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Highland Science Department Percentage Yield. Highland Science Department Percentage Yield Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible. Highland Science Department Percentage Yield

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## Highland Science Department Percentage Yield

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**Highland Science Department**Percentage Yield**Highland Science Department**Percentage Yield Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible.**Highland Science Department**Percentage Yield Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible. -it's rare to produce the same amount of product as predicted by the balanced chemical equation**Highland Science Department**Percentage Yield Percentage Yield: a comparison of the mass of product actually obtained in an experiment to the amount theoretically possible. -it's rare to produce the same amount of product as predicted by the balanced chemical equation Reasons: -side reactions -reaction does not go to completion -loss of product during separation**Highland Science Department**Percentage Yield % yield = actual yield x 100% theoretical yield**Highland Science Department**Percentage Yield e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield.**Highland Science Department**Percentage Yield e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield. G: actual yield of PbS = 189 g theoretical yield of PbS = 239 g U: % yield S: % yield = actual yield x 100% theoretical yield**Highland Science Department**Percentage Yield e.g. 1. If 189 g of lead (II) sulfide was actually obtained in a reaction for which the theoretical yield was 239 g, calculate the percentage yield. G: actual yield of PbS = 189 g theoretical yield of PbS = 239 g U: % yield S: % yield = actual yield x 100% theoretical yield S: % yield = 189 g x 100% 239 g = 79.1 %**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride.**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. G: 2Na(s) + Cl2(g) 2NaCl(s)MM Cl2 mass of Na = 8.30 g Cl = 2 x 35.4 = 70.8 g mass of Cl2 = 19.5 g MM NaCl actual yield of NaCl = 19.5 g Na = 1 x 23.0 = 23.0 Cl = 1 x 35.4 = 35.4 58.4 g**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. G: 2Na(s) + Cl2(g) 2NaCl(s)M.M. Cl2 mass of Na = 8.30 g Cl = 2 x 35.4 = 70.8 g mass of Cl2 = 19.5 g M.M. NaCl actual yield of NaCl = 19.5 g Na = 1 x 23.0 = 23.0 Cl = 1 x 35.4 = 35.4 U: percentage yield 58.4 g S: 1: limiting reactant = mass ÷ lowest # M.M. 2: limiting limiting NaCl NaCl reactant reactant moles mass mass moles 3: % yield = actual mass x 100% theoretical mass**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. S: Na amount = 8.30g = 0.361 mol Cl2 amount = 14.0g = 0.198 mol**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. S: Na amount = 8.30g = 0.361 mol = 1.82 0.198 mol Cl2 amount = 14.0g = 0.198 mol = 1 0.198 mol**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. S: Na amount = 8.30g = 0.361 mol = 1.82 < 2 0.198 mol limiting Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess 0.198 mol**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. S: Na amount = 8.30g = 0.361 mol = 1.82 < 2 0.198 mol limiting Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess 0.198 mol theoretical = 8.30 g Na mass NaCl = 8.30 x 58.4 g NaCl 23.0 = 21.1 g**Highland Science Department**Percentage Yield e.g. 2. If 8.30 g of sodium and 14.0 g of chlorine are heated together, a total of 19.5 g of sodium chloride is isolated. Determine the percentage yield of sodium chloride. S: Na amount = 8.30g = 0.361 mol = 1.82 < 2 0.198 mol limiting Cl2 amount = 14.0g = 0.198 mol = 1 = 1 excess 0.198 mol theoretical = 8.30 g Na mass NaCl = 8.30 x 58.4 g NaCl 23.0 = 21.1 g % yield = 19.5 g x 100% 21.1 g = 92.4 %

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