The mole concept

IB Chemistry SL· Reactivity 1.1: Introducing stoichiometry· 6 min read

1. Definition of the Mole and Avogadro's Constant★☆☆☆☆⏱ 10 min

📘 Definition

Mole

n (unit: mol)

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)

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
$2 \times 1.505 \times 10^{23} = 3.01 \times 10^{23} \text{ H atoms}$

2. Molar Mass Calculations★☆☆☆☆⏱ 10 min

📘 Definition

Molar Mass

M (unit: g mol⁻¹)

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

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 ( $C_{6} H_{12} O_{6}$ ). 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
$n = \frac{36.0 \text{ g}}{180 \text{ g mol}^{-1}} = 0.200 \text{ mol}$

3. Interconverting Mass, Moles and Particles★★☆☆☆⏱ 15 min

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.

Mass → particles: calculate moles from mass, then multiply by $N_{A}$
Particles → mass: calculate moles by dividing particles by $N_{A}$ , then multiply by $M$

📐 Worked Example

What is the mass of $1.204 \times 1 0^{24}$ atoms of carbon? $A_{r} (C) = 12.0$ , $N_{A} = 6.02 \times 1 0^{23} mol^{- 1}$ .

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
$m(C) = n \times M = 2.00 \text{ mol} \times 12.0 \text{ g mol}^{-1} = 24.0 \text{ g}$

✓ Quick check

Test your understanding:

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
Reveal answer
A) 0.5 mol —
Correct! M(H₂O) = 18 g mol⁻¹, so $n = 9/18 = 0.5$ mol

4. Mole Ratios for Empirical Formula★★☆☆☆⏱ 15 min

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.

Convert percentage composition to mass (assume 100 g total sample, so percentages become grams)
Calculate moles of each element with $n = m / M$
Divide all mole values by the smallest mole value to get the simplest ratio
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
$CH_2O$

5. Common Pitfalls

Wrong move:

Forgetting to multiply by the number of target atoms per formula unit when counting particles

Why:

You only calculate the number of molecules/formula units, not the number of individual atoms/ions

Correct move:

After finding moles of the compound, multiply by the number of target atoms per formula unit before multiplying by $N_{A}$

Wrong move:

Using mass in kilograms instead of grams with $n = m / M$

Why:

Molar mass is defined in g mol⁻¹, so mismatched units give wrong mole values

Correct move:

Always convert mass to grams before calculating moles, and confirm M is in g mol⁻¹

Wrong move:

Rounding mole ratios early when calculating empirical formula

Why:

Intermediate rounding leads to incorrect whole number ratios (e.g. 1.33 rounded to 1)

Correct move:

Keep decimals for intermediate ratios; multiply 1.33 by 3, 1.5 by 2 etc. to get whole numbers at the final step

Wrong move:

Confusing moles (amount of substance) with mass

Why:

The mole is a unit of count, not mass. 1 mol of lead has far more mass than 1 mol of carbon

Correct move:

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

6. Quick Reference 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	$N = n \times N_{A}$

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.

2022 · 1
Mole to particle conversion
2023 · 2
Molar mass calculation
2021 · 1
Empirical formula mole ratio

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.