IB DP Chemistry Structure 1.4: Counting particles by mass: the mole – Revision Notes

These notes condense Structure 1.4, “Counting particles by mass: the mole”, and follow the IB Diploma Programme Chemistry guide for first assessment in 2025 (published in February 2023). Statements Structure 1.4.1 to Structure 1.4.6 are all included. Structure 1.4 has no AHL material, so the notes serve SL and HL alike, for the course examined from 2025 onwards. If a method here is unfamiliar, the study guide explains it in full.

Test yourself afterwards with the Structure 1.4 practice set. The overview of Structure 1, the DP Chemistry course page, a checklist to print and our short diagnostics are also there when you need them.

Key definitions

Term	Meaning	Units
Mole (mol)	SI unit for amount of substance; one mole holds NA elementary entities	mol
Avogadro constant, NA	Number of entities per mole: 6.02 × 10²³ to 3 s.f.	mol⁻¹
Elementary entity	The particle you are counting: atom, molecule, ion, electron or a stated group of particles	none
Relative atomic mass, Ar	Average mass of an element’s atoms relative to one-twelfth of a ¹²C atom	none
Relative formula mass, Mr	Sum of the Ar values in a formula	none
Molar mass, M	Mass of one mole of a substance	g mol⁻¹
Empirical formula	Lowest whole-number ratio between the atoms present	none
Molecular formula	Exact atom count for every element in a single molecule	none
Molar concentration, [X]	Amount of solute per unit volume of solution	mol dm⁻³
Avogadro’s law	At equal temperature and pressure, equal gas volumes contain equal numbers of molecules	none

Formulas

Relationship	Rearranged	Notes
N = n × NA	n = N ÷ NA	N is the number of the specified entity
n = m / M	m = n × M; M = m / n	m in g, M in g mol⁻¹
n = C × V	C = n / V; V = n / C	V in dm³ (cm³ ÷ 1000)
g dm⁻³ = mol dm⁻³ × M	mol dm⁻³ = g dm⁻³ ÷ M	conversion the guide requires
V(gas A) / V(gas B) = n(A) / n(B)	—	same temperature and pressure only

The guide notes that NA, n = m/M and n = CV are given in the data booklet, and that its two-decimal-place Ar values should be used in calculations.

Method in steps

Counting entities (1.4.1)

1. Identify the entity asked for (atoms of one element, ions, molecules, electrons).
2. Multiply the amount of the substance by the number of those entities per formula unit.
3. Multiply by NA.

Mass, moles, particles (1.4.3)

1. Find M by adding Ar values; multiply through every bracket and every water of crystallisation.
2. n = m / M.
3. Use the formula to scale to the entity you need, then × NA.
4. Going backwards: ÷ NA, then × M.

Empirical formula from percentages (1.4.4)

1. Treat each percentage as grams in a 100 g sample.
2. Divide each by Ar.
3. Divide by the smallest result.
4. Multiply through if a value ends near .5 (× 2), .33 or .67 (× 3), .25 or .75 (× 4).
5. For the molecular formula: multiple = M ÷ empirical formula mass, then multiply every subscript.

Concentration (1.4.5)

1. Convert volumes to dm³.
2. Use n = CV, rearranged as needed.
3. For g dm⁻³, multiply mol dm⁻³ by M.
4. On dilution, n stays the same: C₁V₁ = C₂V₂.
5. Ion concentrations: multiply [salt] by the number of each ion in the formula.

Gas volumes (1.4.6)

1. Write the balanced equation with state symbols.
2. Use the mole ratio as a volume ratio for gases only.
3. Find which gas runs out and which is in excess.
4. Final volume = gaseous products + leftover reactant gas. Liquids and solids add nothing.

Small worked reminders

Ions from an amount. 0.0200 mol of sodium carbonate, Na₂CO₃, contains 2 × 0.0200 = 0.0400 mol of Na⁺ ions, which is 0.0400 × 6.02 × 10²³ = 2.41 × 10²² sodium ions.

Mass to concentration, both units. 5.55 g of calcium chloride, CaCl₂ (M = 40.08 + 2 × 35.45 = 110.98 g mol⁻¹), is made up to 500.0 cm³.

n = 5.55 / 110.98 = 0.0500 mol
[CaCl₂] = 0.0500 / 0.5000 = 0.100 mol dm⁻³
[Cl⁻] = 2 × 0.100 = 0.200 mol dm⁻³
mass concentration = 5.55 / 0.5000 = 11.1 g dm⁻³

Particles to mass. To find the mass of a given number of molecules: divide by NA to get moles, then multiply by M. Never multiply the number of particles by M directly.

Must-know distinctions

• Ar and Mr versus M. Same number, but Ar and Mr have no units and M is in g mol⁻¹.
• Empirical versus molecular formula. Ethene has molecular formula C₂H₄ and empirical formula CH₂. An ionic compound such as NaCl has only an empirical formula.
• Amount of substance versus number of particles. Amount is in moles; the number of particles is a pure number (often about 10²² to 10²⁴ for lab samples).
• “Mole of oxygen”. Always state O or O₂. 1 mol of O₂ contains 2 mol of O atoms.
• mol dm⁻³ versus g dm⁻³. Convert using M. Square brackets mean mol dm⁻³.
• Volume of solution versus volume of solvent. C uses the final volume of solution.
• Avogadro’s law versus molar volume. Avogadro’s law gives volume ratios without any numerical molar volume. Molar volume for an ideal gas belongs to Structure 1.5.
• Gas versus liquid products. Water formed below its boiling point is a liquid and is left out of the gas volume.

Quick self-test

Use NA = 6.02 × 10²³ mol⁻¹ and Ar values: H 1.01, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, Fe 55.85.

1. How many electrons are in 0.0500 mol of ammonium ions, NH₄⁺?
2. Calculate Mr of magnesium hydroxide, Mg(OH)₂.
3. Calculate the amount, in mol, in 3.40 g of ammonia, NH₃.
4. An oxide of iron contains 69.94% iron by mass. Deduce its empirical formula.
5. A hydrocarbon has empirical formula C₂H₅ and molar mass 58.14 g mol⁻¹. Deduce its molecular formula.
6. Convert a sodium hydroxide concentration of 0.125 mol dm⁻³ into g dm⁻³.
7. Calculate the amount of solute in 15.0 cm³ of a 0.200 mol dm⁻³ solution.
8. 2CO(g) + O₂(g) → 2CO₂(g). What volume of oxygen reacts with 30 cm³ of carbon monoxide, and what volume of carbon dioxide forms (same conditions)?
9. State the units of Ar and of molar mass.
10. Calculate [Na⁺] in a 0.040 mol dm⁻³ solution of sodium phosphate, Na₃PO₄.

Answers

1. Each NH₄⁺ has 7 + 4 − 1 = 10 electrons. n(e⁻) = 10 × 0.0500 = 0.500 mol, so 0.500 × 6.02 × 10²³ = 3.01 × 10²³ electrons.
2. 24.31 + 2 × (16.00 + 1.01) = 58.33.
3. M = 14.01 + 3 × 1.01 = 17.04 g mol⁻¹; n = 3.40 / 17.04 = 0.200 mol.
4. O = 100 − 69.94 = 30.06%. Fe: 69.94 / 55.85 = 1.252; O: 30.06 / 16.00 = 1.879. Ratio 1 : 1.50, so × 2 gives Fe₂O₃.
5. Empirical formula mass = 2 × 12.01 + 5 × 1.01 = 29.07; 58.14 / 29.07 = 2, so C₄H₁₀.
6. M(NaOH) = 22.99 + 16.00 + 1.01 = 40.00 g mol⁻¹; 0.125 × 40.00 = 5.00 g dm⁻³.
7. n = 0.200 × 0.0150 = 3.00 × 10⁻³ mol.
8. Ratio CO : O₂ : CO₂ = 2 : 1 : 2, so 15 cm³ of O₂ reacts and 30 cm³ of CO₂ forms.
9. Ar has no units; molar mass is in g mol⁻¹.
10. Three Na⁺ per formula unit: 3 × 0.040 = 0.12 mol dm⁻³.

Where marks are usually lost

• Giving Ar or Mr a unit, or giving molar mass without g mol⁻¹.
• Multiplying the amount of a compound by NA when the question asks for atoms or ions of one kind, so the answer is short by a factor.
• Rounding mid-calculation (for example, moles to 2 s.f.) which shifts the last digit of the final answer.
• Rounding a mole ratio of 1 : 1.33 or 1 : 1.5 to 1 : 1 in an empirical formula.
• Using the empirical formula mass as the molar mass, or dividing the wrong way round when finding the multiple.
• Forgetting to convert cm³ to dm³ before using n = CV.
• Using the volume of water added rather than the total volume of solution.
• Writing [Cl⁻] equal to [CaCl₂] instead of twice it.
• Including liquid water in a final gas volume, or forgetting the unreacted excess gas.
• Leaving the answer with more significant figures than the data justify; match the data, usually 3 s.f.

Official syllabus

Source document: Diploma Programme Chemistry guide (International Baccalaureate Organization), for first assessment in 2025, published in February 2023. This page follows its Structure 1.4 statements.