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:
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:
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:
[latex]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}[/latex]
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be
[latex]\text{percent yield} = (\frac{0.392 \;\text{g Cu}}{0.5072 \;\text{g Cu}}) \times 100[/latex]
[latex]= 77.3\%[/latex]
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?
Check Your Answer[1]
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 Principle