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.
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.
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)
Molar Mass (M)
Stoichiometric Coefficients
Limiting & Excess Reagents
Theoretical & Percentage Yield
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.
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.
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.
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.
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.
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.
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 |
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 |
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.
Practice Problems
Summary — Amount of Chemical Change
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.