0:01 hi everyone in this video we're
0:04 continuing to discuss reaction yields
0:06 our learning objectives are to explain
0:08 the concepts of theoretical yield and
0:10 limiting reagents we touched on limiting
0:12 reagents last time so here we're going
0:14 to focus on the theoretical yield we're
0:16 going to derive the theoretical yield
0:18 for reaction under specified conditions
0:20 and then we'll use it to calculate the
0:23 percent yield for a given
0:25 reaction so the amount of product that
0:27 can be produced by a reaction when we
0:29 calculate it based upon the reaction stochiometry
0:30 stochiometry
0:32 is called the theoretical yield of the
0:35 reaction right so assuming that we use
0:37 up all of our limiting reagent we
0:39 determine how much product that could
0:43 form that would be our theoretical yield
0:45 in reality the amount of product that we
0:48 obtain is What's called the actual yield
0:50 and it is usually less than the
0:52 theoretical yield for a number of
0:54 reasons right there could be competing
0:56 side reactions meaning that the
0:58 reactants could be doing some other
1:00 reaction that's not represent presented
1:02 uh in your chemical equation there could
1:04 be an incomplete reaction meaning that
1:07 not all of your limiting reagent is
1:10 actually used up for whatever reason uh
1:12 it could be difficult to recover the
1:14 product so for example the product that
1:16 you make would be mixed in with the
1:19 reagents right the reactants and if you
1:21 aren't able to separate them then you'll
1:23 only recover a portion of what you
1:25 actually make in terms of the
1:28 product so the extent to which reactions
1:30 theoretical yield is ad achieved is
1:32 commonly expressed as something called
1:35 the percent yield and this is calculated
1:38 by taking the ratio of the actual yield
1:39 over the theoretical yield and
1:41 multiplying by
1:44 100% so when we calculate the percent
1:47 yield it's important that the actual
1:49 yield and the theoretical yield have the
1:51 same units right so if they're both in
1:54 grams then grams would cancel and then
1:56 our percentage would simply be a
1:59 percent so for example if we have a
2:02 reaction in which we are able to recover
2:07 75 gam of product but based upon our
2:09 limiting reagent and how much product
2:11 that should have made right we would
2:14 have expected 100 grams so in this case
2:17 our actual yield is 75 grams our
2:19 theoretical yield is 100 gram if we want
2:22 to calculate the percent yield we would
2:25 take this ratio and then multiply by 100
2:27 and we would get
2:31 75% so this 75 % tells us that we were
2:35 able to to uh recover 75% of the
2:38 theoretical yield here we have that upon
2:53 1.2745 N2 gram of copper metal so this
2:56 is one of our products was obtained
3:00 according to the equation what is the %
3:02 yield so let's Orient ourselves to the
3:06 problem a little bit we can see that the
3:07 problem told us that we have copper
3:11 sulfate and excess zinc metal so this
3:14 excess tells me that the zinc is the
3:17 excess reagent and so the copper sulfate
3:20 must be our limiting reagent so the
3:22 amount of product that we could form it
3:24 needs to be based upon how much limiting
3:27 reagent we have we're also told the
3:30 amount of copper that is recovering in
3:34 the reaction so that's 0.392 G this is our
3:36 our actual
3:39 actual
3:41 yield right this is how much was
3:43 actually formed in the reaction so if we
3:45 want to calculate the percent yield
3:47 right we said in the previous slide that
3:49 our percent yield would be the actual
3:54 yield divided by the theoretical yield
3:58 and then multiplied by
4:01 100% so we already have the actual yield
4:02 which is given in the problem and we
4:04 need to calculate what the theoretical
4:09 yield would be so our theoretical
4:12 yield is going to be based upon using up
4:15 all of our limiting reagent the copper
4:17 sulfate so the problem told us that we
4:23 had 1. 1274 G of copper
4:25 sulfate just as we've done in previous
4:28 problems we know that our stochiometric
4:30 factor is in moles so first we're going
4:33 to need to convert GS of copper sulfate to
4:33 to
4:37 moles so 1 Mo of copper
5:10 copper and then finally we need our
5:12 theoretical yield to be in the same unit
5:14 as our actual right this was provided to
5:16 us in grams so we're going to need to
5:20 convert from moles of copper to grams of
5:22 copper so here again we're going to need
5:25 a molar mass I want moles of copper to
5:28 cancel so I'm going to put one mole of
5:30 copper in the denominator and then the
5:34 mass for that one mole is
5:36 63.5 5
5:39 G so moles of copper cancel and I'm left
5:44 with grams of copper and this should be
5:48 0.507 Gams of copper so this is my
5:50 theoretical yield we're trying to
5:53 calculate the percent yield so my percent
5:54 percent
5:57 yield is going to be the actual yield so
6:00 that was given in our problem as 0.392
6:02 0.392
6:05 G over the theoretical yield this is
6:08 what we calculated as 0.507
6:10 0.507
6:13 G now the grams will cancel so that's
6:14 what we want and we're multiplying this
6:18 by 100% to make it a
6:19 percentage and when you do this
6:26 77.3% this means that we were able to recover
6:28 recover
