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2 To access a customizable version of this book, as well as other interactive content, visit AUTHOR CK12 Editor CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 2013 CK-12 Foundation, The names CK-12 and CK12 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License ( licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: December 11, 2013

3 Concept 1. Percent Yield CONCEPT 1 Percent Yield Lesson Objectives Define theoretical and actual yield. Explain the difference between theoretical and actual yield. Calculate percent yield (or reaction efficiency). Introduction The amount of product that should be formed when the limiting reactant is completely consumed is called the theoretical yield of that product. This is the maximum amount of the product that could form from the quantities of reactants used. In actual practice, this theoretical yield is seldom obtained because of side reactions, failure of the reaction to go to completion, and other complications. The actual amount of product produced in a laboratory or industrial reaction is called the actual yield. The actual yield is almost always less than the theoretical yield. The actual yield is often expressed as a percentage of the theoretical yield. This is called the percent yield. Theoretical Yield Compared to Actual Yield When we calculate the amount of product that can be produced from limiting reactants, we are determining the maximum theoretical amount of product we could obtain from the reaction. In other words, the theoretical yield is the maximum amount obtained when all of the limiting reactant has reacted in the balanced chemical equation. Think about our sandwich maker. If our sandwich maker was given 10 slices of bread and 12 cheese slices, how many cheese sandwiches could he make with 2 slices of bread and 2 cheese slices? The sandwich maker could actually only make 5 sandwiches because there are only 10 slices of bread and 5 2 = 10. Thus, the theoretical yield of sandwiches is 5. The same idea can be translated into chemical equations. When Kerry-Sue worked in the laboratory with solutions of silver nitrate and potassium sulfide, she worked with the equation that is given below: 2 AgNO 3(aq) + K 2 S (aq) Ag 2 S (s) + KNO 3(aq) When Kerry-Sue added 5.00 g of the silver nitrate with 2.50 g potassium sulfide, what was her theoretical yield? Notice that this is a limiting reagent problem because we are given two reactant amounts and we have to determine which of these reactants runs out first. We have to use the steps from the previous lesson in this section as well. We start with our first step to calculate the moles of each given reactant before we continue to determine the limiting reactant and then the mass of product or the theoretical yield. 2 AgNO 3(aq) + K 2 S (aq) Ag 2 S (s) + KNO 3(aq) 1

4 Step 1: Determine the number of moles of each reactant in the chemical equation. mols AgNO 3 = 5.00 g 170. g/mol = mol AgNO 3 mols K 2 S = 2.50 g 110. g/mol = mol K 2S Step 2: Determine the moles of AgNO 3 that are required to consume all the K 2 S. mols of AgNO 3 required to react with all the K 2 S = ( mol K 2 S) ( ) 2 mol AgNO3 = mol AgNO 3 1 mol K 2 S Step 3: Compare moles of AgNO 3 found in step 2 with actual amount given in step 1. If step 2 <step 1, AgNO 3 is limiting reagent. If step 2 >step 1, AgNO 3 is in excess. We know that we have mol of AgNO 3 (Step 1) We know that we need mol of AgNO 3 (Step 2) to use up the entire K 2 S. Since step 2 >step1, and AgNO 3 is the limiting reagent. This means that K 2 S is the species in excess. Now that we know which is limiting (AgNO 3 ), we can go on to answer the question: How much Ag 2 S(s) can be formed? What is the theoretical yield? 2AgNO 3 + K 2 S Ag 2 S + KNO 3 mols AgNO 3 = 5.00 g 170. g.mol = mol AgNO 3 x mol Ag 2 S mol AgNO 3 = 1 mol Ag 2S 2 mol AgNO 3 x = mol Ag 2 S grams Ag 2 S = (mols)(molar mass) = ( mol Ag 2 S)(248g/mol) = 3.64 g Ag 2 S Therefore, when Kerry-Sue did her experiment in the lab, her theoretical yield would have been 3.64 g of Ag 2 S(s). Alternatively using dimensional analysis: 5.00 g AgNO 3 1 mol AgNO g AgNO 3 1 mol Ag 2S 2 mol AgNO g Ag 2S 1 mol Ag 2 S = 3.64 g Ag 2S Sample Question #1: Magnesium hydroxide reacts with hydrochloric acid according to the reaction below. Janet was working to determine the theoretical yield of magnesium chloride given that she started with 3.65 g of magnesium hydroxide and 4.24 g of hydrochloric acid. The balanced equation is given below: Mg(OH) 2 + 2HCl MgCl 2 + 2H 2 O 2

5 Concept 1. Percent Yield mols Mg(OH) 2 = 3.65 g 58.3 g/mol = mol Mg(OH) 2 mols HCl = 4.24 g = mol HCl 36.5 g/mol Step 2: Determine the moles of Mg(OH) 2 that are required to consume all the HCl. ( ) 1 mol Mg(OH)2 mols Mg(OH) 2 = (0.116 mol HCl) = mol Mg(OH) 2 2 mol HCl Step 3: Compare moles of Mg(OH) 2 found in step 2 with actual amount given in step 1. If step 2 >; step 1, Mg(OH) 2 is limiting reagent. If step 2 <step 1, Mg(OH) 2 is in excess. We know that we have mol of Mg(OH) 2 (Step 1) We know that we need mol of Mg(OH) 2 (Step 2) to use up the entire HCl. Since step 2 <step1, Mg(OH) 2 is in excess. This means that HCl is the limiting reagent. Now that we know which is limiting (HCl), we can go on to answer the question: How much MgCl 2(aq) can be formed? What is the theoretical yield? Mg(OH) 2(s) + 2 HCl (aq) MgCl 2(aq) + 2 H 2 O (L) 4.24 g HCl 1 mol HCl 36.5 g HCl 1 mol MgCl g MgCl 2 = 5.54 g MgCl 2 2 mol HCl 1 mol MgCl 2 Therefore, Janet knows that the theoretical yield is 5.54 g for magnesium chloride given the above conditions for this reaction. Sample question #2: Fire extinguishers that produce foam use aluminum hydroxide, Al(OH) 3, and CO 2 both to smother a fire. The foam that you normally see is the aluminum hydroxide. In the reaction below, calculate the amount of aluminum hydroxide produced from g of baking soda and grams of aluminum sulfate. Solution: 6 NaHCO 3 + Al 2 (SO 4 ) 3 3 Na 2 SO 4 + 2Al(OH) CO 2 mols NaHCO 3 = g 84.1 g/mol = 5.95 mol NaHCO 3 mols Al 2 (SO 4 ) 3 = g 342 g/mol = mol Al 2(SO 4 ) 3 Step 2: Determine the moles of NaHCO 3 that are required to consume all the Al 2 (SO 4 ) 3. ( ) 6 mol NaHCO3 mols NaHCO 3 = (0.731 mol Al 2 (SO 4 ) 3 ) = 4.39 mol NaHCO 3 1 mol Al 2 (SO 4 ) 3 Step 3: Compare moles of NaHCO 3 found in step 2 with actual amount given in step 1. If step 2 >step 1, NaHCO 3 is limiting reagent. If step 2 <step 1, NaHCO 3 is in excess. 3

6 We know that we have 5.95 mol of NaHCO 3 (Step 1) We know that we need 4.39 mol of NaHCO 3 (Step 2) to use up the entire Al 2 (SO 4 ) 3. Since step 2 <step1, NaHCO 3 is in excess. This means that Al 2 (SO 4 ) 3 is the limiting reagent. Now that we know the limiting reactant is Al 2 (SO 4 ) 3, we can go on to answer the question: How much Al(OH) 3 can be formed? What is the theoretical yield? 6 NaHCO 3 + Al 2 (SO 4 ) 3 3 Na 2 SO 4 + 2Al(OH) CO 2 We don t actually have to go all the way back to grams of aluminum sulfate for this calculation because we already converted the grams of aluminum sulfate to moles in an earlier part of the problem mol Al 2 (SO 4 ) 3 2 mol Al(OH) 3 1 mol Al 2 (SO 4 ) g Al(OH) 3 1 mol Al(OH) 3 = 114 g Al(OH) 3 Therefore, the theoretical yield of aluminum hydroxide foam from the fire extinguisher is g. If you look at these calculations, there is nothing new. The only new part, to date, in this lesson has been the definition of the theoretical yield, or the amount that is calculated on the product side of the equation. Stop for a moment and think of a situation where you are in the lab and conditions are not good. Have you ever had one of those days when things are just not going your way, where it just seems like conditions just aren t working for you? The same holds true for reactions. In other words, theoretical yields are the optimum yields. These are the yields that are calculated if conditions allow all 100% of the reactant to reacts. If, on the other hand, anything were to happen to jeopardize this, the actual yield will differ from the theoretical yield. The actual yield is the actual amount that is obtained from the experiment and is almost always less than the theoretical yield. Now let s go back to the sandwich maker for a moment. What if, while making his sandwiches, his sister comes in and takes a piece of bread off the table and eats it. What does this do to the actual yield, the actual number of sandwiches the sandwich maker has produced? Does he still have 5 sandwiches? No, because during the process of the reaction, or his sandwich making, the reaction conditions changed and the actual yield became less (down to 4 sandwiches) than the theoretical yield. The actual yield is massed out at the end of the experiment and thus is provided as a data value. We often calculate how close this actual yield is to the theoretical yield. The percentage of the theoretical yield that is actually produced (actual yield) is known as the percent yield. The Yield Efficiency Can be Found by Calculating Percent Yield The efficiency of a chemical reaction can be measured in terms of their various yields. The efficiency of a chemical reaction is determined by the percent yield. The percent yield is found using the following formula. %yield = actual yield theoretical yield 100 Remember the actual yield is measured and the theoretical yield is calculated. Let s look at a sample question. In the reaction shown below, a student was able to produce a actual yield of 5.12 g of calcium sulfate from 4.95 g of sulfuric acid and excess calcium hydroxide. What was her percent yield? 4

7 Concept 1. Percent Yield H 2 SO 4 +Ca(OH) 2 CaSO H 2 O We can calculate the theoretical yield by regular stoichiometric means. The method employed below is the factorlabel method g H 2 SO 4 1 mol H 2SO g H 2 SO 4 1 mol CaSO 4 1 mol H 2 SO g CaSO 4 1 mol CaSO 4 = 6.88 g CaSO 4 Once we have found the theoretic yield, we can use the actual yield given in the problem to find the percent yield. %yield = actual yield 5.12 g 100 = 100 = 74.4% theoretical yield 6.88 g Sample question #1: Potassium hydrogen phthalate, KHC 8 H 4 O 4, is a compound used quite frequently in acid-base chemistry for a procedure known as standardization. Standardization has to do with the process of determining the concentration of a standard solution. In a certain experiment, g of potassium hydroxide is mixed with g of KHP. If 5.26 g of the product, K 2 C 8 H 4 O 4, is produced, what is the percent yield? KOH + KHC 8 H 4 O 4 K 2 C 8 H 4 O 4 + H 2 O Solution: mols KOH = g 56.1 g/mol = mol KOH mols KHC 8H 4 O 4 = 12.5 g 204 g/mol = mol KHC 8H 4 O 4 Step 2: Determine the moles of KOH that are required to consume all the KHC 8 H 4 O 4. ) mols KOH required to react with all the KHC 8 H 4 O 4 = ( mol KHC 8 H 4 O 4 )( 1 mol KOH 1 mol = mol KOH KHC 8 H 4 O 4 Step 3: Compare moles of KOH found in step 2 with actual amount given in step 1. If step 2 >step 1, KOH is limiting reagent. If step 2 <step 1, KOH is in excess. We know that we have mol of KOH (Step 1) We know that we need mol of KOH (Step 2) to use up the entire KHC 8 H 4 O 4. Since step 2 <step1,koh is the limiting reagent. This means that KHC 8 H 4 O 4 is in excess. Now that we know the limiting reactant is KOH, we can go on to answer the question: How much K 2 C 8 H 4 O 4 can be formed? What is the theoretical yield? mol KOH 1 mol K 2C 8 H 4 O 4 1 mol KOH g K 2C 8 H 4 O 4 1 mol K 2 C 8 H 4 O 4 = 5.98 g K 2 C 8 H 4 O 4 Therefore, the theoretical yield of K 2 C 8 H 4 O 4 is 5.98 g. Remember the actual yield given in the question was 5.26g. Now to calculate the percent yield. 5

8 %yield = actual yield 5.26 g 100 = 100 = 88.0% theoretical yield 5.98 g Sample question #2: Zeanxanthin is a compound that has, as its claim to fame, the responsibility to cause the colors of the maple leaf to change in the fall. It has the formula C 38 H 56 O 2. In a combustion reaction of 0.95 g of zeanxanthin with excess oxygen, 2.2 g of carbon dioxide was produced. The other product was water. What is the percent yield of CO 2? Solution: C 38 H 56 O O 2 38 CO H 2 O mols C 38 H 56 O 2 = 0.95 g 544 g/mol = mol C 38H 56 O 2 x mol CO mol C 38 H 56 O 2 = 38 mol CO 2 1 mol C 38 H 56 O 2 x = mol CO 2 grams CO 2 = (0.066 mol)(44 g/mol) = 2.9 g CO 2 (theoretical yield) %yield = actual yield 2.2 g 100 = 100 = 76% theoretical yield 2.9 g Lesson Summary The percent yield of a reaction is the actual amount of product that is produced. The theoretical yield is the amount of product that is produced under ideal conditions. Review Questions 6 1. Is it possible for the actual yield to be more than the theoretical yield? 2. What happens when competing reactions occur when performing an experiment in the lab? 3. If the actual yield is 4.5 g but the theoretical yield was 5.5 g, what is the percent yield for this data? 4. Solid aluminum and sulfur come together in a reaction to produce 7.5 g of aluminum sulfide. If 5.00 g of each solid react together, what is the percent yield? Remember to balance the reaction. Al + S Al 2 S 3 a. 32.0% b. 53.4% c. 96.2% d %

9 Concept 1. Percent Yield 5. In her experiment, Gerry finds she has obtained 3.65 g of lead(ii) iodide. She knows her reaction was lead (II) nitrate that reacted completely with potassium iodide to produce lead (II) iodide and potassium nitrate. The potassium iodide produced is a brilliant yellow colored precipitate. Pb(NO 3 ) 2 (aq) + KI(aq) PbI 2 (s) + KNO 3 (aq) Gerry began with 5.00 g of potassium iodide. What was her percent yield? Remember to balance your equation first! a. 26.3% b. 36.0% c. 52.6% d. 72.0% 6. If a percentage yield in the reaction below was found to be 78.3%and the actual yield was 1.01 g, what was the original mass of the limiting reagent, oxygen. a g b g c g d g 2S + 3O 2 2SO 3 7. Bromine pentafluoride can be produced from a reaction between liquid bromine and fluorine gas. If 3.25 g of fluorine reacts with 2.74 g of bromine to produce 4.83 g of bromine pentafluoride, what is the percent yield of the product? 8. Ammonia can react with nitrogen in a reaction that is similar to a combustion reaction. The products, however, are nitrogen monoxide and water rather than carbon dioxide and water. In the reaction between ammonia and oxygen, 15 g of each reactant are placed in a container and 10.5 g of nitrogen monoxide was produced. What is the percent yield of the nitrogen monoxide? Further Reading / Supplemental Links Website with lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 16-6 is on Percent Yield. Vocabulary theoretical yield The amount obtained when all of the limiting reactant has reacted in the balanced chemical equation. actual yield The actual amount that is obtained from the experiment and is always less than the theoretical yield. percent yield % yield = actual yield theoretical yield 100 7

10 yield efficiency The percent yield of the reaction compared to the optimal yield. 8