Percent Yield
Step 1. You must have a balanced chemical equation for the reaction.
Step 2. From the weights of the reactants and reagents, calculate the moles of each.
Step 3. Determine which of the reactants or reagents is the limiting reagent.
Step 4. From the moles of limiting reagent and the balanced equation, calculate the
number of moles of product that could theoretically be produced.
Step 5. From the theoretical moles of product, calculate the theoretical yield in grams.
Step 6. Percent yield = [actual yield (grams) / theoretical yield (grams)] * 100%
For example, a student is oxidizing cyclohexene with potassium permanganate in
aqueous solution. She uses 8.2 g of cylcohexene, 12.6 g of KmnO4, and 100 mL of
water. After the final distillation, she obtains 5.2 grams of cis-1,2-cyclohexanediol.
What is her percent yield?
The balanced chemical equation.
+ 2 KMnO4 + 4 H2O
3
+ 2 MnO2 + 2 KOH
3
HO
8.2 g
82 g/mol
0.10 mol
12.6 g
158 g/mol
0.080 mol
OH
100 mL
18 g/mole
5.6 mol
0.10 moles of cyclohexene will react with 0.067 moles of KMnO4, according to the
balanced equation. The permanganate is in excess and the cyclohexene is the limiting
reagent.
According to the balanced equation, the number of moles of cis-1,2-cyclohexanediol
produced is equal to the number of moles of cyclohexene used. So, 0.10 moles of
cyclohexene will theoretically yield 0.10 moles of the diol.
The molecular weight of cis-1,2-cyclohexanediol is 116 g/mol; 0.10 moles would weigh
11.6 grams. This is the theoretical yield, in grams.
The student’s percent yield is therefore = [ 5.2 g / 11.6 g ] * 100% = 45%.
Note the correct number of significant figures in the answer!