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

This study guide teaches Structure 1.4, “Counting particles by mass: the mole”, and is aligned to the IB Diploma Programme Chemistry guide for first assessment in 2025 (published in February 2023). Every understanding from Structure 1.4.1 to Structure 1.4.6 is taught. Structure 1.4 has no AHL material, so all of it is common to SL and HL. It applies to the course examined from 2025 onwards on that guide.

Pair it with the revision notes and practice questions. For planning, use the overview of Structure 1, the DP Chemistry course page, the checklist you can print and our diagnostics.

What this unit covers

Guide statement	Skills required	Level
Structure 1.4.1	Use the mole and the Avogadro constant to convert an amount n into a number of specified entities, and back	SL and HL
Structure 1.4.2	Work out relative formula mass Mr from relative atomic masses Ar	SL and HL
Structure 1.4.3	Move between mass in grams, amount in moles and number of particles using molar mass in g mol⁻¹	SL and HL
Structure 1.4.4	Convert percentage composition by mass into an empirical formula and the reverse; find a molecular formula from the empirical formula and the molar mass	SL and HL
Structure 1.4.5	Solve problems connecting molar concentration with the amount dissolved and the solution volume; use square brackets; convert between g dm⁻³ and mol dm⁻³	SL and HL
Structure 1.4.6	Use Avogadro’s law with mole ratios to solve problems on gas volumes	SL and HL

The guide suggests about 7 teaching hours for this sub-topic at both levels.

Structure 1.4.1 The mole and the Avogadro constant

Atoms are too small to count singly, so chemists count them in batches called the mole (symbol mol), which is the SI unit for amount of substance. A mole is defined as holding exactly as many elementary entities as the value of the Avogadro constant, NA. To three significant figures, NA = 6.02 × 10²³ mol⁻¹. The guide notes that NA is given in the data booklet, and that its unit is mol⁻¹.

An elementary entity is whatever you choose to count: an atom, molecule, ion or electron, or a stated group of particles such as a formula unit of an ionic compound. You must always say which entity you mean. “One mole of oxygen” is ambiguous. One mole of O₂ molecules contains two moles of O atoms, so the two readings differ by a factor of two.

The conversion is a single multiplication or division:

number of entities = n × NA
n = number of entities ÷ NA

Worked example 1: counting ions and atoms in a salt

A sample contains 0.0400 mol of aluminium sulfate, Al₂(SO₄)₃. Use NA = 6.02 × 10²³ mol⁻¹. Find (a) the number of sulfate ions, (b) the number of oxygen atoms, (c) the total number of ions.

(a) each formula unit has 3 SO₄²⁻ ions
    n(SO₄²⁻) = 3 × 0.0400 = 0.120 mol
    number   = 0.120 × 6.02 × 10²³ = 7.22 × 10²² sulfate ions

(b) each formula unit has 3 × 4 = 12 O atoms
    n(O) = 12 × 0.0400 = 0.480 mol
    number = 0.480 × 6.02 × 10²³ = 2.89 × 10²³ oxygen atoms

(c) each formula unit has 2 Al³⁺ + 3 SO₄²⁻ = 5 ions
    n(ions) = 5 × 0.0400 = 0.200 mol
    number  = 0.200 × 6.02 × 10²³ = 1.20 × 10²³ ions

The trick is to scale the amount by the count per formula unit before multiplying by NA.

Structure 1.4.2 Relative atomic mass and relative formula mass

Atomic masses are compared on a relative scale based on carbon-12. The relative atomic mass, Ar, of an element is the weighted average mass of its atoms on a scale where one ¹²C atom has a mass of exactly 12. The relative formula mass, Mr, of a compound is the sum of the Ar values of all the atoms in its formula. Because both are ratios of masses, Ar and Mr have no units.

The guide says calculations should use the data booklet’s Ar values, which are quoted to two decimal places. In the examples on this page, use: H 1.01, C 12.01, N 14.01, O 16.00, Mg 24.31, S 32.07, Cl 35.45, Fe 55.85, Ag 107.87.

Worked example 2: Mr of an ionic salt and a hydrate

(a) Iron(III) sulfate, Fe₂(SO₄)₃:

Fe: 2 × 55.85        = 111.70
S:  3 × 32.07        =  96.21
O: 12 × 16.00        = 192.00
Mr                   = 399.91

(b) Magnesium chloride hexahydrate, MgCl₂·6H₂O. The dot means the six water molecules are part of the formula, so add them:

MgCl₂: 24.31 + 2 × 35.45      =  95.21
6H₂O:  6 × (2 × 1.01 + 16.00) = 108.12
Mr                            = 203.33

A bracket multiplies everything inside it, not just the last atom.

Structure 1.4.3 Molar mass

The molar mass, M, is the mass of one mole of a substance. It has the same number as Mr but carries the unit g mol⁻¹.

The link between mass m (in g), amount n (in mol) and molar mass M is:

n = m / M

The guide notes this relationship is given in the data booklet. Combined with NA, it gives a three-step chain:

mass (g)  --÷ M-->  amount (mol)  --× NA-->  number of entities
mass (g)  <--× M--  amount (mol)  <--÷ NA--  number of entities

Worked example 3: mass to atoms

How many nitrogen atoms are in 1.80 g of urea, CO(NH₂)₂?

M = 12.01 + 16.00 + 2 × (14.01 + 2 × 1.01) = 60.07 g mol⁻¹
n(urea) = 1.80 / 60.07 = 0.02997 mol
each molecule has 2 N atoms, so n(N) = 2 × 0.02997 = 0.0599 mol
number of N atoms = 0.0599 × 6.02 × 10²³ = 3.61 × 10²² atoms

Carry unrounded values between steps and round at the end. To go from particles to mass, run the chain backwards: divide by NA, then multiply by M.

Structure 1.4.4 Empirical and molecular formulae

An empirical formula gives the lowest whole-number ratio between the atoms present. A molecular formula gives the exact atom count for every element in a single molecule, so it is always the empirical formula scaled up by a whole number (sometimes 1). Ionic compounds have only empirical formulae.

Percentage composition to empirical formula

Method:

1. Assume 100 g of compound, so each percentage becomes a mass in grams.
2. Divide each mass by the element’s Ar to get moles.
3. Divide every value by the smallest.
4. If any value is close to x.5, x.33 or x.25, multiply all values by 2, 3 or 4 to reach whole numbers.

Step 4 needs judgement: 1.003 is treated as 1, but 1.33 is not.

Formula to percentage composition

Go the other way by dividing each element’s total mass in the formula by Mr and multiplying by 100. For MgCl₂·6H₂O, the percentage by mass of water is 108.12 / 203.33 × 100 = 53.17%.

Worked example 4: empirical then molecular formula

A compound contains 41.39% carbon, 3.48% hydrogen and 55.13% oxygen by mass. Its molar mass is 116.08 g mol⁻¹. Find its empirical and molecular formulae.

            C               H              O
mass/g      41.39           3.48           55.13
÷ Ar        41.39/12.01     3.48/1.01      55.13/16.00
moles       3.446           3.446          3.446
÷ smallest  1.00            1.00           1.00

Empirical formula: CHO. Empirical formula mass = 12.01 + 1.01 + 16.00 = 29.02.

multiple = molar mass / empirical formula mass = 116.08 / 29.02 = 4.00
molecular formula = (CHO) × 4 = C₄H₄O₄

The molecular formula is C₄H₄O₄ (check: its M is 116.08).

If you are given experimental masses of each element instead of percentages, skip step 1.

Structure 1.4.5 Molar concentration

A solution’s molar concentration depends on how much solute (in mol) is present and on the volume of the whole solution (not the volume of solvent added). Concentration in mol dm⁻³ is found from:

n = C × V      (n in mol, C in mol dm⁻³, V in dm³)

The guide notes this is in the data booklet. Divide a volume in cm³ by 1000 to get dm³.

The guide requires square brackets for molar concentration: [AgNO₃] = 0.100 mol dm⁻³ means the concentration of silver nitrate is 0.100 mol dm⁻³. You must also handle mass concentration in g dm⁻³ and convert:

concentration in g dm⁻³ = concentration in mol dm⁻³ × M
concentration in mol dm⁻³ = concentration in g dm⁻³ ÷ M

For an ionic solute, the ion concentration follows the formula. If [CaCl₂] = 0.150 mol dm⁻³, then [Ca²⁺] = 0.150 mol dm⁻³ and [Cl⁻] = 0.300 mol dm⁻³.

Worked example 5: making and diluting a solution

4.25 g of silver nitrate, AgNO₃, is dissolved in water and made up to 250.0 cm³. Find the concentration in mol dm⁻³ and in g dm⁻³. Then 25.00 cm³ of this solution is diluted to 100.0 cm³. Find the new [AgNO₃].

M(AgNO₃) = 107.87 + 14.01 + 3 × 16.00 = 169.88 g mol⁻¹
n = 4.25 / 169.88 = 0.0250 mol
V = 250.0 / 1000 = 0.2500 dm³
[AgNO₃] = 0.0250 / 0.2500 = 0.100 mol dm⁻³
in g dm⁻³: 0.100 × 169.88 = 17.0 g dm⁻³

dilution: n taken = 0.100 × 0.02500 = 2.50 × 10⁻³ mol
new [AgNO₃] = 2.50 × 10⁻³ / 0.1000 = 0.0250 mol dm⁻³

Dilution never changes the amount of solute, only the volume it is spread through.

Structure 1.4.6 Avogadro’s law

Avogadro’s law: if two gases are at the same temperature and pressure, equal volumes of them hold the same number of molecules. So at fixed temperature and pressure, the volume ratio of gases equals their mole ratio in the balanced equation. No molar volume is needed; that belongs to Structure 1.5, which links Avogadro’s law to ideal gases.

Only gases count (solid and liquid volumes are negligible), and all volumes must be measured under the same conditions.

Worked example 6: reacting gas volumes with an excess

40 cm³ of hydrogen sulfide is mixed with 100 cm³ of oxygen and ignited. After cooling to the starting temperature and pressure, the water is liquid.

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l)

Find the total volume of gas left.

mole ratio H₂S : O₂ = 2 : 3, so 40 cm³ H₂S needs 40 × 3/2 = 60 cm³ O₂
O₂ available = 100 cm³, so O₂ is in excess by 100 − 60 = 40 cm³
SO₂ formed: ratio H₂S : SO₂ = 2 : 2, so 40 cm³ SO₂
water is liquid, so it adds no gas volume
final gas volume = 40 cm³ O₂ + 40 cm³ SO₂ = 80 cm³

State symbols matter here: if the question had said the water stayed as a gas, you would add another 40 cm³.

Common errors

• Writing units on Ar or Mr. They have none; molar mass has g mol⁻¹.
• Forgetting to scale by the atoms or ions per formula unit before multiplying by NA.
• Rounding a mole ratio such as 1.33 to 1 instead of multiplying through by 3.
• Leaving volume in cm³ when concentration is in mol dm⁻³.
• Counting a liquid product, or ignoring an excess reactant gas, in gas volume problems.

Where to go next

Move on to the revision notes, then the practice questions. A summary of the whole component is in the notes for Structure 1.

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.