Finding Your Percentage And Theoretical Yields
Part One: Percentage Yield
For Teaching Purposes, let's use the following:
For the balanced equation 2Al+3S=>Al2S3,
if the reaction of 47.9 grams of S produces 31.6 grams of Al2S3, what is the percent yield?
Step One: Figure Out How Many Moles Of Limiting Reagent You Have
Step One: Find the molar mass of the limiting reagent (given!)
S * 1 = 32 * 1 = 32 grams/moles
Step Two: Divide the number of moles by the molar mass of the limiting reagent
47.9 grams = 1.49 moles
32 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
1.49 (1 mole of Al2S3 / 3 moles of S) = 0.496 moles
Step Three: Multiply Your Result By The Molar Mass of the Product
Step One: Find the molar mass of the product
Al * 2 = 27 * 2 = 54
S * 3 = 32 * 3 = 96
54 + 96 = 150 grams/mole
Step Two: Multiply the molar mass of the product by the amount of moles of the limiting reagent
0.496 moles * (150g/m) = 74.4 grams (Note: This is your theoretical yield)
Step Four: Solve For Percent Yield
Percent yield = actual yield (given) * 100
theoretical yield (just found)
Percent yield = 31.6 grams * 100
74.4 grams
Percent yield = 42.47%
Part Two: Actual Yield
For Teaching Purposes, Let's Use the Following:
For the balanced equation C2H4+3O2=>2CO2+2H2O, if the reaction of 37.8 grams of C2H4 produces a 51.6% yield, how many grams of CO2 would be produced ?
Step One: Figure Out How Many Moles of the Limiting Reagent You Have
Step One: Find the molar mass of the limiting reagent (given!)
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 simply 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