Revision NotesExam QuestionsFlashcardsPast PapersMock Exams
IBChemistryDPSLRevision NotesHow Much, How Fast & How Far?How Much? The Amount of Chemical ChangePercentage Yield Calculations

Percentage Yield Calculations (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

DP Chemistry: SLDP Chemistry: HL

Percentage yield calculations

Theoretical and experimental yield

  • The theoretical yield is the maximum amount of product that can be formed from a given amount of reactants, assuming perfect conversion with no losses

    • Percentage yield calculations depend on the limiting reactant

    • The limiting reactant determines how much product can be made

      • The excess reactant is left over.

    • So, identifying the limiting reactant is essential for calculating a correct theoretical yield.

  • The experimental yield (or actual yield) is the amount of product you actually collect in the lab

  • The difference is often due to:

    • Side reactions

    • Loss of product during transfer or purification

    • Incomplete reaction

Calculating percentage yield

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

  • The formula for percentage yield is:

percentage yield = fraction numerator actual space yield over denominator theoretical space yield end fractionx 100

  • For example:

    • The theoretical yield of a reaction is 6.5 g, but only 5.8 g of product is collected:

percentage yield = fraction numerator 5.8 over denominator 6.5 end fraction × 100 = 89.2%

Examiner Tips and Tricks

Ensure the actual and theoretical yields have the same unit (e.g. grams) before applying the formula

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 of moles = fraction numerator 6.5 space straight g over denominator 65.38 space straight g space mol to the power of negative 1 end exponent end fraction

number of moles = 0.0994 almost equal to 0.10 mol

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

  • The question states that an excess of copper(II) sulfate solution is added

    • This means that Zn is the limiting reactant

  • The ratio of Zn (s) to Cu (s) is 1:1

  • Therefore, 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 yield = space fraction numerator 4.8 space straight g over denominator 6.4 space straight g end fraction x 100 = 75%

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards

    Did this page help you?

    Previous:Limiting & Excess ReactantsNext:Atom Economy
    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry Content Creator

    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

    Richard Boole

    Reviewer: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.