# Highland Science Department Percentage Yield. Highland Science Department Percentage Yield Percentage Yield: a comparison of the mass of product actually.

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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 gx 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) + Cl 2(g)  2NaCl (s) MM Cl 2 mass of Na = 8.30 gCl = 2 x 35.4 = 70.8 g mass of Cl 2 = 19.5 gMM NaCl actual yield of NaCl = 19.5 gNa = 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) + Cl 2(g)  2NaCl (s) M.M. Cl 2 mass of Na = 8.30 gCl = 2 x 35.4 = 70.8 g mass of Cl 2 = 19.5 gM.M. NaCl actual yield of NaCl = 19.5 gNa = 1 x 23.0 = 23.0 Cl = 1 x 35.4 = 35.4 U:percentage yield58.4 g S:1: limiting reactant =mass÷lowest # M.M. 2: limiting  limiting  NaCl  NaCl reactantreactant molesmass massmoles 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 Cl 2 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 Cl 2 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 mollimiting Cl 2 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 mollimiting Cl 2 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 mollimiting Cl 2 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 gx 100% 21.1 g = 92.4 %

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