Lesson Video: Percentage Yield | Nagwa Lesson Video: Percentage Yield | Nagwa

# Lesson Video: Percentage Yield Chemistry • First Year of Secondary School

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

17:43

### Video Transcript

In this video, we will learn how to calculate the yield of a reaction in terms of a percentage. To be able to do this, we will first learn how to identify which reactant is the limiting reactant and how to calculate the theoretical yield. Lastly, we will investigate what factors in a reaction have an effect on the percentage yield.

Let’s look at percentage yield in a very broad context before we discuss it in depth. Most of the things we use in everyday life are the results of a manufacturing process involving chemical reactions. The paints, plastics, metals around us, the fertilizers farmers use, and the food additives and pharmaceutical drugs that we consume were all initially produced or synthesized in large factories using chemical reactions.

The chemical industry is big business. And much research takes place to synthesize these chemicals quickly and in a cost-effective manner. Measuring the output or yield of these reactions as a percentage of the ideal yield is an important factor. In the real world, however, the amount of substance produced in a chemical reaction often differs from the maximum amount that can be formed under certain conditions or the ideal amount that can be formed under a certain set of conditions.

Let’s look at the equation that scientists use to describe the actual output from a chemical reaction versus the maximum possible or ideal output. And then we will put this in terms of a percentage. The output of a reaction can be described by taking the actual output or yield divided by the theoretical output or yield multiplied by 100 percent if we want to express the answer in terms of a percent. Note that sometimes the percentage sign is lift off in this equation.

The actual yield, also called the experimental yield, is the amount of product obtained in reality after a reaction. It is usually measured in moles or grams. And if the product is a solid, it can be weighed out in the lab in terms of grams to get the actual yield. The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants. It is also usually expressed in moles or grams and is determined by calculation.

The percentage yield is always a positive value. It is usually less than 100 percent, but occasionally it is greater than 100 percent. If it is greater than 100 percent, we know that something unexpected occurred in the reaction and that there is an impurity or solvent remaining with the product. The presence of impurity or solvent make the mass of product higher than the maximum possible value, which is 100 percent.

So to calculate percentage yield, we need two values: actual yield and theoretical yield. We said that actual yield is a measured amount and theoretical yield a calculated amount. So to calculate percentage yield, we would first need to calculate the theoretical yield. But to determine theoretical yield, we first need to determine which reactant is limiting. But what exactly is a limiting reactant, otherwise known as a limiting reagent?

The reaction below is for the reaction of iron with sulfur to produce iron(II) sulfide. The stoichiometric coefficients show us that one atom of iron reacts with one atom of sulfur to produce one formula unit of iron(II) sulfide or that one mole of iron reacts with one mole of sulfur to produce one mole of product. However, the diagram shows that sometimes the reactants are not present in the correct ratio according to the balanced equation.

The diagram shows us that when iron and sulfur react, one of the iron atoms does not have a sulfur atom to react with. We can see that iron is not completely used up. We say that iron is in excess and is the excess reactant in this example. On the other hand, sulfur in this particular example is completely used up. All of the sulfur has reacted. We call sulfur the limiting reactant or limiting reagent. At the end of a reaction, the limiting reactant is always completely used up. And some of the excess reactant will remain left over or unreacted.

It is important to identify which reactant is limiting because it is the limiting reaction which determines how much product can form. The limiting reactant in this example, sulfur, directly determined how much product, iron(II) sulfide, could form. In other words, the limiting reactant determines the theoretical yield.

Let’s have a look at how to identify which reactant is limiting. Imagine that two moles of propane, C3H8, reacts with seven moles of oxygen gas to produce carbon dioxide gas and water. The balanced equation is shown. There are two methods to determine which reactant is limiting.

In method one, there are two steps. Step one is to calculate the moles of reactant one needed if all of the given moles of reactant two are used up and step two to compare the calculated moles of reactant one needed with its given number of moles. From the balanced equation, we know that one mole of propane reacts with five moles of oxygen. We can calculate the moles of propane needed if all the moles of oxygen are used up by writing 𝑥 moles of propane reacts with seven moles of O2.

We can now solve for 𝑥 by saying five 𝑥 equals one multiplied by seven. We can make 𝑥 the subject of the formula. 𝑥 equals one times seven divided by five. And we get an answer of 𝑥 equals 1.4 moles of propane needed to use up all the seven moles of oxygen.

In step two, we need to compare the moles of reactant one, propane, needed with its given number of moles. And we can see that the number of moles of propane needed, 1.4 moles, is less than the number of moles of propane given, which is two moles. So we say that propane is in excess because we have present more propane than we need. We say that propane is the excess reactant. And by default, oxygen is the limiting reactant. Let’s write E next to propane in the equation and L next to oxygen for excess and limiting.

There is an alternative way to do method one. And that is for step one to rather calculate the moles of reactant two needed if all the given moles of reactant one are used up and for step two to compare the calculated moles of reactant two needed with its given number of moles. So let’s erase the numbers underlined in pink and swap them. Where there was a one, there is now a two. And where there was a two, there is now a one.

The difference is that here we would take the given moles of propane to solve for the moles of oxygen needed to use up the given moles of propane. We would do the exact same calculation as before and would get an answer of 10 moles of oxygen needed to use up the two moles of propane. If we then compare the moles of oxygen needed, 10 moles, with the moles of oxygen given, seven moles, we see that we need more moles than we are given. So all seven moles of oxygen will be used up in the reaction, and oxygen is thus the limiting reactant. And by default, propane is in excess or the excess reactant. And this confirms what we saw in the first version of method one.

In method two, we need to calculate the amount of product formed from the amounts of each reactant separately and then compare the two amounts we have calculated for the product to find the smaller amount. Let’s make carbon dioxide the product in question. We will relate the moles of propane to the moles of carbon dioxide and the moles of oxygen to the moles of carbon dioxide to complete step one.

Here we do not have sentence form, but a ratio form to save space. One mole of propane reacts to produce three moles of carbon dioxide. So the two given moles of propane will produce 𝑥 moles of carbon dioxide. Solving as we did previously, we get 𝑥 equals six moles of carbon dioxide that can be produced from the two moles of propane. And we saw that five moles of oxygen reacts to produce three moles of carbon dioxide. So taking the seven given moles of oxygen, we can solve for 𝑥, the relative number of moles of carbon dioxide produced from the seven moles of oxygen. Solving again as previously, we get an answer of 𝑥 equals 4.2 moles of carbon dioxide that can be produced from the seven moles of oxygen.

The next step is to compare these two values to see which is the smaller amount. And we see that the amount of moles of CO2 that can be produced from the available oxygen is less than the moles of CO2 that can be produced from the available propane. This again confirms that oxygen is the limiting reactant and that propane is present in excess amounts.

The 4.2 moles of CO2 is the theoretical yield or the maximum possible yield of CO2 that can form. We know this because this amount of moles is produced from the moles of limiting reagent oxygen. And we said it is the limiting reagent or reactant which determines the moles of product which can form.

So far, we have learned what a limiting reactant is, how to identify a limiting reactant by calculation, and how it can determine the maximum amount of product which can form, in other words the theoretical yield. Let’s now do an example of calculating the percentage yield.

The equation shows that 31.5 grams of zinc reacted with excess copper sulfate solution to form 28.1 grams of copper and zinc sulfate in solution. We can take the mass of copper given as the actual yield obtained in reality. And because copper sulfate is in excess, we automatically know that zinc is the limiting reactant.

We need to calculate the molar mass of zinc then its number of moles. And from this, we can calculate the number of moles of product, which is the theoretical yield in terms of moles. And we can use the molar mass of copper to work out its mass. And this is its theoretical yield in grams. The two starred values are the values we will need to calculate the percentage yield. So if we take the molar mass of zinc from the periodic table, 65.38 grams per mole, we can determine its number of moles, putting in our mass and our molar mass values, which gives us 0.482 moles of zinc.

We know that zinc produces copper in a ratio one is to one moles from the balanced equation. So 0.482 moles of zinc would produce 0.482 moles of copper product. So we can now calculate the mass of copper product formed from 0.482 moles of copper. Plugging in our mole value and molar mass value for copper from the periodic table, we get 30.63 grams of copper that can be produced from the 31.5 grams of zinc. This is the theoretical yield of copper in terms of grams.

Knowing our actual yield and theoretical yield of copper, we can now calculate its percentage yield by taking actual yield divided by theoretical yield multiplied by 100 percent. We can put in our values. And we get an answer of 91.7 percent copper. What does this mean? This means that 91.7 percent of copper which could have formed was actually formed. The remaining 8.3 percent was either not produced or it wasn’t collected. In reality, an answer of 91.7 percent makes sense because the numerator is just slightly smaller than the denominator. This is often the case. Often the percentage yield is less than 100 percent, although occasionally it can be higher than 100 percent.

Let’s have a look at factors which affect the percentage yield. Some of these factors are listed here. The first four factors contribute to percentage yields being less than 100 percent. And the last factor causes percentage yields to be greater than 100 percent or at least greater than the realistic actual yield.

In an incomplete reaction, not all of the reactants are used up and converted into product. In reversible reactions, some of the product reforms the initial reactants. When side reactions occur, less of the desired product can form. During product recovery at the end of a reaction, sometimes not all of the formed product is collected. This may be due to a variety of reasons. For example, in recrystallization processes, sometimes not all of the product dissolved in a solvent recrystallizes to form proper crystals, which can then be filtered off. Another example is when transferring product from one piece of glassware to another. Sometimes some particles of the product remain on the first piece of glassware and are impossible to collect.

For a contaminated or wet product, sometimes the actual yield is higher than the theoretical yield, resulting in a percentage yield larger than 100 percent. This is because an impurity or solvent, represented by blue on the bar graph, present with a product increases or inflates its actual yield higher than is realistic.

A percentage yield less than 100 percent, though not ideal, is fine. For example, the Haber process in industry is a reversible reaction and has very low percentage yield, sometimes in the region of 20 percent when the reaction is carried out at high temperature. However, a percentage yield greater than 100 percent always indicates that something has gone wrong. So next time you get a yield higher than 100 percent, don’t celebrate too soon.

Let’s recap the main points we have learned in this lesson. Firstly, actual yield is the amount of product formed and collected in reality. The theoretical yield is the calculated maximum amount of product that can form from the given amounts of reactants. We learned that percentage yield is the actual yield divided by the theoretical yield multiplied by 100 percent and that the limiting reactant directly determines the theoretical yield of product which can form.

And finally, we learned about factors which affect the percentage yield. And these include reaction completion, side reactions, reversibility of reactions, and the recovery and purity of the product.

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