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Percentage Yield Calculations (DP IB Chemistry) : Revision Note

Philippa Platt

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DP Chemistry: HLDP Chemistry: SL

Percentage Yield Calculations

Percentage yield

  • In a lot of reactions, not all reactants react to form products which can be due to several factors:

    • Other reactions take place simultaneously

    • The reaction does not go to completion

    • Products are lost during separation and purification

  • The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

percentage space yield space equals space fraction numerator actual space yield over denominator theoretical space yield end fraction space cross times space 100

  • The actual yield is the number of moles or mass of product obtained experimentally

  • The theoretical yield is the number of moles or mass obtained by a reacting mass calculation

Worked Example

In an experiment to displace copper from copper(II) sulfate, 6.5 g of zinc was added to an excess of copper(II) sulfate solution. The resulting copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g.

Calculate the percentage yield of copper.

Answer:

Step 1: The balanced symbol equation is:

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

Step 2: Calculate the amount of zinc reacted in moles

number space of space moles space equals space fraction numerator 6.5 space straight g over denominator 65.4 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 0.10 space mol

Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:
   

Since the ratio of Zn (s) to Cu (s) is 1:1 a maximum of 0.10 moles can be produced

Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

mass =   mol  x  M

mass =   0.10 mol x 63.55 g mol-1

mass =   6.4 g (2 sig figs)

Step 5: Calculate the percentage yield of copper

percentage space yield space equals space fraction numerator 4.8 space straight g over denominator 6.4 space straight g end fraction space cross times space 100 space equals space 75 percent sign

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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    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