Theoretical, Actual, and Percentage Yield
Now that you have learned how to find the limiting reagent and how to use the limiting reagent, we can move on to understanding the theoretical yield. To find the theoretical yield you must follow this formula: theoretical yield = actual yield/theoretical yield x 100 Actual yield is the quantity of a product found to be formed in a chemical reaction. To find the percentage yield you follow this formula: percentage yield = molar mass of product/molar mass of reactant x mass of reactant
Find the percentage yield Example 1: For the balanced equation below, if the reaction of 47.9 grams of S produces 31.6 grams of Al2S3, what is the percent yield?
2Al+3S=>Al2S3,
step 1: find the molar massS * 1 = 32 * 1 = 32 grams/moles
step 2: multiply your result by the molar ratio of the product to the limiting reagent 1.49 (1 mole of Al2S3 / 3 moles of S) = 0.496 moles
step 3 (broken down into two parts) :first - find the molar mass of the productAl * 2 = 27 * 2 = 54
S * 3 = 32 * 3 = 96
54 + 96 = 150 grams/mole
second - multiply the molar mass of the product by the amount of moles of the limiting reagent 0.496 moles * (150g/m) = 74.4 grams
step 4: solve for the percentage yield
percentage yield = molar mass of product/molar mass of reactant x mass of reactant
31.6 grams/74.4 grams * 100
percentage yield =42.47%
Now we find the actual yield
For the balanced equation below, if the reaction of 37.8 grams of C2H4 produces a 51.6% yield, how many grams of CO2 would be produced ?
C2H4+3O2=>2CO2+2H2O
Step One: find how many moles of the limiting reagents you have
Step One: Find the molar mass of the limiting reagent
C * 2 = 12 * 2 = 24
H * 4 = 1 * 4 = 4
24 + 4 = 28 grams/mole
Step Two: Divide the number of grams of the limiting reagent by its molar mass
37.8 grams = 1.35 moles
28 g/m
Step Two: Determine How Many Moles Of Product Would Be Formed
Multiply Your Result By The Molar Ratio of Product to Limiting Reagent
1.35 moles * (2 moles of CO2 / 1 mole of C2H4) = 2.7 moles
Step Three: Multiply Your Result By The Molar Mass Of The Product
Step One: Find The Molar Mass Of The Product
C * 1 = 12 * 1 = 12
O * 2 = 16 * 2 = 32
12 + 32 = 44 grams/mole
Step Two: Multiply Your Molar Mass by Your Amount of Moles of The Limiting Reagent
2.7 moles * (44 g/m) = 118.8 grams (Note: This is your theoretical yield)
Step Four: Solve For The Actual Yield
you can do this by plugging in your numbers into the formula below:
Percent yield (given) = actual yield (solve for) * 100
theoretical yield (just found)
51.6 = (actual yield) * 100
118.8 grams
actual yield = 61.3 grams
take a quiz on percentage yield !
Find the percentage yield Example 1: For the balanced equation below, if the reaction of 47.9 grams of S produces 31.6 grams of Al2S3, what is the percent yield?
2Al+3S=>Al2S3,
step 1: find the molar massS * 1 = 32 * 1 = 32 grams/moles
step 2: multiply your result by the molar ratio of the product to the limiting reagent 1.49 (1 mole of Al2S3 / 3 moles of S) = 0.496 moles
step 3 (broken down into two parts) :first - find the molar mass of the productAl * 2 = 27 * 2 = 54
S * 3 = 32 * 3 = 96
54 + 96 = 150 grams/mole
second - multiply the molar mass of the product by the amount of moles of the limiting reagent 0.496 moles * (150g/m) = 74.4 grams
step 4: solve for the percentage yield
percentage yield = molar mass of product/molar mass of reactant x mass of reactant
31.6 grams/74.4 grams * 100
percentage yield =42.47%
Now we find the actual yield
For the balanced equation below, if the reaction of 37.8 grams of C2H4 produces a 51.6% yield, how many grams of CO2 would be produced ?
C2H4+3O2=>2CO2+2H2O
Step One: find how many moles of the limiting reagents you have
Step One: Find the molar mass of the limiting reagent
C * 2 = 12 * 2 = 24
H * 4 = 1 * 4 = 4
24 + 4 = 28 grams/mole
Step Two: Divide the number of grams of the limiting reagent by its molar mass
37.8 grams = 1.35 moles
28 g/m
Step Two: Determine How Many Moles Of Product Would Be Formed
Multiply Your Result By The Molar Ratio of Product to Limiting Reagent
1.35 moles * (2 moles of CO2 / 1 mole of C2H4) = 2.7 moles
Step Three: Multiply Your Result By The Molar Mass Of The Product
Step One: Find The Molar Mass Of The Product
C * 1 = 12 * 1 = 12
O * 2 = 16 * 2 = 32
12 + 32 = 44 grams/mole
Step Two: Multiply Your Molar Mass by Your Amount of Moles of The Limiting Reagent
2.7 moles * (44 g/m) = 118.8 grams (Note: This is your theoretical yield)
Step Four: Solve For The Actual Yield
you can do this by plugging in your numbers into the formula below:
Percent yield (given) = actual yield (solve for) * 100
theoretical yield (just found)
51.6 = (actual yield) * 100
118.8 grams
actual yield = 61.3 grams
take a quiz on percentage yield !