hi everyone in this video we're continuing to discuss reaction yields our learning objectives are to explain the concepts of theoretical yield and limiting reagents we touched on limiting reagents last time so here we're going to focus on the theoretical yield we're going to derive the theoretical yield for reaction under specified conditions and then we'll use it to calculate the percent yield for a given reaction so the amount of product that can be produced by a reaction when we calculate it based upon the reaction stochiometry is called the theoretical yield of the reaction right so assuming that we use up all of our limiting reagent we determine how much product that could form that would be our theoretical yield in reality the amount of product that we obtain is What's called the actual yield and it is usually less than the theoretical yield for a number of reasons right there could be competing side reactions meaning that the reactants could be doing some other reaction that's not represent presented uh in your chemical equation there could be an incomplete reaction meaning that not all of your limiting reagent is actually used up for whatever reason uh it could be difficult to recover the product so for example the product that you make would be mixed in with the reagents right the reactants and if you aren't able to separate them then you'll only recover a portion of what you actually make in terms of the product so the extent to which reactions theoretical yield is ad achieved is commonly expressed as something called the percent yield and this is calculated by taking the ratio of the actual yield over the theoretical yield and multiplying by 100% so when we calculate the percent yield it's important that the actual yield and the theoretical yield have the same units right so if they're both in grams then grams would cancel and then our percentage would simply be a percent so for example if we have a reaction in which we are able to recover 75 gam of product but based upon our limiting reagent and how much product that should have made right we would have expected 100 grams so in this case our actual yield is 75 grams our theoretical yield is 100 gram if we want to calculate the percent yield we would take this ratio and then multiply by 100 and we would get 75% so this 75 % tells us that we were able to to uh recover 75% of the theoretical yield here we have that upon reaction of 1.2745 N2 gram of copper metal so this is one of our products was obtained according to the equation what is the % yield so let's Orient ourselves to the problem a little bit we can see that the problem told us that we have copper sulfate and excess zinc metal so this excess tells me that the zinc is the excess reagent and so the copper sulfate must be our limiting reagent so the amount of product that we could form it needs to be based upon how much limiting reagent we have we're also told the amount of copper that is recovering in the reaction so that's 0.392 G this is our actual yield right this is how much was actually formed in the reaction so if we want to calculate the percent yield right we said in the previous slide that our percent yield would be the actual yield divided by the theoretical yield and then multiplied by 100% so we already have the actual yield which is given in the problem and we need to calculate what the theoretical yield would be so our theoretical yield is going to be based upon using up all of our limiting reagent the copper sulfate so the problem told us that we had 1. 1274 G of copper sulfate just as we've done in previous problems we know that our stochiometric factor is in moles so first we're going to need to convert GS of copper sulfate to moles so 1 Mo of copper sulfate is equal to we're able to uh yield one mole of copper and then finally we need our theoretical yield to be in the same unit as our actual right this was provided to us in grams so we're going to need to convert from moles of copper to grams of copper so here again we're going to need a molar mass I want moles of copper to cancel so I'm going to put one mole of copper in the denominator and then the mass for that one mole is 63.5 5 G so moles of copper cancel and I'm left with grams of copper and this should be 0.507 Gams of copper so this is my theoretical yield we're trying to calculate the percent yield so my percent yield is going to be the actual yield so that was given in our problem as 0.392 G over the theoretical yield this is what we calculated as 0.507 G now the grams will cancel so that's what we want and we're multiplying this by 100% to make it a percentage and when you do this calculation this gives us 77.3% this means that we were able to recover 77.3% of of the theoretical yield