Chapter 8: Stoichiometry of Chemical Reactions

# 8.4 Reaction Yields

### Learning Outcomes

• Explain the concepts of theoretical yield and limiting reactants/reagents
• Derive the theoretical yield for a reaction under specified conditions
• Calculate the percent yield for a reaction

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding section involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this chapter, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

## Limiting Reactant

Consider another food analogy, making grilled cheese sandwiches (Figure 8.4.1):

$\text{1 slice of cheese}+\text{2 slices of bread}\rightarrow\text{1 sandwich}$

Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

${\text{H}}_{2}\text{(}s\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow\text{2 HCl}\text{(}g\text{)}$

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of $\ce{H2}$ and 2 moles of $\ce{Cl2}$. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.

An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield

$\text{mol HCl produced}=3\text{ mol H}_{2}\times\dfrac{2\text{ mol HCl}}{1\text{ mol H}_{2}}=\text{6 mol HCl}$

Complete reaction of the provided chlorine would produce

$\text{mol HCl produced}=2\text{ mol Cl}_{2}\times\dfrac{\text{2 mol HCl}}{\text{1 mol Cl}_{2}}=\text{4 mol HCl}$

The chlorine will be completely consumed once 4 moles of $\ce{HCl}$ have been produced. Since enough hydrogen was provided to yield 6 moles of $\ce{HCl}$, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 8.4.2).

### Example 8.4.1: Identifying the Limiting Reactant

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

$\text{3Si}\text{(}s\text{)}+2{\text{N}}_{2}\text{(}g\text{)}\rightarrow{\text{Si}}_{3}{\text{N}}_{4}\text{(}s\text{)}$

Which is the limiting reactant when 2.00 g of $\ce{Si}$ and 1.50 g of $\ce{N2}$ react?

Show Solution

Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

$\text{mol Si}=2.00\cancel{\text{g Si}}\times \dfrac{\text{1 mol Si}}{28.09\cancel{\text{g Si}}}=\text{0.0712 mol Si}$

$\text{mol}{\text{ N}}_{2}=1.50\cancel{\text{g}{\text{ N}}_{2}}\times \dfrac{\text{1 mol}{\text{ N}}_{2}}{28.09\cancel{\text{g}{\text{ N}}_{2}}}=\text{0.0535 mol}{\text{ N}}_{2}$

The provided $\ce{Si:N2}$ molar ratio is:

$\dfrac{\text{0.0712 mol Si}}{\text{0.0535 mol}{\text{ N}}_{2}}=\dfrac{\text{1.33 mol Si}}{\text{1 mol}{\text{ N}}_{2}}$

The stoichiometric $\ce{Si:N2}$ ratio is:

$\dfrac{\text{3 mol Si}}{\text{2 mol}{\text{ N}}_{2}}=\dfrac{\text{1.5 mol Si}}{\text{1 mol}{\text{ N}}_{2}}$

Comparing these ratios shows that $\ce{Si}$ is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

$\text{mol Si}_{3}{\text{N}}_{4}\text{ produced}=\text{0.0712 mol Si}\times \dfrac{1\text{ mol }{\text{Si}}_{3}{\text{N}}_{4}}{\text{3 mol Si}}=0.0237\text{ mol }{\text{Si}}_{3}{\text{N}}_{4}$

while the 0.0535 moles of nitrogen would produce

$\text{mol}{\text{ Si}}_{3}{\text{N}}_{4}\text{ produced}=\text{0.0535 mol}{\text{ N}}_{2}\times \dfrac{\text{1 mol}{\text{ Si}}_{3}{\text{N}}_{4}}{\text{2 mol}{\text{ N}}_{2}}=\text{0.0268 mol}{\text{ Si}}_{3}{\text{N}}_{4}$

Since silicon yields the lesser amount of product, it is the limiting reactant.

## Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

$\text{percent yield}=\dfrac{\text{actual yield}}{\text{theoretical yield}}\times 100\%$

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

### Example 8.4.2: Calculation of Percent Yield

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

${\text{CuSO}}_{4}\text{(}aq\text{)}+\text{Zn}\text{(}s\text{)}\rightarrow\text{Cu}\text{(}s\text{)}+{\text{ZnSO}}_{4}\text{(}aq\text{)}$

What is the percent yield?

Show Solution

The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous section, as shown here:

$1.274\cancel{\text{g}{\text{ CuSO}}_{4}}\times \dfrac{1\cancel{\text{mol}{\text{ CuSO}}_{4}}}{159.62\cancel{\text{g}{\text{ CuSO}}_{4}}}\times \dfrac{1\cancel{\text{mol Cu}}}{1\cancel{\text{mol}{\text{ CuSO}}_{4}}}\times \dfrac{63.55\text{g Cu}}{1\cancel{\text{mol Cu}}}=\text{0.5072 g Cu}$

Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

$\text{percent yield}=\left(\dfrac{\text{actual yield}}{\text{theoretical yield}}\right)\times 100$
$\begin{array}{c}\\ \text{percent yield}=\left(\dfrac{\text{0.392 g Cu}}{\text{0.5072 g Cu}}\right)\times 100\\ =77.3\%\end{array}$

### Green Chemistry and Atom Economy

The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at the EPA’s “Green Chemistry” website). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

$\text{atom economy}=\dfrac{\text{mass of product}}{\text{mass of reactants}}\times 100\%$

Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 8.4.3). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.

### Key Concepts and Summary

When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.

#### Key Equations

• $\text{percent yield}=\left(\dfrac{\text{actual yield}}{\text{theoretical yield}}\right)\times 100$

### Try It

1. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?
2. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield?
3. Toluene, $\ce{C6H5CH3}$, is oxidized by air under carefully controlled conditions to benzoic acid, $\ce{C6H5CO2H}$, which is used to prepare the food preservative sodium benzoate, $\ce{C6H5CO2Na}$. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid? $2{\text{C}}_{6}{\text{H}}_{5}{\text{CH}}_{3}+3{\text{O}}_{2}\rightarrow 2{\text{C}}_{6}{\text{H}}_{5}{\text{CO}}_{2}\text{H}+2{\text{H}}_{2}\text{O}$
4. Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for \$5? Explain why or why not. Find the current price of gold at https://www.cnn.com/business/markets/commodities $\text{(}\text{1 troy ounce}=\text{31.1 g}\text{)}$
Show Selected Solutions
1. Write the balanced chemical equation and determine the number of moles of each reactant available.
The reaction is:$2\text{Na}+{\text{Cl}}_{2}\rightarrow 2\text{NaCl}$
$\text{Moles of Na}=\text{8 g Na}\times \frac{\text{1 mol}}{\text{23.0 g}}=\text{0.3 mol Na}$
$\text{Moles of}{\text{ Cl}}_{2}=\text{8 g}\times \frac{\text{1 mol}}{\text{70.9 g}}=\text{0.1 mol}{\text{ Cl}}_{2}$The stoichiometric ratio is 2 mol $\ce{Na}$: 1 mol $\ce{Cl2}$; since the reactants are provided in a 0.3:0.1 or 3:1 ratio, $\ce{Na}$ is present in excess and $\ce{Cl2}$ is the limiting reactant.
2. $\text{Percent yield}=\frac{\text{25 g}}{\text{81 g}}\times 100\%=31\%$
3. $\begin{array}{l} 100\cancel{\text{g}{\text{ C}}_{6}{\text{H}}_{5}{\text{CH}}_{3}}\times \frac{1\cancel{\text{mol toluene}}}{92.13\cancel{\text{g}{\text{C}}_{6}{\text{H}}_{5}{\text{CH}}_{3}}}\times \frac{1\cancel{\text{mol benzoic acid}}}{1\cancel{\text{mol toluene}}}\times \frac{122.1\cancel{\text{g benzoic acid}}}{1\cancel{\text{mol benzoic acid}}}\\ \times \frac{\text{1 kg}}{1000\cancel{\text{g benzoic acid}}}=\text{1.325 kg benzoic acid}\text{(}\text{theoretical yield}\text{)}\end{array}$
$\text{percent yield}=\frac{\text{1.21 g}}{\text{1.325 g}}\times 100\%=91.3\%$
4. $\text{mass Au}=\frac{1\times {10}^{12}\cancel{\text{atoms Au}}}{6.022\times {10}^{23}\cancel{\text{atoms}{\text{ mol}}^{-1}}}\times 196.97\cancel{\text{g}{\text{ mol}}^{-1}}=3.27\times {10}^{-10}\text{g}$ This amount cannot be weighted by ordinary balances and is worthless.

## Glossary

actual yield: amount of product formed in a reaction

excess reactant: reactant present in an amount greater than required by the reaction stoichiometry

limiting reactant: reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated

percent yield: measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield

theoretical yield: amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry