## 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