Percentage Yield

OUTCOME QUESTION(S): C11-3-15 STOICHIOMETRY Vocabulary & Concepts Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction. Include: heat of reaction Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data. Include: theoretical yield, experimental yield Vocabulary & Concepts  Actual Yield

- 5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g) ? HAVE 75.0 g 150.0 L - USE 58.9 g 191 L 16.1 g (excess) C3H8 - excess reactant O2 - limiting reactant 150 L O2 1 mol O2 3 mol CO2 22.4 L CO2 = 90 L CO2 22.4 L O2 5 mol O2 1 mol CO2 Our calculations suggest the reaction will use ALL 150 L of O2 and only 58.9 g of C3H8 to make 90 L of CO2 BUT…will it actually produce 90 L when the experiment is run?

Pressure (of gas reactants) Human error Reactions are influenced by external factors which effect the amount of yield produced: Concentration State (s, l, g, aq) Temperature Pressure (of gas reactants) Human error Nothing is 100% efficient Poor recovery /technique Impurities in reactants Miscalculation Unexpected side reaction

actual yield x 100 percent yield = theoretical yield Theoretical yield: amounts of product calculated from the limiting reagent. Actual yield: (also experimental yield), amount produced during the conducted experiment. Percentage yield: yield ratio (actual to theoretical) expressed as a percentage. actual yield theoretical yield percent yield = x 100

5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g) ? HAVE 75.0 g 150.0 L USE 58.9 g 191 L 150 L O2 1 mol O2 3 CO2 22.4 L CO2 = 90 L CO2 22.4 L O2 5 O2 1 mol CO2 Theoretical Yield An experiment is run producing 74 L of CO2 gas. What is the % yield? Actual Yield actual yield 74 L percent yield = x 100 = 82 % theoretical yield 90 L

5. 00 g of KClO3 is heated and decomposes to yield 1. 78 g of O2 5.00 g of KClO3 is heated and decomposes to yield 1.78 g of O2. What is the % yield? Actual Yield 2 KClO3  2 KCl + 3 O2 5.00 g ? g mol mol 5.00 g KClO3 1 mole 3 O2 32.0 g O2 122.5 g 2 KClO3 1 mole O2 = 1.96 g O2 Theoretical Yield actual yield 1.78 g x 100 = 90.8 % percent yield = theoretical yield 1.96 g

Be careful rearranging this equation – don’t forget about the 100%... 2 KClO3  2 KCl + 3 O2 How much O2 would be produced if the percentage yield was 78.5%? Ac. yield 1.96 g 78.5 % = % yield = x 100 Th. yield 1.53 g = Actual yield Be careful rearranging this equation – don’t forget about the 100%...

O2 - excess reactant H2 - limiting reactant What is the % yield if 58 g H2O are produced by combining 60.0 g O2 and 7.0 g H2? O2 (g) + 2 H2 (g) 2 H2O (g) HAVE 60.0 g 7.0 g USE 7.5 g 60.0 g O2 1 mole O2 2 H2 2.0 g H2 = 7.5 g H2 32.0 g O2 1 O2 1 mole H2 O2 - excess reactant H2 - limiting reactant This is a Limiting Reactant question, BUT it doesn’t ask to find excess – so just “pick one” way and identify the LR

O2 (g) + 2 H2 (g) 2 H2O (g) = 63 g H2O 58 g percent yield = x 100 ? HAVE 60.0 g 7.0 g USE 7.5 g 7.0 g H2 1 mole H2 2 H2O 18.0 g H2O 2.0 g H2 2 H2 1 mole H2O = 63 g H2O Actual Yield: 58 g Theoretical Yield 58 g actual yield percent yield = x 100 = 92 % theoretical yield 63 g

CAN YOU / HAVE YOU? C11-3-15 STOICHIOMETRY Vocabulary & Concepts Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction. Include: heat of reaction Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data. Include: theoretical yield, experimental yield Vocabulary & Concepts  Actual Yield