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Percent Yield. Percent yield describes the efficiency of a chemical reaction 100 free throws Theoretically… Actually…

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Presentation on theme: "Percent Yield. Percent yield describes the efficiency of a chemical reaction 100 free throws Theoretically… Actually…"— Presentation transcript:

1 Percent Yield

2 Percent yield describes the efficiency of a chemical reaction 100 free throws Theoretically… Actually…

3 Chemical reactions also have theoretical and actual outcomes Reasons for not producing theoretical amounts: – Reactants or products might stick to container

4 Reasons for not producing theoretical amounts: – Reactants or products might evaporate

5 Reasons for not producing theoretical amounts: – Other unintended products might form

6 Reasons for not producing theoretical amounts: – Purification process might reduce reactant amounts

7 Because of these problems, scientists need to know how to gage the yield of a chemical reaction

8 Theoretical yield Maximum amount of product that can be produced from a given amount of reactant N 2 + 3 H 2 → 2 NH 3

9 Actual yield Amount of product produced when a chemical reaction is carried out in an experiment

10 Popcorn If there are 100 kernels in a bag… – What is the theoretical yield? – What is the actual yield?

11 Percent yield Ratio of the actual yield to the theoretical yield shown as percent

12 Percent yield If there are 105 kernels in a bag of popcorn and 13 were left unpopped, what is the percent yield of that bag? 92/105 = 87.6 %

13 Now apply this to chemistry… Solid silver chromate (Ag₂CrO₄) forms when potassium chromate (K₂CrO₄) is added to a solution containing 0.500 g of silver nitrate (AgNO₃). Determine the theoretical yield of Ag₂CrO₄. Write out the equation and balance it!

14 2AgNO₃ + K₂CrO₄ = Ag₂CrO₄ +2KNO₃ Change grams to moles then use mole ratios to find out how much product it will produce 0.500 g / 169.9 g = 0.00294 mol 0.00294 x 1 mol Ag₂CrO₄ = 0.00147 mol Ag₂CrO₄ 2 mol AgNO₃ 0.500 g?

15 2AgNO₃ + K₂CrO₄ = Ag₂CrO₄ +2KNO₃ Change moles to grams 0.00147 mol Ag₂CrO₄ x 331.7 g = 0.488 g Ag₂CrO₄ 0.500 g? Theoretical yield

16 If the experiment is performed and the actual yield is 0.455 g Ag₂CrO₄, what is the percent yield? 0.455 g 0.488 g x 100 = 93.2% Ag₂CrO₄

17 Popcorn time!!!


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