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Whenever we do experiments, the actual result is a little different from the result we predicted. In chemistry, this discrepancy is compared by calculating the percent yield. In this lesson, we will define percent yield and go over a few examples.

What Is Percent Yield?

If you’ve ever cooked a meal from a recipe, you will have noticed they often come with serving amounts, or a number that states how many people the recipe can feed. Sometimes, however, this number can be off, and you’ll wind up with either less or more food than you anticipated.

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A number of things can account for this difference – food spillage prior to cooking, leaving ingredients too long on the stovetop, using the wrong measuring cup, and so forth.The same thing can happen whenever we perform experiments in the chemistry lab to make a compound. We first make calculations for how much of a compound we’ll end up with. However, these calculations are made under ideal conditions.

They do not account for experimental errors or personal errors of the experimenter. In the end, your prediction is often different from what you have actually made.In chemistry, we have theoretical yield, which is the amount of the product calculated from the limiting reactant. The limiting reactant is the reactant in the chemical reaction which limits the amount of product that can be formed. The actual yield is the actual amount produced when the experiment or reaction is carried out.The discrepancy between the theoretical yield and the actual yield can be calculated using the percent yield, which uses this formula:

To use this formula for percent yield, you need to make sure that your actual yield and theoretical yield are in the same units. If the actual yield is in grams, then theoretical yield also needs to be in grams. If theoretical yield is in moles, then the actual yield also needs to be in moles. You will need to perform the conversions if you are given one measurement unit and you need the other.

Example 1

Let’s look at an example. Magnesium carbonate (MgCO3) decomposed to form 15 grams of MgO in the actual experiment. If the theoretical yield is 19 grams, what is the percent yield of MgO?

Problem 4 Chemical Reaction


In this problem, you need to calculate the percent yield of magnesium oxide. To do this, you need to know the actual and theoretical yields of magnesium oxide. Both these values are already given to you in the question, so the only thing you need to do is to plug these values in the percent yield formula:

Problem 4 Final Answer


Example 2

Sometimes, you will also need to calculate the theoretical yield from the given chemical reaction.

To do this, you will need to convert the amount of your limiting reactant into the amount of your final product. Let’s take a look at how this is done with this problem.If the reaction of 30 grams of calcium carbonate (CaCO3) produces 15 grams of calcium oxide (CaO), what is the percent yield for the following reaction?

Problem 1 Chemical Reaction


The problem has already given you the actual yield of 15 grams. So, in order to find the percent yield, you need to calculate the theoretical yield.

You look at your chemical reaction, and you see that your one and only limiting reactant is calcium carbonate. You can take your given amount of 30 grams of calcium carbonate and convert it into the amount of calcium oxide that would be formed under ideal conditions. That’s your theoretical yield. Take a moment and refresh your memory and your conversion skills if you need to before continuing.You end up with this calculation.

You get a theoretical yield of 16.8 grams of calcium oxide.Now that you have your theoretical yield, you can now simply plug your values into the formula for percent yield to find your answer.

Example 3

Let’s take a look at one final example.If 45 g of glucose (C6H12O6) reacts with an excess of oxygen and 55 g of carbon dioxide (CO2) is produced, what is the percent yield of CO2?

Problem 3 Chemical Reaction


Before we can find the percent yield of CO2, we need to know both the actual and the theoretical yields of CO2.

We know the actual yield, but we first need to calculate the theoretical yield.This problem has two reactants, so which one is the limiting reactant? In this case, the phrase ‘an excess of oxygen’ means we don’t need to do any calculations to know that C6H12O6 is the limiting reactant. So we convert the 45 grams of glucose to grams of CO2.

Problem 3 Theoretical Yield


Now that we have both the theoretical and actual yields of CO2, we can calculate the percent yield.

The question tells us that the actual yield is 55 g of CO2. We now plug the actual yield and theoretical yield into the formula:

Problem 3 Final Answer


Lesson Summary

When chemical reactions happen in real life, they do not occur in ideal conditions, so the actual yield of products produced by chemical reactions is different from the calculated or theoretical yield. This difference between the actual and theoretical yield is quantified by the percent yield.In chemistry, we have theoretical yield, which is the amount of the product calculated from the limiting reactant. The limiting reactant is the reactant in the chemical reaction which limits the amount of product that can be formed.

The actual yield is the actual amount produced when the experiment or reaction is carried out.In calculating the percent yield, we need to calculate the theoretical yield based on the limiting reactant. If there is more than one reactant, this is the reactant that produces a smaller amount of product. After we have determined the limiting reactant, we can ensure that our calculation of theoretical yield is correct.The calculation of the percent yield in chemical reactions is very important because it can help us determine and be conscious of, and possibly correct, the conditions that made the actual yield and theoretical yield differ.

Learning Outcomes

Once you are finished, you should be able to:

  • State what percent yield represents
  • Write the formula for percent yield
  • Recall why knowing the percent yield for a chemical reaction is important

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