Percent Yield.

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Presentation transcript:

Percent Yield

Theoretical Yield The maximum amount of product that can be produced from a given amount of reactant Theoretical yield is calculated using stoichiometry

Actual Yield The amount of product actually produced when the chemical reaction is carried out in an experiment

Percent Yield A comparison of the actual and theoretical yield In general, the higher the yield, the better the results are from the experiment. % Yield = actual yield (experiment) theoretical yield (calculation) × 100

Steps Identify what is given in the problem. One product and one reactant (go to step 2) One product and two reactants (go to step 3) The product given is the actual yield, calculate the theoretical yield using stoichiometry and the reactant given The product given is the actual yield, calculate the limiting reactant and that becomes your theoretical yield Calculate the percent yield using the equation

Determine the theoretical yield of Ag2CrO4 if 0 Determine the theoretical yield of Ag2CrO4 if 0.500 g of AgNO3 is used to react with K2CrO4. Also if 0.455 g of Ag2CrO4 is obtained from an experiment, calculate the percent yield. 2 AgNO3 + 1 K2CrO4 1 Ag2CrO4 + 2 KNO3 0.500g AgNO3 1 mol AgNO3 1 mol AgCrO4 331.74 g AgCrO4 169.88 g AgNO3 2 mol AgNO3 1 mol AgCrO4 % Yield = Actual 0.455 g AgCrO4 Theoretical 0.488 g AgCrO4 ×100 = ×100 = 93.2% = 0.488 g AgCrO4

CCl4 was prepared by reacting 100. 0 g of CS2 and 100. 0 g Cl2 CCl4 was prepared by reacting 100.0 g of CS2 and 100.0 g Cl2. Calculate the theoretical yield and percent yield if 65.0 g of CCl4 was obtained from the reaction. 1 CS2 + 3 Cl2 1 CCl4 + 1 S2Cl2 100.0g CS2 1 mol CS2 1 mol CCl4 153.8 g CCl4 76.15 g CS2 1 mol CS2 1 mol CCl4 = 202.0 g CCl4 100.0g Cl2 1 mol Cl2 1 mol CCl4 153.8 g Ccl4 70.90 g Cl2 3 mol Cl2 1 mol CCl4 = 72.31 g CCl4 L.R. = theoretical % Yield = Actual 65.0 g CCl4 Theoretical 72.31 g CCl4 ×100 = ×100 = 89.9%

1 MgBr2 + 2 AgNO3 1 Mg(NO3)2 + 2 AgBr Silver bromide (AgBr) was prepared by reacting 200.0 g of magnesium bromide and 100.0 g of silver nitrate. Calculate the theoretical and percent yield if 100.0 g of silver bromide was obtained from the reaction. 1 MgBr2 + 2 AgNO3 1 Mg(NO3)2 + 2 AgBr 200.0g MgBr2 1 mol MgBr2 2 mol AgBr 187.77 g AgBr 184.11 g MgBr2 1 mol MgBr2 1 mol AgBr =408.0g AgBr = 110.5 g AgBr 100.0g AgNO3 1 mol AgNO3 2 mol AgBr 187.77 g AgBr 169.88 g AgNO3 2 mol AgNO3 1 mol AgBr L.R. = theoretical % Yield = Actual 100.0 g AgBr Theoretical 110.5 g AgBr ×100 = ×100 = 90.5%