PERCENT YIELD Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product that can be formed. calculation (Stoichiometry) Actual yield The amount of product obtained when the reaction takes place. (In lab) Percent yield percent yield = actual yield (g) x 100 theoretical yield (g)

Calculating Percent Yield You prepared cookie dough to make 5 dozen cookies. The phone rings and while you are 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

PRACTICE PROBLEMS If 8.87 grams of As2O3 react with excess carbon, how many grams of As will be produced? 2 As2O3 + 3 C → 3CO2 + 4 As

8.87 g As2O3 x 1mol As2O3 x 4 mol As x 75 gAs 198g As2O3 2mol As2O3 1mol As 6.72 g

B) In lab, a student collects 5. 33 grams of As. What is the % yield B) In lab, a student collects 5.33 grams of As. What is the % yield? % yield = 5.33 g X 100 6.72 g 79.3 %

2) How many grams of chlorine will be produced by the decomposition of 585 grams of NaCl? 2 NaCl → 2Na + Cl2 585 g NaCl x 1 mol NaCl x 1mol Cl2 X 71 g Cl2 58.5 g NaCl 2mol NaCl 1 mol Cl2 355g

B) What is the % yield of chlorine if 318 grams of Cl2 are collected B) What is the % yield of chlorine if 318 grams of Cl2 are collected? % yield = 318 g x 100 355 g 89.6 %

3) Calculate the % yield of W if 56. 9 g of WO3 produces 41. 4 g of W 3) Calculate the % yield of W if 56.9 g of WO3 produces 41.4 g of W. WO3 + 3 H2 → W + 3 H2O

56.9 g WO3 x 1 mol WO3 x 1 mol W x 184 g 232 g WO3 1mol WO3 1 mol W 45.1 g (theoretical) % yield = 41.4 g x100 45.1 g 91.8%

Practice 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 30.0 g O2 x 1 mol O2 x 2 mol CO x 28 g CO 32.00 g O2 1 mol O2 1 mol CO = 52.5 g CO (theoretical) percent yield 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

Practice 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 17.0 g NH3 2.0 g H2 3 mol H2 1 mol NH3 = 28.3 g NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.5 % 28.3 g NH3