Percentage Yield.

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Presentation transcript:

Percentage Yield

What is Yield? It’s the amount of product that is produced in a chemical reaction  actual yield. What you get in a chemical reaction may not be what you were supposed to get. The theoretical yield tells you how much product you should get according to the balanced equation.

Factors affecting Yield Splattering when heating a solution Filtering may not get all of the product Using impure/contaminated reactants Not letting wet solids dry properly Use of equipment (old or faulty) Side chemical reactions (eg. When you burn Mg in the air. It reacts with O2. However, air also has nitrogen, so Mg may react with N2 too!)

Percentage Yield The comparison between the actual yield and theoretical yield of a product in a reaction. Percentage yield = Actual Yield x 100% Theoretical Yield

Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g) Example 1 Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas, as seen below: Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g) Calculate the theoretical yield and percentage yield of ZnCl2 if 1.541 g of ZnCl2 (aq) is produced when 0.999 g of HCl (aq) is used.

2. Find # moles of product, using mole ratio Balanced Eq’n Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g) Given Mass (g) Molar Mass (g/mol) 0.999 g 1.541 g (actual) 36.46 g/mol 136.29 g/mol 1. Find # moles reactant nHCl = 0.999 g = 0.0274 mol 36.46 g/mol 2. Find # moles of product, using mole ratio nZnCl2 = 0.0274 mol HCl x __________ = 1 mol ZnCl2 2 mol HCl 0.0137 mol

3. Find mass of product mZnCl2 = n x M = 0.0137 mol x 136.29 g/mol = 1.86 g  theoretical yield 4. Calculate % yield % yield = actual yield x 100% theoretical yield = 1.541 g x 100% 1.86 g = 82.8%

Example 2 In an experiment, 16.1 g of iron sulfide, FeS (s), is added to oxygen. 14.1 g of iron(III) oxide, Fe2O3 (s) is produced. Calculate the theoretical yield and percentage yield of Fe2O3 (s). 4FeS (s) + 7O2 (g) → 2Fe2O3 (s) + 4SO2 (g)

2. Find # moles of product, using mole ratio Balanced Eq’n 4FeS (s) + 7O2 (g) → 2Fe2O3 (s) + 4SO2 (g) Given Mass (g) Molar Mass (g/mol) 16.1 g 14.1 g (actual) 87.92 g/mol 159.7 g/mol 1. Find # moles reactant nFeS = 16.1g = 0.183 mol 87.92 g/mol 2. Find # moles of product, using mole ratio nFe2O3 = 0.183 mol FeS x __________ = 2 mol Fe2O3 4 mol FeS 0.0915 mol

3. Find mass of product mFe2O3 = n x M = 0.0915 mol x 159.7 g/mol = 14.61 g  theoretical yield 4. Calculate % yield % yield = actual yield x 100% theoretical yield = 14.1 g x 100% 14.61 g = 96.5%