Percentage Yield and Actual Yield
We have been calculating the theoretical yield, the amount of product that would be produced in an ideal situation. However, we don't live in an ideal world so the amount of product that will actually be produced is different from the theoretical yield, this amount is known as the actual yield. The difference between the theoretical yield and the actual yield is known as the percentage yield.
Calculating the percentage yield and actual yield is quite simple. We must use the equation below:
Percentage Yield = (Actual Yield/ Theoretical Yield) * 100
Lets take a look at an example.
For the balanced equation shown below, if the reaction of 88.8 grams of C6H6S2 produces a 57.9% yield, how many grams of SO3 would be produced ?
2C6H6S2+15O2=>12CO+6H2O+4SO3
We are asked to find the actual yield.
In the problems we will look at the limited reagent will be given; however this will not always be the case, so you will have to find limiting reagent.
The first step is finding the theoretical yield. Lets find the molar mass.
C = 12
H = 1
S = 32
O = 16
12 * 6 = 72
1 * 6 = 6
32 * 2 = 64
72 + 6 + 64 = C6H6S2 has 142 grams/mole
32 * 1 = 32
16 * 3 = 48
32 + 48 = SO3 has 80 grams/mole
Now we need to look at the coefficient of the limiting reagent and the coefficient of the product. It says that for every 2 moles of C6H6S2 there are 4 moles of SO3.
So lets multiply the molar mass of the limiting reagent and the product by their coefficients.
142 * 2 = 284
80 * 4 = 320
Now we have to divide the grams of the product by the grams of the limiting reagent:
320/284 = 1.126
Now we have to multiply this by the given mass of the limiting reagent.
1.126 * 88.8 = 100 grams
So now we have the theoretical yield. The next and final step is plugging the theoretical yield, and the precentage yield (given to us in the original question) into the formula.
Percentage Yield = (Actual Yield/ Theoretical Yield) * 100
57.9 = (Actual Yield/100) * 100
Now use algebra.
57.9 = (Actual Yield/100) * 100
100
100 * .579 = (Ay/100) *100
57.9 = Ay
Actual Yield = 57.9 grams
Sometimes you will see problems where the actual yield is given to you and you are asked to find the percentage yield.
For instance if the example said:
For the balanced equation shown below, if the reaction of 88.8 grams of C6H6S2 produces a 57.9 grams of SO3, what is the percent yield?
2C6H6S2+15O2=>12CO+6H2O+4SO3
After finding the theoretical yield you just plug in the actual yield and the theoretical into the percentage yield equation.
Percentage Yield = (57.9/100) * 100
Percentage Yield = (.579) * 100
Percentage Yield = 57.9 %