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5.05 Theoretical & Percent Yields

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Chimie

Université du Kentucky

4.7 (640 notes) | 65K étudiants inscrits

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This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

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À partir de la leçon

Week 5

This unit will delve into the quantitative relationships we can determine from a balanced chemical equation to determine the relative amounts of substances needed to react or the amount of products formed. We will also explore limiting reagents and percent yield to address practical aspects of chemical reactions because there may be leftover reactants and/or incomplete formation of products.

5.01 Stoichiometry of Chemical Reactions: Mole to Mole 8:19

5.01a How many Moles of water are produced1:11

5.01b How many Moles of CO2 are produced 1:47

5.02 Stoichiometry of Chemical Reactions: Mass Relationships11:02

5.02a How many grams of magnesium are required 2:27

5.02b How many grams of magnesium do we need 4:34

5.03 Limiting and Excess Reagents11:18

5.04 Limiting and Excess Reagents: Calculations 7:57

5.04a How many grams of H2S can be formed 4:13

5.04b What mass of excess reactant will remain 6:31

5.05 Theoretical & Percent Yields 9:56

5.05a Methyl Salicylate is prepared5:48

5.05b A student needs 625 g of zinc sulfide5:32

5.06 Solution Concentration 15:16

5.06a Find the mass percentage for a solution prepared 1:56

5.06b Find the volume percentage for a solution prepared 1:51

5.06c Find the mass/volume percentage for a solution prepared 2:39

5.06d What is the molarity of a solution made 2:23

5.06e How many grams of LiCl do we need 2:44

5.07 Dilution of Solutions7:16

5.07a What is the concentration of the solution 1:45

5.08 Solution Stoichiometry 9:06

5.08a What volume of HBr is required 5:03

Enseigné par

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

In this module, we'll talk about theoretical and percent yields.

By the end of this module, you should be able to calculate a percent yield based on

both the theoretical and experimental yield of a reaction.

So, up until now, we've assumed that the amount of product we produce would be

the maximum amount possible based on the amount of reactants.

The reality is that we cannot produce as much product as we actually calculate.

If we go back to our sandwich analogy, what we know is that sometimes we drop

something on the floor, somebody steals a piece of turkey from our sandwich, and

we can't make as many as we had originally planned based on our supply.

For a reaction, sometimes what happens is, it simply doesn't go to completion for

a variety of reasons that are beyond the scope of this course.

Other times we will lose some substance if we're producing a gas or

using a gas as a reactant.

Sometimes we'd lose some of that in the transfer process.

We can also have situations of unrecovered product, either because it spilled or

because we have a, say,

solid stuck to a filter paper that we can't remove from that.

So lots of different things can happen to affect our percent yield.

And this is a big issue when you start to look in industry,

because even reducing the per the waste and increasing the percent yield by

a small amount can make a dramatic difference in the bottom line.

So the calculations we've been doing up until now have been for

theoretical yield or the maximum amount of product possible.

For our example using the snowman, we determined that our yield,

our theoretical yield or calculated amount was 3, because that was the maximum of

snowmen that we could build from the parts that we had.

We would still have some excess reagents, but

we were limited by the number of snowmen heads that we had.

While the theoretical yield gives us the maximum amount we can possibly make,

the actual yield is what we really get when we run the experiment.

This is sometimes also called the experimental yield because it's

what results when we actually do the reaction.

What we do know is that it will always be less than or it could be equal to, but

more than likely it's going to be less than, the theoretical yield.

And if we go back to our sandwich analogy, if I had the supplies to make six, and

someone stole a piece of turkey off the counter,

then I'm only going to be able to make five now.

And so, my actual, or experimental yield, is less than my theoretical yield.

So, when we do calculations, we're calculating the theoretical yield.

The experimental or

actual yield has to actually be measured after we do the real experiment.

So, we can actually quantify the difference in these two values by using

the percent yield.

And it basically shows us the efficiency of a reaction.

Manufacturers are very concerned about percent yield,

because increasing the percent yield even by a small amount can be

a large savings cost in the amount of reactants and

less waste that they have to deal with after the experiment is complete.

To find the percent yield, we get the actual yield, or

what we actually did in our experiment.

So this is the experimental yield,

what we actually measured out after we did the reaction.

And our theoretical yield, and this is the calculated value.

And so this is either based on simply the amount of one reactant and

knowing the other's an excess.

Or we might also combine this type of problem with a limiting an excess

reagent problem where we have to actually first identify the limiting reagent to

determine the theoretical yield of a particular reaction.

We do want this value to be as close to 100% as possible.

That doesn't always happen, and

sometimes the values are very low, even in the 20 and 30% yield.

But for those particular reactions, that may be outstanding.

It depends on what you're comparing it to.

But we want to get as close to 100% as possible.

What that's telling us is that we've produced an amount of

product that is very similar to the amount we calculated we could get based on

the amount of our reactants.

The question we want to ask is, can this percent yield be greater than 100%?

Well, the answer is no.

Because if it was greater than 100%, what that would mean is

that the actual yield has a greater value than the theoretical yield.

And we've already determined that the calculated value is

the maximum amount that can be formed, or produced.

So if you get a value greater than 100%, there are two things you should check.

One, make sure you've done your calculations correctly, and two,

if you've done the actual experiment, to look back through and see if

there's something else in that sample that you're actually weighing along with it.

Such as a wet sample, so you're actually weighing the mass of water and

your product rather than just the product alone.

So let's look at an example with the percent yield.

What is the percent yield if the theoretical yield for a particular set of

conditions is 0.750 grams and the actual yield is 0.606 grams of zinc sulfide.

So we're given the balanced chemical equation.

And we can look at the information given.

We're given the actual yield, which is 0.606 grams.

We're given the theoretical, which is 0.750 grams.

Now I simply need to calculate the percent yield,

which will be equal to 0.606 divided by 0.750, times 100.

And when I do that,

I end up with a percent yield of 80.8%.

This tells me I got 80.8% of what I actually expected to get.

Not a bad return on my reaction.

Let's look at a problem that uses percent yield in a slightly different way.

If we start with 42 grams of potassium chlorate,

what is our theoretical yield of potassium chloride in grams?

That's the first question.

The next part is,

if the reaction runs at a 87.3% yield, what was the actual yield of the reaction?

We're given a balanced chemical equation.

And what we need to do is find our theoretical yield first of the KCl.

So, just like any other stoichiometry problem,

we start with our 42.0 grams of KClO3.

Then we use our molar mass of KClO3, which is 122.55

grams per mole of KClO3 to get to moles of our reactant.

Then we use our mole to mole ratio which is given in the balanced equation.

We see that we have two moles of KClO3 to two moles of KCl,

because that's the product that we're interested in.

And then we can figure out the mass of the KCl that is produced

in the reaction by converting from moles of KCl to grams of KCl.

And again we'll use the molar mass,

which is 74.55 grams per mole of KCl.

Now, we can do the calculation to find out the actual theoretical yield.

And when we do that, we find that we get 25.5 grams of KCl.

Remember, this our theoretical yield or

the maximum amount we can produce, because this is the calculated value.

Now, for the second part of the equation, we're given the percent yield,

we know our theoretical yield and we're looking to find the actual yield.

And so you're actually going to get an opportunity to solve the second half of

this problem.

So here we are, we're given the same mass of KCIO3.

We just calculated the theoretical yield.

Now you can find the actual yield,

since we're given in this case the percent yield.

We're still going to use the same formula to solve this problem,

we're just going to have some different information going into it.

The equation for percent yield equals actual over

theoretical times 100.

In this case, we know our percent yield is 87.3%.

We don't know our actual yield, that's what's being asked for in the problem, but

we do know our theoretical yield, which is

25.5 grams of the KCl times 100.

Now, we can divide both sides by 100 and get 0.873, and

then multiply both sides by 25.5 and get our actual yield in units of grams of KCl.

And what we find is we have 14.4 grams of KCl as our actual yield.

This as also known as the experimental,

because it's exactly how much we got when we did the real experiment.

In the next module, we're going to look at solution concentration.

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