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STOICHIOMETRY

STOICHIOMETRY. Percent yield. Using your book, look up and write the definitions for these in your notes:. 1. percent yield 2. actual yield 3. theoretical yield. Here are OCP’s definitions.

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STOICHIOMETRY

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  1. STOICHIOMETRY Percent yield

  2. Using your book, look up and write the definitions for these in your notes: 1. percent yield 2. actual yield 3. theoretical yield

  3. Here are OCP’s definitions actual yield: how much product you did recover in the experiment. The actual yield is given in the problem or measured at the end of the lab. (Think of it as how many points you earned on a test.)

  4. Here are OCP’s definitions theoretical yield: how much product you should have been able to recover based on the amounts of the reactants. You have to calculate this using stoichiometry and dimensional analysis. (Think of it as the number of points you should have been able to earn on a test.)

  5. Here are OCP’s definitions Percent yield: the comparison of how much product was made to how much product should have been made multiplied by 100. Let’s consider you last test. The theoretical yield was 38 points (what you should have earned) and let’s say you scored 30 points (actual yield). Your percent yield is …….

  6. The theoretical yield was 38 points (what you should have earned) and let’s say you scored 30 points (actual yield). Your percent yield is: 30 x 100 = 78.9% 38

  7. Look at Example 13 on pg. 229 • There are two measurements in the problem. Place each value under it’s substance in the equation. • CaCO3 CaO + CO2

  8. Look at Example 13 on pg. 229 • There are two measurements in the problem. Place each value under it’s substance in the equation. • CaCO3 CaO + CO2 24.8 g 13.1 g Since 13.1 grams belongs to the product, it is the actual yield.

  9. Look at Example 13 on pg. 229 To calculate the theoretical yield, look at the solution for this problem in the book. The 24.8 grams of CaCO3 (which is the reactant) was converted to grams of CaO (the product we have an actual yield for). The answer to this conversion (13.9 g) is the theoretical yield.

  10. Look at Example 13 on pg. 229 To calculate the percent yield, divide the actual yield given in the problem (13.1 grams CaO) by the calculated theoretical yield (13.9 grams). Always multiple by 100. The answer is 94.2%

  11. Circle up into your groups to solve the problem on the next slide. Group jobs *Reader *Periodic table expert *Calculator *Researcher-looks at example 13 to help guide on this problem. EVERYONE WRITES IN THEIR NOTES!!!

  12. Circle up into your groups to solve the following problem: • When iron and an ore of antimony are heated together, pure antimony and ferrous sulfide are the products. Sb2S3 (s) + Fe (s)  2Sb(s) + FeS (s) When 15.0 grams of Sb2S3 reacts with excess iron, 9.84 grams of Sb are produced. Calculate the percent yield.

  13. And the answer is:

  14. And the answer is: 91.1% Did you get it right? Let’s see how I did it…..

  15. Sb2S3 + 3Fe  2Sb + 3FeS 15.0 g 9.84 g • Since 9.84 grams is the mass of our product, it is the actual yield. Save it till the end. 2. 15.0 grams is the measurement we want to change to the mass of Sb (theortical yield).

  16. Sb2S3 + 3Fe  2Sb + 3FeS 15.0 g 9.84 g 15.O g Sb2S3 x 1 mol Sb2S3 x2 molesSb x 121.8 gSb 1 339.8 g Sb2S3 1 mole Sb 1 mole Sb = 10.75 grams = 10.8 grams (sf) (theoretical yield) Percent yield = 9.84 grams Sb x 100 = 91.1% (sf) 10.8 grams Sb

  17. Homework Time Do problems #19-21 from the 11-3 worksheet showing your work in DA and rounding your answer to the correct number of sig. figs. Due tomorrow. Don’t forget about the problems from last Friday (11-3 worksheet, #1,3,15,16). These are also due tomorrow. Both assignments can be down on a separate sheet of paper.

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