7.3 Percentage Yield

Theoretical Yield and Actual Yield The amount of product that is predicted by stoichiometry is called the theoretical yield The amount of product that is obtained in an experiment is called the actual yield Factors that can affect yield: competing reactions experimental design and technique impure reactants

Calculating Percentage Yield Percentage yield = Actual yield x 100% Theoretical yield

Example 1 Consider: N2(g) + 3H2(g) → 2NH3(g) When 7.5 × 101 g of nitrogen reacts with sufficient hydrogen, the theoretical yield of ammonia is 9.10 g. If 1.72 g of ammonia is obtained by experiment, what is the % yield of the reaction?

Solution Percentage yield = Actual yield x 100% Theoretical yield = 1.72 g x 100% 9.10 g Therefore the percentage yield of the reaction is 18.9%

Example 2 Consider: CaCO3(s) → CaO(s) + CO2(g) This reaction proceeds with a 92.4% yield of calcium oxide. How many grams of CaO can the chemist expect to obtain if 12.4 g of CaCO3 is heated?

Solution Start by converting mass of reactant to moles: n mol of CaCO3 = 12.4 g of CaCO3 100.086 g/mol = 0.124 mol of CaCO3 Then find the moles of product using mole ratios: 1 mol CaO : 1 mol CaCO3 n mol CaO : 0.124 mol CaCO3 There is a theoretical yield of 0.124 moles of CaO

Solution continued… Then find the theoretical yield in grams: mass of CaO = (0.124 mol CaO)(56.077 g/mol) = 6.95 g actual yield = theoretical yield x % yield = 6.95 g CaO x 92.4% = 6.42 g CaO Therefore the chemist can expect to obtain 6.42 g of calcium oxide if 12.4 g of calcium carbonate is heated in a reaction that proceeds with a 92.4% yield.

Percentage Purity Impure reactants often cause percentage yields less than 100% Percentage purity of a sample describes what proportion, by mass, of the sample is composed of a specific compound or element

Example 3 Suppose that you have a 13.9 g sample of impure iron pyrite, FeS2. You heat the sample in air to produce iron(III) oxide, Fe2O3, and sulfur dioxide, SO2. 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g) If you obtain 8.02 g of iron(III) oxide, what was the percentage of iron pyrite in the original sample? Assume that the reaction proceeds to completion. That is, all the available iron pyrite reacts completely.

Solution Start by converting mass of product to moles: n mol of Fe2O3 = 8.02 g of Fe2O3 160 g/mol = 0.0502 mol of Fe2O3 Then find the moles of reactant using mole ratios: 4 mol FeS2 : 2 mol Fe2O3 n mol FeS2: 0.0502 mol Fe2O3 There were 0.100 moles of FeS2 reacted

Solution continued… Then convert the moles of reactant to grams: mass of FeS2 = (0.100 mol FeS2)(120 g/mol) = 12.0 g FeS2 Then find the percentage purity: percentage purity = theoretical mass x 100% sample mass = 12.0 g x 100% 13.9 g = 86.3% Therefore the percentage purity of the FeS2 is 86.3%

Homework P 270 #1-4 McGraw Hill P 339 #1 – 10 Nelson