2015 AP Chemistry free response 2a (part 2/2) and b | Chemistry | Khan Academy

All right, now let's tackle, in the last video we did the first part of Part A. Now let's do the second part of Part A.

So the second part of Part A, they say calculate the number of moles of ethine that would be produced if the dehydration reaction went to completion. Well, this is the dehydration reaction right over here, and they're telling us— they're telling us that we start with 0.2 g of ethanol.

So if we start with 0.200000 gram of ethanol, and if we figure out how many moles of ethanol that is, well, for every mole of ethanol, if we have the reaction go to— if the dehydration reaction goes to completion, for every mole of ethanol that we start with, we're going to have a mole of ethine and a mole of water.

So if we can just figure out how many moles of ethanol this is, then we'll say okay, if this were to completely react, we would have that many moles of ethine. So let's figure this out.

So we have ethanol. For ethanol, we are starting with 0.2000 g, and we want to convert this to moles. So we want to multiply this times—we want grams in the denominator to cancel out with this grams and moles in the numerator.

So, one mole of ethanol has a mass of how many grams? Well, they tell us that earlier on in the problem. They say ethanol molar mass is 46.1 g per mole. So, ethanol molar mass of 46.1 g per mole or another way of thinking about it: one mole would have a mass of 46.1 g.

And so if we do this, we are going to get 0.2 over 46.1, and then grams cancel with grams, and that's going to be how many moles we have. This is going to be equal to—we have three significant digits that we're going to be dealing with, three significant digits divided by three significant digits.

So, all right, let me clear this. So, 0.2—I could write 0.2000, but it's going to— for the calculator's purposes, it's the same thing. Divided by 46.1 is equal to, and let's see, we have three significant digits here. So, 0.264 divided by 0.0434 is equal to—and then I want three significant figures here.

So, 0.434. So, 0.434 moles. So, what we've seen so far: if the ethanol were— the reaction— if we have this many moles of ethanol, well if we—and if they completely react, well then we should end up with that many moles of ethine.

And so if we could say dehydration—let me write it this way— if the dehydration reaction goes to completion, every mole of ethanol would be converted to a mole of ethine and a mole of water. I shouldn't write the shorthand there: a mole of ethine and a mole of water.

So, 0.434 moles of ethanol would yield that same number of moles—0.434 moles of ethine. So, if the reaction went to completion, that's how many moles you would produce.

What we actually measure is a smaller number than that, so the reaction didn't go fully to completion.

So now, let's tackle Part B. Let's tackle Part B, and they ask us to calculate the percent yield of ethine in the experiment. Well, the percent yield is going to be how much we got divided by how much we would ideally have gotten if the reaction went fully to completion.

So, yield is going to be equal to how much we actually got, which we figured out in Part One—so 0.264 moles—over what we would have ideally gotten if the reaction went to completion, so divided by 0.434 moles.

And this is going to be equal to—so 0.264 divided by 0.434 is equal to—and let's see, we have three significant digits here. So, I could say 60.8% yield. So I could write this as—actually, I’m rounding, so I could say it's 60.8% yield.

And there you go.