1. Stoichiometry Chapter 9 Percent Yield Stoich ppt _5 Percent Yield

2. Stoichiometry After this presentation, you should understand:  Calculations using balanced chemical equations: for example, for a given mass of a reactant, calculate the volume of product.  How to identify the theoretical yield and the actual yield of a reactions.  How to calculate the percent yield of a reaction

3. Percent Yield Percent yield = actual yield (measured) x 100 theoretical yield (calculated)

4. Time to Practice! Grab your calculator and periodic table.

5. Calculating Percent Yield Ex: When potassium chromate (K 2 CrO 4 ) is added to a solution containing 0.500 g of silver nitrate (AgNO 3 ), solid silver chromate (Ag 2 CrO 4 ) and potassium nitrate (KNO 3 ) are formed. a.Determine the theoretical yield of the silver chromate precipitate. b.If 0.455 g of silver chromate is obtained, calculate the percent yield.

6. To Solve 1.What do we know? a.mass of silver nitrate = 0.500 g AgNO 3 b.percent yield = ? % yield of Ag 2 CrO 4 2.Write a balanced equation for the reaction 2 AgNO 3 + 1 K 2 CrO 4 ---  1 Ag 2 CrO 4 + 2 KNO 3 0.500 ga)? g b)%Yield

7. 2 AgNO 3 + 1 K 2 CrO 4 ---  1 Ag 2 CrO 4 + 2 KNO 3 Convert grams to moles of AgNO 3 : 0.500 g AgNO 3 x 1 mol AgNO 3 = 2.94 x 10 -3 mol AgNO 3 1 169.9 g AgNO 3 Use the appropriate mole ratio to convert mol AgNO 3 to mol Ag 2 CrO 4 : 2.94 x 10 -3 mol AgNO 3 x 1 mol Ag 2 CrO 4 = 1.47 x 10 -3 mol Ag 2 CrO 4 1 2 mol AgNO 3

8. Calculate the mass of Ag 2 CrO 4 (theoretical yield) by multiplying mol Ag 2 CrO 4 by the molar mass From last slide: 2.94 x 10 -3 mol AgNO 3 x 1 mol Ag 2 CrO 4 = 1.47 x 10 -3 mol Ag 2 CrO 4 1 2 mol AgNO 3 1.47 x 10 -3 mol Ag 2 CrO 4 x 331.7 g Ag 2 CrO 4 = 0.488 g Ag 2 CrO 4 1 1 mol Ag 2 CrO 4 Theoretical Yield

9. Calculate the percent yield of Ag 2 CrO 4 Divide actual yield by theoretical yield and multiply times 100: Percent yield = actual yield (measured) x 100 theoretical yield (calculated) Percent yield = 0.455 g Ag 2 CrO 4 x 100 = 93.2 % Ag 2 CrO 4 0.488 g Ag 2 CrO 4 Given in the problem From mole- mole ratios