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Theoretical and Actual Yields

Key Terms

(Excess reagent, limiting reagent)
Theoretical and actual yields
Percentage or actual yield

Skills to develop

Theoretical and Actual Yields

Reactants not completely used up are called excess reagent, and the reactant completely reacts is called limiting reagent. This concept has been illustrated for the reaction:

2 Na + Cl2 = 2 NaCl,

Amounts of products calculated from the complete reaction of the limiting reagent is called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield. 


                 actual yield
percent yield = ----------------- x 100
                theoretical yield

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction.

Example 1

Methyl alcohol can be produced in a high-pressure reaction CO(g) + 2 H2(g) = CH3OH(l) If 6.1 metric tones of methyl alcohol is obtained from 1.2 metric tones of hydrogen reacting with excess amount of CO, estimate the theoretical and the percentage yield?
Hint...
To calculate the theoretical yield, consider the reaction 
 CO(g) + 2 H2(g) = CH3OH(l)
 28.0    4.0        32.0 (stoichiometric masses in g, kg, or tons)

             32.0 CH3OH
1.2 tons H2 ------------ = 9.6 tones CH3OH
               4.0 H2
Thus, the theoretical yield from 1.2 metric tons (1.2x106 g) of hydrogen gas is 9.6 tons.
The actual yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is 
          6.1 tones
% yield = --------- x 100 = 64 %
          9.6 tones

Discussion...
Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. A loss during the recovery process causes even lower actual yield.

Example 2

A solution containing silver ion, Ag+, has been treated with excess of chloride ions Cl-. When dried, 0.1234 g of AgCl has recovered. Assume the percentage yield to be 98.7%, how many grams of silver ions were present in the solution?
Hint...
The reaction and relative masses of reagents and product are: 
Ag+(aq) + Cl-(aq) = AgCl(s)
107.868 + 35.453 = 143.321
The calculation, 
               107.868 g Ag+
0.1234 g AgCl --------------- = 0.09287 g Ag+.
               143.321 g AgCl
shows that 0.1234 g dry AgCl comes from 0.09287 g Ag+ ions. Since the actual yield is only 98.7%, the actual amount of Ag+ ions present is therefore, 
0.09287 g Ag+
-------------- = 0.09409 g Ag+
    0.987

Discussion...
One can also calculate the theoretical yield of AgCl from the percentage yield of 98.7% to be, 
0.1234 g AgCl
------------- = 0.1250 g AgCl
   0.987
From 0.1250 g AgCl, the amount of Ag+ present is also 0.09409 g.

Skill Developing Problems

  1. In an analytical experiment, you are asked to determine the amount of iodide ion I+ in 10.00 mL of a solution that does not contain any other ions that will form a precipitate with silver ions. You have learned that Ag+ ions precipitate all the iodide ions in a solution. In performing the experiment, shall you treat AgNO3 as the excess reagent or limiting reagent? Molar mass or atomic weight: Ag, 107.868; I, 126.904 (You should know where to find them).

    Skill -
    Apply the concept of excess and limiting reagent for work. You can add AgNO3 slowly until the clear portion of the solution gives no precipitate when a drop of AgNO3 solution is added. This indicates that all the I- ions are consumed.

  2. From the 10.00 mL iodide solution, you have added AgNO3 solution or solid. How do you know that you have added excess amount of AgNO3 to precipitate the iodide ions?

    Skill -
    Excess reagent can be tested for its presence, and limiting reagent can be tested for its absence.

  3. In an analytical experiment, 0.1234 g of AgI was obtained from a 10.00-mL solution with excess silver nitrate. How much (in g) iodide ions are present?

    Skill -
    Calculate the amount of limiting reagent from the amount of products.

  4. In an analytical experiment, 0.1234 g of AgI was obtained from a 10.00-mL solution with excess silver nitrate. What is the iodide concentration (mol/L or M) in the solution?

    Skill -
    Calculate the concentration when the amount of solute is known. The concept of concentration will be covered in the unit dealing with solution, but you should be able to convert see the following relationship. 0.1234 g AgI = 0.0005256 mol = 0.5256 mili-mol AgI or Ag or I.

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