Percentage Yields

So we know how to do theorectical yield, which is pretty much what is produced in the ideal situation, the difference between percentage yield, is that percentage yield is the actual yield divided by the theorectical yield.
Real life example
Rachel makes cars for a living. She is going on vacation in a week and does not want to order more parts before she leaves. She has 2 car bodies and 12 tires, because of this she can only make two cars. Then one of the cars gets into a wreck on its test drive, so only one car is made. The theorectical yield is 2 cars, but the actual yield is only 1 car, and so the percentage yield is 1 over 2 or 1/2 or 50%.

Percentage Yield =     Mass of Actual Yield       x   100% 
                                Mass of Theoretical Yield

Okay so lets do an example
For the balanced equation shown below, if the reaction of 9.61 grams of C2H3O2Cl produces a 67.6% yield, how many grams of Cl2 would be produced ?

4C2H3O2Cl+7O2 Yields 8CO2+6H2O+2Cl2

1. First you must locate what you are looking for, for in these equations they can be asking for different things
In this equation they are looking for the amount of Cl2 that will be produced in the experiment
2. Locate the grams of the first reactent, in this equation is 9.61 and make the grams of Cl2 (what you are looking for) as x
3. Use a division sign underneath the grams and put the GFW of the two compounds. (If you have forgotton how to locate GFW Click here)
At this point the equation should look like this
Compound 1         + Compound 2 = Result 1 + Result 2 + Result 3
9.61g/94.45 GFW                                                        x/70.9 GFW
Then when you divide the first compound that will give you .101746 as the molarity
Just so you now the compound in any of these equations on the left hand side of the yield sign is considered the limiting reagent
5. Find the ratio between the coiffients of the Compounds
4:2 or 1:1/2
6. Multiply the ratio by the molarity
1/2(.101746) = .050508734
7. Multiply that number given to you by the GFW of the result which in this case is 70.9
70.9(.50508734) = 3.6068957
Now comes the tricky part
8. Since the equation for percentage yield is actual yield/predicted yield = percentage
We have the percentage which is 67% and the theorectical yield which is 3.6068957, then we put it in the equation 
*Mind you because the percentage is in percentage, you have to divide by 100*
Actual = x
Theorectical = 3.6068957
Percentage = .67 
    x                            =  .67
3.6068957                        1
Then cross multiply .67(3.6068957) = 2.4
The actual yield of the experiment will give you 2.4 grams of Cl2
Lets try Another one
2) For the balanced equation shown below, if the reaction of 49.7 grams of O2 produces 51.2 grams of SO2, what is the percent yield?

4FeS2+11O2 Yields 2Fe2O3+8SO2

1. First Locate what you are looking for, in this case it is the percentage
Note that the grams that are produced in this equation are the actual yield and so first we are just looking for the predicted yield
2. Find the molarity of the compound on the left side of the equation
grams/ GFW or 49.7 grams/32 GFW = 1.553125 as the molarity
3. Multiply the molarity by the ratio between its compound and the compound produced
11:8 or 1:8/11
8/11(1.553125) = 1.129545455
4. Multiply that number by the GFW of the compound produced
GFW of SO2 = 64
64(1.129545455) = 72.29090909
That number is the predicted yield
5. Set up the equation
Actual yield/ predicted yield = percentage yield x 100
51.2/72.29090909 = .708249497
.708249497 x 100 = 70.8249497 % or 70%
Thats how you do percentage yield
Click on ET to try some practice problems
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