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Published byGerald Snow Modified over 9 years ago
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Percentage Yield of a Chemical Reaction
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Let’s look at your last Chemistry Test You scored 32/40. What’s your % grade? (32/40) * 100% = 80% What is the “theoretical” grade on this test? (theoretical = highest possible grade) 40 What was your “actual” grade? 32
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What has this to do with Chemistry? Theoretical yield of a chemical reaction is predicted by stoichiometry. The amount of product obtained by the chemist is the actual yield.
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Actual yields are often less than theoretical yields due to competing (side) reactions loss of product due to poor lab technique chemical equilibrium (See y’all next year!) impure reactants
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Actual yields can also be greater than theoretical yields due to an impure or contaminated product a solid product that hasn’t been sufficiently dried
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Calculating Percentage Yield % yield=actual yield* 100% theoretical yield Use same units for actual yield as for the theoretical yield. ie. grams/grams; mol/mol, etc
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Sample Problem 1 Determine the % yield when 1.7 g of NH 3 are produced from the reaction of 7.5 g N 2 with sufficient H 2, according to: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Steps to solve this problem: 1.Determine theoretical yield using stoich. in units of grams. 2. Calculate % yield.
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Theoretical yield is 1 mol N 2 (g):2 NH 3 (g) 7.5 g ↓(/28.0 g/mol) 0.27 mol (x 2/1) 0.54 mol ↓x 17.0 g/mol 9.1 g NH 3 is the theoretical yield
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% =(actual/theoretical) * 100% = (1.7g/9.1g) * 100% =19% (to two sf) Does this answer make sense?
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Sample Problem 2 Calcium carbonate, CaCO 3, thermally decomposes to produce CaO and CO 2 according to CaCO 3 (s) CaO(s) + CO 2 (g) If the reaction proceeds with a 92.5% yield, what volume, at SATP, of CO 2 can be expected if 12.4 g CaCO 3 is heated?
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1 mol CaCO 3 (s):1 mol CO 2 (g) 12.4 g ↓/ 100.1 g.mol 0.123 mol (x 1/1) 0.123 mol ↓ x 24.0 L/mol 2.97 L theoretical ↓ x 0.925 yield 2.75 L actual yield at SATP
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p 262 PP 31 – 33 p 264 PP 34 – 37 Homework... there’s more...
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Percentage Purity Impure reactants are often the cause of less than 100% yield. For example, the reactant you massed is only 70% pure. What will this do to the % yield? Yield will be 70%.
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Sample Problem 3 Iron pyrite (fool’s gold) has the formula FeS 2. When a 13.9 g sample of impure iron pyrite is heated in the presence of oxygen, O 2, 8.02 g of Fe 2 O 3 is produced according to: 4 FeS 2 (s) + 11 O 2 (g) 2 Fe 2 O 3 (s) + 8 SO 2 (g) What is the % purity of the iron pyrite sample?
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4 FeS 2 (s) + 11 O 2 (g) 2 Fe 2 O 3 (s) + 8 SO 2 (g) 4 mol FeS 2 :2 mol Fe 2 O 3 8.02 g ↓/159.7 g/mol 0.100 mol (X4/2) 0.0502 mol ↓*120.0 g/mol 12.0 g FeS 2 This represents the mass of pure FeS 2. % purity= (12.0 g/13.9 g) * 100% = 86.3% is the purity of iron pyrite
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Homework PP #38, 39, 40 on p 269 SR #1 – 4 on p 270 Get started on Ch 7 review problems.
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