It said that if you get a percent yield over 100 it'sometimes because you made a mistake in the lab, which makes sense to me, but I've heard that there can be other, more scientific reasons.... how does that work?

Nice question! The theoretical yield is maximum 100% by definition. You can run an experiment correctly and still get >100% although It's not quite a "mistake" as you say: your final product will often include impurities (things other than your pure desired compound) so it will seem like you made more than you could have gotten because you also weigh the impurities. The other, simillar, reason is that your final product is not completely dry, in which the "impurity" would be water.

In Step 3, Calculation of Percent Yield the equation shows 1.82/2.15 as equaling 83.9%. I believe it should say 84.7%, If I'm incorrect in this I blame Texas Instruments the producer of my calculator.

You are correct. The answer is not 83.9, but 84.6. It's 84.6 instead of 84.7 because of minor rounding issues. Kudos to Texas Instruments.

Where do you get the actual yield from? Thanks:)

If speaking in terms of doing a lab or experiment, the actual yield comes from the actual result of the lab (hence the name).

On a standardized test, how would you distinguish differences between a solely stoichiometric problem and a limiting reagent problem?

The amount of the reactants would be given to you.

For the percent yield equation, must the equation be in grams or can it be done in moles as well?

You can use whatever units you wish, provided the actual yield and the theoretical yield are expressed in the same units.

why are there no videos for the hard stochiometry for unit 6?

```They want to challenge us.```

So there isn't a way to find the actual yield without doing an experiment?

Theoretical yield is what you think should happen:, and is a calculated amount derived through stoichiometry. Actual yield is what you observed to happen or you have been told happened: and is derived by measurement in an experiment or manufacturing process. Expected yield is the amount that has been consistently reported, when control measures are used. So there is another way to figure yield as expected yield, which would include the parameters and control methods used to increase yield in a real life manufacture of a chemical process. We can use theoretical yield or expected yield to calculate percent yield.

How would you express the actual yield if a side reaction occurs?

In general, the theoretical yield is calculated assuming no side reactions will occur (this is almost never actually the case! But still that is the assumption that is usually made). In the case that there is a side reaction, you would calculate the actual yield based on how much pure product you were able to make. This means you need to be able to either separate your product from the side products (or leftover reactants), or you have some analytical method to analyze the purity of your product.

Could anyone give more detail about why the actual yield is almost always lower than theoretical? What exactly is meant by side reactions and purification steps?

The theoretical yield assumes that all of your reactants (100% of them) react together in the desired reaction to produce your products. But real life is more messy than idealized math solutions and mistakes happen. Your actual yield is almost always going to be less than your theoretical yield because you do not obtain the entirety of your product. One reason is that you can simply have unreacted reactants which do not produce products. This can be remedied by increasing the contact reactants have with each other such as by better (or constant) stirring. Another reason if that often you are transferring solutions to and from glassware for entire reactions and products can simply be spilled by accident. Or they can be left in the glassware attached to tiny scratches on the inside of the glassware where you are unable to collect it. This is mendable by having good quality glassware and being careful as a chemist when transferring solutions. Another common frustration is the occurrence of side reactions, or undesired reactions which creates a product that you did not wish to have. This can actually create a scenario where your percent yield is actually higher than your theoretical yield because it is contaminated with impurities. These can possibly be fixed by being extra cautious to not expose your reaction to air since chemicals like water vapor in the air can react with your reaction to create impurities. That being said if you performed your reaction well and produced a pure sample your percent yield will be less than 100%. An excellent percent yield would be somewhere in the 90% range. 80% is considered good. Around 50% is considered adequate. Hope that helps.

In step 2 method 1, how did you go from having 1.74 mol of Al to 0.67 mol of Al?

The *theoretical molar ratio* is (2 mol Al)/(3 mol Cl₂). Divide numerator and denominator by 3, and the ratio becomes (0.67 mol Al)/(1 mol Cl₂).

Main content

Course: Chemistry library > Unit 3

Lesson 3: Limiting reagent stoichiometry

Limiting reactant and reaction yields

Google Classroom

Limiting reactant and theoretical yield

It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make?

Assuming that hot dogs and buns combine in a

1 : 1

ratio, we can make four complete hot dogs. Once we run out of buns, we'll have to stop making complete hot dogs. In other words, the hot dog buns limit the number of complete hot dogs we can produce.

In much the same way, a reactant in a chemical reaction can limit the amounts of products formed by the reaction. When this happens, we refer to the reactant as the limiting reactant (or limiting reagent). The amount of a product that is formed when the limiting reactant is fully consumed in a reaction is known as the theoretical yield. In the case of our hot dog example, we already determined the theoretical yield (four complete hot dogs) based on the number of hot dogs buns we were working with.

Enough about hot dogs, though! In the next example, we'll see how to identify the limiting reactant and calculate the theoretical yield for an actual chemical reaction.

Example 1: Using the limiting reactant to calculate theoretical yield

2.80 g

sample of

Al (s)

reacts with a

4.15 g

sample of

Cl_{2} (g)

according to the equation shown below.

2 Al (s) + 3 Cl_{2} (g) \to 2 AlCl_{3} (s)

What is the theoretical yield of $AlCl_{3}$ ‍ in this reaction?

To solve this problem, we first need to determine which reactant,

Al

Cl_{2}

, is limiting. We can do so by converting both reactant masses to moles and then using one or more mole ratios from the balanced equation to identify the limiting reactant. From there, we can use the amount of the limiting reactant to calculate the theoretical yield of

AlCl_{3}

Step 1: Convert reactant masses to moles

Let's start by converting the masses of

Al

and

Cl_{2}

to moles using their molar masses:

\begin{aligned} 2.80 g Al & \times \frac{1 mol Al}{26.98 g Al} = 1.04 \times 10^{- 1} mol Al \\ 4.15 {g Cl}_{2} & \times \frac{1 {mol Cl}_{2}}{70.90 {g Cl}_{2}} = 5.85 \times 10^{- 2} {mol Cl}_{2} \end{aligned}

Step 2: Find the limiting reactant

Now that we know the quantities of

Al

and

Cl_{2}

in moles, we can determine which reactant is limiting. As you'll see below, there are multiple ways to do so, each of which uses the concept of the mole ratio. All of the methods give the same answer, though, so you can choose whichever approach you prefer!

Method 1: For the first method, we'll determine the limiting reactant by comparing the mole ratio between

Al

and

Cl_{2}

in the balanced equation to the mole ratio actually present. In this case, the mole ratio of

Al

and

Cl_{2}

required by balanced equation is

\frac{moles of Al}{{moles of Cl}_{2}} (required) = \frac{2}{3} = 0.6 \overset{―}{6}

and the actual mole ratio is

\frac{moles of Al}{{moles of Cl}_{2}} (actual) = \frac{1.04 \times 10^{- 1}}{5.85 \times 10^{- 2}} = 1.78

Since the actual ratio is greater than the required ratio, we have more

Al

than is needed to completely react the

Cl_{2}

. This means that the

{Cl}_{2}

must be the limiting reactant. If the actual ratio had been smaller than the required ratio, then we would have had excess

{Cl}_{2}

, instead, and the

Al

would be limiting.

Method 2: For the second method, we'll use the mole ratio between

Al

and

Cl_{2}

to determine how much

Cl_{2}

we would need to fully consume

1.04 \times 10^{- 1}

moles of

Al

. Then, we'll compare the answer to the amount of

Cl_{2}

we actually have to see if

Cl_{2}

is limiting or not. The number of moles of

Cl_{2}

required to react with

1.04 \times 10^{- 1}

moles of

Al

1.04 \times 10^{- 1} mol Al \times \frac{3 {mol Cl}_{2}}{2 mol Al} = 1.56 \times 10^{- 1} {mol Cl}_{2}

According to our earlier calculations, we have

5.85 \times 10^{- 2}

moles of

Cl_{2}

, which is less than

1.56 \times 10^{- 1}

moles. Again, this means that the

Cl_{2}

is limiting. (Note that we could have done a similar analysis for

Al

instead of

Cl_{2}

, and we would have arrived at the same conclusion.)

Method 3: For the third and final method, we'll use mole ratios from the balanced equation to calculate the amount of

AlCl_{3}

that would be formed by complete consumption of

Al

and

Cl_{2}

. The reactant that produces the smallest amount of

AlCl_{3}

must be limiting. To start, let's calculate how much

AlCl_{3}

would be formed if the

Al

was completely consumed:

1.04 \times 10^{- 1} mol Al \times \frac{2 {mol AlCl}_{3}}{2 mol Al} = 1.04 \times 10^{- 1} {mol AlCl}_{3}

Then, let's calculate the amount of

AlCl_{3}

that would be formed if the

Cl_{2}

was completely consumed:

5.85 \times 10^{- 2} {mol Cl}_{2} \times \frac{2 {mol AlCl}_{3}}{3 {mol Cl}_{2}} = 3.90 \times 10^{- 2} {mol AlCl}_{3}

Since the

Cl_{2}

produces a smaller amount of

AlCl_{3}

than the

Al

does, the

Cl_{2}

must be the limiting reactant.

Step 3: Calculate the theoretical yield

Our final step is to determine the theoretical yield of

{AlCl}_{3}

in the reaction. Remember that the theoretical yield is the amount of product that is produced when the limiting reactant is fully consumed. In this case, the limiting reactant is

Cl_{2}

, so the maximum amount of

{AlCl}_{3}

that can be formed is

5.85 \times 10^{- 2} {mol Cl}_{2} \times \frac{2 {mol AlCl}_{3}}{3 {mol Cl}_{2}} = 3.90 \times 10^{- 2} {mol AlCl}_{3}

Note that we had already calculated this value while working through Method 3! Since a theoretical yield is typically reported with units of mass, let's use the molar mass of

AlCl_{3}

to convert from moles of

AlCl_{3}

to grams:

3.90 \times 10^{- 2} {mol AlCl}_{3} \times \frac{133.33 {g AlCl}_{3}}{1 {mol AlCl}_{3}} = 5.20 {g AlCl}_{3}

Percent yield

As we just learned, the theoretical yield is the maximum amount of product that can be formed in a chemical reaction based on the amount of limiting reactant. In practice, however, the actual yield of product—the amount of product that is actually obtained—is almost always lower than the theoretical yield. This can be due to a number of factors, including side reactions (secondary reactions that form undesired products) or purification steps that lower the amount of product isolated after the reaction.

The actual yield of a reaction is typically reported as a percent yield, or the percentage of the theoretical yield that was actually obtained. The percent yield is calculated as follows:

Percent yield = \frac{actual yield}{theoretical yield} \times 100 %

Based on this definition, we would expect a percent yield to have a value between 0% and 100%. If our percent yield is greater than 100%, that means we probably calculated something incorrectly or made an experimental error. With all this in mind, let's try calculating the percent yield for a precipitation reaction in the following example.

Example 2: Calculating percent yield

A students mixes

25.0 mL

0.314 M

BaCl_{2}

with excess

AgNO_{3}

, causing

AgCl

to precipitate. The balanced equation for the reaction is shown below.

BaCl_{2} (a q) + 2 AgNO_{3} (a q) \to 2 AgCl (s) + Ba (NO_{3})_{2} (a q)

If the student isolates $1.82 g$ ‍ of $AgCl (s)$ ‍, what is the percent yield of the reaction?

To solve this problem, we'll need to use the given information about the limiting reactant,

BaCl_{2}

, to calculate the theoretical yield of

AgCl

for the reaction. Then, we can compare this value to the actual yield of

AgCl

to determine the percent yield. Let's walk through the steps now:

Step 1: Find moles of the limiting reactant

To determine the theoretical yield of

AgCl

, we first need to know how many moles of

BaCl_{2}

were consumed in the reaction. We're given the volume (

0.0250 L

) and molarity (

0.314 M

) of the

BaCl_{2}

solution, so we can find the number of moles of

BaCl_{2}

by multiplying these two values:

\begin{aligned} {Moles BaCl}_{2} & = M_{{BaCl}_{2}} \times liters soln \\ = \frac{0.314 {mol BaCl}_{2}}{1 L soln} \times .0250 L soln \\ = 7.85 \times 10^{- 3} {mol BaCl}_{2} \end{aligned}

Step 2: Determine the theoretical yield (in grams)

Now that we have the quantity of

BaCl_{2}

in moles, we can find the theoretical yield of

AgCl

in grams:

7.85 \times 10^{- 3} {mol BaCl}_{2} \times \frac{2 mol AgCl}{1 {mol BaCl}_{2}} \times \frac{143.32 g AgCl}{1 mol AgCl} = 2.25 g AgCl

Step 3: Calculate the percent yield

Finally, we can calculate the percent yield of

AgCl

by dividing the actual yield of

AgCl

(

1.82 g

) by our theoretical yield from Step 2:

\begin{aligned} % yield & = \frac{actual yield}{theoretical yield} \times 100 % \\ = \frac{1.82 g AgCl}{2.25 g AgCl} \times 100 % \\ = 80.9 % \end{aligned}

Summary

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed. As we saw in Example 1, there are many different ways to determine the limiting reactant, but they all involve using mole ratios from the balanced chemical equation.

The amount of product that can be formed based on the limiting reactant is called the theoretical yield. In reality, the amount of product actually collected, known as the actual yield, is almost always smaller than the theoretical yield. The actual yield is usually expressed as a percent yield, which specifies what percentage of the theoretical yield was obtained.

This article was adapted from the following articles:

"Chemical Reactions" from UC Davis ChemWiki, CC BY-NC-SA 3.0
"Stoichiometry and Balancing Reactions" from UC Davis ChemWiki, CC BY-NC-SA 3.0
"Reaction Stoichiometry" from Boundless Chemistry, CC BY-SA 4.0

The modified article is licensed under a CC BY-NC-SA 4.0 license.

Additional references:

Zumdahl, S. S., and Zumdahl, S. A. (2014). The Concept of Limiting Reactant. In Chemistry (9th ed., pp. 114–123). Belmont, CA: Brooks/Cole.

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micah.ariel.snow
Posted 8 years ago. Direct link to micah.ariel.snow's post “It said that if you get a...”
It said that if you get a percent yield over 100 it'sometimes because you made a mistake in the lab, which makes sense to me, but I've heard that there can be other, more scientific reasons.... how does that work?
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(52 votes)
Answer
- Matt B
  Posted 8 years ago. Direct link to Matt B's post “Nice question! The theore...”
  Nice question! The theoretical yield is maximum 100% by definition. You can run an experiment correctly and still get >100% although It's not quite a "mistake" as you say: your final product will often include impurities (things other than your pure desired compound) so it will seem like you made more than you could have gotten because you also weigh the impurities. The other, simillar, reason is that your final product is not completely dry, in which the "impurity" would be water.
  Comment on Matt B's post “Nice question! The theore...”
  (121 votes)
Jonny Cartee
Posted 9 years ago. Direct link to Jonny Cartee's post “In Step 3, Calculation of...”
In Step 3, Calculation of Percent Yield the equation shows 1.82/2.15 as equaling 83.9%. I believe it should say 84.7%, If I'm incorrect in this I blame Texas Instruments the producer of my calculator.
Button navigates to signup pageComment on Jonny Cartee's post “In Step 3, Calculation of...”
(29 votes)
Answer
- DJ Daba
  Posted 5 years ago. Direct link to DJ Daba's post “You are correct. The answ...”
  You are correct. The answer is not 83.9, but 84.6.
  It's 84.6 instead of 84.7 because of minor rounding issues.
  
  Kudos to Texas Instruments.
  Button navigates to signup page
  (23 votes)
brookeleath
Posted 8 years ago. Direct link to brookeleath's post “Where do you get the actu...”
Where do you get the actual yield from? Thanks:)
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(13 votes)
Answer
- cmc18
  Posted 8 years ago. Direct link to cmc18's post “If speaking in terms of d...”
  If speaking in terms of doing a lab or experiment, the actual yield comes from the actual result of the lab (hence the name).
  Button navigates to signup page
  (4 votes)
SpamShield2.0
Posted 8 years ago. Direct link to SpamShield2.0's post “On a standardized test, h...”
On a standardized test, how would you distinguish differences between a solely stoichiometric problem and a limiting reagent problem?
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(7 votes)
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- mga0002
  Posted 8 years ago. Direct link to mga0002's post “The amount of the reactan...”
  The amount of the reactants would be given to you.
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  (13 votes)
Marvyn Greco
Posted a year ago. Direct link to Marvyn Greco's post “why are there no videos f...”
why are there no videos for the hard stochiometry for unit 6?
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(6 votes)
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- Soh (Sine) Cah (Cosine) Toa (Tangent)
  Posted a year ago. Direct link to Soh (Sine) Cah (Cosine) Toa (Tangent)'s post “```They want to challenge...”
  They want to challenge us.
  Comment on Soh (Sine) Cah (Cosine) Toa (Tangent)'s post “```They want to challenge...”
  (13 votes)
SpamShield2.0
Posted 8 years ago. Direct link to SpamShield2.0's post “How would you express the...”
How would you express the actual yield if a side reaction occurs?
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(4 votes)
Answer
- yuki
  Posted 8 years ago. Direct link to yuki's post “In general, the theoretic...”
  In general, the theoretical yield is calculated assuming no side reactions will occur (this is almost never actually the case! But still that is the assumption that is usually made). In the case that there is a side reaction, you would calculate the actual yield based on how much pure product you were able to make. This means you need to be able to either separate your product from the side products (or leftover reactants), or you have some analytical method to analyze the purity of your product.
  Button navigates to signup page
  (15 votes)
kathrynjade777
Posted 7 years ago. Direct link to kathrynjade777's post “For the percent yield equ...”
For the percent yield equation, must the equation be in grams or can it be done in moles as well?
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(7 votes)
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- RogerP
  Posted 7 years ago. Direct link to RogerP's post “You can use whatever unit...”
  You can use whatever units you wish, provided the actual yield and the theoretical yield are expressed in the same units.
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  (7 votes)
waissene
Posted 3 years ago. Direct link to waissene's post “Could anyone give more de...”
Could anyone give more detail about why the actual yield is almost always lower than theoretical? What exactly is meant by side reactions and purification steps?
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(4 votes)
Answer
- Richard
  Posted 3 years ago. Direct link to Richard's post “The theoretical yield ass...”
  The theoretical yield assumes that all of your reactants (100% of them) react together in the desired reaction to produce your products. But real life is more messy than idealized math solutions and mistakes happen. Your actual yield is almost always going to be less than your theoretical yield because you do not obtain the entirety of your product.
  
  One reason is that you can simply have unreacted reactants which do not produce products. This can be remedied by increasing the contact reactants have with each other such as by better (or constant) stirring.
  
  Another reason if that often you are transferring solutions to and from glassware for entire reactions and products can simply be spilled by accident. Or they can be left in the glassware attached to tiny scratches on the inside of the glassware where you are unable to collect it. This is mendable by having good quality glassware and being careful as a chemist when transferring solutions.
  
  Another common frustration is the occurrence of side reactions, or undesired reactions which creates a product that you did not wish to have. This can actually create a scenario where your percent yield is actually higher than your theoretical yield because it is contaminated with impurities. These can possibly be fixed by being extra cautious to not expose your reaction to air since chemicals like water vapor in the air can react with your reaction to create impurities.
  
  That being said if you performed your reaction well and produced a pure sample your percent yield will be less than 100%. An excellent percent yield would be somewhere in the 90% range. 80% is considered good. Around 50% is considered adequate.
  
  Hope that helps.
  Button navigates to signup page
  (9 votes)
Gray
Posted 7 years ago. Direct link to Gray's post “So there isn't a way to f...”
So there isn't a way to find the actual yield without doing an experiment?
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(6 votes)
Answer
- George Karkanis
  Posted 7 years ago. Direct link to George Karkanis's post “Theoretical yield is what...”
  Theoretical yield is what you think should happen:,
  and is a calculated amount derived through stoichiometry.
  Actual yield is what you observed to happen or you have been told happened:
  and is derived by measurement in an experiment or manufacturing process.
  Expected yield is the amount that has been consistently reported, when control measures are used.
  So there is another way to figure yield as expected yield, which would include the parameters and control methods used to increase yield in a real life manufacture of a chemical process.
  We can use theoretical yield or expected yield to calculate percent yield.
  Button navigates to signup page
  (3 votes)
Emma Salgado
Posted 7 years ago. Direct link to Emma Salgado's post “In step 2 method 1, how d...”
In step 2 method 1, how did you go from having 1.74 mol of Al to 0.67 mol of Al?
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(4 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “The *theoretical molar ra...”
  The theoretical molar ratio is (2 mol Al)/(3 mol Cl₂).
  Divide numerator and denominator by 3, and the ratio becomes (0.67 mol Al)/(1 mol Cl₂).
  Comment on Ernest Zinck's post “The *theoretical molar ra...”
  (7 votes)