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The Mole & The Avogadro Constant (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

The Mole & the Avogadro Constant

  • The Avogadro constant (NA or L) is the number of particles equivalent to the relative atomic mass or molecular mass of a substance

    • The Avogadro constant applies to atoms, molecules, ions and electrons

  • The value of NA is 6.02 x 1023 g mol-1

  • The mass of a substance with this number of particles is called a mole (mol)

    • The mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.00 g of 12C

  • One mole of any element is equal to the relative atomic mass of that element in grams

    • One mole of carbon, that is if you had 6.02 x 1023 atoms of carbon in your hand, would have a mass of 12 g

    • One mole of water would have a mass of (2 x 1 + 16) = 18 g

  • The number of moles or particles can be calculated easily using a formula triangle:

Formula triangle diagram linking moles, particles and Avogadro's constant 

Formula triangle diagram linking moles, particles and Avagadro's constant
The moles and particles formula triangle – cover the one you want to find and follow the directions in the triangle

Worked Example

Determine the number of atoms, molecules and the relative mass of 1 mole of:

  1. Na

  2. H2 

  3. NaCl

Answer 1

  • The relative atomic mass of Na is 23.0

  • Therefore, 1 mol of Na has a mass of 23.0 g mol-1

  • 1 mol of Na will contain 6.02 x 1023 atoms of Na (Avogadro’s constant)

Answer 2

  • The relative atomic mass of H is 1.0

  • Since there are 2 H atoms in H2, the mass of 1 mol of H2 is (2 x 1.0) = 2.0 g mol-1

  • 1 mol of H2 will contain 6.02 x 1023 molecules of H2

  • Since there are 2 H atoms in H2, 1 mol of H2 will contain 2 x 6.02 x 1023  = 1.204 x 1024 H atoms

Answer 3

  • The relative atomic mass of Na and Cl is 23.0 and 35.5 respectively

  • Therefore, 1 mol of NaCl has a mass of (23.0 + 35.5) = 58.5 g mol-1

  • 1 mol of NaCl will contain 6.02 x 1023 molecules of NaCl

  • Since there is one Na and one Cl atom in NaCl, 1 mol of NaCl will contain 2 x 6.02 x 1023  = 1.204 x 1024 atoms in total

1 mole of

Number of atoms

Number of molecules

Relative mass
(g mol-1)

Na

6.02 x 1023

-

23.0

H2

1.204 x 1024 

6.02 x 1023

2.0

NaCl

1.204 x 1024 

6.02 x 1023

58.5

Examiner Tips and Tricks

You only have to be able to work with Avogadro's constant, you will not be expected to recall the value.

Other mole calculations

  • The mole is an integral part of other chemical calculations:

Moles, mass and Mr

  • The number of moles of any chemical can be calculated from the mass of the chemical combined with the Periodic Table

  • With this information, the number of moles can be calculated by using a formula triangle:

Formula triangle diagram linking moles, mass and molar mass 

Formula triangle diagram linking moles, mass and molar mass
The moles and mass formula triangle – cover the one you want to find and follow the directions in the triangle

Worked Example

  1. What is the mass of 0.250 moles of zinc?

  2. How many moles are in 2.64 g of sucrose, C12H11O22  (Mr = 342.3)?

Answers:

  1. 0.250 moles of zinc

  • From the periodic table, the relative atomic / molar mass of Zn is 65.4 g mol-1

  • Mass = moles x molar mass

  • Mass = 0.250 mol x 65.4 g mol-1 = 16.4 g

  1. 2.64 g of sucrose:

  • The molar mass of sucrose is 342.3 g mol-1

  • Moles = mass ÷ molar mass

  • Moles = 2.64 g ÷ 342.3 g mol-1 = 7.71 x 10-3 mol

Moles, concentration and volume

  • The number of moles of any chemical can be calculated using the concentration and volume of the chemical

  • With this information, the number of moles can be calculated by using a formula triangle:

Formula triangle diagram linking moles, concentration and volume 

Formula triangle diagram linking moles, concentration and volume
The concentration-moles formula triangle – cover the one you want to find and follow the directions in the triangle

  • A common application of the moles = concentration x volume formula is titration calculations, e.g. Required Practical 1

Worked Example

Calculate the concentration, mol dm-3, of the solution formed when 80.0 g of sodium hydroxide is dissolved in 500 cm3 of water.

Answer:

  • The relative formula mass of NaOH is:

    • 23.0 + 16.0 + 1.0 = 40.0 g mol-1

  • So, 80 g of sodium hydroxide is:

    • Moles = fraction numerator m a s s over denominator M subscript r end fraction

    • Moles = fraction numerator 80.0 over denominator 40.0 end fraction = 2 moles

  • The volume in dm3 is:

    • 500 over 1000 = 0.5 dm3

  • The concentration in mol dm-3 is:

    • Concentration = fraction numerator m o l e s over denominator v o l u m e space open parentheses d m cubed close parentheses end fraction

    • Concentration = fraction numerator 2 space m o l e s over denominator 0.5 space d m cubed end fraction = 4 mol dm-3

Examiner Tips and Tricks

Make sure that you use the correct units for moles, concentration, volume calculations.

These calculations have units of mol dm-3 for concentration and dm3 for volume but questions will typically use cm3.

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    Author
    Reviewer
    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.