Stoichiometry Solving Problems Involving Mass/Mole ...

Stoichiometry Solving Problems Involving Mass/Mole Relationships Theoretical Yield and Percent Yield Page [1 of 2]

Here what we're going to do is use the ideas that we've used up until now, the idea of using the balanced reaction to calculate the amount of product we expect and stuff like that, and introduce two new concepts, the theoretical yield and the percent yield. Now the theoretical yield is as it sounds. How much stuff do you expect to get? And obviously it's going to depend on how much stuff you started with. How many gin and tonics do you expect to be able to make is obviously going to depend on how much gin you have, how much tonic you have, and more importantly which one are you going to run out of first? Because that's going to determine how many gin and tonics you make.

So let's look at a chemical problem where we're going to apply that exact same idea. And the reaction that I'd like to look at is the reaction of ethanol, which is in purple, and acetic anhydride, which is in blue, to form what we're interested in, which is ethyl acetate plus acetic acid, which is in red. And these things are interesting because ethanol is grain alcohol. That's the alcohol in our gin. Ethyl acetate is a common form of nail polish remover, not the only kind, but one of the two kinds of nail polish removers. And then acetic acid is the acid in vinegar.

And so here's the problem. How many grams of ethyl acetate, again the thing in green, should we expect from a reaction of 20 grams of ethanol and 60 grams of acetic anhydride? And this is actually a balanced reaction. In other words, 1 mole of ethanol plus 1 mole of acetic anhydride forms 1 mole of ethyl acetate and 1 mole of acetic acid.

Now, as an aside, this happens to be exactly the same reaction that takes you from morphine to heroin. So in other words, morphine plus acetic anhydride goes to heroin plus acetic acid. And if you were a police officer, how would you detect where they have labs that are making heroin from morphine? And the answer is you'd train dogs to smell the vinegar. The smell of vinegar is indicative of the fact that either somebody is making heroin out of morphine, or somebody is baking French bread.

Anyway, so here's the balanced reaction expressed in a molecular formula terminology. Again, this is the ethanol, color-coded. This is the acetic anhydride going to ethyl acetate and acetic acid. And it's 1 mole plus 1 mole goes to 1 mole plus 1 mole.

So how do we approach this problem? Remember. To use the balanced reaction, things have to be in moles. Moles, moles, moles. And in order to figure out how much stuff we're going to make, we have to figure out first what are we going to run out of first? That's a limiting reagent problem. So let's first calculate the limiting reagent. And to do that we first have to figure out how many moles of ethanol we started with. That was one of the two reactants that we started with. We started with 20 grams. The molar mass of ethanol is 46 grams per mole. So we multiply this through and we get 4.35 moles. And this is moles of ethanol. So let's go ahead and write moles of ethanol. And then let's do the same for acetic anhydride. We started with 60 grams. And the molar mass of acetic anhydride is 102 grams per mole. So we multiply through and we get .588 mole. And this is acetic anhydride.

And now let's use method two for reasons that will become clear later on. You could use method one. If you'd rather use method one, go ahead, but I'm going to show you something bonus that you get from method two. So if we calculate how many moles of ethyl acetate we would have produced, assuming that ethanol is the limiting reagent, it would be 4.35 moles of ethanol times 1 mole of ethyl acetate for every mole of ethanol. This is 1 mole of product formed, which is ethyl acetate, divided by 1 mole of ethanol, which is the reagent. And we get that we would produce .435 moles of ethyl acetate. That's assuming that ethanol was the limiting reagent.

Now we'll do the second part of method two, which is calculate how much ethyl acetate we're going to make assuming that acetic anhydride is the limiting reagent. It's the same idea. Now it's 1 mole of ethyl acetate formed for every 1 mole of acetic anhydride used. And so we would form, theoretically, .588 moles of ethyl acetate starting with .588 moles of acetic anhydride. Now remember method two says whichever one you would form the least amount of product, that's your limiting reagent. So that means ethanol is the limiting reagent.

Now the nice thing about method two is you get a bonus. And the bonus is you've also, at the same time, calculated the theoretical yield of ethyl acetate, because we've determined how much is the maximum amount of ethyl acetate we can make from our reagents. And that turned out to be .435 moles of ethyl acetate. So we've sort of inadvertently calculated also the theoretical yield in moles of ethyl acetate. And it is .435 moles. If you wanted to calculate the theoretical yield of ethyl acetate in grams, we'd just multiply by the molar mass of ethyl acetate in grams, and that gives us 38.3 grams of ethyl acetat e. So this is of ethyl acetate.

Stoichiometry Solving Problems Involving Mass/Mole Relationships Theoretical Yield and Percent Yield Page [2 of 2]

The interesting thing is that most chemical reactions don't give all of the product that you expected. Why is that? Well not all reactions, even though we write them stoichiometrically, and it looks like you should get exactly what you predicted, the reality is that a lot of reactions don't necessarily go only to one single product. Maybe they give a couple of products, but those are minor products. So you don't get everything you expected.

Similarly, when you're trying to purify your product, sometimes you lose a little. If you get a distillation, if you get crystallization, you might lose a little. And sometimes you're just plain sloppy. Suppose you spill part of your product. Well you still have to report that you didn't get as much as you thought you were going to get. And so the quantity of product that you actually get when you do the experiment, that's what we call the actual yield. And the actual yield, if you think about it, is always going to be less than the theoretical yield, because the theoretical yield is, if you got everything that you should have gotten, that would be the theoretical yield. And the actual yield is going to be less because of losses. All of the suggestions that I gave for why you wouldn't get everything that you expected are because of losses. You spilled some. The reaction didn't go to completion. The reaction gave some side products that really you weren't interested in. Let's define a quantity called the percent yield. And the percent yield is the actually yield. So again, this is the experimentally-determined, go?into-the-lab-and-measure-it yield, divided by the theoretical yield, which is something that we calculated based on what we put into the pot. And we multiply by 100%.

Note that because of the compound that we're talking about for the actual yield, in other words, the actual yield of ethyl acetate in our problem divided by the theoretical yield of ethyl acetate in our problem, because it's the same compound in the numerator and the denominator, and it will always be the case when you're doing the percent yield calculation, you could have actual yield in grams and theoretical yield in grams. Or you could do actual yield in moles and theoretical yield in moles. And you'd get exactly the same answer, because these of course are related by the molar mass, and there would be a term of molar mass in both the numerator and the denominator. So you could do it either way. And what we're going to do is we're going to do it in grams, but it doesn't make much difference.

In order to use this equation, what do you need? Well you need some actual data. So suppose we went in the lab and we took 20 grams of ethanol and we took 60 grams of acetic anhydride. And we reacted them. And we got ethyl acetate. And we purified it, and got actually a clean sample. Suppose we only isolated 34.2 grams of ethyl acetate. What is the percent yield? Well we use our percent yield equation. Actual yield was what we determined experimentally, 34.2 grams. Theoretical yield we calculated, remember, based on the limiting reagent, which was ethanol, 38.3 grams. And we work this out and the percent yield is 89.3%.

What have we done? What we've done is indicated a concept of percent yield. And we talked also about theoretical yield. Remember. Theoretical yield depends on the limiting reagents. So you have to calculate the limiting reagent first. And remember to calculate a limiting reagent; things have to be in moles. And then finally, we've introduced the concept of percent yield, which tells you how well you did. How well you did compared to what you should have been able to get in the laboratory. And obviously people that are very skilled are going to get closer to the theoretical yield. Whereas sloppy chemists are going to get very low percent yields.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download