Percentage Yield

Theoretical Yield – amount of product that is predicted using stoichiometry Actual Yield – amount of product that is obtained in an experiment Percent Yield – compares the mass of product obtained by experiment with the mass of product determined by stoichiometric calculations

Reasons For Low Yield Sources of error. (Experimental procedures) Impurities in reagents used (different grades of chemicals) Side reactions (other products formed) Reactions are reversible

Reasons for High Yield 1. Sources of error (experimental procedures)

Solving Percent Yield Problems First, determine the theoretical yield (how much you should get) using stoichiometry Second using the experimental/ actual yield, use the formula. % yield = Experimental yield x 100% theoretical yield

Example 1: In a particular experiment 10 Example 1: In a particular experiment 10.0 g of sugar should be produced but only 0.664 g is produced. What is the percentage yield? % yield = x 100 = x 100 = 6.64 % Experimental yield theoretical yield 0.664 g 10.0g

Step 1: Write out the balanced chemical equation Example 2: Aluminium reacts with oxygen to form aluminium oxide. If 635g of Aluminium oxide is obtained from reacting 1150g of aluminium, what is the percentage yield? Step 1: Write out the balanced chemical equation Al + O2  Al2O3 4Al + 3O2  2Al2O3

Step 2: Convert given mass into moles (n=m/M) Part 1: Find the theoretical yield through stoichmetry (how much you should have got) Step 2: Convert given mass into moles (n=m/M) Moles of Al = 1150g 27.0g/mol = 42.59 mol of Al

Step 3: Use moles of given substance to find moles of required substance (mole to mole ratio) 4Al + 3O2  2Al2O3 x mols of Al2O3 = 2 mol of Al2O3 42.59 mol of Al 4mol of Al = 21.31 mols of Al2O3

Step 4: Convert moles of required substance to the mass (m=n x M) mass of Al2O3 = 21.31 mol of Al2O3 X 102g/mol = 2173 g of Al2O3 This is your theoretical yield

Part 2: Calculate Percent Yield Step 6: Using experimental and theoretical yield find percent yield % yield = x 100 = x 100 = 29.2 % Experimental yield theoretical yield 635 g 2173g

Step 1: Write out the balanced chemical equation Example 3: Iron (III) oxide reacts with carbon monoxide to produce carbon dioxide and iron. If 300g of iron is produced when 425 of iron ore is used. What is the percentage yield? Step 1: Write out the balanced chemical equation Fe2O3 + 3CO  2Fe + 3CO2 m= 425g m= ?

Step 2: Convert given mass into moles (n=m/M) Part 1: Find the theoretical yield through stoichmetry (how much you should have got) Step 2: Convert given mass into moles (n=m/M) Moles of Fe2O3 = 425g 159.7g/mol = 2.66 mol of Fe2O3

Step 3: Calculate the number of mols of the required substance (mol to mole ratio) Fe2O3 + 3CO  2Fe + 3CO2 x mols of Fe = 2 mol of Fe 2.66 mols of Fe2O3 1mol of Fe2O3 = 5.32 mols of Fe

Step 4: Convert moles of required substance to the mass mass of Fe = 5.32 mols of Fe X 55.85g/mol of Fe = 297g of Fe

Part 2: use percent yield formula Step 6: Using experimental and theoretical yield find percent yield % yield = x 100 = x 100 = 101 % Experimental yield theoretical yield 300 g 297g

Homework p. 319 #51-60