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Lesson Explainer: Percentage Yield Chemistry • 10th Grade

In this explainer, we will learn how to identify the limiting reagent and calculate the percentage yield of desired products based on the actual and theoretical yield.

A chemical equation can provide lots of useful information about a reaction. It can tell us about the reactants, the products formed, and the molar ratio among everything involved in the reaction. Consider the following chemical reaction between sodium hydroxide and hydrochloric acid: NaOH+HClNaCl+HO2

By looking at this equation, we can instantly tell that 1 mole of NaOH reacts with 1 mole of HCl to form 1 mole of NaCl and of HO2. By using the molar mass for each substance in the reaction, we can see that 40 g of NaOH and 36.5 g of HCl should produce 58.5 g of NaCl. The mass of the desired product, which in this case is NaCl, is known as the yield of the reaction. Calculating the mass from the balanced chemical equation is known as the theoretical yield of the reaction.

Definition: Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants.

If we were to actually perform this reaction in a lab using the same masses of NaOH and HCl, we might find that only 46.8 g of NaCl is produced. This is clearly not the 58.5 g of product we were expecting from our theoretical yield calculation. The actual mass of a product made during a chemical reaction is called the actual yield of the reaction.

Definition: Actual Yield

The actual yield is the amount of product that is obtained from carrying out a chemical reaction.

We can express the actual mass of product obtained compared with the theoretical mass by calculating the percentage yield:percentageyieldactualyieldtheoreticalyield=×100%.

Example 1: Defining What a Theoretical Yield Is

Which of the following statements best describes the theoretical yield?

  1. The theoretical yield is the same as the actual yield multiplied by 100.
  2. The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants.
  3. The theoretical yield is the amount of product that is obtained experimentally from the reaction.
  4. The theoretical yield is the result of adding the percentage yield to the actual yield.

Answer

The yield of a chemical reaction is the amount of product made. We can calculate the theoretical yield of a reaction by assuming that all the reactants are changed into products. This suggests that statement B best describes theoretical yield. However, to be certain let’s look at the other possible answers. Statement A suggests that the theoretical yield can be calculated by multiplying the actual yield by 100. The actual yield is the amount of product obtained when performing the reaction. This is not correct, as the theoretical yield is not based on the actual yield, but instead the maximum amount of product that can be obtained. As we have just seen, what is being described in statement C is the actual yield of the reaction, and not the theoretical yield. Statement D refers to the equation used to calculate the percentage yield. In this equation, the percentage yield is calculated by dividing the actual yield by the theoretical yield and not by adding the percentage yield to the actual yield as suggested in the statement. We can therefore say with certainty that statement B best describes theoretical yield.

A percentage yield should always be between 0 and 100%. The smaller the percentage yield, the less product that was obtained. Going back to our reaction of sodium hydroxide with hydrochloric acid, we determined a theoretical yield of 58.5 g for sodium chloride. The actual yield obtained from the reaction was 46.8 g. Using this information, we can calculate the percentage yield for this reaction:percentageyieldactualyieldtheoreticalyieldpercentageyieldgramsgramspercentageyield=×100%=46.858.5×100%=80%.

Therefore, in our experiment, we obtained an 80% yield of sodium chloride.

Equation: Percentage Yield

Percentageyieldactualyieldtheoreticalyield=×100%

Example 2: Calculating the Percentage Yield for a General Reaction

A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was its percent yield? Give your answer to 1 decimal place.

Answer

This question is asking us to calculate the percent yield for an unknown chemical reaction.

To calculate the percentage yield of a reaction, we require two pieces of information: the theoretical yield (mass) of the product and the actual yield (mass) of the product. Both of these pieces of information are given in the question. We are told that the student obtains an actual yield of 25 grams but that the theoretical yield should be 81 grams.

To calculate the percent yield, we can use the following equation: percentageyieldactualyieldtheoreticalyield=×100%.

We can now substitute in the values for the actual and theoretical yields: percentageyieldgramsgramspercentageyield=2581×100%=30.8641%.

Remember to round the final value to 1 decimal place, which gives us a percentage yield of 30.9%.

There are many reasons why a percentage yield is less than 100%:

  • The reaction may be incomplete and not all the reactants have reacted.
  • Reactions that are reversible may result in both products and reactants being present in the final sample.
  • The experimental procedure itself can also lead to a loss of product. Separation processes such as filtration and transferring substances between beakers can also result in loss of product.
  • It is possible that unwanted side reactions may occur during the experiment leading to undesired products being made instead.
  • Impurities in the reactants themselves can lead to lower yields. These impurities mean that the starting mass of the reactant is not accurate but can also be the cause of side reactions and impurities in the final product.

In practice, all these reasons mean that obtaining a percentage yield of 100% is very difficult. Normally, a percentage yield of 90% or greater is considered a very good yield.

Example 3: Identifying Which Factor Does Not Affect Percentage Yield

Which of these is not a factor that may affect percentage yield?

  1. The amount of gas produced
  2. The reaction being reversible
  3. The loss of product during separation or handling
  4. Unwanted products formed from side reactions
  5. The purity of the reactants

Answer

This question asks us to identify which of the listed factors does not affect the percentage yield. In practice, a percentage yield is always less than 100%. Let’s imagine that we are performing an experiment. We have measured out a certain mass of both of our reactants. We assume that the reactants are pure and do not contain any unwanted substances. However, any impurities can make the actual mass of our reactants lower than our measured mass. This might already result in a lower percentage yield and suggests that statement E is indeed a factor that affects the yield.

We now perform our experiment. During this process, there might be other reactions taking place that we did not expect or want. These side reactions could lead to unwanted products being formed, which again will give us a lower yield, making statement D a factor. What about if the reaction is reversible? This means that any product formed could be converted back into a reactant and so our final product may also contain some of the starting reactant. This will definitely decrease the actual yield we obtain and therefore lower the percentage yield. Therefore, statement B is a factor as well.

The final process in our experiment is to extract and purify our product. This is likely to involve transferring the product between glassware and separating it using techniques such as filtration. During these processes, some products may be left behind or lost. This will decrease our actual yield and so affect the percentage yield, making statement C a factor.

This leaves us with statement A and suggests that this factor does not affect the percentage yield, but let’s check. Statement A suggests that the amount of gas produced affects the percentage yield. This is not true; whether the reaction produces lots or little amounts of gas will not affect the amount of product being produced. It will also have no effect on the reactants or the products when being handled or separated. So this makes statement A the correct answer.

Knowing the percentage yield of a reaction is very important in industry. The reactants and resources needed for large-scale chemical reactions are very expensive and so a reaction that produces a very low yield is not ideal for a chemical company. Yields for chemical reactions are often reported in scientific journals and there is much research being done to change existing reactions or find alternative ones that give much higher yields.

A percent yield claiming to be over 100% might result for two main reasons. One is an error with the calculation, but the other is the nature of the product itself. Any impurities in the product will increase its mass and give the impression of a much higher yield. These impurities could be water from not drying the product fully, or contamination during the reaction.

Example 4: Reasons Why a Percentage Yield Is over 100%

Which of the following explains why a reaction yield may appear to be above 100%?

  1. There are two or more reactions that occur simultaneously so that some reactants are converted to products.
  2. All reactions are complete; therefore, all reactants are converted to products.
  3. The reagents are of high purity.
  4. The product of the reaction contains impurities.

Answer

To answer this question, we need to decide which of the given statements provides a valid reason for a reaction yield being over 100%. A yield over 100% means that our actual yield is greater than the theoretical yield. The theoretical yield is the amount of product made if all the reactants are converted into product. So if statement B were true, then our actual yield would equal the theoretical yield and the reaction yield would equal 100% but would not be above; so this rules out statement B. Statement A refers to other reactions that may occur during an experiment. While this would affect the reaction yield, the side reactions would make unwanted products and so would actually decrease our yield to less than 100%, not above it. Statement C refers to the purity of the reagents. A high purity of reagents means there is a low amount of impurities. Impurities in the reagents would mean that the actual amount of reagent we have measured is not correct. This would result in our actual yield always being lower than the theoretical yield which would lower the reaction yield.

If our actual yield is greater than the theoretical yield, then the mass of our final product is greater than it should be. This suggests the product contains other undesired products. This could be water from not drying the sample properly, or any reactants that have not been removed. These impurities could therefore be a good reason for a reaction yield being above 100% and so statement D is the correct answer.

So far, we have looked at reactions with a molar ratio of 11, but what about when the ratio is different or when the masses of reactants do not equal their molar mass? To calculate the theoretical yield, we will need to calculate the number of moles of each substance.

Sulfuric acid reacts with potassium hydroxide to produce potassium sulfate and water according to the following equation: HSO+2KOHKSO+2HO24242

If an experiment using 2.46 g of KOH produced 2.98 g of KSO24, then what would the percentage yield for this reaction be? Firstly, we need to check if the given chemical equation is balanced, which it is. This chemical equation tells us that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide to form 1 mole of potassium sulfate and 2 moles of water. The actual yield of the reaction has been given, but we still need to calculate the theoretical yield of the desired product, potassium sulfate.

We can use the given mass of KOH to calculate the number of moles of KOH at the start of the reaction. Recall that the number of moles can be calculated using the formula 𝑛=𝑚()𝑀(/),ggmol where 𝑛 is the number of moles, 𝑚 is the mass, and 𝑀 is the molar mass. We can therefore calculate the number of moles of KOH as 𝑛=2.46()56(/)=0.0439286.ggmolmoles

We can see from the balanced equation that the molar ratio between KOH and KSO24 is 21. In other words, the number of moles of KSO24 is half that of KOH. So we need to divide the number of moles of KOH by 2 to determine the number of moles of KSO24: numberofmolesofKSOmoles24=0.04392862=0.0219643.

Now that we have the number of moles of KSO24, we can multiply this by the molar mass of KSO24 (174 g/mol) to give a theoretical yield of 3.82 g. Having determined the theoretical yield of KSO24 in this reaction, we can calculate the percentage yield: percentageyieldactualyieldtheoreticalyieldpercentageyieldgramsgramspercentageyield=×100%=2.463.82×100%=78%.

Therefore, the reaction produced KSO24 with a percentage yield of 78%, a fairly decent yield.

Example 5: Calculating the Percentage Yield for an Industrial Reaction

In an industrial reaction, iron oxide reacts with carbon to make iron and carbon monoxide: FeO+4C3Fe+4CO34 [Fe=56g/mol, O=16g/mol]

Calculate, to the nearest whole number, the percentage yield if 2 kg of FeO34 produces 800 g of iron.

Answer

In this question, we have been given the mass of the reactants and the actual yield of iron (Fe) and have been asked to calculate the percentage yield. In order to do this, we need to calculate the theoretical yield of iron. First, we need to check that the given chemical equation is balanced, and it is. The next step is to calculate the number of moles of reactant we have. In the question, we have only been given the mass of FeO34, so let’s calculate its number of moles. Taking the molar mass as 232 g/mol, and remembering to convert from kilograms to grams, then the number of moles of FeO34 is 𝑛=2000()232(/)=8.62.ggmolmoles

From the balanced chemical equation, we can see that 1 mole of FeO34 produces 3 moles of iron. We can now calculate the number of moles of iron by multiplying the value above by 3: numberofmolesofFemolesmoles=8.62×3=25.86.

Multiplying the number of moles of iron by its molar mass (56 g/mol) gives a theoretical yield of 1‎ ‎448.16 g. Now that we have calculated the theoretical yield of iron, we can use the actual yield given in the question to determine the percentage yield: percentageyieldactualyieldtheoreticalyieldpercentageyieldgramsgramspercentageyield=×100%=8001448.16×100%=55%.

The percentage yield of this reaction is 55%.

We have seen that in order to calculate the theoretical yield of a reaction, we must know the molar ratio between the reactants and the products. However, what about if the number of moles of two reactants is different? Which value do we use to calculate the theoretical yield? For problems such as this, it is necessary to first determine the limiting reagent of the reaction. The limiting reagent is the reactant that gets used up first during a reaction. Once all the limiting reagent has reacted, the reaction cannot continue. In other words, the reaction is limited by the amount of this reactant, which is why we call it the limiting reagent. The reactant that is not limiting is then said to be in excess.

When calculating the theoretical yield of a reaction, it is the number of moles of the limiting reagent that need to be used. This makes sense as once all of the limiting reagent has been used up, there is none of it left to continue the reaction.

In an example reaction, 1 mole of A reacts with 2 moles of B. However, if in our experiment we have 0.5 moles of A and 0.9 moles of B, then we do not have the correct 12 molar ratio—we should have 1 mole of B. This means that reactant B is the limiting reagent and will get used up first. Once all of B has been used, the reaction will stop, but there will still be some of reactant A left. Reactant A was therefore in excess.

Definition: Limiting Reagent

The limiting reagent is the reactant that is first to be completely used up during a chemical reaction.

Consider the reaction between phosphorus trioxide and hydrochloric acid: PO+HClHPO+PCl46333

In an experiment, a student reacts 1.75 g of PO46 with 1.16 g of HCl to produce 0.61 g of HPO33. How would we determine the percentage yield for this reaction? Let’s start as before and make sure the chemical equation is balanced. This time, the given equation is not balanced, so we need to determine the coefficients to make it balanced: PO+6HCl2HPO+2PCl46333

Having correctly balanced the chemical equation, we can see that 1 mole of PO46 reacts with 6 moles of HCl to produce 2 moles of our desired product, HPO33. The next step is to calculate the number of moles of reactant we have, but which reactant should we use? We will start by calculating the number of moles of both. Taking the molecular mass of PO46 and HCl to be 220 g/mol and 36.5 g/mol, respectively, results in 0.0080 moles of PO46 and 0.0318 moles of HCl.

Remember, from our balanced equation, that 6 moles of HCl reacts with 1 mole of PO46. However, from our calculation it is clear that we do not have a 16 ratio; it is actually closer to a 14 ratio. This means that we have an excess of PO46 and so the limiting reactant is HCl. In order to calculate our theoretical yield, we will therefore need to use the number of moles of HCl, the limiting reagent.

We can now calculate the number of moles of HPO33. The balanced equation tells us that 6 moles of HCl produces 2 moles of HPO33 and so we need to divide the number of moles of HCl by 3: numberofmolesofHPOmoles33=0.03183=0.0106.

Finally, we can multiply the number of moles of HPO33 by its molecular mass (82 g/mol) to obtain a theoretical yield of 0.87 g. Having determined the theoretical yield, we can now calculate the percent yield for the reaction: percentageyieldactualyieldtheoreticalyieldpercentageyieldgramsgramspercentageyield=×100%=0.610.87×100%=70%.

In our experiment, we produced HPO33 with a yield of 70%.

Example 6: Calculating the Percentage Yield for the Reaction of Sodium Aluminum Hydride with Lithium Chloride

20 g of NaAlH4 reacts with 10 g of LiCl to produce 8.3 g of LiAlH4: NaAlH+LiClLiAlH+NaCl44 [Na=23g/mol, Cl=35.5g/mol, Al=27g/mol, H=1g/mol, Li=7g/mol]

What is the percentage yield for this reaction to the nearest whole number?

Answer

In this question, we have been given the masses of the two reactants and the mass of the desired product and are asked to calculate the percentage yield for the reaction. As always, the first step is to look at the given chemical equation and check it is correctly balanced. Having confirmed the chemical equation is indeed balanced, we can see there is a 11 ratio between all the reactants and products.

Having been given the masses for both reactants, we need to determine which one is the limiting reagent. To do this, we need to calculate the number of moles of each: ForNaAlHggmolmolesForLiClggmolmoles4𝑛=20()54(/)=0.370𝑛=10()42.5(/)=0.235

Clearly, we do not have a 11 ratio of reactants here. Since the small number of moles is lower for LiCl, then this is our limiting reagent, while the NaAlH4 is in excess.

Having determined the limiting reagent, we can now calculate the number of moles of our desired product, LiAlH4. As the molar ratio between LiCl and LiAlH4 is 11, the number of moles of LiAlH4 equals the number of moles of LiCl, 0.235 moles. Multiplying the number of moles of LiAlH4 by its molar mass (38 g/mol) gives us the theoretical yield of 8.93 g.

We can now calculate the percentage yield for the reaction: percentageyieldactualyieldtheoreticalyieldpercentageyieldgramsgramspercentageyield=×100%=8.38.93×100%=92.945%.

Rounding this value to the nearest whole number gives a percentage yield of 93%.

We can summarize the procedure needed to calculate the percentage yield of a reaction to the following steps:

  1. Write a balanced chemical equation for the reaction.
  2. Determine the number of moles for each reactant.
  3. Use the number of moles and the balanced equation to determine the limiting reagent in the reaction.
  4. Determine the number of moles of the desired product.
  5. Calculate the theoretical yield (mass) of the desired product.
  6. Use the theoretical and actual yield to calculate the percentage yield.

We will now conclude this explainer with a summary of the key points about percentage yield and limiting reagents.

Key Points

  • The yield of a reaction is the amount of desired product made.
  • The theoretical yield of a reaction is the maximum amount of desired product that can be made, assuming all reactants are converted to products.
  • The actual yield of a reaction is the amount of desired product obtained from an experiment.
  • The percentage or reaction yield can be calculated using percentageyieldactualyieldtheoreticalyield=×100%.
  • The limiting reagent of a reaction is the reactant that is first to be completely used up during a reaction.
  • The theoretical yield of a reaction should be calculated from the number of moles of the limiting reagent.

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