The Mole and Avogadro Constant

Learning Objectives (OCR Spec 2.1.3)

• Define the mole and the Avogadro constant
• Relate moles, mass and molar mass for elements and compounds
• Calculate the number of particles from moles, and vice versa
• Use molar mass to convert between mass and moles confidently
• Apply n = m/M in multi-step problems
• Understand the historical definition and modern SI definition of the mole

Why Chemists Use Moles

Atoms and molecules are far too small to count individually. A teaspoon of water (about 5 g) contains approximately 1.7 × 10²³ molecules — a number so vast it would take all human beings who have ever lived more than a trillion years to count them one per second. Instead, chemists count particles in bundles called moles, much as a shopkeeper counts eggs in dozens or a printer counts paper in reams (500 sheets).

The mole is the bridge between the microscopic world (individual atoms and molecules, where we use amu and relative masses) and the macroscopic world (grams and dm³ that we actually measure in the lab).

Definition of the Mole

The historical IUPAC definition:

The mole (mol) is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12.

That number is the Avogadro constant, symbol L (or N_A**):

$L = 6.02 \times 10^{23} \text{ mol}^{-1}$L=6.02×1023 mol−1

You should memorise this value; it will be needed in calculations and is given in the OCR data booklet.

The 2019 Redefinition

In May 2019, the SI definition was updated: the mole is now defined as containing exactly 6.022 140 76 × 10²³ elementary entities. The mass of 12 g of carbon-12 is no longer the reference — instead, Avogadro's number itself is fixed by definition. This is part of the same SI redefinition that fixed Planck's constant to define the kilogram.

For A-Level purposes the practical meaning is unchanged: 1 mole = 6.02 × 10²³ particles. The new definition is just more precise and philosophically cleaner.

The word "particles" is deliberately generic. A mole of anything is 6.02 × 10²³ of that thing:

Substance	1 mole contains
Carbon	6.02 × 10²³ atoms
Water	6.02 × 10²³ molecules
Sodium chloride	6.02 × 10²³ formula units
Electrons	6.02 × 10²³ electrons
Ideas (in principle)	6.02 × 10²³ ideas

How Big Is Avogadro's Number?

To appreciate the scale:

• One mole of golf balls would fill a volume roughly equal to the entire Earth
• Stacked in a square, one mole of rice grains would cover India 1000 m deep
• If you had a mole of pounds sterling, you could give every person on Earth 80 trillion pounds each

And yet just 18 g of water (a tablespoon) contains this many molecules.

Molar Mass (M)

Molar mass is the mass in grams of one mole of a substance. Numerically, molar mass equals the relative atomic or molecular mass, but with units of g mol⁻¹:

• Ar(Na) = 23.0, so M(Na) = 23.0 g mol⁻¹
• Mr(H₂O) = 18.0, so M(H₂O) = 18.0 g mol⁻¹
• Mr(CaCO₃) = 100.1, so M(CaCO₃) = 100.1 g mol⁻¹
• Mr(NaOH) = 40.0, so M(NaOH) = 40.0 g mol⁻¹
• Mr(KMnO₄) = 158.0, so M(KMnO₄) = 158.0 g mol⁻¹
• Mr(Fe₂(SO₄)₃) = 399.9, so M(Fe₂(SO₄)₃) = 399.9 g mol⁻¹

Note that Ar/Mr are dimensionless while molar mass has units. This is a subtle but important distinction.

The Master Equation

The most important equation in Chemistry A-Level:

$n = \frac{m}{M}$n=Mm​

Where:

• n = amount of substance in moles (mol)
• m = mass in grams (g)
• M = molar mass (g mol⁻¹)

flowchart LR
    A[Mass m in g] -- divide by M --> B[Moles n]
    B -- multiply by M --> A
    B -- multiply by L --> C[Number of particles]
    C -- divide by L --> B

Worked Example 1: Mass to Moles

Calculate the number of moles in 40.0 g of sodium hydroxide, NaOH.

• M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹
• n = m/M = 40.0 / 40.0 = 1.00 mol

Worked Example 2: Moles to Mass

What mass of CaCl₂ contains 0.250 mol?

• M(CaCl₂) = 40.1 + 2(35.5) = 111.1 g mol⁻¹
• m = n × M = 0.250 × 111.1 = 27.8 g

Worked Example 3: Moles to Particles

How many molecules are in 0.500 mol of carbon dioxide?

• Number of particles = n × L = 0.500 × 6.02 × 10²³ = 3.01 × 10²³ molecules

Worked Example 4: Particles to Moles

A sample contains 1.50 × 10²⁴ atoms of iron. How many moles is this?

• n = N/L = (1.50 × 10²⁴) / (6.02 × 10²³) = 2.49 mol

Worked Example 5: Mass to Particles (Two Steps)

How many molecules are in 9.00 g of water?

Step 1: n = m/M = 9.00 / 18.0 = 0.500 mol Step 2: N = n × L = 0.500 × 6.02 × 10²³ = 3.01 × 10²³ molecules

Worked Example 6: Atoms of a Specific Element in a Compound