National 5ChemistrySQARevision NotesChemical Changes & StructureFormulae & Reacting QuantitiesCalculations Involving the Mole

Calculations Involving the Mole (SQA National 5 Chemistry): Revision Note

Exam code: X813 75

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

Moles & mass

  • In chemistry, we need a way to measure the amount of a substance

    • We do this using a unit called the mole

  • A mole represents a specific number of particles, but for calculations, we are interested in its mass

Gram formula mass (GFM)

  • The gram formula mass (GFM) is the mass of one mole of a substance, measured in grams (g)

  • We calculate the GFM by adding up the relative atomic masses (RAM) of all the atoms in the chemical formula

  • You can find the relative atomic masses (RAM) for all the elements on page 7 of the SQA Data Booklet

Worked Example

What is the mass of one mole of the following compounds:

  1. Sodium chloride, NaCl

  2. Magnesium bromide, MgBr2

  3. Aluminium oxide, Al2O3

  4. Potassium nitrate, KNO3

  5. Calcium hydroxide, Ca(OH)2

[5]

Answer

  • NaCl

Atoms: 1 x Na, 1 x Cl

RAMs: Na = 23, Cl = 35.5

Mass of Na = 1 x 23 = 23

Mass of Cl = 1 x 35.5 = 35.5

Total GFM: 23 + 35.5 = 58.5 [1 mark]

  • MgBr2

Atoms: 1 x Mg, 2 x Br

RAMs: Mg = 24.5, Br = 80

Mass of Mg = 1 x 24.5 = 24.5

Mass of Br = 2 x 80 = 160

Total GFM: 24.5 + 160 = 184.5 [1 mark]

  • Al2O3

Atoms: 2 x Al, 3 x O

RAMs: Al = 27, O = 16

Mass of Al = 2 x 27 = 54

Mass of O = 3 x 16 = 48

Total GFM: 54 + 48 = 102 [1 mark]

  • KNO3

Atoms: 1 x K, 1 x N, 3 x O

RAMs: K = 39, N = 14, O = 16

Mass of K = 1 x 39 = 39

Mass of N = 1 x 14 = 14

Mass of O = 3 x 16 = 48

Total GFM: 39 + 14 + 48 = 101 [1 mark]

  • Ca(OH)2

Atoms: 1 x Ca, 2 x O, 2 x H (Remember the 2 outside the bracket applies to both the O and H)

RAMs: Ca = 40, O = 16, H = 1

Mass of Ca = 1 x 40 = 40

Mass of O = 2 x 16 = 32

Mass of H = 2 x 1 = 2

Total GFM: 40 + 32 + 2 = 74 [1 mark]

The mole triangle

  • The number of moles (n), the mass in grams (mass), and the Gram Formula Mass (GFM) are all related.

  • We can use a triangle to help us with the calculations

The mole triangle

Triangle diagram for moles calculation: top section for mass in grams, bottom left for moles, bottom right for gram formula mass.
To use the triangle cover up the quantity you want to calculate, and the triangle shows you the formula to use.

Worked Example

Calculating moles

  1. How many moles are in 8g of methane (CH4)?

[2]

  1. How many moles are in 3.4 g of ammonia, NH3?

[2]

Answer 1

  • Step 1: Calculate GFM of CH4:

C + (4 x H)

12 + (4 x 1) = 16 [1 mark]

  • Step 2: Use the mole triangle to find the number of moles (n)

n = fraction numerator mass space open parentheses straight g close parentheses over denominator GFM end fraction

n = 8 over 16 = 0.5 moles [1 mark]

Answer 2

  • Step 1: Calculate the GFM of NH3

1 x N, 3 x H

14 + 3 = 17 [1 mark]

  • Step 2: Use the mole triangle to find the number of moles (n)

n = fraction numerator mass space open parentheses straight g close parentheses over denominator GFM end fraction

n = fraction numerator 3.4 over denominator 17 end fraction= 0.2 moles [1 mark]

Worked Example

Calculating mass

  1. What is the mass of 0.5 moles of carbon dioxide, CO2?

  2. What is the mass of 0.2 moles of aluminium hydroxide, Al(OH)3?

Answer 1

  • Step 1: Calculate GFM of CO2:

C + (2 x O)

GFM = 12 + (2 x 16) = 44 [1 mark]

  • Step 2: Calculate the mass of CO2 using the mole triangle:

mass = n x GFM

mass = 0.5 x 44

mass = 22 g [1 mark]

Answer 2

  • Step 1: Calculate GFM of Al(OH)3 formula for

Al + (3 x O) + (3 x H)

GFM = 27 + 48 + 3 = 78 [1 mark]

  • Step 2: Calculate the mass of Al(OH)3 using the mole triangle:

mass = n × GFM

mass = 0.2 × 78

mass = 15.6 g [1 mark]

Examiner Tips and Tricks

Always show your workings in calculations as it's easier to check for errors and you may pick up credit if you get the final answer wrong.

Reacting mass calculations

  • A balanced chemical equation is like a recipe

    • It tells you the ratio of moles of reactants that are needed and products that are made

  • We can use this ratio to calculate the mass of a substance in a reaction if we know the mass of another

The 3-Step Method

  • To solve these problems, we always follow the same three steps:

  1. Calculate moles of the 'known':

    • Use the mass given in the question and the GFM to find the number of moles of your starting substance.

n = fraction numerator mass space open parentheses straight g close parentheses over denominator GFM end fraction

  1. Use the mole ratio:

    • Look at the balancing numbers in the balanced equation to find the ratio between your 'known' substance and the substance you want to find out about (the 'unknown')

  2. Calculate mass of the 'unknown':

    • Use the new number of moles and the GFM of the 'unknown' substance to calculate its mass

mass = n × GFM

Worked Example

The equation for magnesium burning is:

2Mg (s) + O2 (g) ⟶ 2MgO (s)

Calculate the mass of magnesium oxide, MgO, that can be made by completely burning 6.0 g of magnesium, Mg.

(Use RAMs from page 7 of the Data Booklet: Mg = 24.5, O = 16)

[3]

Answer:

  1. Calculate moles of the 'known' (Mg)

RAM of Mg = 24.5

n = fraction numerator mass space open parentheses straight g close parentheses over denominator GFM end fraction

n = fraction numerator 6 over denominator 24.5 end fraction = 0.245 moles of Mg [1 mark]

  1. Use the mole ratio

    • From the balanced equation, find the ratio between Mg and MgO

    • 2Mg : 2MgO

      • Simplify the ratio to 1 : 1

    • This means 0.245 moles of Mg will produce 0.245 moles of MgO [1 mark]

  1. Calculate mass of the 'unknown' magnesium oxide

GFM of MgO = 24.5 + 16 = 40.5 g

mass = n × GFM

mass = 0.245 × 40.5

mass = 9.92 g of MgO [1 mark]

Worked Example

The equation for the decomposition of aluminium oxide is:

2Al2O3 ⟶ 4Al + 3O2

Calculate the maximum possible mass of aluminium, Al, in tonnes, that can be produced from 51 tonnes of aluminium oxide, Al2O3.

(Use RAMs from page 7 of the Data Booklet: Al = 27, O = 16)

[3]

Answer:

  1. Calculate moles of the 'known', Al2O3

GFM of Al2O3 = (2 x 27) + (3 x 16) = 54 + 48 = 102

n = fraction numerator mass space open parentheses straight g close parentheses over denominator GFM end fraction

n =fraction numerator 51 space 000 space 000 over denominator 102 end fraction = 500 000 moles of Al2O3 [1 mark]

  1. Use the mole ratio

    • From the balanced equation, find the ratio between Al2O3 and Al.

      • 2Al2O3 : 4Al

      • Simplify the ratio to 1 : 2

    • This means 500 000 moles of Al2O3 will produce (500 000 x 2) = 1 000 000 moles of Al [1 mark]

  1. Calculate mass of the 'unknown', Al

RAMof Al = 27

mass = n × GFM

mass = 1 000 000 × 27

mass = 27 000 000 g = 27 tonnes of Al [1 mark]

Examiner Tips and Tricks

In multi-step calculations, a very common mistake is rounding the answer from one step before using it in the next.

This can create a "rounding error", making your final answer slightly inaccurate and potentially costing you the final mark.

Only round the final answer that you write on the answer line. If the question doesn't specify, rounding to one or two decimal places is usually fine.

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    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry Content Creator

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

    Richard Boole

    Reviewer: Richard Boole

    Expertise: Chemistry Content Creator

    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.