Understand Amount of Chemical Change

SECTION 1

Historical Context & Motivation

For most of human history, chemistry was a practical art rather than a precise science. Alchemists mixed substances by feel, measuring ingredients by handfuls rather than by formula. The transformation from guesswork to prediction required answering one deceptively simple question: how much? How much iron can a lump of ore yield? How much oxygen does a candle consume? Answering these questions demanded new concepts—conservation of mass, fixed proportions, and ultimately the mole—that form the backbone of quantitative chemistry.

1774

Lavoisier & Conservation of Mass

Antoine Lavoisier conducted careful combustion experiments in sealed vessels, demonstrating that total mass is conserved in every chemical reaction. This principle made quantitative chemistry possible.

1799

Proust & Definite Proportions

Joseph Proust showed that a given compound always contains the same elements in the same mass ratio, establishing the law of definite proportions.

1803

Dalton's Atomic Theory

John Dalton proposed that atoms of each element have a characteristic mass, allowing chemists to compute relative atomic masses and write formulas for compounds.

1811

Avogadro's Hypothesis

Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain the same number of particles, linking macroscopic volume to particle count.

1971

The Mole Becomes an SI Unit

The 14th General Conference on Weights and Measures formally adopted the mole as the SI unit for amount of substance, defined via Avogadro's number (6.022 × 10²³).

These developments converge on a single challenge that remains central to IB Chemistry today: given a balanced equation, can you predict the exact masses, volumes, and amounts of every reactant and product? That is the question Reactivity 2.1 equips you to answer.

SECTION 2

Core Principles & Definitions

Quantitative chemistry rests on a handful of interconnected ideas. Understanding these will let you move confidently between mass, moles, number of particles, and gas volume—all the quantities the IB expects you to handle.

The Mole (mol)

One mole is exactly 6.022 × 10²³ particles (Avogadro's constant, N_A). It bridges the atomic scale to the laboratory scale, much like "a dozen" bridges individual eggs to a carton.

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 read from the periodic table. For H₂O: M = 2(1.01) + 16.00 = 18.02 g mol⁻¹.

Stoichiometry

The mole ratios given by the coefficients in a balanced equation. In 2 H₂ + O₂ → 2 H₂O, the ratio is 2 : 1 : 2, telling you exactly how many moles of each substance react or form.

Limiting & Excess Reagents

The limiting reagent is the reactant that runs out first, controlling the maximum amount of product. The other reactant(s) are in excess.

Molar Volume of a Gas

At STP (0 °C, 100 kPa), one mole of any ideal gas occupies approximately 22.7 dm³. This lets you convert directly between moles of gas and volume.

✦ KEY TAKEAWAY

Think of a balanced equation like a recipe: "2 cups flour + 1 cup sugar → 2 batches of cookies." The recipe tells you the ratio, not the absolute amount. Doubling or halving all ingredients keeps the ratio intact. The mole is the chemist's "cup"—a standard counting unit that makes the recipe work at any scale.

SECTION 3

Visual Explanation — The Mole Conversion Map

The diagram below shows the central role of the mole in quantitative chemistry. Starting from any measurable quantity—mass, number of particles, concentration, or gas volume—you can always convert to moles first, then use stoichiometry to reach any other quantity. This "mole map" is the single most important problem-solving tool in Reactivity 2.1.

The mole conversion map. Start from any outer box (mass, particles, gas volume, or concentration) and convert to the central moles hub using the formula on each arrow. Then apply the stoichiometric ratio from the balanced equation to find moles of any other substance.

Notice that every conversion passes through moles. When a problem asks you to go from the mass of reactant A to the volume of gas B, you follow three steps: convert mass to moles (÷ M), apply the mole ratio from the balanced equation, then convert moles to volume (× 22.7 dm³ mol⁻¹ at STP). Keeping this map in mind prevents you from ever feeling lost during a stoichiometry problem.

SECTION 4

Mathematical Framework

All stoichiometric calculations rely on a small set of equations. Master these, and every IB problem in Reactivity 2.1 reduces to picking the right formula and plugging in values.

MOLES FROM MASS

n = m / M

Where n = amount in mol, m = mass in g, M = molar mass in g mol⁻¹.

MOLES FROM PARTICLE COUNT

n = N / Nₐ

Where N = number of particles and Nₐ = 6.022 × 10²³ mol⁻¹ (Avogadro's constant).

MOLES FROM GAS VOLUME (STP)

n = V / 22.7

Where V = volume in dm³ at STP (0 °C, 100 kPa). The IB data booklet value for molar volume at STP is 22.7 dm³ mol⁻¹.

MOLES FROM SOLUTION CONCENTRATION

n = c × V

Where c = concentration in mol dm⁻³ and V = volume of solution in dm³. Remember to convert cm³ to dm³ by dividing by 1000.

⚠️ IB Exam Tip

The IB defines STP as 0 °C and 100 kPa. Some older textbooks use 1 atm (101.3 kPa) and quote 22.4 dm³ mol⁻¹. On the IB exam, always use 22.7 dm³ mol⁻¹ unless told otherwise.

SECTION 5

Limiting Reagents & Percentage Yield

In real experiments, reactants are rarely provided in perfect stoichiometric amounts. One reactant typically runs out before the others, and it is this limiting reagent that determines the maximum amount of product—the theoretical yield. The remaining reactant that is not fully consumed is called the excess reagent.

When 2 mol N₂ and 4 mol H₂ are mixed, H₂ needs 4/3 mol N₂ (less than the 2 mol available), so H₂ is limiting and determines the amount of NH₃ produced.

To identify the limiting reagent, convert the given mass or volume of each reactant to moles, then divide each by its stoichiometric coefficient. The reactant with the smallest quotient is the limiting reagent.

PERCENTAGE YIELD

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

The theoretical yield is calculated from the limiting reagent. The actual yield is what you actually collect in the lab. Losses occur due to side reactions, incomplete reactions, or transfer losses.

ATOM ECONOMY

% atom economy = (M of desired product / Σ M of all products) × 100

Atom economy measures how much of the reactant atoms end up in the desired product. A reaction with 100% atom economy produces no by-products. This is important for green chemistry.

SECTION 6

Worked Example — Stoichiometry with a Limiting Reagent

Problem: 5.40 g of aluminium reacts with 25.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid. The equation is:

2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)

Determine the limiting reagent and calculate the maximum mass of H₂ produced.

Stoichiometry with Limiting Reagent

Step 1 — Calculate moles of AlM(Al) = 26.98 g mol⁻¹. Using n = m / M: n(Al) = 5.40 / 26.98 = 0.200 mol

n(Al) = 0.200 mol

Step 2 — Calculate moles of HClConvert volume: 25.0 cm³ = 0.0250 dm³. Using n = c × V: n(HCl) = 2.00 × 0.0250 = 0.0500 mol

n(HCl) = 0.0500 mol

Step 3 — Identify the limiting reagentDivide each amount by its coefficient in the balanced equation: Al: 0.200 / 2 = 0.100 HCl: 0.0500 / 6 = 0.00833 HCl has the smaller quotient, so HCl is the limiting reagent.

Limiting reagent: HCl

Step 4 — Calculate moles of H₂ producedFrom the equation, the mole ratio HCl : H₂ is 6 : 3, which simplifies to 2 : 1. n(H₂) = 0.0500 × (3/6) = 0.0250 mol

n(H₂) = 0.0250 mol

Step 5 — Convert moles of H₂ to massM(H₂) = 2 × 1.01 = 2.02 g mol⁻¹. Using m = n × M: m(H₂) = 0.0250 × 2.02 = 0.0505 g

Maximum mass of H₂ = 0.0505 g

SECTION 7

Strengths, Limitations & Common Pitfalls

Stoichiometric calculations are powerful, but they rest on several assumptions. Understanding where those assumptions break down will keep you from making errors on the exam and in the lab.

Strengths and limitations of stoichiometric predictions

Strengths	Limitations
Predicts exact amounts of reactants consumed and products formed from a balanced equation.	Assumes the reaction goes to completion; many reactions reach equilibrium and stop short.
Works for any phase—solids, liquids, gases, and solutions—using appropriate conversion formulas.	Gas volume formula (n = V / 22.7) assumes ideal gas behaviour, which breaks down at high pressures or low temperatures.
Allows calculation of atom economy and percentage yield, useful for evaluating efficiency.	Does not account for practical losses such as transfer losses, side reactions, or impurities in reagents.
Applies universally to all balanced chemical equations, from combustion to acid–base reactions.	Requires a correctly balanced equation; an incorrect equation will produce wrong predictions.

⚠️ COMMON PITFALLS

Three frequent mistakes in IB exams: (1) forgetting to convert cm³ to dm³ before using n = c × V, (2) using the wrong molar volume (22.4 instead of the IB value 22.7 dm³ mol⁻¹ at STP), and (3) reading the wrong coefficients because the equation is unbalanced. Always balance first, convert to moles, then apply the ratio.

SECTION 8

Connection to Advanced Topics

Stoichiometry is the foundation, but IB Chemistry builds several more sophisticated ideas on top of it. The table below previews how the skills you have learned in Reactivity 2.1 connect to later topics.

From stoichiometry to advanced IB topics

Reactivity 2.1 Skill	Advanced Application	Where in IB
Mole ratios from balanced equations	Equilibrium constant expressions (K) use mole-based concentrations or partial pressures	Reactivity 3 – Equilibrium
Limiting reagent identification	Titration calculations require identifying the point where the analyte is fully consumed	Reactivity 2.3 – Acid–base
Percentage yield	Evaluating green chemistry and industrial efficiency; multi-step synthesis yield	Reactivity 2.1 (HL) & IA
n = c × V for solutions	Electrochemistry and Faraday's laws use moles of electrons transferred	Reactivity 3.2 – Electron transfer

At Higher Level, you will also encounter back-titration and multi-step reaction sequences where you chain stoichiometric calculations together. Every one of these extensions begins with the same question: How many moles? If you can answer that confidently, you are well prepared for the rest of the course.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

Explain why a balanced chemical equation is necessary before performing any stoichiometric calculation. What information do the coefficients provide?

PROBLEM 2 — BASIC CALCULATION

Calculate the mass of carbon dioxide produced when 12.0 g of carbon reacts completely with excess oxygen: C(s) + O₂(g) → CO₂(g). (M: C = 12.01, O = 16.00)

PROBLEM 3 — INTERMEDIATE

In the reaction 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g), 4.60 g of sodium is added to excess water. Calculate the volume of hydrogen gas produced at STP. (M: Na = 22.99)

PROBLEM 4 — APPLIED

A student reacts 3.24 g of Zn with 50.0 cm³ of 0.800 mol dm⁻³ HCl: Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g). Identify the limiting reagent, calculate the theoretical yield of H₂ in grams, and determine the percentage yield if 0.0720 g of H₂ is collected. (M: Zn = 65.38, H = 1.01)

PROBLEM 5 — CRITICAL THINKING

A reaction has two possible pathways producing the same desired product but different by-products. Pathway A: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Pathway B: 2 C₃H₈ + 7 O₂ → 6 CO + 8 H₂O (with CO as an undesired by-product). Compare the atom economy of these two pathways with respect to CO₂ (Pathway A) and CO (Pathway B). Discuss why atom economy alone is not sufficient to evaluate the "greenness" of a reaction.

SUMMARY

Lesson Summary

The mole is the central unit linking the atomic world to the laboratory. You convert between mass (n = m / M), number of particles (n = N / N_A), gas volume at STP (n = V / 22.7), and solution concentration (n = c × V) to reach moles. Once in moles, the stoichiometric coefficients of a balanced equation provide the ratio needed to move between substances.

When reactants are not in perfect ratio, the limiting reagent determines the maximum product—the theoretical yield. The percentage yield compares what you actually collect to this maximum, while atom economy evaluates how efficiently atoms end up in the desired product. Together, these tools let you predict, measure, and evaluate the amount of chemical change in any reaction.