# The mole concept

> IB Chemistry SL · Reactivity 1: Stoichiometry and Periodicity
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u4-the-mole-concept/

This foundational sub-topic introduces the mole, the standard counting unit for chemical particles. You will learn to convert between mass, moles and number of particles, and master core calculations for all stoichiometry problems.

**Prerequisites:** [Basic atomic structure and relative atomic mass](https://www.owlsprep.com/study/ib-chemistry-sl-u1-atomic-structure/)

## Learning objectives

- Define the mole and Avogadro's constant correctly
- Calculate molar mass from relative atomic masses
- Interconvert mass, moles and number of particles
- Apply the mole concept to calculate empirical formula ratios

## Definition of the Mole and Avogadro's Constant

**Mole** — The SI base unit of amount of substance, defined as the amount of substance that contains as many elementary particles as there are atoms in 12 grams of carbon-12 (¹²C)

*Notation:* n (unit: mol)

*Example:* One mole of water contains 6.02 × 10²³ water molecules

Avogadro's constant ($N_A$) is the fixed number of particles per mole of any substance:

$$N_A = 6.02 \times 10^{23} \text{ mol}^{-1}$$

**Worked example:** How many hydrogen atoms are in 0.25 mol of H₂O?

1. First calculate the total number of H₂O molecules, using $N = n \times N_A$:
2. $$0.25 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 1.505 \times 10^{23} \text{ H}_2\text{O molecules}$$
3. Each H₂O molecule contains 2 hydrogen atoms, so multiply by 2:
4. $$2 \times 1.505 \times 10^{23} = 3.01 \times 10^{23} \text{ H atoms}$$

## Molar Mass Calculations

**Molar Mass** — The mass per mole of a substance, numerically equal to the sum of relative atomic masses of all atoms in the chemical formula

*Notation:* M (unit: g mol⁻¹)

*Example:* Molar mass of H₂O = (2×1.0) + 16.0 = 18.0 g mol⁻¹

The core relationship between mass ($m$, in grams), moles ($n$) and molar mass ($M$) is:

$$n = \frac{m}{M}$$

This rearranges to give $m = n \times M$ (find mass from moles) and $M = \frac{m}{n}$ (find molar mass from mass and moles).

**Worked example:** Calculate the amount (in mol) of 36.0 g of glucose ($C_6H_{12}O_6$). Use $A_r$: C=12.0, H=1.0, O=16.0.

1. First calculate the molar mass of glucose by summing relative atomic masses:
2. $$M(C_6H_{12}O_6) = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 180 \text{ g mol}^{-1}$$
3. Substitute into $n = \frac{m}{M}$:
4. $$n = \frac{36.0 \text{ g}}{180 \text{ g mol}^{-1}} = 0.200 \text{ mol}$$

## Interconverting Mass, Moles and Particles

All mole calculations link two core relationships: $n = \frac{m}{M}$ (mass-moles) and $N = n \times N_A$ (moles-particles), where $N$ is the number of particles. You can combine these to convert between any two of the three quantities.

1. Mass → particles: calculate moles from mass, then multiply by $N_A$
2. Particles → mass: calculate moles by dividing particles by $N_A$, then multiply by $M$

**Worked example:** What is the mass of $1.204 × 10^{24}$ atoms of carbon? $A_r(C) = 12.0$, $N_A = 6.02 × 10^{23} \text{ mol}^{-1}$.

1. First calculate moles of carbon:
2. $$n(C) = \frac{\text{Number of atoms}}{N_A} = \frac{1.204 \times 10^{24}}{6.02 \times 10^{23} \text{ mol}^{-1}} = 2.00 \text{ mol}$$
3. Then calculate mass from moles:
4. $$m(C) = n \times M = 2.00 \text{ mol} \times 12.0 \text{ g mol}^{-1} = 24.0 \text{ g}$$

**Check your understanding**

Test your understanding:

1. How many moles are in 9.0 g of H₂O?

   - A) 0.5 mol
   - B) 1.0 mol
   - C) 2.0 mol
   - D) 18 mol

   *Why:* Correct! M(H₂O) = 18 g mol⁻¹, so $n = 9/18 = 0.5$ mol

## Mole Ratios for Empirical Formula

The mole concept is used to find empirical formula, the simplest whole number ratio of atoms of each element in a compound, from percentage composition or mass data.

1. Convert percentage composition to mass (assume 100 g total sample, so percentages become grams)
2. Calculate moles of each element with $n = m/M$
3. Divide all mole values by the smallest mole value to get the simplest ratio
4. Multiply by whole numbers if needed to get integer ratios

**Worked example:** A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Find its empirical formula. $A_r$: C=12, H=1, O=16.

1. Assume 100 g sample, so masses: C = 40.0 g, H = 6.7 g, O = 53.3 g
2. Calculate moles of each element:
3. $$n(C) = 40.0/12 = 3.33 \text{ mol}, n(H) = 6.7/1 = 6.7 \text{ mol}, n(O) = 53.3/16 = 3.33 \text{ mol}$$
4. Divide all values by the smallest mole value (3.33):
5. $$C: 3.33/3.33 = 1, H: 6.7/3.33 ≈ 2, O: 3.33/3.33 = 1$$
6. The simplest whole number ratio is 1:2:1, so empirical formula is:
7. $$CH_2O$$

## Common pitfalls

- **Wrong:** Forgetting to multiply by the number of target atoms per formula unit when counting particles
  - Why it fails: You only calculate the number of molecules/formula units, not the number of individual atoms/ions
  - Correct: After finding moles of the compound, multiply by the number of target atoms per formula unit before multiplying by $N_A$
- **Wrong:** Using mass in kilograms instead of grams with $n = m/M$
  - Why it fails: Molar mass is defined in g mol⁻¹, so mismatched units give wrong mole values
  - Correct: Always convert mass to grams before calculating moles, and confirm M is in g mol⁻¹
- **Wrong:** Rounding mole ratios early when calculating empirical formula
  - Why it fails: Intermediate rounding leads to incorrect whole number ratios (e.g. 1.33 rounded to 1)
  - Correct: Keep decimals for intermediate ratios; multiply 1.33 by 3, 1.5 by 2 etc. to get whole numbers at the final step
- **Wrong:** Confusing moles (amount of substance) with mass
  - Why it fails: The mole is a unit of count, not mass. 1 mol of lead has far more mass than 1 mol of carbon
  - Correct: Always use $n = m/M$ to relate moles and mass, and remember 1 mol of any substance has the same number of particles, not the same mass

## Cheatsheet

| Quantity | Symbol | Unit | Core Relationship |
| --- | --- | --- | --- |
| Amount of substance | $n$ | mol | - |
| Avogadro's constant | $N_A$ | mol⁻¹ | $N = n \times N_A$ |
| Molar mass | $M$ | g mol⁻¹ | $n = \frac{m}{M}$ |
| Mass of sample | $m$ | g | $m = n \times M$ |
| Number of particles | $N$ | unitless | $N = n \times N_A$ |

## What's next

Mastering the mole concept is the foundation for all quantitative chemistry and stoichiometry topics in IB Chemistry SL. Every subsequent calculation topic, from reacting mass problems to solution stoichiometry, gas laws and titration calculations, relies on the core relationships between mass, moles and number of particles you learned here. Solid understanding of this sub-topic will make all more advanced stoichiometry problems much easier to solve, and it appears in every paper of the IB Chemistry SL exam.

- [Reacting masses and volumes](https://www.owlsprep.com/study/ib-chemistry-sl-u4-reacting-masses-and-volumes/)
- [Periodic table classification of elements](https://www.owlsprep.com/study/ib-chemistry-sl-u4-periodic-table-classification-of-elements/)
- [Periodic trends in atomic properties](https://www.owlsprep.com/study/ib-chemistry-sl-u4-periodic-trends-in-atomic-properties/)

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