Suggest how the atom economy of the whole process could be improved. [3]

Solutions and mark-scheme guidance · Set R2B

R2B.1 C

HCl: 0.10 ÷ 2 = 0.05; Mg: 0.20 ÷ 1 = 0.20. HCl is limiting and gives 0.10 ÷ 2 = 0.05 mol H₂.

R2B.2 B

Fe₂O₃: 5.0 ÷ 1 = 5.0; CO: 6.0 ÷ 3 = 2.0, so CO is limiting. It uses 2.0 mol Fe₂O₃, leaving 5.0 − 2.0 = 3.0 mol.

R2B.3 B

H₂SO₄ + 2NaOH: 0.10 mol H₂SO₄ needs 0.20 mol NaOH, so with only 0.10 mol NaOH the NaOH runs out. In A the amounts are exactly stoichiometric; HNO₃ reacts 1 : 1, so in C and D NaOH is in excess or exactly matched.

R2B.4 D

Theoretical yield = (7 ÷ 28) × 46 g; percentage = 6 ÷ [(7 ÷ 28) × 46] × 100 = (6 × 28 × 100) ÷ (7 × 46).

R2B.5 A

Once the limiting reactant has been consumed, adding more of the excess reactant forms no more product: a horizontal line. Only 47 % chose it.

R2B.6 [4]

(i) n(CuSO₄) = 0.0800 × 0.200 = 0.0160 mol AND n(Fe) = 3.26 ÷ 55.85 = 0.0584 mol ✔; ratio 1 : 1, so CuSO₄ is the limiting reactant ✔. The second mark is not awarded without the mole calculation.

(ii) Theoretical mass of Cu = 0.0160 × 63.55 = 1.02 g ✔; percentage yield = 0.872 ÷ 1.02 × 100 = 85.5 % ✔ (85.6 % if 1.017 g is carried). The report found this generally well done.

R2B.7 [2]

n(Na₂O) = 5.00 ÷ 61.98 = 0.0807 mol = theoretical n(Na₂O₂); theoretical mass = 0.0807 × 77.98 = 6.291 g ✔. Percentage yield = 5.50 ÷ 6.291 × 100 = 87.4 % ✔ (or 0.0705 ÷ 0.0807 × 100).

R2B.8 [2]

Expected yield = 2.83 × 60.06 ÷ 46.03 = 3.69 g ✔; percentage yield = 1.72 ÷ 3.69 × 100 = 46.6 % ✔ (46.5 % by amounts: 0.0286 ÷ 0.0615). The scheme awards [0] for 60.8 %, the simple ratio of final to starting mass — the error the report says left many with no marks.

R2B.9 [3]

M(C₂H₅Cl) = 64.52 g mol⁻¹. Route 1: 64.52 ÷ (28.06 + 36.46) × 100 = 100 % ✔ (addition). Route 2: 64.52 ÷ (30.08 + 70.90) × 100 = 63.9 % ✔. Route 1 produces less waste: all reactant atoms end up in the product, whereas route 2 forms HCl as a by-product ✔ (route 2 also gives further substitution products).

R2B.10 [3]

(a) M(TiCl₄) = 47.87 + 4(35.45) = 189.67; 2M(Mg) = 48.62; total 238.29 g ✔. Atom economy = 47.87 ÷ 238.29 × 100 = 20.1 % ✔. (b) Use the by-product: electrolyse the MgCl₂ to regenerate magnesium and chlorine, which can be recycled ✔.

Review · Reactivity 2.1

6Misconceptions, the examiner’s view, and the question types

Six question types cover the sub-topic.

7Quick check

8Summary and knowledge organiser

Essential knowledge

• Coefficients give the mole ratio; balancing conserves atoms (and charge in ionic equations) by changing coefficients only.
• State symbols: (s), (l), (g), (aq). Water is (l); dissolved salts, acids and alkalis are (aq); precipitates are (s).
• Every stoichiometry calculation: given quantity → amount → mole ratio → amount → required quantity.
• n = m/M; n = cV (V in dm³); n = V/22.7 dm³ mol⁻¹ at STP; gas volume ratio = mole ratio at the same T and p.
• Limiting reactant: smallest amount ÷ coefficient; it fixes the theoretical yield. Excess left = present − reacted.
• Percentage yield = experimental ÷ theoretical × 100 %. Atom economy = M(desired) ÷ ΣM(reactants) × 100 %; low atom economy means more waste.

Examination checklist

• Write and check the balanced equation before any calculation.
• Convert cm³ to dm³; use Ar values to two decimal places.
• Show amount, ratio and answer on separate lines so error carried forward can be awarded.
• Never express one substance's mass as a percentage of another's.
• Give a unit and appropriate significant figures with every final answer.

Knowledge organiser