Percent Yield - Online Tutor, Practice Problems & Exam Prep

Stoichiometry helps determine the theoretical amount of product from given reactants, while percent yield measures the actual success of a chemical reaction. Percent yield is calculated using the formula: $\mathrm{Percent}\mathrm{Yield}=\frac{\mathrm{Actual}}{Yield}/\frac{\mathrm{Theoretical}}{Yield}100$. Yields are categorized as excellent (≥90%), very good (≥80%), good (≥70%), or poor (<40%). Actual yield is always less than theoretical yield due to inefficiencies in reactions.

Percent Yield

Now, stoichiometry is a way of us determining theoretically on paper how much product we could make if we are given some amount of starting material. But when we do the experiment in real life and obtain a certain amount of product, it's the percent yield that determines how successful we were. Percent yield determines how successful the scientist was in creating their desired product. The higher the percent yield, then the higher the efficiency of a chemical reaction. So, just like you want the highest possible percentages on an exam, we want the highest possible percentages in percent yield, that will show us being successful in creating our product. We can describe percent yield values as excellent, very good, good, and poor. If you have a percent yield that is equal to or greater than 90%, you could call that an excellent yield. If you have an amount that's equal to or greater than 80 percent, then that would be very good. If you were good, that means you would be equal to or greater than 70%. And then you'd have a poor yield if your percent yield was less than 40%.

Now, with percent yield comes the percent yield formula. And the percent yield formula equals actual yield over theoretical yield times 100. Remember, we have our purple box here. That means that this is a formula you need to commit to memory because oftentimes it's not given on your formula sheet. Now, we know that the percent yield measures how successful we are in terms of carrying this reaction out in real life. We know that theoretical yield is what we do as calculations on paper when they give us multiple amounts of given, and we figure out how much product we're making. Actual yield, though, actual yield is the amount of pure product actually created when the experiment is done in the laboratory. Often, what you do on paper, you'll find that when you do the experiment in real life, things don't exactly match up. We're going to say here that the units used in the formula are based on the units of the actual yield. So, let's assume that our actual yield is in grams of our product, but our theoretical yield that we calculated is in moles. You'd have to change those theoretical yield units to match the actual yield unit. So you have to change moles to grams.

We're going to say here that no chemical reaction is 100% efficient. So you'll never get a percent yield that is 100%. There's always going to be something that happens. You may spill some of your compound, lose some randomly, or outside forces might play a part. So you'll never achieve a 100% yield. Because of this, that means that your actual yield is always less than your theoretical yield. So, just remember, we have percent yield, we have actual yield, and we have theoretical yield. Together, they give us a good insight into how efficient our chemical reaction is.

Percent Yield Example 1

Here is our example question that says, consider the following balanced chemical reaction. So we have 2 moles of C₆H₆ reacting with 15 moles of oxygen gas to produce 12 moles of carbon dioxide gas, plus 6 moles of water as a liquid. It says, if a 2.6 gram sample of C₆H₆ reacted with excess oxygen to produce 1.25 grams of water, what is the percent yield of water? Alright. They want us to determine the percent yield. We know percent yield equals actual yield over theoretical yield times 100. If we read back on the question, they are telling us that we produced 1.25 grams of water. How can we determine if we produced something, or made something, made product? The only way we would know for sure is if we did the reaction in real life and obtained that amount. That 1.25 grams represents our actual yield. So really, what we have to do is calculate our theoretical yield. And remember, our theoretical yield can be determined using stoichiometry. So if we look here, it says step 1, map out the portion of the stoichiometric chart you will use. Well, excess means we ignore. They are telling us that we have 2.6 grams of this reacted. That's the amount they gave to us. So that represents our grams of given. So the math that we're going to have to work out is grams of given, convert that into moles of given, then we're going to have to convert those moles of given to moles of unknown. What's our unknown? Well, to figure out the units that we need, remember, your theoretical yield must have the same units as your actual yield. Our actual yield is grams of water. So we have to go all the way to grams of water. So we're gonna have here moles of unknown, which is water, and then finally end here in grams of unknown. Alright. So that's the stoichiometric chart, at least the portion of it that we need to utilize to get our theoretical yield. Alright.
So we're going to take the given quantity and change it into moles of given. We have 2.6 grams of benzene. We have 6 carbons and 6 hydrogens. When you look on the periodic table and take their atomic masses, add up the 6 carbons and the 6 hydrogens, you'll have a combined mass of 78.108 grams per 1 mole of C₆H₆. Next, we're going to say here, step 3, do a mole to mole comparison to convert moles of given into moles of unknown. To do that, remember, we look at the coefficients of the balanced equation. We're going to go from moles of C₆H₆ to moles of water. And in our balanced equation, it is a 2 to 6 ratio. Then it says, if necessary, convert the moles of unknown into desired units, into the final desired units. Since our actual yield is using grams of water, our theoretical yield needs to go to grams of water. One mole of water, taking into account the 2 hydrogens and the 1 oxygen, has a combined mass of 18.016 grams. Then if we multiply everything out and divide by what's on the bottom, we get 1.799 grams of water. This here represents our theoretical yield. So plug that into our answer. And then step 5, we're already doing it. We're plugging in the actual yield and the theoretical yield, and that'll help us define our percent yield. So 1.251.799×100 gives us 69.48% as my percent yield for this particular question. So just remember, parts of this are pretty familiar. They're incorporating things we've learned about stoichiometry and limiting reagents, and now just tacking on actual yield. With that information, we're able to find our percent yield for this particular question.

What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO₄?  

Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)

• A. 63.1 %
• B. 90.7 %
• C. 79.2 %
• D. 67.1 %
• E. 50.0 %

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?  

2 NH₃ + OCl^– → N₂H­₄ + Cl^– + H₂O

In this practice question, it says, ammonia, which is represented as NH3, reacts with hypochlorite ion, OCl− to produce hydrazine, which is N2H4. How many grams of hydrogen are produced from 115 grams of ammonia if the reaction has an 81.5% yield. Alright. They're giving us the percent yield because it's 81.5%. Remember, percent yield equals actual yield over theoretical yield times 100. They're giving us 81.5, so that's 81.5 percent here. They are asking me how many grams of hydrazine are produced. So, they're saying how much is produced, and what they're trying to imply here is how much do we make in real life from this experiment. We know the percent yield, so we should actually be able to determine how much of the product we actually make. Alright. So what they're asking us to solve for is actual, so that becomes our x. But I can't solve my actual yield yet because I'm still missing my theoretical yield. How do I figure out my theoretical yield? Well, to figure out the theoretical yield, we use stoichiometry. Here is our grams of given for a reactant, and if I do stoichiometry, I can figure out how much of the product I could make theoretically. So that will represent my theoretical yield. Plug that into the formula and solve for the x variable. Alright. So we're going to have 115 grams of ammonia, and what we're going to do is we're going to change our grams of given into moles of given. So 1 mole of ammonia has 1 nitrogen and 3 hydrogens. The combined mass is 17.034 grams. Remember, we're trying to figure out the grams of hydrazine. So we know we have to go all the way to grams of hydrazine in terms of the stoichiometric chart. So we're going to go from moles of given to our moles of unknown. Remember, at this point, to go from moles of given to moles of unknown, we use the jump and we do a mole-to-mole comparison. For every 2 moles of this, there is 1 mole of this. So these moles cancel out. Now, finally, I'm going to say for every 1 mole of hydrazine, we need its combined mass of the 2 nitrogens and the 4 hydrogens. When we look at the periodic table and add them all up together, its combined mass is 32.052. Here, that's going to give me 10.82 grams of hydrazine. So take that number and plug it over here. That is our theoretical yield. So now this becomes a simple type of algebraic question where we just have to solve for the x variable. So I'm going to bring it over here. So it's 81.5 equals a 100 and x are both numerators, so it's a 100x divided by 10.82. Multiply both sides by 10.82, so these cancel out here. So when I do that, that's going to give me 881.83 equals 100x, divide both sides now by 100, and when I divide both sides by 100, I'm going to see that x equals 8.82 grams of hydrazine, roughly. Now remember, we know that our answer here will be in grams of hydrazine because they asked us to find grams of hydrazine here in the question. So just remember we're dealing with the percent yield formula. Reading the question carefully tells you which components of the formula you have and which one you have to solve. The wording was a little bit elusive, a little bit cloudy, but realize they were just asking for the actual yield. Find the theoretical, plug it in, find your percent yield, plug it in, solve for actual yield, and you'll have your answer.

The reduction of iron (III) oxide creates the following reaction: 

Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g) 

If the above reaction only went to 75% completion, how many moles of Fe₂O₃ were require to produce 0.850 moles of Fe? 

• A. 0.963 mol
• B. 2.27 mol
• C. 0.425 mol
• D. 0.567 mol
• E. 1.70 mol

Here’s what students ask on this topic:

What is the formula for calculating percent yield?

The formula for calculating percent yield is:

$\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100$

In this formula, the actual yield is the amount of product obtained from the experiment, while the theoretical yield is the amount of product predicted by stoichiometric calculations. The result is then multiplied by 100 to convert it to a percentage.

Why is percent yield important in chemistry?

Percent yield is important in chemistry because it measures the efficiency of a chemical reaction. It indicates how successful a scientist was in producing the desired product. A higher percent yield means a more efficient reaction, which is crucial for both academic research and industrial applications. It helps in evaluating the practicality of a reaction and in optimizing conditions to maximize product yield. Additionally, it provides insight into potential losses and inefficiencies in the experimental process.

What are the categories of percent yield values?

Percent yield values are categorized as follows:

• Excellent: ≥90%
• Very Good: ≥80%
• Good: ≥70%
• Poor: <40%

These categories help in assessing the efficiency of a chemical reaction. An excellent yield indicates a highly efficient reaction, while a poor yield suggests significant inefficiencies or losses during the process.

Why is it impossible to achieve a 100% percent yield in a chemical reaction?

It is impossible to achieve a 100% percent yield in a chemical reaction because no reaction is perfectly efficient. Various factors contribute to this, such as incomplete reactions, side reactions, loss of product during transfer, and measurement inaccuracies. Additionally, external factors like temperature, pressure, and human error can also affect the yield. Therefore, the actual yield is always less than the theoretical yield, making a 100% yield unattainable.

How do you convert theoretical yield units to match actual yield units?

To convert theoretical yield units to match actual yield units, you need to ensure both yields are in the same unit of measurement. For example, if the actual yield is given in grams and the theoretical yield is in moles, you must convert the theoretical yield from moles to grams. This is done using the molar mass of the product:

$\text{Theoretical Yield (g)}=\text{Theoretical Yield (mol)}\times \text{Molar Mass (g/mol)}$

This ensures consistency in units, allowing for accurate calculation of percent yield.