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4.3: Percent Yield

  • Page ID
    51455
  • Learning Objectives

    • Differentiate between actual yield, theoretical yield, and percent yield
    • Calculate percent yield

    Percent Yield

    Percent Yield is defined as the actual yield divided by the theoretical yield times 100.

    \[\text{Percent Yield} = \left ( \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \right ) \times 100\%\]

    There are many reasons why the actual yield of a chemical reaction may be less than the theoretical yield, and these will be taken up during later Chapters of the course. Here are some reasons, most of which deal with topics that will be covered in the second semester of this course.

    1. Equilibria between products and reactants, where the limiting reagent is not completely consumed (Chapter 15)

    \[Ca_3PO_4(s) \leftrightharpoons 3Ca^{+2} + 2PO_4^{-3}\]

    1. Kinetics, simply speaking, the reaction may be very slow and not over (Chapter 14).
    2. Formation of Intermediates. A chemical equation does not represent the mechanism but the overall balance of mass. The mechanism tells how the reaction actually proceeds (section 14.6), and there are often intermediate chemical species that are not completely consumed. This is exemplified in the following reaction.

    \[H_{2}(g) + 2ICl(g) \rightarrow I_{2}(g) + 2HCl{g}\]

    It is highly unlikely that this reaction takes place in one step where three molecules simultaneously collide. A more plausible way for the reaction to occur is in two steps as below, where HI is an intermediate.

    \[\begin{align} H_{2} + ICl & \rightarrow HCl + \textcolor{red}{HI} \\ \textcolor{red}{HI} + ICl & \rightarrow HCl + I_{2} \end{align}\]

    In the first step HCl and the intermediate HI are formed from the bimolecular collision of H2 and ICl. In the second step, the HI (intermediate) collides with another ICl to form HCl and I2. If you add the two steps the intermediate cancels out and you get the balanced equation.

    \[H_{2}(g) + 2ICl(g) \rightarrow I_{2}(g) + 2HCl{g}\]


    Percent Yield Problem

    Phosphoric Acid can be synthesized from Phosphorous, Oxygen and water according to the following equations

    \[4P + 5O_2+ 6H_2O \rightarrow 4H_3PO_4\]

    What is the percent yield if 50.3 g of phosphoric acid is created from 20.0 g phosphorous, 15.0 g of water and 30.0 g of oxygen?

    First we set up the problem and it is suggested that we write down the given quantities under each species of the balanced equation (as that makes it easier to see the given information)

    \[\underset{m=20.0g \\ fw = 30.98g/mol}{4P} + \underset{m=30.0g \\fw=32.00g/mol}{5O_2}+ \underset{m=15.0g \\ fw=18.01g/mol}{6H_2O} \rightarrow \underset{Actual Yield=50.3g \\ fw=97.995g/mol \\ Theoretical Yield =?}{4H_3PO_4}\]

    After calculating the theoretical yield (based on the complete consumption of the limiting reagent) we calculate the percent yield. This problem is solved in video \(\PageIndex{1}\).

    Video \(\PageIndex{1}\): 4'40" YouTube solving a percent yield problem (https://youtu.be/BCOdyr1QKj0).

    Exercise \(\PageIndex{1}\)

    A chemist decomposes 1.006 g of NaHCO3 and obtains 0.434 g of Na2CO3. What are the theoretical yield and the actual yield? What is the percent yield?

    2NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g)

    Answer

    theoretical yield = 0.6345 g; actual yield = 0.434 g

    percent yield = 68.4%

    Contributors and Attributions

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • November Palmer & Ronia Kattoum (UALR)
    • anonymous