# 9.4 Reaction Yields

## Learning Objectives

By the end of this section, you will be able to:

• Explain the concepts of theoretical yield
• Derive the theoretical yield for a reaction under specified conditions
• Calculate the percent yield for a reaction

## 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:

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

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 9.4a

#### 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:

$\text{CuSO}_4(aq) + \text{Zn}(s) \longrightarrow \text{Cu}(s) + \text{ZnSO}_4(aq)$

What is the percent yield?

#### 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 module, as shown here:

$1.274 \;\rule[0.5ex]{3.75em}{0.1ex}\hspace{-3.75em}\text{g CuSO}_4 \times \frac{1 \;\rule[0.25ex]{4em}{0.1ex}\hspace{-4em}\text{mol CuSO}_4}{159.62 \;\rule[0.25ex]{3em}{0.1ex}\hspace{-3em}\text{g CuSO}_4} \times \frac{1 \;\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol Cu}}{1 \rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol CuSO}_4} \times \frac{63.55 \;\text{g Cu}}{1 \;\rule[0.5ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol Cu}} = 0.5072 \;\text{g Cu}$

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

$\text{percent yield} = (\frac{\text{actual yield}}{\text{theoretical yield}}) \times 100$

$\text{percent yield} = (\frac{0.392 \;\text{g Cu}}{0.5072 \;\text{g Cu}}) \times 100$

$= 77.3\%$

### Exercise 9.4a

What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?

$\text{CCl}_4 + 2\text{HF} \longrightarrow \text{CF}_2 \text{Cl}_2 + 2\text{HCl}$

### 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”. 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:

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

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 9.4a). 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 Equations

• $\text{percent yield} = (\frac{\text{actual yield}}{\text{theoretical yield}}) \times 100$

## Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards from “General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)​.

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