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ASChemistryOCRRevision Notes2. Foundations in ChemistryAmount of SubstancePercentage Yield & Atom Economy

Percentage Yield & Atom Economy (OCR AS Chemistry A): Revision Note

Exam code: H032

Richard Boole

Written by: Richard Boole

Reviewed by: Philippa Platt

Updated on

Percentage yield calculations

  • In many reactions, not all reactants are fully converted into products

    • This can happen for several reasons:

      • The reaction may not go to completion

      • Side reactions may occur

      • Some product may be lost during separation or purification

  • The percentage yield tells you how much product you actually obtained compared to the maximum theoretical amount:

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

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

  • The theoretical yield is the mass or number of moles of product expected from a reacting mass calculation using the balanced chemical equation

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:

  1. The balanced symbol equation is:

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

  1. Calculate the amount of zinc in moles:

n(Zn) = 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

n(Zn) = 0.10 mol

  1. Use the molar ratio to calculate the maximum amount of copper produced:

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

    • Therefore, a maximum of 0.10 moles can be produced

  2. Calculate the maximum (theoretical) mass of copper produced:

mass = mol x M

mass = 0.10 mol x 63.55 g mol-1

mass = 6.4 g (2 sig figs)

  1. Calculate the percentage yield of copper:

percentage yield = fraction numerator 4.8 space straight g over denominator 6.4 space straight g end fraction x 100

percentage yield = 75%

Atom economy calculations

What is atom economy?

  • Atom economy tells you how many of the atoms in the reactants end up in the desired product

    • Any atoms not used in the main product are considered waste

  • It is found directly from the balanced equation by calculating the Mr of the desired product

Atom economy = fraction numerator molecular space mass space of space desired space product over denominator total space molecular space mass space of space ALL space products end fraction x 100

  • In addition reactions, there is only one product

    • Therefore, the atom economy is 100%

  • In substitution or elimination reactions, some atoms form by-products

    • This reduces the atom economy

  • Atom economy can also be calculated using mass, instead or Mr:

Atom economy = fraction numerator space mass space of space desired space product over denominator total space mass space of space ALL space products end fraction x 10

Worked Example

Qualitative atom economy

Ethanol can be produced by various reactions, such as:

  • Hydration of ethene:

C2H4 + H2O → C2H5OH

  • Substitution of bromoethane:

C2H5Br + NaOH → C2H5OH + NaBr

Explain which reaction has a higher atom economy.

Answer:

  • Hydration of ethene has a higher atom economy

    • The hydration of ethene has a higher atom economy (100%) because it produces only one product

    • The substitution reaction produces an unwanted by-product, NaBr, which reduces atom economy

Worked Example

Quantitative atom economy

The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron.

Fe2O3 + 3CO → 2Fe + 3CO2

Calculate the atom economy for this reaction, assuming that iron is the desired product.

(Ar / Mr data: Fe2O3 = 159.6, CO = 28.0, Fe = 55.8, CO2 = 44.0)

Answer:

  1. Write the equation:

Atom economy = fraction numerator molecular space mass space of space desired space product over denominator total space mass space of space ALL space products end fraction x 100

  1. Substitute values and evaluate:

Atom economy = fraction numerator 2 cross times 55.8 over denominator 159.6 plus open parentheses 3 cross times 28.0 close parentheses end fraction x 100

Atom economy = 45.8%

Examiner Tips and Tricks

Careful: Sometimes a question may ask you to show your working when calculating atom economy.

In this case, even if it is an addition reaction and it is obvious that the atom economy is 100%, you will still need to show your working.

Benefits of high atom economy

  • Chemists use percentage yield to measure how much of the theoretical product is actually obtained

    • A high yield suggests efficient conversion of reactants to products

  • For example, the Haber process has an estimated percentage yield of around 15% per pass

    • This is due to a compromise between pressure, temperature, and safety

    • However, because unreacted materials are recycled, the overall conversion of reactants reaches around 97%

  • However, percentage yield does not account for waste products

    • A reaction may have a high yield but low atom economy, meaning a lot of waste is still produced

  • Atom economy measures how much of the total mass of reactants ends up as the desired product:

Atom economy = fraction numerator molecular space mass space of space desired space product over denominator sum space of space molecular space masses space of space ALL space products end fraction x 100

Atom economy and Green Chemistry

  • Chemists often have multiple options for making a compound. Choosing a route with high atom economy is a key principle of Green Chemistry

Diagram showing green chemistry principles: prevent waste, atom economy, less hazardous synthesis, safer chemicals, safer solvents, energy efficiency.
The twelve principles of green chemistry

  • A route with fewer steps reduces waste, energy use, and cost

  • High atom economy means:

    • Less waste

    • Greater sustainability

    • Reduced use of raw/finite resources

Example: Ibuprofen synthesis

  • The original 1960s Boots method was a 6-step process

    • Even with 90% atom economy per step, the overall atom economy was 53%

  • The modern method is a 3-step process

    • Same step efficiency gives overall atom economy = 73%

  • Selecting routes with high atom economy and fewer steps improves:

    • Environmental impact

    • Long-term sustainability

    • Industrial efficiency

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    Author
    Reviewer
    Richard Boole

    Author: 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.

    Philippa Platt

    Reviewer: 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