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GCSEChemistry: Combined ScienceAQARevision Notes3. Quantitative Chemistry3.2 Calculating Mass of SubstancesAmounts of Substances in Equations

Amounts of Substances in Equations (AQA GCSE Chemistry: Combined Science)

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Calculating masses from balanced equations

Higher tier only

  • Chemical / symbol equations can be used to calculate:
    • The moles of reactants and products 
    • The mass of reactants and products
  • To do this:
    • Information from the question is used to find the amount in moles of the substances being considered
    • Then, the ratio between the substances is identified using the balanced chemical equation
    • Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses

Worked example

Quantitative chemistry example 1:

Magnesium undergoes combustion to produce magnesium oxide.

The overall reaction that is taking place is shown in the equation below.

2Mg (s) + O2 (g)  ⟶ 2 MgO (s) 

Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction:

Ar(O) = 16   Ar(Mg) = 24

Answer:

  • Step 1 - calculate the moles of magnesium 
    • Moles = open parentheses mass over M subscript r close parentheses = open parentheses 6 over 24 close parentheses = 0.25
  • Step 2 - use the molar ratio from the balanced symbol equation
    • 2 moles of magnesium produce 2 moles of magnesium oxide
    • The ratio is 1 : 1
    • Therefore, 0.25 moles of magnesium oxide is produced
  • Step 3 - calculate the mass of magnesium oxide
    • Mass = moles x Mr = 0.25 moles x (24 + 16) = 10 g

Worked example

Quantitative chemistry example 2:

In theory, aluminium could decompose as shown in the equation below.

2Al2O3  ⟶  4Al +  3O2 

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

Ar(O) = 16   Ar(Al) = 27

Answer:

  • Step 1 - calculate the moles of aluminium oxide 
    • Mass = 51 tonnes x 106 = 51 000 000 g
    • Moles = open parentheses mass over M subscript r close parentheses = open parentheses 51000000 over 102 close parentheses = 500 000
  • Step 2 - use the molar ratio from the balanced symbol equation
    • 2 moles of aluminium oxide produces 4 moles of aluminium 
    • The ratio is 1 : 2
    • Therefore, 2 x 500 000 = 1 000 000 moles of aluminium is produced
  • Step 3 - calculate the mass of aluminium 
    • Mass = moles x Mr = 1 000 000 moles x 27 = 27 000 000 g
    • Mass in tonnes = open parentheses 27000000 over 10 to the power of 6 close parentheses = 27 tonnes

Examiner Tip

  • When you see any of the following phrases in a question, make sure you work methodically through the process of calculating moles, using molar ratios and calculating mass
    • Calculate the mass...
    • Calculate the minimum mass...
    • Calculate the maximum mass...
  • As long as you are consistent, it doesn't matter whether you work in grams or tonnes or any other mass unit as the reacting masses will always be in proportion to the balanced equation

Balancing equations using reacting masses

  • If the masses of reactants and products of a reaction are known then we can use them to write a balanced equation for that reaction
  • This is done by converting the masses to moles and simplifying to find the molar ratios

Worked example

A student reacts 1.2 g of carbon with 16.2 g of zinc oxide. The resulting products are 4.4 g of carbon dioxide and 13 g of zinc.

Determine the balanced equation for the reaction.

Answer: 

  • Step 1: Write the unbalanced chemical equation:
    • _C + _ZnO → _CO2 + _Zn
  • Step 2: Write down the masses of each substance:
    • C = 12
    • ZnO = 65 + 16 = 81
    • CO2 = 12 + (2 x 16) = 44
    • Zn = 65
  • Step 3: Calculate the moles of each substance:
    • C = 1.2 / 12 = 0.1
    • ZnO = 16.2 / 81 = 0.2
    • CO2 = 4.4 / 44 = 0.1
    • Zn = 13 / 65 = 0.2
  • Step 4: Deduce the whole number ratio of substances:
    • C : ZnO : CO2 : Zn
    • 0.1 : 0.2 : 0.1 : 0.2
    • 1 : 2 : 1 : 2
  • Step 5: Apply the whole number ratio to the unbalanced equation:
    • 1C + 2ZnO → 1CO2 + 2Zn
    • C + 2ZnO → CO2 + 2Zn

Examiner Tip

These questions look hard but they are actually quite easy to do, as long as you follow the steps and organise your work neatly.

Remember the molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.

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    Stewart

    Author: Stewart

    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 Exam Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.