Brought to you by Coach Cox PERCENT YIELD. WHAT IS PERCENT YIELD? Theoretical Yield – the maximum amount of product that can be produced from a given.

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Presentation transcript:

Brought to you by Coach Cox PERCENT YIELD

WHAT IS PERCENT YIELD? Theoretical Yield – the maximum amount of product that can be produced from a given amount of reactant. This is determined mathematically using stoichiometry calculations Actual Yield – the amount of product actually produced when a chemical reaction is carried out in the lab Percent Yield – ratio of the actual yield to the theoretical yield expressed as a percent (%)

PERCENT YIELD FORMULA Percent Yield = Actual Yield x 100 Theoretical Yield It is on your formula chart!

PERCENT YIELD EXAMPLE 1 In lab, you burned 3.2 g of Mg and produced 4.8 g of MgO according to the given reaction. Your calculations showed that you should have been able to make 5.3 g of MgO. What is your percent yield ? Mg (s) + O 2(g)  MgO (s)

PERCENT YIELD EXAMPLE 1 ANSWER Actual Yield: 4.8 g MgO Theoretical Yield: 5.3 g MgO Percent yield = Actual Yield x 100 Theoretical Yield 4.8 g MgO x 100 = 90.6% Yield MgO 5.3 g MgO

PERCENT YIELD EXAMPLE 2 During a lab experiment, 24.8 g of calcium carbonate, CaCO 3, is decomposed and 13.1 grams of calcium oxide, CaO, is produced. CaCO 3(s)  CaO (s) + CO 2(g) a.What is the actual yield of CaO? _________________ b.What is the theoretical yield of CaO? _________________ c.What is the percent yield of CaO?_________________

PERCENT YIELD EXAMPLE 2 ANSWER a.What is the actual yield of CaO? 13.1 g CaO b.What is the theoretical yield of CaO? g CaO 24.8 g CaCO 3 x 1 mol CaCO 3 x 1 mol CaO x g CaO = g CaO g CaCO 3 1 mol CaCO 3 1 mol CaO c.What is the percent yield of CaO?94.2% Yield CaO 13.1 g CaO x 100 = 94.2% Yield CaO 13.9 g CaO

PERCENT YIELD EXAMPLE 3 During a lab experiment, 1.87 g of aluminum reacts with an excess of copper (II) sulfate, (CuSO 4 ), and 4.65 g or copper is produced. What is the percent yield? 2 Al + 3 CuSO 4  Al 2 (SO 4 ) Cu

PERCENT YIELD EXAMPLE 3 ANSWER Actual Yield: 4.65 g Cu Theoretical Yield: 6.61 g Cu 1.87 g Al x 1 mol Al x 3 mol Cu x g Cu = 6.61 g Sb g Al 2 mol Al 1 mol Cu Percent yield = Actual Yield x 100 Theoretical Yield 4.65 g MgO x 100 = 70.3% Yield Cu 6.61 g MgO

PERCENT YIELD EXAMPLE 4 An ore of antimony, (Sb 2 S 3 ), reacts with excess iron according to the given reaction. If 15 g of Sb 2 S 3 reacts and the percent yield of Sb is 91.5%, what mass of Sb is actually produced ? Sb 2 S 3(s) + 3 Fe (s)  2 Sb (s) + 3 FeS (s)

PERCENT YIELD EXAMPLE 4 ANSWER Actual Yield: ? g Sb Theoretical Yield: g Sb 15 g Sb 2 S 3 x 1 mol Sb 2 S 3 x 2 mol Sb x g Sb = g Sb g Sb 2 S 3 1 mol Sb 2 S 3 1 mol Sb Percent yield = Actual Yield x 100 Theoretical Yield ? g MgO x 100 = 91.5% Yield Sb g MgO Actual Yield = 9.84 g Sb

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