- 127 Views
- Uploaded on
- Presentation posted in: General

Highland Science Department Percentage Yield

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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) + Cl2(g) 2NaCl(s)MM Cl2

mass of Na = 8.30 gCl = 2 x 35.4 = 70.8 g

mass of Cl2 = 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) + Cl2(g) 2NaCl(s)M.M. Cl2

mass of Na = 8.30 gCl = 2 x 35.4 = 70.8 g

mass of Cl2 = 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: limitinglimitingNaCl NaCl

reactantreactant molesmass

massmoles

3: % yield = actual mass x 100%

theoretical mass

Highland Science Department

Percentage Yield

S:Na amount=8.30g = 0.361 mol

Cl2 amount =14.0g = 0.198 mol

Highland Science Department

Percentage Yield

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

S:Na amount=8.30g = 0.361 mol= 1.82 < 2

0.198 mollimiting

Cl2 amount =14.0g = 0.198 mol= 1 = 1 excess

0.198 mol

Highland Science Department

Percentage Yield

S:Na amount=8.30g = 0.361 mol= 1.82 < 2

0.198 mollimiting

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

S:Na amount=8.30g = 0.361 mol= 1.82 < 2

0.198 mollimiting

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 gx 100%

21.1 g

= 92.4 %