Theoretical and Actual Yields

Key Terms

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

Learning Objectives

• Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction and explain why.
• Calculate theoretical yields of products formed in reactions that involve limiting reagents.
• Evaluate percentage or actual yields from known amounts of reactants

Theoretical and Actual Yields

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

\[\mathrm{2 Na + Cl_2 \rightarrow 2 NaCl} \nonumber \]

Amounts of products calculated from the complete reaction of the limiting reagent are 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.

\(\mathrm{percent\: yield = \dfrac{actual\: yield}{theoretical\: yield}\times100}\)

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 \(\PageIndex{1}\)

Methyl alcohol can be produced in a high-pressure reaction

\(\mathrm{CO_{\large{(g)}} + 2 H_{2\large{(g)}} \rightarrow CH_3OH_{\large{(l)}}}\)

If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess amount of \(\ce{CO}\), estimate the theoretical and the percentage yield?

Solution

To calculate the theoretical yield, consider the reaction

\(\begin{alignat}{2}
\ce{&CO_{\large{(g)}} +\, &&2 H_{2\large{(g)}} \rightarrow \, &&CH_3OH_{\large{(l)}}}\\
&\:28.0 &&\:4.0 &&\:\:\:32.0 \hspace{45px}\ce{(stoichiometric\: masses\: in\: g,\: kg,\: or\: tons)}
\end{alignat}\)

\(\mathrm{1.2\: tons\: H_2 \times\dfrac{32.0\: CH_3OH}{4.0\: H_2}= 9.6\: tons\: CH_3OH}\)

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

\(\mathrm{\%\: yield =\dfrac{6.1\: tons}{9.6\: tons}\times 100 = 64 \%}\)

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. Losses during the recovery process of the product will cause an even lower actual yield.

Example \(\PageIndex{2}\)

A solution containing silver ion, \(\ce{Ag+}\), has been treated with excess of chloride ions \(\ce{Cl-}\). When dried, 0.1234 g of \(\ce{AgCl}\) was recovered. Assuming 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:

\(\begin{alignat}{2}\ce{
Ag^+_{\large{(aq)}}} &+ \mathrm{Cl^-_{\large{(aq)}}} &&\rightarrow \ce{AgCl_{\large{(s)}}} \\
107.868 &+ 35.453 &&= 143.321
\end{alignat}\)

The calculation,

\(\mathrm{0.1234\: g\: AgCl \times \dfrac{107.868\: g\: Ag^+}{143.321\: g\: AgCl}= 0.09287\: g\: Ag^+}\)

shows that 0.1234 g dry \(\ce{AgCl}\) comes from 0.09287 g \(\ce{Ag+}\) ions. Since the actual yield is only 98.7%, the actual amount of \(\ce{Ag+}\) ions present is therefore

\(\mathrm{\dfrac{0.09287\: g\: Ag^+}{0.987}= 0.09409\: g\: Ag^+}\)

DISCUSSION

One can also calculate the theoretical yield of \(\ce{AgCl}\) from the percentage yield of 98.7% to be

\(\mathrm{\dfrac{0.1234\: g\: AgCl}{0.987}= 0.1250\: g\: AgCl}\)

From 0.1250 g \(\ce{AgCl}\), the amount of \(\ce{Ag+}\) present is also 0.09409 g.

Skill Developing Problems

1. In an analytical experiment, you are asked to determine the amount of iodide ion \(\ce{I+}\) 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 \(\ce{Ag+}\) ions precipitate all the iodide ions in a solution. In performing the experiment, shall you treat \(\ce{AgNO3}\) as the excess reagent or limiting reagent? Molar mass or atomic weight: \(\ce{Ag}\), 107.868; \(\ce{I}\), 126.904 (You should know where to find them).

Hint: \(\ce{AgNO3}\) is the excess reagent

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

2. From the 10.00 mL iodide solution, you have added \(\ce{AgNO3}\) solution or solid. How do you know that you have added an excess amount of \(\ce{AgNO3}\) to precipitate the iodide ions?

Hint: Test for excess or limiting reagent.

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 \(\ce{AgI}\) was obtained from a 10.00-mL solution with excess silver nitrate. How much (in g) iodide ions are present?

Hint: 0.05670 g \(\ce{Ag}\) and 0.06670 g \(\ce{I}\)

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

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

Hint: 0.05256 mol/L

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 \(\ce{AgI}\) = 0.0005256 mol = 0.5256 mili-mol \(\ce{AgI}\) or \(\ce{Ag}\) or \(\ce{I}\).

Contributors and Attributions

• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)