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

- [Voiceover] Alright I'll tackle. In the last video we did the first part of part A, now, so the second part of part A. So, the second part of part A, they say, "Calculate the number of moles of ethene 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 0.200 grams of ethanol, so, if we start with 0.200 grams 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 ethene, 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 completely react, we would have that many moles of ethene". So, let's figure this out. So if we say, so, we have ethanol, so for ethanol, we are starting with zero .200 grams, and we wanna convert this to moles, so we wanna 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 is, as a mass of how many grams, well, they tell us that earlier on the problem, they say, "Ethanol, molar mass 46.1 grams per mol", so, ethanol, molar mass of 46.1 grams per mol, or another way to thinking about it, one mole would have mass of 46.1 grams of 46.1 grams, and so if we do this, we are going to get 0.200 over 46.1, and then grams cancel with the grams, and that going to be how many moles we have, and this is going to be equal to, we have three significant digits that we're gonna be with three significant digits divided by three significant digits, so this is going to be all right, so we clear this, so .2, I could write 00, but it's going to, for the calculator's purposes this is the same thing, divided by 46.1 is equal to 0.0043, I want three significant figures here, so 43 I'm gonna round up, 434, .00434 so 0.00434 moles. So, we've seen so far, if the ethanol were (mumbling) the reaction, if we have this many moles of ethanol, well, and if they completely react, well, then we should end up with that many moles of ethene, and so if we have, if it we get a dehydration, I'm gonna write it this way, if dehydration dehydration reaction goes to completion goes to completion every every mole of ethanol would be converted to a mole of ethene and a mole of water. I shouldn't write the shorthand there, a mole, if I'm writing it out, mole of and a mole water. So 0.00434 mole of ethanol would yield, that same number of moles, 0.00434 moles of ethene. So, if the reaction went to completion that how many moles it 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, "Calculate he percent yield of ethene in the experiment". Well, the percent yield is going to be how much we got, divide by how much we would ideally have gotten if the reaction went fully to completion. So yield yield is going to be equal to how much we actually got, which we figured out in part one, so 0.00264 moles over what we actually got, i'm sorry, what we actually got over what we'd have ideally gotten if the reaction went to completion, so divided by 0.00434 moles, and this is going to be equal to this is going to be equal to so, 00264 divided by 00434, is equal to, and let's see, we have three significant digits here, so 60, I can say .608, or I could say 60.8% yield. So, I'm gonna write this as actually this is gonna, I'm rounding, so I can say 60.8% yield, and there you go.