Percentage Yield

Define the following terms: yield, theoretical yield, actual yield, percentage yield. Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield Actual yield x 100% Theoretical Yield (stoichiometric calculation)

The reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. Not all product is recovered (e.g. spattering) Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk) A side reaction occurs (e.g. MgO vs. Mg3N2) The reaction does not go to completion

Sample problem Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100% x 100% 96.7% =

Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100% x 100% 17.8% =

Challenging question 2H2 + O2  2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100% x 100% 92.9% =

More Percent Yield Questions The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2 What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

More Percent Yield Questions Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?

Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? Actual yield is given: 12.3 g O2 Next, calculate theoretical yield # g O2= 1 mol H2O 18.02 g H2O x 1 mol O2 2 mol H2O x 32 g O2 1 mol O2 x 14.0 g H2O 12.43 g = Finally, calculate % yield actual theoretical 12.3 g O2 12.43 g O2 = % yield = x 100% x 100% 98.9% =

Q2 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3  2KCI + 3O2 Actual yield is given: 107 g O2 Next, calculate theoretical yield # g O2= 1 mol KClO3 122.55 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 300 g KClO3 117.5 g = Finally, calculate % yield actual theoretical 107 g O2 117.5 g O2 = % yield = x 100% x 100% 91.1% =

Q3 What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS Actual yield is given: 220 g FeS Next, calculate theoretical yield 1 mol FeS 1 mol Fe x 87.91 g FeS 1 mol FeS x # g FeS= 3.00 mol Fe 263.7 g = Finally, calculate % yield actual theoretical 220 g O2 263.7 g O2 = % yield = x 100% x 100% 83.4% =

First, determine limiting reagent 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 1 mol FeS2 119.97 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x 159.7 g Fe2O3 1 mol Fe2O3 x 300 g FeS2 199.7 g Fe2O3 = 1 mol O2 32 g O2 x 2 mol Fe2O3 11 mol O2 x 159.7 g Fe2O3 1 mol Fe2O3 x 200 g O2 181.48 g Fe2O3 = actual theoretical 143 g Fe2O3 181.48 g Fe2O3 = % yield = x 100% x 100% 78.8% =