Stoichiometric relationships

IB Chemistry HL· Unit 5: Stoichiometric Relationships, R2: How much / how fast / how far?· 6 min read

1. The Mole and Molar Mass★☆☆☆☆⏱ 15 min

📘 Definition

Mole

mol

The SI base unit for amount of substance. 1 mole of any substance contains exactly $6.022 \times 1 0^{23}$ elementary entities (atoms, molecules, ions), equal to Avogadro's constant ( $N_{A}$ ).

Example:

1 mol of carbon atoms contains $6.022 \times 1 0^{23}$ C atoms.

The core relationship between amount of substance ( $n$ , in mol), mass ( $m$ , in g) and molar mass ( $M$ , in g mol⁻¹) is:

n = \frac{m}{M}

📐 Worked Example

Calculate the amount of sodium chloride (NaCl) in 12.5 g of pure solid NaCl. Given $M (N a) = 23.0$ g mol⁻¹, $M (C l) = 35.5$ g mol⁻¹.

1
First calculate the molar mass of NaCl:
$M(NaCl) = M(Na) + M(Cl) = 23.0 + 35.5 = 58.5 \text{ g mol}^{-1}$
2
Substitute values into the mole formula to solve for $n$ :
$n = \frac{m}{M} = \frac{12.5 \text{ g}}{58.5 \text{ g mol}^{-1}} = 0.214 \text{ mol}$

Exam tip:

Always include units in your final answer; IB exam markers deduct 1 mark per question for missing units.

2. Empirical and Molecular Formulas★★☆☆☆⏱ 20 min

📘 Definition

Empirical Formula

The simplest whole number ratio of atoms of each element present in a compound.

To find the empirical formula from mass or percentage composition data, follow these four steps:

Divide the mass (or percentage mass) of each element by its molar mass to get moles
Divide each mole value by the smallest mole value from the first step
Multiply all values by a whole number if needed to get whole number ratios
Write the ratio as subscripts to get the empirical formula

📐 Worked Example

A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.

1
Assume 100 g of compound, so masses are 40.0 g C, 6.7 g H, 53.3 g O. Calculate moles of each element:
$n(C) = \frac{40.0}{12.0} = 3.33 \text{ mol}, \quad n(H) = \frac{6.7}{1.0} = 6.7 \text{ mol}, \quad n(O) = \frac{53.3}{16.0} = 3.33 \text{ mol}$
2
Divide all values by the smallest mole value (3.33):
$C: \frac{3.33}{3.33} =1, \quad H: \frac{6.7}{3.33} ≈ 2, \quad O: \frac{3.33}{3.33} = 1$
3
The ratio 1:2:1 is already whole numbers, so the empirical formula is:

To get the molecular formula, you need the actual molar mass of the compound. Calculate the empirical formula mass, divide the actual molar mass by the empirical mass to get a multiplier, then multiply all subscripts by this multiplier.

3. Balanced Equations and Reaction Stoichiometry★★☆☆☆⏱ 20 min

A balanced chemical equation has equal numbers of each atom on both reactant and product sides, following the law of conservation of mass. The coefficients in a balanced equation give the mole ratio of all reactants and products.

📐 Worked Example

2.5 g of calcium carbonate ( $C a C O_{3}$ ) reacts completely with hydrochloric acid (HCl). Calculate the mass of $C O_{2}$ produced.

1
Write and balance the full chemical equation:
$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$
2
Calculate moles of $C a C O_{3}$ , $M (C a C O_{3}) = 100.1$ g mol⁻¹:
$n(CaCO_3) = \frac{2.5}{100.1} ≈ 0.025 \text{ mol}$
3
Use the 1:1 mole ratio from the balanced equation: 1 mol $C a C O_{3}$ produces 1 mol $C O_{2}$ , so $n (C O_{2}) = 0.025$ mol
4
Calculate mass of $C O_{2}$ , $M (C O_{2}) = 44.0$ g mol⁻¹:
$m(CO_2) = n \times M = 0.025 \times 44.0 = 1.1 \text{ g}$

4. Percentage Yield★★★☆☆⏱ 15 min

📘 Definition

Percentage Yield

% yield

A measure of how much product is actually obtained compared to the maximum theoretical amount predicted by stoichiometry, calculated as: $% yield = \frac{actual yield}{theoretical yield} \times 100%$

Example:

An 82% yield means 82% of the expected product was collected from the reaction.

📐 Worked Example

In the reaction of calcium carbonate above, 0.9 g of $C O_{2}$ was collected. Calculate the percentage yield.

1
We already calculated the theoretical yield of $C O_{2}$ as 1.1 g. The actual yield collected is 0.9 g.
2
Substitute into the percentage yield formula:
$\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{0.9}{1.1} \times 100 = 82\%$

Percentage yield is always less than 100% due to practical losses during filtration, transfer or purification, or incomplete reactions.

5. Common Pitfalls

Wrong move:

Forgetting to count all atoms when calculating molar mass, e.g. $M (H_{2} O) = 1 + 16 = 17$ g mol⁻¹

Why:

Misses the two hydrogen atoms in the water molecule, leading to an incorrect molar mass and wrong final answer

Correct move:

Count all atoms in the formula: $M (H_{2} O) = 2 (1.0) + 16.0 = 18.0$ g mol⁻¹

Wrong move:

Using excess reactant moles for product yield calculations instead of limiting reactant moles

Why:

The reaction stops when the limiting reactant is completely consumed; excess reactant does not contribute to forming more product

Correct move:

Always identify the limiting reactant first before calculating any product yields

Wrong move:

Rounding intermediate calculation values too early, leading to inaccurate final answers

Why:

Early rounding introduces cumulative error that can change the final answer enough to lose marks in exams

Correct move:

Keep at least one extra significant figure in intermediate steps, only round the final answer to the correct number of sig figs

Wrong move:

Confusing empirical and molecular formulas when answering exam questions

Why:

Many compounds share the same empirical formula but have different molecular formulas and molar masses

Correct move:

Always confirm what the question asks for, and use the given actual molar mass to find the molecular formula if required

Wrong move:

Using the mass of an impure reactant directly in mole calculations

Why:

Impure samples contain non-reacting contaminants that add mass but do not participate in the reaction

Correct move:

Multiply the total mass of the impure sample by the percentage purity to get the mass of pure reactant before calculating moles

6. Quick Reference Cheatsheet

Concept	Core Relationship	Units
Amount of substance	$n = \frac{m}{M} = \frac{N}{N _{A}}$	$n$ : mol, $m$ : g, $M$ : g mol⁻¹
Empirical formula steps	Mass → divide by M → divide by smallest → whole number ratio	None
Percentage yield	$% yield = \frac{actual yield}{theoretical yield} \times 100$	%
Mole ratio	Equal to coefficient ratio in balanced equation	None
Molecular formula	$Multiplier = \frac{M _{actual}}{M _{empirical}}$	None

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Empirical formula calculation from data
2024 · Paper 2
Percentage yield from reaction data
2023 · Paper 1
Mole ratio calculation from balanced equation

Going deeper

syllabusOfficial IB Chemistry HL 2025 SyllabusFull syllabus for stoichiometric relationships

What's Next

Stoichiometry is the foundation of all quantitative chemistry, so the skills you learned here will be applied in every subsequent topic in IB Chemistry HL, from thermochemistry enthalpy calculations to organic reaction yield calculations. Mastering these core calculations now will prevent common errors in more complex topics later. Next, you will extend these ideas to gas stoichiometry and solution stoichiometry, where you apply the mole concept to gas volumes and solution concentrations, both commonly tested in Paper 1 and Paper 2 exams. You will also use these skills in your practical internal assessment when calculating yields from your own experiments.