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Percent Yield

Percent Yield. Brought to you by Coach Cox. 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

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Percent Yield

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  1. Percent Yield Brought to you by Coach Cox

  2. 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 (%)

  3. Percent Yield Formula • Percent Yield = Actual Yield x 100 Theoretical Yield It is on your formula chart! 

  4. 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) + O2(g)  MgO(s)

  5. 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

  6. Percent yield example 2 • During a lab experiment, 24.8 g of calcium carbonate, CaCO3, is decomposed and 13.1 grams of calcium oxide, CaO, is produced. CaCO3(s) CaO(s) + CO2(g) • What is the actual yield of CaO? _________________ • What is the theoretical yield of CaO? _________________ • What is the percent yield of CaO? _________________

  7. Percent yield example 2 Answer • What is the actual yield of CaO? 13.1 g CaO • What is the theoretical yield of CaO? 13.90 g CaO 24.8 g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56.077 g CaO = 13.895 g CaO 100.086 g CaCO3 1 mol CaCO3 1 mol CaO • What is the percent yield of CaO? 94.2% Yield CaO 13.1 g CaOx 100 = 94.2% Yield CaO 13.9 g CaO

  8. Percent yield Example 3 • During a lab experiment, 1.87 g of aluminum reacts with an excess of copper (II) sulfate, (CuSO4), and 4.65 g or copper is produced. What is the percent yield? 2 Al+ 3 CuSO4  Al2(SO4)3 + 3 Cu

  9. Percent yield example 3 answer • Actual Yield: 4.65 g Cu • Theoretical Yield: 6.61 g Cu 1.87 g Al x 1 molAl x 3 molCu x 63.546 g Cu= 6.61 g Sb 26.98 g Al 2 molAl 1 molCu • Percent yield = Actual Yield x 100 Theoretical Yield 4.65 g MgO x 100 = 70.3% Yield Cu 6.61 g MgO

  10. Percent yield example 4 • An ore of antimony, (Sb2S3), reacts with excess iron according to the given reaction. If 15 g of Sb2S3 reacts and the percent yield of Sb is 91.5%, what mass of Sb is actually produced? Sb2S3(s) + 3 Fe(s)  2Sb(s) + 3FeS(s)

  11. Percent yield example 4 Answer • Actual Yield: ? g Sb • Theoretical Yield: 10.75 g Sb 15 g Sb2S3x 1 mol Sb2S3x 2 molSbx 121.76 g Sb= 10.75 g Sb 339.718 g Sb2S31 mol Sb2S31 molSb • Percent yield = Actual Yield x 100 Theoretical Yield ? g MgOx 100 = 91.5% Yield Sb 10.75 g MgO Actual Yield = 9.84 g Sb

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