The Mole, Avogadro Constant and Molar Mass — AQA A-level Chemistry

SPEC 3.1.2.1-3.1.2.2 - AMOUNT OF SUBSTANCE

Relative Mass, The Mole & Avogadro Constant

Understanding relative atomic mass, molecular mass, and the mole concept for quantitative chemistry

Why Do We Need Relative Mass?

Atoms are incredibly small and light. A single carbon atom has a mass of approximately 1.99 × 10⁻²³ g. Working with such tiny numbers is impractical for everyday chemistry calculations.

The Problem

We need a convenient way to compare atomic masses and work with measurable quantities of substances in the laboratory.

The solution is to use a relative scale based on a standard reference atom, and to define a quantity called the mole that links atomic-scale masses to laboratory-scale masses.

In This Lesson

You will learn how to calculate relative atomic and molecular masses, understand what the mole represents, and convert between mass, moles, and number of particles.

Relative Atomic Mass (A_r)

Definition

Relative atomic mass (A_r) is the weighted mean mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.

A_{r} = \frac{mean mass of one atom of an element}{\frac{1}{12} \times mass of one atom of ^{12} C}

Key Points

A_r is a ratio so has no units

Carbon-12 has A_r = 12.00 exactly (by definition)

A_r is a weighted mean accounting for all naturally occurring isotopes

Calculating A_r from Isotopic Data

Most elements exist as a mixture of isotopes. The relative atomic mass is the weighted average of these isotopic masses.

Formula

A_{r} = \frac{\sum ( isotopic mass \times abundance )}{100}

From Mass Spectra

The m/z value gives the isotopic mass; the peak height gives relative abundance. Convert heights to percentages before calculating A_r.

Worked Example01

Chlorine has two naturally occurring isotopes: ³⁵Cl (75.77%) and ³⁷Cl (24.23%). Calculate the relative atomic mass of chlorine.

1

Write the formula:

A_{r} = \frac{\sum ( mass \times abundance )}{100}

2

Substitute:

A_{r} = \frac{( 35 \times 75.77 ) + ( 37 \times 24.23 )}{100}

3

Calculate:

A_{r} = \frac{2651.95 + 896.51}{100} = \frac{3548.46}{100} = 35.48

AnswerA_r(Cl) = 35.5 (to 3 s.f.)

Worked Example02

A mass spectrum of neon shows peaks at m/z = 20, 21, and 22 with relative heights of 90.48, 0.27, and 9.25 respectively. Calculate the relative atomic mass of neon.

1

Check abundances sum to 100: 90.48 + 0.27 + 9.25 = 100 (already percentages)

2

Apply:

A_{r} = \frac{( 20 \times 90.48 ) + ( 21 \times 0.27 ) + ( 22 \times 9.25 )}{100}

3

Calculate:

A_{r} = \frac{1809.6 + 5.67 + 203.5}{100} = \frac{2018.77}{100} = 20.19

AnswerA_r(Ne) = 20.2 (to 3 s.f.)

Relative Molecular Mass (M_r)

Definition

Relative molecular mass (M_r) is the weighted mean mass of a molecule compared to 1/12th of the mass of an atom of carbon-12.

M_{r} = \frac{mean mass of one molecule}{\frac{1}{12} \times mass of one atom of ^{12} C}

In practice: M_r is calculated by adding up the relative atomic masses of all atoms in the molecular formula.

Formula

M_{r} = \sum (number of atoms \times A_{r} of each element)

Terminology

Relative formula mass is used for ionic compounds (which don't form discrete molecules). The calculation method is identical. Both use the symbol M_r and have no units.

Worked Example03

Calculate the relative molecular mass of sulfuric acid, H₂SO₄.

A_r values: H = 1.0, S = 32.1, O = 16.0

1

Identify atoms: H₂SO₄ contains 2 × H, 1 × S, 4 × O

2

Multiply: H: 2 × 1.0 = 2.0 | S: 1 × 32.1 = 32.1 | O: 4 × 16.0 = 64.0

3

Sum: M_r = 2.0 + 32.1 + 64.0 = 98.1

AnswerM_r(H₂SO₄) = 98.1

Worked Example04

Calculate the relative formula mass of calcium hydroxide, Ca(OH)₂.

A_r values: Ca = 40.1, O = 16.0, H = 1.0

1

Expand: Ca(OH)₂ contains 1 × Ca, 2 × O, 2 × H (subscript applies to brackets)

2

Multiply: Ca: 40.1 | O: 2 × 16.0 = 32.0 | H: 2 × 1.0 = 2.0

3

Sum: M_r = 40.1 + 32.0 + 2.0 = 74.1

AnswerM_r(Ca(OH)₂) = 74.1

The Mole & Avogadro Constant

Definition

The mole (mol) is the SI unit for amount of substance. One mole contains exactly 6.022 × 10²³ particles (atoms, molecules, ions, electrons, etc.).

Avogadro Constant

The Avogadro constant (N_A) = 6.022 × 10²³ mol⁻¹

This is the number of particles in exactly one mole of any substance.

It is defined as the number of atoms in exactly 12 g of carbon-12.

Why This Number?

The Avogadro constant was chosen so that one mole of carbon-12 atoms has a mass of exactly 12 grams. This creates a direct link between the relative atomic mass scale and measurable masses in grams.

Molar Mass & Key Relationships

Molar Mass

Molar mass (M) is the mass of one mole of a substance. It is numerically equal to M_r but has units of g mol⁻¹.

Key Insight

If you weigh out the relative atomic mass of any element in grams, this amount will always contain the same number of atoms (6.022 × 10²³). The same applies to molecules: weighing out M_r in grams gives one mole of entities.

EntitiesFormulaM_rMolar Mass / g

Oxygen atomsO16.016.0

Oxygen moleculesO₂32.032.0

Sodium ionsNa⁺23.023.0

Sodium fluorideNaF42.042.0

The Mole Equation

n = \frac{m}{M}

SymbolMeaningUnitnAmount (number of moles)molmMass of substancegMMolar massg mol⁻¹

Calculating Number of Particles

To convert between moles and number of particles, use the Avogadro constant:

Formula

Number of particles = n \times N_{A}

1 Mole = 6.022 × 10²³ Particles

Be Specific About Particles

It is easy to confuse moles of atoms and moles of molecules. Always state the formula when working out moles:

1 mol of H₂O = 6.022 × 10²³ molecules

1 mol of H₂O = 1.204 × 10²⁴ hydrogen atoms (2 per molecule)

1 mol of NaCl = 6.022 × 10²³ Na⁺ ions AND 6.022 × 10²³ Cl⁻ ions

Example

"10 moles of hydrogen" is ambiguous — it could mean 10 moles of H atoms or 10 moles of H₂ molecules (which contains twice as many atoms). Always write the formula!

Worked Example05

Calculate the number of moles in 4.40 g of carbon dioxide (CO₂).

A_r values: C = 12.0, O = 16.0

1

Calculate M_r: M_r = 12.0 + (2 × 16.0) = 44.0

2

Molar mass: M = 44.0 g mol⁻¹

3

Apply n = m/M:

n = \frac{4.40}{44.0} = 0.100 mol

Answern = 0.100 mol

Worked Example06

Calculate the mass of 0.250 mol of magnesium sulfate (MgSO₄).

A_r values: Mg = 24.3, S = 32.1, O = 16.0

1

Calculate M_r: M_r = 24.3 + 32.1 + (4 × 16.0) = 120.4

2

Use m = n × M: m = 0.250 × 120.4 = 30.1 g

Answerm = 30.1 g

Worked Example07

How many molecules are present in 9.00 g of water (H₂O)?

A_r: H = 1.0, O = 16.0. N_A = 6.022 × 10²³ mol⁻¹

1

Calculate M_r: M_r = (2 × 1.0) + 16.0 = 18.0

2

Calculate moles:

n = \frac{9.00}{18.0} = 0.500 mol

3

Number of molecules = 0.500 × 6.022 × 10²³ = 3.01 × 10²³

Answer3.01 × 10²³ molecules

Practice

Question 1: Copper has two isotopes: ⁶³Cu (69.17%) and ⁶⁵Cu (30.83%). Calculate the relative atomic mass of copper.

A_{r} = \frac{( 63 \times 69.17 ) + ( 65 \times 30.83 )}{100} = \frac{6361.66}{100} = 63.6

AnswerA_r(Cu) = 63.6

Practice

Question 2: Calculate the relative formula mass of ammonium nitrate, NH₄NO₃.

A_r: N = 14.0, H = 1.0, O = 16.0

NH₄NO₃ contains: 2 × N, 4 × H, 3 × O

M_r = (2 × 14.0) + (4 × 1.0) + (3 × 16.0) = 28.0 + 4.0 + 48.0 = 80.0

AnswerM_r(NH₄NO₃) = 80.0

Practice

Question 3: A sample of ethanol (C₂H₅OH) contains 1.20 × 10²⁴ molecules. Calculate the mass of the sample.

A_r: C = 12.0, H = 1.0, O = 16.0. N_A = 6.022 × 10²³ mol⁻¹

Step 1: M_r(C₂H₅OH) = 24.0 + 6.0 + 16.0 = 46.0

Step 2: n = (1.20 × 10²⁴) ÷ (6.022 × 10²³) = 1.99 mol

Step 3: m = 1.99 × 46.0 = 91.5 g

Answerm = 91.5 g

Summary

Key Definitions

TermDefinition

A_rWeighted mean mass of an atom relative to 1/12th of carbon-12

M_rSum of A_r values in a molecular/formula unit

MoleAmount containing 6.022 × 10²³ particles

N_AAvogadro constant = 6.022 × 10²³ mol⁻¹

Molar massMass of 1 mole (numerically = M_r, units g mol⁻¹)

Key Equations

A_{r} = \frac{\sum ( mass \times abundance )}{100}

n = \frac{m}{M} Particles = n \times N_{A}