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IGCSEChemistryCambridge (CIE)Revision Notes3. StoichiometryThe Mole & the Avogadro ConstantReacting Masses

Reacting Masses (Cambridge (CIE) IGCSE Chemistry)

Revision Note

Caroline Carroll

Written by: Caroline Carroll

Reviewed by: Stewart Hird

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Reacting masses

Extended 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

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:

  1. Calculate the moles of magnesium 

    • Moles = begin mathsize 14px style mass over M subscript r end style = begin mathsize 14px style 6 over 24 end style = 0.25

  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

  3. Calculate the mass of magnesium oxide

    • Mass = moles x Mr = 0.25 moles x (24 + 16) = 10 g

Worked Example

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:

  1. Calculate the moles of aluminium oxide 

    • Mass = 51 tonnes x 106 = 51 000 000 g

    • Mr of aluminium oxide = (2 x 27) + (3 x 16) = 102

    • Moles = mass over M subscript r = fraction numerator 51 space 000 space 000 over denominator 102 end fraction = 500 000

  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

  3. Calculate the mass of aluminium 

    • Mass = moles x Mr = 1 000 000 moles x 27 = 27 000 000 g

    • Mass in tonnes = fraction numerator 27 space 000 space 000 over denominator 10 to the power of 6 end fraction = 27 tonnes

Examiner Tips and Tricks

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

Limiting reactants

  • A chemical reaction stops when one of the reactants is used up

  • The reactant that is used up first is the limiting reactant, as it limits the duration and hence the amount of product that a reaction can produce

    • The one that is remaining is the excess reactant

    • The limiting reagent is the reactant which is not present in excess in a reaction

  • The amount of product is therefore directly proportional to the amount of the limiting reactant added at the beginning of a reaction

Determining the limiting reactant

  • In order to determine which reactant is the limiting reagent in a reaction, we have to consider the amounts of each reactant used and the molar ratio of the balanced chemical equation

  • When performing reacting mass calculations, the limiting reagent is always the number that should be used, as it indicates the maximum possible amount of product that can form

    • Once all of a limiting reagent has been used up, the reaction cannot continue

  • The steps are:

    1. Convert the mass of each reactant into moles by dividing by the molar masses

    2. Write the balanced equation and determine the molar ratio

    3. Look at the equation and compare the moles

Worked Example

9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na2S.

Which reactant is in excess and which is the limiting reactant?

Relative atomic masses (Ar):     Na = 23;   S = 32

Answer:

  1. Calculate the moles of each reactant

    • Moles = mass over M subscript r

    • Moles Na = fraction numerator 9.2 over denominator 23 end fraction = 0.40

    • Moles S = fraction numerator 8.0 over denominator 32 end fraction = 0.25

  2. Write the balanced equation and determine the molar ratio

    • 2Na + S → Na2S

    • So, the molar ratio of Na : S is 2 : 1

  3. Compare the moles

    • To react completely 0.40 moles of Na requires 0.20 moles of S

    • Since there are 0.25 moles of sulfur:

      • S is in excess

      • Therefore, Na is the limiting reactant

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    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.

    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.