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Master Percentage Yield and Atom Economy Calculations
Dive into the world of reaction efficiency with our comprehensive guide on percentage yield and atom economy. Learn to assess chemical reactions, optimize processes, and understand their real-world impact.

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Now Playing:Percentage yield and atom economy – Example 0a
Intros
  1. Why do chemists do chemical reactions?
  2. Why do chemists do chemical reactions?
    Practical problems in chemical reactions.
  3. Why do chemists do chemical reactions?
    Atom economy and percentage yield.
Examples
  1. Find the atom economy and percentage yield of chemical reactions.
    Water can be produced by reaction of hydrogen and oxygen gas according to this equation:

    2 H2+_2 + O2 _2 \, →  \, 2 H2_2O

    1. What is the atom economy of this reaction?

    2. If the reaction was performed and 150 g was the theoretical yield of water and the actual yield was only 92 g, what is the percentage yield?

Percentage yield and atom economy
Jump to:NotesConceptFAQsPrerequisites
Notes
In this lesson, we will learn:
  • The meaning of atom economy, percentage yield and the difference between the terms.
  • How to calculate atom economy and percentage yield from example chemical reactions.
  • To explain the importance of atom economy as a chemist when planning and running chemical processes.

Notes:

  • In Moles, excess and limiting reagents, we learned that in chemical reactions, knowing your excess and limiting reagents is important practical information. When your limiting reagent runs out, any amounts in excess have nothing to react with, so you will stop making your products.

    • Percentage yield and atom economy are two other practical considerations when doing chemical reactions. They are both related to the amount of useful product generated in a reaction, compared to undesirable side products or unreacted starting material.

  • The atom economy of a reaction is the percentage of atomic mass of useful products in a reaction. It is calculated by:

    • Atom Economy Atom \, Economy \, (%) = Atomic mass of useful productsTotal atomic mass of produtcs ∗ \frac{Atomic \, mass \, of \, useful \, products } {Total \, atomic \, mass \, of \, produtcs } \, * \, 100

    A high atom economy means most of what the process makes is useful! In the same way, a low atom economy tells you that a reaction is mostly producing unwanted side products.
    The atom economy is used to show the efficiency of a reaction: does it make a lot of waste products that will require storage and disposal, or is most of the product valuable to us?
    • Reactions with a single product have a 100% atom economy because the only chemical being produced is the desired product.
    • Atom economy is used as an application of the conservation of mass: no atoms are created or destroyed in chemical reactions, they are only re-arranged by breaking and forming substances. We are using the amount of atomic mass as a measure of the reaction efficiency.

  • The percentage yield of a reaction is the mass of products formed as a percentage of how much could have been formed given the mass of reactants used. The equation to calculate percentage yield is:

    • Percentage yield = actual yieldtheoretical yield ∗ Percentage \, yield \, = \, \frac{actual \, yield } {theoretical \, yield} \, * \, 100

    Where:
    • Actual yield is the yield of product obtained in the experiment (in g or moles).
    • Theoretical yield is the yield of product based on the limiting reagent, i.e. the calculations done in Moles, excess and limiting reagents.

    A low percentage yield means that not much of the reactants you used has become products. A high percentage yield therefore means that a lot of the reactant chemicals you used successfully reacted to make the products.

  • Worked example: Calculate the atom economy of a reaction.
    The reaction below shows the production of iron metal by reacting iron oxide (Fe2O3) with carbon.

    • 2Fe2O3 (s) + 3C (s)   \, →  \, 4Fe (s)+ 3CO2 (g)

    This reaction is intended to produce iron metal with CO2 as an unwanted side product. What is the atom economy of this reaction?

    To begin finding atom economy, we need to know what is desirable in the products. We already know that Fe is the desired product and CO2 is undesirable side product.
    Now we need to find the atomic mass of both of these – how much of the product, in terms of atomic mass, is valuable?

    % Atom economy = \, Atom \, economy \, = \, 4 ∗ (55.8 g mol−1 Fe)(4 ∗ (55.8 g mol−1 Fe)) + (3 ∗ (44 g mol CO2)) ∗ \large \frac{4 \, * \, (55.8 \, g \, mol^{-1}\, Fe )} { (4 \, * \, (55.8 \, g \, mol^{-1} \, Fe ) ) \, + \, (3 \, * \, (44 \, g \, mol \, CO_{2})) } \, * \, 100 = 62.8 %

  • Worked example: Calculate the percentage yield of a reaction.
    The reaction below shows the production of iron metal by reacting iron oxide with carbon:

    • 2Fe2O3 (s) + 3C (s)   \, →  \, 4Fe (s) + 3CO2 (g)

    In one run of this reaction, the limiting reagent is Fe2O3. 750g of Fe2O3 was reacted in this run, and an experimental yield of 460 g of Fe metal was obtained.

    What is the theoretical and percentage yield of this run of the reaction?

    To begin, we need to find the theoretical yield. If we only have 750g of Fe2O3 and we are told this is the limiting reagent then we know this is going to run out and this amount dictates how much Fe can possibly be made.

    You can use conversion factors to find the theoretical yield of Fe metal product from the amount of Fe2O3 reactant.

    750g Fe2O3 ∗ g \, Fe_{2}O_{3} \, * \, 1 mol Fe2O3159.6g Fe2O3 ∗ 4 mol Fe2 mol Fe2O3 ∗ 55.8 g Fe1 mol Fe \large \frac{1 \, mol \, Fe_{2}O_{3} } {159.6g \, Fe_{2}O_{3} } \, * \, \frac{4\, mol \, Fe } {2 \, mol \, Fe_{2}O_{3} } \, * \, \frac{55.8 \, g \, Fe } {1 \, mol \, Fe } \, = 524.4  g Fe\, g \, Fe


    This 524.4g Fe tells us the theoretical yield of the reaction. It is the maximum possible amount of product we can get in this reaction, because we only have 750g of the Fe2O3 limiting reagent. In terms of percentage yield, this is your 100% mark.

    Now, we have the 100% mark, and we have an experimental yield of 460g of Fe. This is our actual “score”. Let’s convert 460/524.4 into a percentage.

    Percentage yield = Percentage \, yield \, = \, 460 g Fe524.4 g Fe ∗ \large \frac{460 \, g \, Fe } {524.4 \, g \, Fe} \, * \, 100 = 87.7%

    This is the percentage yield of the reaction.

  • Know the difference between yield and atom economy!
    • Atom economy is about how wasteful the reaction is. If your reaction has low atom economy, it will always be wasteful; most of the product made is simply not valuable to you. This is a chemical problem, not a practical one.
      The unwanted products are a waste of money/resources and an environmental problem because waste has to be stored or disposed of safely.
    • Yield is about how much product was successfully made. If your yield is low, this is probably a practical problem, such as not enough time to react or your conditions may need to be changed.
Concept

Introduction to Percentage Yield and Atom Economy

Percentage yield and atom economy are crucial concepts in chemistry that help assess the efficiency of chemical reactions. Percentage yield compares the actual amount of product obtained to the theoretical maximum, indicating how well a reaction performs in practice. Atom economy, on the other hand, measures the proportion of reactant atoms that end up in the desired product, highlighting the reaction's efficiency in terms of resource utilization. The introduction video provides a comprehensive overview of these concepts, offering visual explanations and practical examples to enhance understanding. By watching this video, students and chemistry enthusiasts can grasp the significance of these measurements in real-world applications. These concepts are essential for evaluating reaction effectiveness, optimizing industrial processes, and developing sustainable chemical practices. Understanding percentage yield and atom economy is fundamental for chemists, chemical engineers, and anyone involved in the design and analysis of chemical reactions, as they provide valuable insights into reaction efficiency and environmental impact.

FAQs

Here are some frequently asked questions about percentage yield and atom economy:

  1. How do you calculate percentage yield?

    Percentage yield is calculated using the formula: (Actual Yield / Theoretical Yield) × 100%. For example, if you expect to produce 50g of a product (theoretical yield) but only obtain 40g (actual yield), the percentage yield would be (40g / 50g) × 100% = 80%.

  2. What does a 50% percent yield mean?

    A 50% percent yield means that the actual amount of product obtained from a reaction is half of the theoretical maximum amount. This could be due to factors such as incomplete reactions, side reactions, or loss during purification.

  3. How do you calculate atom economy?

    Atom economy is calculated using the formula: (Molecular mass of desired product / Total molecular mass of all reactants) × 100%. This measures the efficiency of a reaction in terms of incorporating reactant atoms into the desired product.

  4. What is a good atom economy percentage?

    Generally, an atom economy above 80% is considered good, while above 90% is excellent. Higher atom economy indicates more efficient use of reactants and less waste production. However, the specific context of the reaction should be considered when evaluating atom economy.

  5. Is a high percent yield always good?

    While a high percent yield is generally desirable, it's not always the only factor to consider. Other aspects such as product purity, cost-effectiveness, safety, and environmental impact should also be taken into account. Sometimes, a lower yield might be acceptable if it results in a purer product or uses more sustainable methods.

Prerequisites

Understanding the fundamental concepts of chemistry is crucial when delving into more advanced topics like percentage yield and atom economy. Two essential prerequisite topics that lay the groundwork for this understanding are introduction to chemical reactions and balancing chemical equations.

The study of chemical reactions is fundamental to grasping the concepts of percentage yield and atom economy. When we explore chemical reactions, we learn about the transformation of substances, the rearrangement of atoms, and the formation of new compounds. This knowledge is essential because percentage yield and atom economy are directly related to the outcomes and efficiency of these reactions.

In particular, understanding stoichiometry in chemical reactions is crucial. Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. This concept forms the basis for calculating percentage yield, which compares the actual yield of a reaction to the theoretical yield predicted by stoichiometric calculations.

Moreover, the ability to write and interpret balanced chemical equations is indispensable when studying percentage yield and atom economy. Balanced equations provide the foundation for understanding the conservation of mass in chemical reactions, which is a key principle in calculating both percentage yield and atom economy.

When working with balanced chemical equations, students learn to account for all atoms on both sides of the equation. This skill is directly applicable to atom economy calculations, where we assess the efficiency of a reaction by determining the percentage of atoms from the reactants that end up in the desired product.

By mastering these prerequisite topics, students develop a solid foundation for understanding more complex concepts like percentage yield and atom economy. The ability to write and balance chemical equations allows for accurate predictions of reaction outcomes, while knowledge of stoichiometry enables precise calculations of theoretical yields.

Furthermore, these fundamental concepts help students appreciate the practical implications of percentage yield and atom economy in real-world applications. In industrial processes, for example, optimizing reaction conditions to improve yield and atom economy can lead to more efficient and cost-effective production methods.

In conclusion, a thorough understanding of chemical reactions and balanced equations provides the necessary context for exploring percentage yield and atom economy. These prerequisite topics not only facilitate the learning process but also enhance the overall comprehension of advanced chemical concepts, preparing students for more sophisticated studies in chemistry and its applications.