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ASChemistryAQARevision Notes1. Physical Chemistry1.2 Formulae, Equations & Calculations1.2.4 Reaction Yields

Reaction Yields (AQA AS Chemistry)

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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
    • Reactants or products are lost to the atmosphere
  • 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 yield = fraction numerator bold actual bold space bold yield over denominator bold theoretical bold space bold yield end fraction

  • Where actual yield is the number of moles or mass of product obtained experimentally
  • The predicted yield is the number of moles or mass obtained by calculation

  • You will often have to use the following equation to work out the reacting masses, to calculate the predicted yield

number of mol = fraction numerator bold mass bold space bold of bold space bold a bold space bold substance bold space bold in bold space bold grams bold space stretchy left parenthesis g stretchy right parenthesis over denominator bold molar bold space bold mass bold space stretchy left parenthesis g space mol to the power of negative 1 end exponent stretchy right parenthesis end fraction

  • It is important to be clear about the type of particle you are referring to when dealing with moles
    • Eg. 1 mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two moles of F- ions

Worked example

Calculate % yield using moles

In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution.

The copper was filtered off, washed and dried.

The mass of copper obtained was 4.80 g.

Calculate the percentage yield of copper.

Answer:

  1. Write the balanced symbol equation:

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

  2.  Calculate the number of moles of zinc:

    • n(Zn) = fraction numerator 6.54 space straight g over denominator 65.4 space straight g space mol to the power of negative 1 end exponent end fraction = 0.10 moles

  3. Deduce the number of moles of copper, using the balanced chemical equation:

    • 1 mole of zinc forms 1 mole of copper

      • The ratio is 1 : 1

    • Therefore, n(Cu) = 0.10 moles

  4. Calculate the maximum mass (theoretical yield) of copper:

    • Mass = mol x Mr

    • Mass = 0.10 mol x 63.5 g mol-1

    • Mass = 6.35 g

  5. Calculate the percentage yield of copper:

    • Percentage yield = fraction numerator 4.80 space straight g over denominator 6.35 space straight g end fraction x 100 = 75.6 %

Limiting & Excess Reagents

Limiting & Excess reagents

  • Sometimes, there is an excess of one or more of the reactants (excess reagent)
  • The reactant which is not in excess is called the limiting reagent
  • To determine which reactant is limiting:
    • The number of moles of each reactant should be calculated
    • The ratio of the reactants shown in the equation should be taken into account e.g.

2Na + S → Na2S

  • Here, the ratio of Na : S is 2 : 1, and this should be taken into account when doing calculations
  • Once all of one reactant has been used up, the reaction will stop, even if there are moles of the other reactant(s) leftover
    • The reactant leftover is in excess, the reactant which causes the reaction to stop once it is used up is the limiting reagent

Worked example

Excess & limiting reagent

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?

Answer

  • Step 1: Calculate the moles of each reactant
    • mol(Na) = fraction numerator 9.2 space straight g over denominator 23 space straight g space mol to the power of negative 1 end exponent end fraction = 0.40 mol
    • mol(S) = fraction numerator 8.0 over denominator 32 space straight g space mol to the power of negative 1 end exponent end fraction = 0.25 mol
  • Step 2: Write the balanced equation and determine the molar ratio

2Na + S → Na2S

    • The molar ratio of Na: Na2S is 2:1
  • Step 3: Compare the moles and determine the limiting reagent
    • So, to react completely 0.40 moles of Na require 0.20 moles of S
    • Since there are 0.25 moles of S, then S is in excess
    • Na is therefore the limiting reactant.
    • Once all of the Na has been used up, the reaction will stop, even though there is S left

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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.