Chapter 7: 7.3 Reaction Yields

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Presentation transcript:

Chapter 7: 7.3 Reaction Yields Percentage Yield

Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product calculated using the balanced equation. (Calculated on paper!) Actual yield The amount of product obtained when the reaction takes place. (Actually found in the lab) Percent yield The ratio of actual yield to theoretical yield. percent yield = mass of actual yield (g) x 100% mass of theoretical yield (g)

Calculating Percent Yield To calculate the percent yield, the actual yield and theoretical yield are needed. You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burn and you have to throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies

a) What is the theoretical yield if 5 a) What is the theoretical yield if 5.50 grams of hydrogen react with nitrogen to form ammonia? H2 + N2 -> NH3 3 2 5.50 g H2 1 mole H2 2 moles NH3 17.04 g NH3 1 mol NH3 2.02 g H2 3 moles H2 = 30.9 g NH3 = Theoretical Yield!!!!

b) What is the percent yield if only 20 b) What is the percent yield if only 20.4 grams of ammonia is actually produced in the lab? 20.4 g % Yield = X 100% = 65.96 % yield 30.93 g

Learning Check Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O2(g) 2CO(g) What is the percent yield if 40.0 g CO are produced when 30.0 g O2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

Solution 3) 76.2 % yield theoretical yield of CO percent yield

Learning Check When N2 and 5.00 g H2 are mixed, the reaction produces 16.0 g NH3. What is the percent yield for the reaction? N2(g) + 3H2(g) 2NH3(g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

Solution 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 5.00 g H2 x 1 mol H2 x 2 mol NH3 x 17.03 g NH3 2.016 g H2 3 mol H2 1 mol NH3 = 28.2 g NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.7 % 28.2 g NH3

Reasons for low % Yield Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made. WHY? Experimental design- Can you fix the method? Experimental errors/human error(loss when transferring, or filtering) Impurities in the reactants (most laboratory chemicals are only 80-90% pure) Side reactions