GCSEChemistryAQARevision NotesQuantitative ChemistryYield & Atom EconomyTheoretical Masses of Products

Theoretical Masses of Products (AQA GCSE Chemistry)

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Obtaining calculated masses

Higher tier only

  • We can obtain theoretical masses of products using a balanced equation and a given mass of reactant

How to calculate theoretical mass

  • The process is as follows:
    1. Write out the balanced equation for the reaction(if not already given in the question)
    2. Convert the given mass of reactant(s) into moles, by dividing the masses by the molar masses
    3. Use the coefficients(multipliers) in the equation to deduce the number of moles of the product(s)
    4. Convert the moles of product into mass by multiplying by the molar mass
  • The following example illustrates the process:

Worked example

In an experiment to displace copper from copper(II)sulfate, 6.5g of zinc was added to a solution of copper(II)sulfate.

The balanced equation for the reaction is: 

Zn (s) + CuSO4 (aq)  ⟶ Cu (s) + ZnSO4 (aq)

The copper was filtered off, washed, dried and weighed. the final mass of copper obtained was 4.8 g.

Calculate the percentage yield of copper.

Answer:

  • Step 1: Calculate the moles of zinc that reacted:
    • Moles = mass / Ar 
    • Moles = 6.5 / 65 = 0.1 mol
  • Step 2: Calculate the maximum number of moles of copper that could be formed:
    • From the molar ratio, one mole of zinc reacts to form one mole of copper
    • So, 0.1 moles of zinc will form 0.1 moles of copper
  • Step 3: Calculate the maximum mass of copper that could be formed:
    • Mass = moles x Ar 
    • Mass = 0.1 x 63.5 = 6.35 g
  • Step 4: Calculate the percentage yield of the reaction:
    • Percentage yield = fraction numerator actual space yield over denominator theoretical space yield end fraction x 100
    • Percentage yield = fraction numerator 4.8 over denominator 6.35 end fraction = 76 %

Examiner Tip

The key to success in all calculations is to have a methodical step-by-step approach and show all your workings. Be careful when working out the molar masses not to include coeffiecients. It is very easy to make the mistake of calculating the molar mass of magnesium oxide as 80 g instead of 40 g, when working from an equation:

2Mg (s) + O2 (g) → 2MgO (s)

(Mg = 24 g mol−1; O = 16 g mol−1 )

Remember that molar masses are calculated from chemical formulae, not from equations

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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.