So now you've learned all about theory. But theory is what happens in a perfect world and unless you've been living on a tropical paradise away from civilization for all your life, you know that this is not the case. In fact just the other day I was making Banana toast. I had gotten bored of S'mores and thought I'd take a turn at inventing a snack. Banana toast is simple, it's a piece of toast with a banana sliced in small pieces covering it. I had all the ingredients 10 bananas and 12 slices of bread, so I knew I could make 10 pieces of Banana toast. Except..... When I examined my bananas more closely three of them were bruised and disgusting (I blame Fairway). Therefore I could only use seven bananas, and only make seven pieces of banana toast. This is actual yield, because some of my bananas were bruised I only made 7 pieces of Banana Toast. Actual yield is the amount you can actually produce, not the amount you produce when nothing goes wrong.
Similar to actual yield is percentage yield. Percentage yield is simply the percentage you actually made out of what you would have made without any mishaps. So in the case of the banana toast, I could only make seven pieces, but had my bananas been perfect I could of made 10. Therefore my percentage yield is 7/10 or 70%.
Similar to actual yield is percentage yield. Percentage yield is simply the percentage you actually made out of what you would have made without any mishaps. So in the case of the banana toast, I could only make seven pieces, but had my bananas been perfect I could of made 10. Therefore my percentage yield is 7/10 or 70%.
Now lets apply this to some chemistry:
To find the percentage yield in a chemistry problem you do the following:
1. Find the limiting reagent, if there is only one reactant then this is simple, if not click here to review how to find the limiting reagent.
2. Once you have the limiting reagent calculate the theoretical yield. To review how to find the theoretical yield click here
3. The problem will give you the actual yield, so now you just divide the theoretical yield by the actual yield and multiply by 100 to find the percentage yield.
Examples of finding percentage yield:
For the balanced equation shown below, if the reaction of 20.7 grams of CaCO3 produces 6.81 grams of CaO, what is the percent yield?
CaCO3=>CaO+CO2
In this case you only have one reactant so it is the limiting reagent.
20.7g/100 (GFW of CaCO3)=0.207mol
The ratio is 1:1 so 0.207mol of CaO are produced
0.207mol x 56 (GFW of CaO)=11.592g (theoretical yield)
6.81 grams (actual yield)/ 11.592g (theoretical yield) x 100=58.74 % (percentage yield)
For the balanced equation shown below, if the reaction of 0.112 grams of H2 produces a 73.91% yield. then how many grams of H2O are produced.
Fe3O4+4H2=>3Fe+4H2O
In this case it specifies the reactant so H2 is the limiting reagent.
0.112g/2 (GFW of H2)=0.056mol
The ratio is 4:4 so 0.056mol of CaO are produced
0.056mol x 18 (GFW of H20)=1.008g (theoretical yield)
The difference here is that you have the percentage yield and want to find out the grams so you put what you know in the equation and solve
x grams (actual yield)/ 1.008g (theoretical yield) x 100=73.91 % (percentage yield)
73.91g/100 x 1.008 mol= x grams
x=0.745g of H2O
To find the percentage yield in a chemistry problem you do the following:
1. Find the limiting reagent, if there is only one reactant then this is simple, if not click here to review how to find the limiting reagent.
2. Once you have the limiting reagent calculate the theoretical yield. To review how to find the theoretical yield click here
3. The problem will give you the actual yield, so now you just divide the theoretical yield by the actual yield and multiply by 100 to find the percentage yield.
Examples of finding percentage yield:
For the balanced equation shown below, if the reaction of 20.7 grams of CaCO3 produces 6.81 grams of CaO, what is the percent yield?
CaCO3=>CaO+CO2
In this case you only have one reactant so it is the limiting reagent.
20.7g/100 (GFW of CaCO3)=0.207mol
The ratio is 1:1 so 0.207mol of CaO are produced
0.207mol x 56 (GFW of CaO)=11.592g (theoretical yield)
6.81 grams (actual yield)/ 11.592g (theoretical yield) x 100=58.74 % (percentage yield)
For the balanced equation shown below, if the reaction of 0.112 grams of H2 produces a 73.91% yield. then how many grams of H2O are produced.
Fe3O4+4H2=>3Fe+4H2O
In this case it specifies the reactant so H2 is the limiting reagent.
0.112g/2 (GFW of H2)=0.056mol
The ratio is 4:4 so 0.056mol of CaO are produced
0.056mol x 18 (GFW of H20)=1.008g (theoretical yield)
The difference here is that you have the percentage yield and want to find out the grams so you put what you know in the equation and solve
x grams (actual yield)/ 1.008g (theoretical yield) x 100=73.91 % (percentage yield)
73.91g/100 x 1.008 mol= x grams
x=0.745g of H2O