Stoichiometric Relationships — IB Chemistry HL HL Study Guide

For: IB Chemistry HL candidates sitting IB Chemistry HL.

Covers: all core subtopics for Stoichiometric Relationships including states of matter and changes, mole concept and Avogadro's constant, empirical and molecular formulas, equation balancing and limiting reactant calculations, and ideal gas laws.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Stoichiometric Relationships?

Stoichiometric relationships (commonly shortened to stoichiometry) are the set of quantitative rules and calculations used to relate the amounts of reactants and products in chemical reactions, rooted in the law of conservation of mass. As the first core topic of IB Chemistry HL, it underpins all reaction-based calculations across the syllabus, and is tested in both Paper 1 (multiple choice) and Paper 2 (structured response), often combined with later topics like energetics, kinetics, and organic chemistry in high-mark extended response questions.

2. States of matter and changes

All chemical substances exist in one of three primary states under standard conditions, defined by their particle arrangement and kinetic energy:

• Solid: Particles arranged in a fixed, tightly packed lattice, with minimal kinetic energy. Solids have a fixed shape and fixed volume.
• Liquid: Particles are close together but free to move past one another, with moderate kinetic energy. Liquids have a fixed volume but no fixed shape, taking the shape of their container.
• Gas: Particles are far apart, moving randomly at high speed, with high kinetic energy. Gases have no fixed shape or volume, expanding to fill their container.

State changes are physical (not chemical) changes, where only intermolecular forces are broken, not intramolecular chemical bonds, so total mass of the substance is conserved. Common state changes include melting (solid → liquid), boiling/evaporation (liquid → gas), sublimation (solid → gas, e.g. dry ice, iodine), and reverse processes (freezing, condensation, deposition).

Worked exam-style example: If 15.5 g of solid iodine sublimes completely to iodine gas at room temperature, what mass of iodine gas is produced? Solution: By the law of conservation of mass, mass of gas = mass of solid = 15.5 g. Examiners often include trick options using molar volume to calculate mass for state change questions; remember mass is unchanged regardless of state.

3. Mole concept and Avogadro's constant

The mole (unit symbol: mol) is the SI unit for amount of substance, defined as the amount of substance that contains the same number of elementary entities (atoms, molecules, ions, electrons) as there are atoms in 12.0 g of pure carbon-12 (¹²C). This fixed number is called Avogadro's constant ($N_A$), given in the IB data booklet as $6.02\times 10^{23}{\text{ mol}}^{-1}$.

Two core formulas connect amount in moles to measurable values:

1. Number of particles: $$N=n\times N_A$$ where $N$ = total number of entities, $n$ = amount in mol
2. Mass to mole conversion: $$n=\frac{m}{M}$$ where $m$ = mass in g, $M$ = molar mass of the substance in g mol⁻¹ (equal to the relative atomic/molecular/formula mass of the substance, no unit conversion needed)

Worked exam-style example: Calculate the total number of oxygen atoms in 4.4 g of carbon dioxide (CO₂). Solution:

1. Molar mass of CO₂ = $12.01+(2\times 16.00)=44.01{\text{ g mol}}^{-1}$
2. Moles of CO₂ = $\frac{4.4}{44.01}\approx 0.10\text{ mol}$
3. Each CO₂ molecule has 2 oxygen atoms, so moles of O atoms = $0.10\times 2=0.20\text{ mol}$
4. Number of O atoms = $0.20\times 6.02\times 10^{23}=1.20\times 10^{23}$

Exam tip: Always check if the question asks for number of atoms, ions, or molecules; multiplying by the number of atoms per molecule is a common step students miss.

4. Empirical and molecular formulas

Two types of formulas are used to describe compound composition:

• Empirical formula: The simplest whole-number ratio of atoms of each element in a compound
• Molecular formula: The actual number of atoms of each element in one molecule of a compound, equal to $(\text{empirical formula}{)}_n$, where $n$ is a positive integer.

How to calculate empirical formula:

1. Find the mass (or percentage mass) of each element in the compound
2. Divide each mass by the relative atomic mass of the element to get moles of each element
3. Divide all mole values by the smallest mole value to get a ratio
4. If the ratio is within ±0.1 of a whole number, round to the nearest integer; if you get ratios of ~0.5, multiply all values by 2, ~0.33 multiply by 3, etc, to get whole numbers.

How to calculate molecular formula:

1. Calculate the empirical formula mass (EFM) by adding the relative atomic masses of all atoms in the empirical formula
2. Find $n$ using: $$n=\frac{\text{Molecular mass (M)}}{\text{Empirical formula mass (EFM)}}$$
3. Multiply all subscripts in the empirical formula by $n$ to get the molecular formula.

Worked exam-style example: A compound is 52.2% C, 13.0% H, and 34.8% O by mass, with a molar mass of 46 g mol⁻¹. Find its empirical and molecular formula. Solution:

1. Assume a 100 g sample: masses of C = 52.2 g, H =13.0 g, O=34.8 g
2. Moles: C = $\frac{52.2}{12}=4.35$, H = $\frac{13.0}{1}=13.0$, O = $\frac{34.8}{16}=2.175$
3. Divide by smallest mole value (2.175): C = $\frac{4.35}{2.175}=2$, H = $\frac{13.0}{2.175}\approx 6$, O = $\frac{2.175}{2.175}=1$
4. Empirical formula: ${\text{C}}_2{\text{H}}_6\text{O}$
5. EFM = $(2\times 12)+(6\times 1)+16=46{\text{ g mol}}^{-1}$, so $n=\frac{46}{46}=1$
6. Molecular formula = ${\text{C}}_2{\text{H}}_6\text{O}$ (ethanol)

5. Balancing equations and limiting reactant

All balanced chemical equations follow the law of conservation of mass: the number of atoms of each element on the reactant (left) side equals the number on the product (right) side. To balance an equation:

1. Write the unbalanced equation with correct chemical formulas and state symbols (s, l, g, aq)
2. Balance elements that appear in only one reactant and one product first
3. Balance oxygen and hydrogen last, as they often appear in multiple compounds
4. Check that all atom counts are equal on both sides, and reduce coefficients to the smallest whole-number ratio.

Limiting reactant calculations

The limiting reactant (LR) is the reactant that is completely used up first in a reaction, and limits the maximum amount of product that can be formed. The excess reactant (XS) is the reactant left over after the reaction stops. To identify the limiting reactant:

1. Calculate moles of each reactant
2. Divide each mole value by its stoichiometric coefficient in the balanced equation
3. The smallest resulting value is the limiting reactant; use its moles to calculate the amount of product formed.

Worked exam-style example: 28 g of ethene (${\text{C}}_2{\text{H}}_4$) reacts with 91.25 g of hydrogen chloride (HCl) to form chloroethane (${\text{C}}_2{\text{H}}_5\text{Cl}$) via the reaction: ${\text{C}}_2{\text{H}}_4(g)+\text{HCl}(g)\to {\text{C}}_2{\text{H}}_5\text{Cl}(g)$. Calculate the maximum mass of chloroethane that can be formed. Solution:

1. Moles of ${\text{C}}_2{\text{H}}_4$ = $\frac{28}{28}=1.0$ mol
2. Moles of HCl = $\frac{91.25}{36.5}=2.5$ mol
3. Divide by coefficients: ${\text{C}}_2{\text{H}}_4$: $\frac{1.0}{1}=1.0$, HCl: $\frac{2.5}{1}=2.5$. ${\text{C}}_2{\text{H}}_4$ is the limiting reactant.
4. Moles of ${\text{C}}_2{\text{H}}_5\text{Cl}$ formed = 1.0 mol, mass = $1.0\times 64.5=64.5$ g

Exam tip: Always show your limiting reactant calculation explicitly in Paper 2; even if you get the final answer wrong, you will be awarded method marks for this step.

6. Gas laws — ideal gas equation

An ideal gas is a theoretical model of gas that follows four key assumptions (you must memorize these for IB exams):

1. Gas molecules have negligible volume compared to the total volume of the container
2. There are no intermolecular forces between gas molecules
3. All collisions between molecules and with the container walls are perfectly elastic (no kinetic energy is lost)
4. The average kinetic energy of gas molecules is proportional to the absolute temperature (in Kelvin) of the gas.

The ideal gas equation combines Boyle's Law ($P\propto 1/V$ at constant n, T), Charles' Law ($V\propto T$ at constant n, P), and Avogadro's Law ($V\propto n$ at constant P, T): $$PV=nRT$$ Where:

• $P$ = pressure in Pascals (Pa, 1 atm = 101325 Pa, 1 bar = 100000 Pa)
• $V$ = volume in cubic meters (${\text{m}}^3$, $1{\text{ m}}^3=1000{\text{ dm}}^3=10^6{\text{ cm}}^3$)
• $n$ = amount of gas in mol
• $R$ = gas constant, given in the IB data booklet as $8.31{\text{ J K}}^{-1}{\text{ mol}}^{-1}$
• $T$ = absolute temperature in Kelvin (K = °C + 273)

Worked exam-style example: Calculate the volume occupied by 11.0 g of propane (${\text{C}}_3{\text{H}}_8$) gas at 127°C and 200 kPa pressure. Solution:

1. Convert units: $T=127+273=400$ K, $P=200\text{ kPa}=200000$ Pa, moles of ${\text{C}}_3{\text{H}}_8$ = $\frac{11.0}{44}=0.25$ mol
2. Rearrange ideal gas equation: $V=\frac{nRT}{P}$
3. Substitute values: $V=\frac{0.25\times 8.31\times 400}{200000}=0.004155{\text{ m}}^3=4.16{\text{ dm}}^3$ (3 significant figures)

Exam shortcut: If you use pressure in kPa and volume in dm³, you can use $R=8.31{\text{ kPa dm}}^3{\text{ K}}^{-1}{\text{ mol}}^{-1}$ to avoid unit conversion for volume and pressure; examiners accept this value of R.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using temperature in Celsius instead of Kelvin in ideal gas calculations. Why students do it: They forget gas laws rely on absolute temperature. Correct move: Add 273 to all Celsius values before substituting into the ideal gas equation, and write the unit next to T to double-check.
• Wrong move: Rounding mole ratios to whole numbers too early when calculating empirical formulas (e.g. rounding 1.5 to 2 instead of multiplying all ratios by 2). Why students do it: They rush through calculations to save time. Correct move: Only round if the ratio is within ±0.1 of a whole number; otherwise multiply by 2, 3, etc, to get whole integers.
• Wrong move: Using the excess reactant to calculate product yield instead of the limiting reactant. Why students do it: They skip the limiting reactant identification step. Correct move: Always calculate mole/coefficient ratios for all reactants first to find the LR before calculating product amounts, even if the question does not explicitly ask for the LR.
• Wrong move: Forgetting state symbols in balanced equations. Why students do it: They think state symbols are optional. Correct move: IB requires state symbols for all balanced equations in Paper 2, and you lose 1 mark per missing state symbol. Add state symbols as the final step of balancing every equation.
• Wrong move: Using 22.4 dm³ mol⁻¹ as molar volume at STP. Why students do it: They carry over this value from IGCSE. Correct move: IB uses 22.7 dm³ mol⁻¹ as the molar volume of an ideal gas at STP (273 K, 100 kPa), this value is given in the data booklet.

8. Practice Questions (IB Chemistry HL Style)

Question 1 (Paper 1 multiple choice style)

Which sample contains the greatest number of nitrogen atoms? A) 0.15 mol of ${\text{N}}_2{\text{O}}_3$ B) 0.22 mol of ${\text{N}}_2$ C) 0.30 mol of ${\text{NH}}_3$ D) 0.40 mol of ${\text{NO}}_2$

Worked solution: Calculate moles of N atoms for each option by multiplying moles of compound by number of N atoms per molecule:

• A: $0.15\times 2=0.30$ mol N
• B: $0.22\times 2=0.44$ mol N
• C: $0.30\times 1=0.30$ mol N
• D: $0.40\times 1=0.40$ mol N Correct answer: B

Question 2 (Paper 2 structured style)

A 7.2 g sample of a hydrocarbon undergoes complete combustion to produce 22.0 g of ${\text{CO}}_2$ and 10.8 g of ${\text{H}}_2\text{O}$. a) Calculate the empirical formula of the hydrocarbon. [3 marks] b) If the molar mass of the hydrocarbon is 72 g mol⁻¹, find its molecular formula. [1 mark]

Worked solution: a) 1. Moles of ${\text{CO}}_2=\frac{22.0}{44.0}=0.50$ mol, so moles of C = 0.50 mol, mass of C = $0.50\times 12=6.0$ g 2. Moles of ${\text{H}}_2\text{O}=\frac{10.8}{18.0}=0.60$ mol, so moles of H = $0.60\times 2=1.20$ mol, mass of H = $1.20\times 1=1.2$ g 3. Mole ratio C:H = $0.50:1.20$, divide by 0.50 to get $1:2.4$, multiply by 5 to get whole number ratio $5:12$ Empirical formula: ${\text{C}}_5{\text{H}}_{12}$ b) EFM of ${\text{C}}_5{\text{H}}_{12}=(5\times 12)+12=72$ g mol⁻¹, so $n=1$, molecular formula = ${\text{C}}_5{\text{H}}_{12}$ (pentane)

Question 3 (Paper 2 structured style)

Magnesium reacts with oxygen to form magnesium oxide via the reaction: $2\text{Mg}(s)+{\text{O}}_2(g)\to 2\text{MgO}(s)$. A 4.86 g strip of Mg is placed in a container with $2.24{\text{ dm}}^3$ of ${\text{O}}_2$ gas at STP (273 K, 100 kPa). a) Identify the limiting reactant. [2 marks] b) Calculate the mass of MgO formed. [2 marks]

Worked solution: a) 1. Moles of Mg = $\frac{4.86}{24.3}=0.20$ mol 2. Moles of ${\text{O}}_2=\frac{2.24}{22.7}\approx 0.0987$ mol 3. Divide by coefficients: Mg = $\frac{0.20}{2}=0.10$, ${\text{O}}_2=\frac{0.0987}{1}=0.0987$ ${\text{O}}_2$ has the smaller value, so it is the limiting reactant. b) Moles of MgO = $2\times {\text{moles of O}}_2=2\times 0.0987=0.1974$ mol Mass of MgO = $0.1974\times 40.3=7.96$ g (3 sig figs)

9. Quick Reference Cheatsheet

Core Formulas

Key Rules

1. Balanced equations have equal numbers of each atom on both sides; include state symbols (s, l, g, aq) for all species.
2. Limiting reactant = reactant with the smallest mole/stoichiometric coefficient ratio.
3. Ideal gas assumptions: negligible molecular volume, no intermolecular forces, elastic collisions, average KE ∝ absolute temperature.
4. Temperature conversion: $\text{K}{=}^{\circ }\text{C}+273$.

10. What's Next

Stoichiometric relationships is the foundational topic for all IB Chemistry HL content, so mastering it is critical for success in later units. You will use these calculations in Energetics (Topic 5) to calculate enthalpy changes from reaction masses, in Kinetics (Topic 6) to find reaction rates from concentration changes, in Equilibrium (Topic 7) to calculate equilibrium constants, and in Organic Chemistry (Topic 10) to find percentage yields of synthesis reactions. Examiners frequently combine stoichiometry with these topics in extended response questions in Paper 2 and Paper 3, so you should be comfortable applying these rules to unfamiliar contexts.

To reinforce your learning, you can practise more stoichiometry questions tailored to IB Chemistry HL on the OwlsPrep platform, and you can ask Ollie, our AI tutor, any questions you have about tricky calculations, common mistakes, or exam technique for this topic. You can also move on to our study guide for the next core topic, Atomic Structure, to continue your IB Chemistry HL revision.