# The mole concept

> IB Chemistry HL · S1: Models of the particulate nature of matter
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u1-the-mole-concept/

The mole is the SI base unit for amount of substance, the foundation of all stoichiometric calculations in IB Chemistry. This module covers key definitions and core interconversions you will use in almost every subsequent topic.

**Prerequisites:** [Basic understanding of atomic structure and mass number](https://www.owlsprep.com/study/ib-chemistry-hl-u1-atomic-structure/)

## Learning objectives

- Define the mole and Avogadro's constant as the foundation of chemical amount
- Calculate molar mass of elements, molecules and ionic compounds
- Interconvert between mass, amount in moles and number of particles
- Calculate percentage composition by mass of any compound

## Definition of the mole and Avogadro's constant

**Mole (amount of substance)** — The SI base unit that measures the amount of a substance, defined as containing exactly $6.02214076 \times 10^{23}$ elementary entities (atoms, ions, molecules etc.)

*Notation:* n (unit: mol)

*Example:* One mole of carbon atoms contains ~$6.02 \times 10^{23}$ carbon atoms

Avogadro's constant ($N_A = 6.02 \times 10^{23}$ 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. $$N = 1.5 \times 10^{23} \text{ helium atoms}$$

> **tip**
>
> Always match the number of significant figures in your answer to the given data in the question.

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

## Molar mass calculation

**Molar mass** — 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

*Notation:* M (unit: g mol⁻¹)

*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, $\text{Ca(OH)}_2$

1. Get $A_r$ values from the data booklet: $A_r(\text{Ca}) = 40.08$, $A_r(\text{O}) = 16.00$, $A_r(\text{H}) = 1.01$
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. $$M = 74.10 \text{ g mol}^{-1}$$

## Interconverting mass, moles and particles

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 ($\text{C}_6\text{H}_{12}\text{O}_6$). 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. $$N = 0.0500 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 3.01 \times 10^{22} \text{ molecules}$$

**Check your understanding**

Test your understanding:

1. 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

   *Why:* Correct: $n = 2.0 / 18 = 0.11$ mol, 2 significant figures matching the question data.

## Percentage composition by mass

**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 ($\text{C}_6\text{H}_{12}\text{O}_6$, $M_r = 180.18$)

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. $$\% \text{C} = \frac{72.06}{180.18} \times 100\% = 40.00\%$$

## Common pitfalls

- **Wrong:** Forgetting to multiply all atoms inside brackets by the outer subscript (e.g. counting 1 O in Ca(OH)₂)
  - Why it fails: The outer subscript applies to every atom inside the bracket, not just the last one
  - Correct: Always distribute the outer subscript to all atoms inside the bracket when counting atoms for molar mass calculations
- **Wrong:** Using Avogadro's number to calculate molar mass from mass
  - Why it fails: Avogadro's constant only connects moles to number of particles, not to mass
  - Correct: Use the relationship $n = m/M$ to interconvert mass and moles, with molar mass from the periodic table
- **Wrong:** Writing the unit of molar mass as g instead of g mol⁻¹
  - Why it fails: IB exam mark schemes penalize incorrect units even if the numerical value is correct
  - Correct: Always label molar mass with the correct unit g mol⁻¹
- **Wrong:** Rounding intermediate values in multi-step calculations
  - Why it fails: Premature rounding leads to accumulated error and an incorrect final answer
  - Correct: Keep full unrounded values during calculation, only round the final answer to the required number of significant figures

## 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 | $6.02 \times 10^{23}$ |
| Number of particles | $N$ | - | $N = n \times N_A$ |
| % by mass X | - | % | $\frac{\text{total mass of X}}{M_r\text{(compound)}} \times 100\%$ |

## 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.

- [Ideal gas behaviour](https://www.owlsprep.com/study/ib-chemistry-hl-u1-ideal-gas-behaviour/)
- [Kinetic molecular theory](https://www.owlsprep.com/study/ib-chemistry-hl-u1-kinetic-molecular-theory/)
- [AHL: Advanced mass spectrometry interpretation](https://www.owlsprep.com/study/ib-chemistry-hl-u1-ahl-advanced-mass-spectrometry-interpretation/)

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