Apply Amount of Chemical Change — Apply Reactivity 2.1—How much? The amount of chemical change in problem-solving and explanations - IB Chemistry

SECTION 1

Historical Context & Motivation

For centuries, alchemists mixed substances together hoping for gold, but they had no reliable way to predict how much of each ingredient was needed or how much product would form. Chemistry only became a true science when researchers started measuring masses carefully and discovered that reactions follow fixed, predictable ratios. The question "How much?" is the foundation of all quantitative chemistry, from designing medicines to manufacturing fertilizers.

1774 • Lavoisier and Conservation of Mass

Antoine Lavoisier demonstrated that mass is conserved in chemical reactions by carefully weighing reactants and products in sealed containers. This law of conservation of mass became the bedrock of stoichiometry.

1803 • Dalton's Atomic Theory

John Dalton proposed that elements consist of indivisible atoms with fixed masses, explaining why compounds always form in definite proportions by mass—the law of definite proportions.

1811 • Avogadro's Hypothesis

Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain the same number of particles, laying the groundwork for the mole concept.

1865 • Loschmidt's Number

Josef Loschmidt made the first estimate of the number of molecules in a given volume of gas. This work eventually led to the precise determination of Avogadro's constant (6.022 × 10²³ mol⁻¹).

1971 • The Mole Becomes an SI Unit

The 14th General Conference on Weights and Measures adopted the mole as the seventh SI base unit, formalizing amount of substance as a fundamental measurement in science.

These breakthroughs converged into a single powerful question that IB Chemistry asks you to master: given a balanced equation, how can you calculate the exact amounts of reactants consumed and products formed? This is the heart of Reactivity 2.1, and the skills you develop here will be used in virtually every subsequent topic in the course.

SECTION 2

Core Principles & Definitions

Before you can solve any "how much" problem, you need to be fluent with a handful of foundational ideas. These concepts connect the invisible world of atoms and molecules to the measurable world of grams and litres on your lab bench.

The Mole (n)

A mole is the amount of substance containing exactly 6.022 × 10²³ particles (atoms, molecules, ions, etc.). It bridges particle counts and measurable mass.

Molar Mass (M)

The mass of one mole of a substance, expressed in g mol⁻¹. Numerically equal to the relative atomic or molecular mass from the periodic table.

Stoichiometric Coefficients

The numbers in front of formulas in a balanced equation represent mole ratios. For example, 2H₂ + O₂ → 2H₂O means 2 mol H₂ reacts with 1 mol O₂.

Limiting & Excess Reagents

The limiting reagent is the reactant that runs out first, determining the maximum yield. The other reactant(s) are in excess.

Theoretical & Percentage Yield

Theoretical yield is the maximum product predicted by stoichiometry. Percentage yield = (actual yield ÷ theoretical yield) × 100%.

✦

KEY TAKEAWAY

Think of a balanced equation like a recipe. If a cake recipe calls for 2 eggs and 1 cup of flour, you can't just triple the eggs without also tripling the flour. The mole ratio from the balanced equation is your recipe—it tells you the exact proportions of each ingredient (reactant) and each product. If you run out of eggs first, eggs are your limiting reagent.

SECTION 3

Visual Explanation — The Stoichiometry Roadmap

Every stoichiometry problem follows the same three-step pathway: convert what you're given into moles, use the mole ratio from the balanced equation, and then convert moles into whatever unit the question asks for. The diagram below is your universal roadmap for solving any quantitative chemistry problem.

The stoichiometry roadmap. Every problem starts by converting given information (mass, volume, or concentration) into moles, then uses the balanced-equation mole ratio to find moles of the desired substance, and finally converts back to the required unit.

Notice that moles sit at the centre of the roadmap. No matter what unit a question gives you—grams, litres of gas, or concentration of a solution—you always convert to moles first. The mole ratio from the balanced equation is the bridge that connects one substance to another. After crossing that bridge, you convert moles back into whatever unit the question asks for. Memorise this three-step flow and you will be able to tackle any stoichiometry problem the IB throws at you.

SECTION 4

Mathematical Framework

The following equations form your toolkit for solving amount-of-substance problems. Each one converts between a measurable quantity and moles. You should know all of these by heart for IB Chemistry.

MOLES FROM MASS

n = m ÷ M

n = amount of substance (mol), m = mass of the sample (g), M = molar mass (g mol⁻¹). This is the most commonly used equation in stoichiometry.

MOLES OF GAS AT STP

n = V ÷ 22.7

V = volume of gas (dm³ or L) at STP (standard temperature and pressure: 273.15 K, 100 kPa). The IB data booklet gives the molar volume of an ideal gas at STP as 22.7 dm³ mol⁻¹.

MOLES FROM CONCENTRATION

n = c × V

c = concentration (mol dm⁻³), V = volume of solution (dm³). Be careful with units—if volume is given in cm³, divide by 1000 to convert to dm³.

PERCENTAGE YIELD

% yield = (actual yield ÷ theoretical yield) × 100%

The theoretical yield is calculated from stoichiometry using the limiting reagent. The actual yield is measured experimentally. Percentage yield is always ≤ 100%.

⚠

IB TIP — Units Matter!

The most common error in IB exams is a unit mismatch. Always check: is the volume in cm³ or dm³? The formula n = c × V requires volume in dm³. To convert: 1 dm³ = 1000 cm³. Losing marks to unit errors is completely avoidable—build the habit of writing units at every step.

SECTION 5

Limiting Reagent Analysis — A Deeper Look

In most real-world reactions, reactants are not mixed in perfect stoichiometric proportions. One reactant will be completely consumed before the other, and this is the limiting reagent. The amount of product formed is always determined by the limiting reagent, never the excess reagent. To identify the limiting reagent, calculate the moles of each reactant and compare them using the mole ratio from the balanced equation.

A particle-level view of a limiting reagent scenario. With 2 mol N₂ and only 3 mol H₂, the H₂ is limiting because the reaction requires 3 mol H₂ per mol N₂. Only 1 mol N₂ reacts, leaving 1 mol N₂ in excess. The quick method divides moles by the stoichiometric coefficient for each reactant—the smallest value identifies the limiting reagent.

The quick method shown at the bottom of the diagram is extremely efficient for IB exams. Simply divide the number of moles of each reactant by its stoichiometric coefficient. The reactant with the smallest ratio is the limiting reagent. Use the limiting reagent's moles to calculate the theoretical yield of any product.

SECTION 6

Worked Example — From Start to Finish

Let's work through a complete IB-style problem that combines multiple concepts: mole calculations, limiting reagent identification, theoretical yield, and percentage yield.

ℹ

PROBLEM

Iron(III) oxide reacts with carbon monoxide according to the equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂. A student heats 24.0 g of Fe₂O₃ with 15.0 g of CO. (a) Determine the limiting reagent. (b) Calculate the theoretical yield of iron. (c) If 12.5 g of iron was collected, find the percentage yield.

Solution: Fe₂O₃ + 3CO → 2Fe + 3CO₂

1
Step 1 — Calculate molar masses

Using the periodic table: M(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70 g mol⁻¹. M(CO) = 12.01 + 16.00 = 28.01 g mol⁻¹. M(Fe) = 55.85 g mol⁻¹.
2
Step 2 — Convert masses to moles

n(Fe₂O₃) = 24.0 ÷ 159.70 = 0.1503 mol. n(CO) = 15.0 ÷ 28.01 = 0.5355 mol.
3
Step 3 — Identify the limiting reagent

Divide moles by the stoichiometric coefficient for each reactant. For Fe₂O₃: 0.1503 ÷ 1 = 0.1503. For CO: 0.5355 ÷ 3 = 0.1785. The smaller value is 0.1503 (Fe₂O₃), so Fe₂O₃ is the limiting reagent.

Fe₂O₃ is limiting
4
Step 4 — Calculate theoretical yield of Fe

From the balanced equation, 1 mol Fe₂O₃ produces 2 mol Fe. So n(Fe) = 0.1503 × 2 = 0.3006 mol. Mass of Fe = 0.3006 × 55.85 = 16.79 g (this is the theoretical yield).

Theoretical yield = 16.8 g Fe
5
Step 5 — Calculate percentage yield

% yield = (actual yield ÷ theoretical yield) × 100% = (12.5 ÷ 16.79) × 100% = 74.4%. This means 74.4% of the maximum possible iron was collected. Losses may be due to incomplete reaction, transfer losses, or side reactions.

Percentage yield = 74.4%

SECTION 7

Common Strengths & Pitfalls

Stoichiometric calculations are among the most reliable tools in chemistry, but students frequently lose marks on IB exams due to avoidable errors. The table below highlights the strengths of this approach and the most common pitfalls to watch out for.

Strength	Common Pitfall	How to Avoid It
Balanced equations give exact mole ratios	Using an unbalanced equation, giving wrong ratios	Always balance the equation first, checking each element's atom count
n = m ÷ M works for any solid, liquid, or dissolved substance	Using atomic mass instead of molecular/formula mass	Write out the full formula and add up all atoms. For Fe₂O₃, don't forget the subscripts
Limiting reagent method predicts exact maximum product	Assuming the substance with the smaller mass is limiting	Compare moles ÷ coefficient, not raw masses. A heavier compound can still be limiting
Gas volume calculations are straightforward at STP	Using 22.4 dm³ (old value) instead of 22.7 dm³ (IB value)	The IB uses 22.7 dm³ mol⁻¹ at STP (100 kPa). Always check your data booklet
n = c × V is simple for solution chemistry	Mixing up cm³ and dm³ for volume	Convert cm³ to dm³ by dividing by 1000 before substituting into the formula

Common strengths and pitfalls in stoichiometric calculations

✦

KEY TAKEAWAY

Stoichiometry is like GPS navigation. If you enter the correct starting address (balanced equation and accurate data), the GPS gives you a reliable route (mole ratio) and destination (product amount). But if you enter the wrong starting address—say, an unbalanced equation or the wrong molar mass—the GPS will confidently send you to the wrong place. Garbage in, garbage out. Always double-check your setup before calculating.

SECTION 8

Connection to Advanced Topics

The stoichiometric skills you are building now extend far beyond basic "how much" calculations. In higher-level IB Chemistry and university courses, the same mole-ratio reasoning underpins topics such as enthalpy calculations, equilibrium expressions, electrochemistry, and analytical techniques like titrations. The table below shows how the core ideas from Reactivity 2.1 connect to more advanced content.

Reactivity 2.1 Concept	Advanced Application	Where You'll See It
Mole ratio from balanced equation	Calculating enthalpy change per mole of reaction (ΔH)	Reactivity 1 — Measuring Enthalpy Changes
n = c × V for solutions	Acid–base and redox titration calculations	Reactivity 3 — What are the mechanisms?
Limiting reagent analysis	ICE tables for equilibrium calculations	Reactivity 2.3 — How far? The extent of chemical change
Percentage yield	Atom economy and green chemistry metrics	Structure 1 — Models of the particulate nature of matter

How Reactivity 2.1 connects to advanced IB Chemistry topics

Think of stoichiometry as the grammar of chemistry. Just as you can't write essays without understanding grammar, you can't tackle any quantitative chemistry problem—from calorimetry to electrochemical cell calculations—without fluency in mole conversions and balanced equations. The time you invest mastering these fundamentals will pay dividends throughout the rest of the course.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In the reaction 2Mg + O₂ → 2MgO, a student says: "Since there's a 2 in front of Mg and a 1 in front of O₂, you need twice as much oxygen by mass." Is this statement correct? Explain your reasoning.

Show answer

The statement is incorrect. The coefficients represent mole ratios, not mass ratios. The equation tells us 2 mol Mg reacts with 1 mol O₂, meaning you need twice as many moles of Mg as O₂. Because Mg and O₂ have different molar masses (24.31 g mol⁻¹ vs. 32.00 g mol⁻¹), the mass ratio is 2 × 24.31 : 1 × 32.00 = 48.62 : 32.00, which is approximately 1.5 : 1 by mass—not 2 : 1. Always work with moles, not grams, when using stoichiometric coefficients.

Question 2: BASIC CALCULATION

Calculate the mass of carbon dioxide (CO₂) produced when 10.0 g of calcium carbonate (CaCO₃) decomposes completely: CaCO₃ → CaO + CO₂.

Show answer

First, calculate M(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09 g mol⁻¹ and M(CO₂) = 12.01 + 2(16.00) = 44.01 g mol⁻¹. Find moles: n(CaCO₃) = 10.0 ÷ 100.09 = 0.09991 mol. The mole ratio is 1 : 1, so n(CO₂) = 0.09991 mol. Convert to mass: m(CO₂) = 0.09991 × 44.01 = 4.40 g. This is a straightforward mass-to-moles-to-mass conversion following the stoichiometry roadmap.

Question 3: INTERMEDIATE

25.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid (HCl) is added to 20.0 cm³ of 0.150 mol dm⁻³ sodium hydroxide (NaOH). HCl + NaOH → NaCl + H₂O. Determine the limiting reagent and calculate the number of moles of NaCl produced.

Show answer

Convert volumes to dm³: V(HCl) = 25.0 ÷ 1000 = 0.0250 dm³; V(NaOH) = 20.0 ÷ 1000 = 0.0200 dm³. Calculate moles: n(HCl) = 0.200 × 0.0250 = 0.00500 mol; n(NaOH) = 0.150 × 0.0200 = 0.00300 mol. Since the mole ratio is 1 : 1, compare directly: 0.00500 mol HCl vs. 0.00300 mol NaOH. NaOH has fewer moles and is the limiting reagent. Therefore, n(NaCl) = 0.00300 mol. Note that 0.00200 mol HCl remains unreacted (in excess).

Question 4: APPLIED

In a lab synthesis, a student reacts 5.40 g of aluminium with excess hydrochloric acid: 2Al + 6HCl → 2AlCl₃ + 3H₂. The hydrogen gas produced is collected at STP. What volume of H₂ gas (in dm³) should the student expect? (Use V_m = 22.7 dm³ mol⁻¹.)

Show answer

M(Al) = 26.98 g mol⁻¹. n(Al) = 5.40 ÷ 26.98 = 0.2001 mol. From the balanced equation, 2 mol Al produces 3 mol H₂, so n(H₂) = 0.2001 × (3 ÷ 2) = 0.3002 mol. At STP, V(H₂) = 0.3002 × 22.7 = 6.81 dm³. This problem combines the n = m ÷ M formula, the mole ratio, and the gas volume formula V = n × V_m in a single multi-step calculation—exactly the kind of chain you'll see on IB Paper 2.

Question 5: CRITICAL THINKING

A student performs the reaction: 2KClO₃ → 2KCl + 3O₂. She starts with 12.25 g of KClO₃ and collects 4.20 g of O₂. Calculate the percentage yield. Then suggest two reasons why the yield might be less than 100%, and explain whether a percentage yield greater than 100% would ever be scientifically valid.

Show answer

M(KClO₃) = 39.10 + 35.45 + 3(16.00) = 122.55 g mol⁻¹; M(O₂) = 32.00 g mol⁻¹. n(KClO₃) = 12.25 ÷ 122.55 = 0.09995 mol. From the equation, 2 mol KClO₃ → 3 mol O₂, so n(O₂) = 0.09995 × (3 ÷ 2) = 0.14993 mol. Theoretical mass of O₂ = 0.14993 × 32.00 = 4.798 g. Percentage yield = (4.20 ÷ 4.798) × 100% = 87.5%. Possible reasons for less than 100% yield: (1) Incomplete decomposition—some KClO₃ may not have fully reacted. (2) Gas collection losses—O₂ may have escaped before being collected. As for yields above 100%, this is not scientifically valid and would indicate experimental error, such as water vapour condensing with the product and increasing its apparent mass, or an impure starting material that decomposes more easily.

SUMMARY

Summary — Amount of Chemical Change

Summary

Reactivity 2.1 centres on answering one question: how much? Every stoichiometry problem follows the same pattern. First, convert given quantities into moles using the appropriate formula: n = m ÷ M for masses, n = V ÷ 22.7 for gas volumes at STP, or n = c × V for solutions. Next, use the mole ratio from the balanced equation to find the moles of the target substance. Finally, convert back to the required unit.

When reactants are not in stoichiometric proportions, identify the limiting reagent by dividing each reactant's moles by its coefficient—the smallest value determines the maximum product. Calculate theoretical yield from the limiting reagent and compare it to the actual yield to find percentage yield. These skills form the quantitative backbone of all subsequent IB Chemistry topics—from enthalpy calculations to equilibrium analysis.