The mole concept

IB Chemistry HL· 1.2 The mole concept· 45 min read

1. Definition of the mole and Avogadro's constant★☆☆☆☆⏱ 10 min

📘 Definition

Mole (amount of substance)

n (unit: mol)

The SI base unit that measures the amount of a substance, defined as containing exactly $6.02214076 \times 1 0^{23}$ elementary entities (atoms, ions, molecules etc.)

Example:

One mole of carbon atoms contains ~ $6.02 \times 1 0^{23}$ carbon atoms

Avogadro's constant ( $N_{A} = 6.02 \times 1 0^{23}$ mol⁻¹) is the constant that connects the number of particles to the amount in moles.

📐 Worked Example

How many helium atoms are present in 0.25 mol of helium gas?

1
Recall the relationship between moles and number of atoms
2
$N = n \times N_A$
3
Substitute the values for n and $N_{A}$ :
4
$N = 0.25 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1}$
5
Calculate and round to 2 significant figures:
6
$N = 1.5 \times 10^{23} \text{ helium atoms}$

Exam tip:

Check that the entity is specified (atoms vs molecules) when counting particles, 1 mol of O₂ has 2 mol of O atoms.

2. Molar mass calculation★★☆☆☆⏱ 15 min

📘 Definition

Molar mass

M (unit: g mol⁻¹)

The mass per mole of a substance, numerically equal to its relative atomic mass ( $A_{r}$ ) for elements or relative formula mass ( $M_{r}$ ) for compounds

Example:

Molar mass of H₂O is 18.02 g mol⁻¹

To calculate molar mass of a compound, sum the $A_{r}$ values of all atoms in the compound's formula. $A_{r}$ values are available in the IB data booklet.

📐 Worked Example

Calculate the molar mass of calcium hydroxide, $Ca(OH)_{2}$

1
Get $A_{r}$ values from the data booklet: $A_{r} (Ca) = 40.08$ , $A_{r} (O) = 16.00$ , $A_{r} (H) = 1.01$
2
Count atoms: the subscript 2 outside the bracket multiplies both O and H, so 1 Ca, 2 O, 2 H
3
Sum the masses to get M:
4
$M = (1 \times 40.08) + (2 \times 16.00) + (2 \times 1.01)$
5
$M = 74.10 \text{ g mol}^{-1}$

3. Interconverting mass, moles and particles★★☆☆☆⏱ 20 min

Three core relationships connect all the key quantities:

$n = \frac{m}{M}$ : calculate moles from mass and molar mass
$N = n \times N_{A}$ : calculate number of particles from moles
$n = \frac{N}{N _{A}}$ : calculate moles from number of particles

📐 Worked Example

A 9.01 g sample of glucose ( $C_{6} H_{12} O_{6}$ ). Calculate (a) moles of glucose, (b) number of glucose molecules.

1
First calculate the molar mass of glucose:
2
$M = (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18 \text{ g mol}^{-1}$
3
Part (a): Calculate moles using $n = \frac{m}{M}$ :
4
$n = \frac{9.01 \text{ g}}{180.18 \text{ g mol}^{-1}} = 0.0500 \text{ mol}$
5
Part (b): Calculate number of molecules using $N = n \times N_{A}$ :
6
$N = 0.0500 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 3.01 \times 10^{22} \text{ molecules}$

✓ Quick check

Test your understanding:

How many moles are in 2.0 g of H₂O (M = 18 g mol⁻¹)?
- 0.11 mol
- 0.090 mol
- 9.0 mol
- 36 mol
Reveal answer
0.11 mol —
Correct: $n = 2.0/18 = 0.11$ mol, 2 significant figures matching the question data.

4. Percentage composition by mass★★★☆☆⏱ 15 min

📘 Definition

Percentage composition by mass

The percentage of the total mass of a compound contributed by each element, used to find empirical formulas

The formula for percentage by mass of an element X is:

\% \text{ by mass of X} = \frac{(\text{number of X atoms} \times A_r(\text{X}))}{M_r(\text{compound})} \times 100\%

📐 Worked Example

Calculate the percentage by mass of carbon in glucose ( $C_{6} H_{12} O_{6}$ , $M_{r} = 180.18$ )

1
Calculate total mass of carbon in one mole of glucose:
2
$\text{Total mass C} = 6 \times 12.01 = 72.06$
3
Substitute into the percentage formula:
4
$\% \text{C} = \frac{72.06}{180.18} \times 100\% = 40.00\%$

5. Common Pitfalls

Wrong move:

Forgetting to multiply all atoms inside brackets by the outer subscript (e.g. counting 1 O in Ca(OH)₂)

Why:

The outer subscript applies to every atom inside the bracket, not just the last one

Correct move:

Always distribute the outer subscript to all atoms inside the bracket when counting atoms for molar mass calculations

Wrong move:

Using Avogadro's number to calculate molar mass from mass

Why:

Avogadro's constant only connects moles to number of particles, not to mass

Correct move:

Use the relationship $n = m / M$ to interconvert mass and moles, with molar mass from the periodic table

Wrong move:

Writing the unit of molar mass as g instead of g mol⁻¹

Why:

IB exam mark schemes penalize incorrect units even if the numerical value is correct

Correct move:

Always label molar mass with the correct unit g mol⁻¹

Wrong move:

Rounding intermediate values in multi-step calculations

Why:

Premature rounding leads to accumulated error and an incorrect final answer

Correct move:

Keep full unrounded values during calculation, only round the final answer to the required number of significant figures

6. Quick Reference Cheatsheet

Quantity	Symbol	Unit	Relationship
Amount of substance	$n$	mol	$n = \frac{m}{M} = \frac{N}{N _{A}}$
Molar mass	$M$	g mol⁻¹	Numerically equal to $M_{r}$
Avogadro's constant	$N_{A}$	mol⁻¹	$6.02 \times 1 0^{23}$
Number of particles	$N$		$N = n \times N_{A}$
% by mass X		%	$\frac{total mass of X}{M _{r} (compound)} \times 100%$

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 · P1
Mole to particle interconversion
2024 · P2
Molar mass and percentage composition
2023 · P1
Counting atoms in a mole sample

What's Next

The mole concept is the foundational building block for all stoichiometric calculations in IB Chemistry, from empirical and molecular formula determination to limiting reactant calculations, titrations and gas laws. Every calculation involving reacting masses or yields relies on the core interconversions you learned here, so mastering this topic now will save you time and effort later. Common exam questions range from multiple-choice interconversions to multi-part calculation questions in paper 2, so ensure you can apply these relationships consistently. Next, you will build on this foundation to find empirical formulas from percentage composition data, then move on to more complex stoichiometry problems.