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*Revision Materials and Past Papers* 1 Atomic Structure 2 Amounts of Substance 3 Bonding 4 Energetics 5 Kinetics 6 Chemical Equilibria & Kc 7 Redox Equations 8 Thermodynamics 9 Rate Equations 10 Kp (Equilibrium Constant) 11 Electrode Potentials & Cells 12 Acids and Bases 13 Periodicity 14 Group 2: Alkaline Earth Metals 15 Group 7: The Halogens 16 Period 3 Elements & Oxides 17 Transition Metals 18 Reactions of Ions in Aqueous Solution 19 Intro to Organic Chemistry 20 Alkanes 21 Halogenoalkanes 22 Alkenes 23 Alcohols 24 Organic Analysis 25 Optical Isomerism 26 Aldehydes & Ketones 27 Carboxylic Acids & Derivatives 28 Aromatic Chemistry 29 Amines 30 Polymers 31 Amino Acids, Proteins & DNA 32 Organic Synthesis 33 NMR Spectroscopy 34 Chromatography RP1–RP12 Required Practicals

1.2 Amounts of Substance

1.2.1 Relative Atomic Mass and Relative Molecular Mass 1.2.2 The Mole and the Avogadro Constant 1.2.3 The Ideal Gas Equation 1.2.4 Empirical and Molecular Formulae 1.2.5 Balanced Equations and Atom Economy

The Mole and Avogadro Constant

Specification Reference Physical Chemistry, Amount of Substance 3.1.2.2

Quick Notes

  • Avogadro constant (NA) = 6.02 × 1023
    the number of particles in one mole (you don’t need to memorise the value).
  • A mole is a fixed number of particles and can apply to any species — atoms, ions, molecules, electrons, etc.
  • Key Equations:
    • Moles = mass ÷ Mr
    • Concentration (mol dm−3) = moles ÷ volume (dm3)

Full Notes

The Mole and Avogadro Constant

A mole is a standard unit for measuring amounts in chemistry.

One moles worth of something contains 6.02 × 1023 particles of that something, this number is the Avogadro constant (NA).

Avogadro constant (NA) = 6.02 × 1023 (the number of particles in 1 mole).

You are not expected to recall the exact value in exams.

Examples:

Calculations Using the Mole

Formula 1: Moles from Mass

AQA A‑Level Chemistry formula card: moles equals mass divided by molar mass.
Worked Example: Moles from Mass

How many moles in 18 g of water (H2O)?

  1. Mr(H2O) = 2(1.0) + 16.0 = 18.0
  2. n = mass ÷ Mr = 18 ÷ 18.0 = 1 mol

Worked Example 2: Mass from Moles

What is the mass of 2 moles worth of methane (CH4)?

  1. Mr(CH4) = 4(1.0) + 12.0 = 16.0
  2. If n = mass ÷ Mr then mass = n × Mr
  3. mass = 2 × 16.0 = 32 g

Using the Avogadro Constant

AQA A‑Level Chemistry formula card: number of particles equals moles times Avogadro constant.

where NA = 6.02 × 1023

Worked Example

How many molecules in 0.5 mol of O2?

  1. Number of molecules = 0.5 × NA
  2. = 0.5 × (6.02 × 1023) = 3.01 × 1023

Photo of Matt
Matt’s exam tip

Matt’s exam tip - when working with the Avagadro constant, don’t be surprised if answers seem seem very big (when finding numbers of particles) or very small (finding masses of individual particles). Remember the Avagadro constant is enormous!

Concentration of Solutions

Concentration is the amount of a substance (in moles) per 1 dm3 of solution.

Units: mol dm−3

Moles from Concentration and Volume

AQA A‑Level Chemistry formula card: moles equals concentration times volume in dm3 n = C × V.

To convert cm3 to dm3: Volume (dm3) = Volume (cm3) ÷ 1000

Worked Example

What is the concentration of a solution made by dissolving 0.1 mol of NaOH in 250 cm3 of solution?

  1. Volume = 250 ÷ 1000 = 0.250 dm3
  2. Concentration = 0.1 ÷ 0.250 = 0.4 mol dm−3

Rearranging the Formula

Make sure you are comfortable using rearranged forms of the above formulas.

Quantity Formula Units
Moles from mass n = mass ÷ Mr mol
Mass from moles mass = n × Mr g
Number of particles particles = n × NA particles
Moles from solution n = C × V mol
Concentration C = n ÷ V mol dm−3

Summary