The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Analysis of how much of a compound is produced in any given reaction is an important part of cost control.
If the actual yield of C 6 H 5 Br is 63.6 g, what is the percent yield? Use the following reaction: C 4 H 9 OH + NaBr + H 2 SO 4 C 4 H 9 Br + NaHSO 4 + H 2 O If 15.0 g of C 4 H 9 OH react with 22.4 g of NaBr and 32.7 g of H 2 SO 4 to yield 17.1 g of C 4 H 9 Br, what is the percent yield of this reaction? 5) For the balanced equation shown below, if the reaction of 24.4 grams of C2H3Cl produces a 68.0% yield, how many grams of CO2 would be produced? 23.358 4C2H3Cl+11O2=8CO2+6H2O+2Cl2 Theoretical Yield = (352/250). (24.4) = 34.35 moles Actual Yield = 2335.8/100 = 23.358 grams Click here to go back to Percentage Yield (% Yield).
Percent Yield
Chemical reactions in the real world do not always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are often losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.
To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage:
[text{Percent Yield} = frac{text{Actual Yield}}{text{Theoretical Yield}} times 100%]
Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than (100%) because of the reasons indicated earlier. However, percent yields greater than (100%) are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction.
Example (PageIndex{1})
Percent Yield Skill Practice 34 Practice Percent Yield Answers
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below.
[2 ce{KClO_3} left( s right) rightarrow 2 ce{KCl} left( s right) + 3 ce{O_2} left( g right)]
In a certain experiment, (40.0 : text{g} : ce{KClO_3}) is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed, the oxygen gas is collected, and its mass is found to be (14.9 : text{g}). What is the percent yield for the reaction?
Solution
First, we will calculate the theoretical yield based on the stoichiometry.
Step 1: List the known quantities and plan the problem.
Known
- Given: Mass of (ce{KClO_3} = 40.0 : text{g})
- Molar mass (ce{KClO_3} = 122.55 : text{g/mol})
- Molar mass (ce{O_2} = 32.00 : text{g/mol})
Unknown
- Theoretical yield (ce{O_2} = ? : text{g})
Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
[text{g} : ce{KClO_3} rightarrow text{mol} : ce{KClO_3} rightarrow text{mol} : ce{O_2} rightarrow text{g} : ce{O_2} nonumber]
Step 2: Solve.
Percent Yield Skill Practice 34 Free
[40.0 : text{g} : ce{KClO_3} times frac{1 : text{mol} : ce{KClO_3}}{122.55 : text{g} : ce{KClO_3}} times frac{3 : text{mol} : ce{O_2}}{2 : text{mol} : ce{KClO_3}} times frac{32.00 : text{g} : ce{O_2}}{1 : text{mol} : ce{O_2}} = 15.7 : text{g} : ce{O_2} nonumber]
The theoretical yield of (ce{O_2}) is (15.7 : text{g}).
Step 3: Think about your result.
The mass of oxygen gas must be less than the (40.0 : text{g}) of potassium chlorate that was decomposed.
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
Step 1: List the known quantities and plan the problem.
Known
- Actual yield (= 14.9 : text{g})
- Theoretical yield (= 15.7 : text{g})
Unknown
- Percent yield (= ? %)
[text{Percent Yield} = frac{text{Actual Yield}}{text{Theoretical Yield}} times 100% nonumber]
Use the percent yield equation above.
Step 2: Solve.
[text{Percent Yield} = frac{14.9 : text{g}}{15.7 : text{g}} times 100% = 94.9% nonumber]
Step 3: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under (100%).
Summary
Theoretical yield is calculated based on the stoichiometry of the chemical equation. The actual yield is experimentally determined. The percent yield is determined by calculating the ratio of actual yield/theoretical yield.
Contributors and Attributions
CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.