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Percent yield

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Percent yield

  1. 1. Percent Yield April 14, 2010
  2. 2. <ul><li>In-Class Assignment </li></ul><ul><li>Read the section entitled Percent Yield on page 293 in your textbook. </li></ul><ul><li>Define the following terms in your notes: theoretical yield, actual yield and percent yield. Then, write the equation for percent yield. </li></ul><ul><li>Write down the following statements, and then answer. Give at least two reasons why the actual yield in a chemical reaction often falls short the theoretical yield. Give one real-life example illustrating how a company’s actual yield of products falls short of their theoretical yield. </li></ul><ul><li>*You may use the Internet to find your answers. </li></ul>
  3. 3. <ul><li>Read the sample problem below. </li></ul>Sample Problem: If the actual yield equals 15.0 g and the theoretical yield is 20.0 g, calculate the percent yield. Step 1: Write the equation for percent yield. Step 2: Substitute the given values for actual yield and theoretical yield. Step 3: Multiply your decimal by 100 to obtain a percent. % yield = x 100% x 100% actual theoretical 15.0 g 20.0 g = 75.0% =
  4. 4. Read the more challenging sample problem below. Sample Problem: 16 g of H 2 reacts with excess O 2 to produce 138 g of H 2 O. What is the percent yield for this chemical reaction? Step 1: Write the balanced chemical equation (if it isn’t already given to you). 2H 2 + O 2  2H 2 O Step 2: Determine the actual and theoretical yield. The actual yield is given in the problem. Actual yield = 138 g of H 2 O
  5. 5. The theoretical yield is calculated (use Mass- Mass Stoichiometry from your arrow sheet). Continued… Step 3: Calculate the percent yield. The theoretical yield is 143 g of H 2 O. 16 g H 2 % yield = x 100% x 100% 1 mol H 2 2.02 g H 2 x 2 mol H 2 O 2 mol H 2 x 18.02 g H 2 O 1 mol H 2 O x 143 g = actual theoretical 138 g H 2 O 1 4 3 g H 2 O = 96.7% =
  6. 6. <ul><li>Now, try these percent yield questions on your own: </li></ul><ul><li>If the actual yield equals 4.75 g and the theoretical yield is 5.00 g, then calculate the percent yield. Show your work. </li></ul><ul><li>If the percent yield equals 90.0% and the theoretical yield is 1.0 g, calculate the actual yield. </li></ul><ul><li>When 300 g of potassium chlorate, KClO 3 , is heated, 107 g of oxygen, O 2 , is produced according to the following equation: </li></ul><ul><li>2 KClO 3  2 KCI + 3 O 2 </li></ul><ul><li>Calculate the percent yield. </li></ul>
  7. 7. <ul><li>Continued… </li></ul><ul><li>The electrolysis of water forms H 2 and O 2 . </li></ul><ul><li>2 H 2 O  2 H 2 + O 2 </li></ul><ul><li>What is the % yield of O 2 if 12.3 g of O 2 is produced from the decomposition of 14.0 g H 2 O? </li></ul><ul><li>Iron pyrites (FeS 2 ) reacts with oxygen according to the following equation: </li></ul><ul><li>4 FeS 2 + 11 O 2  2 Fe 2 O 3 + 8 SO 2 </li></ul><ul><li>If 300 g of iron pyrites is burned in 200 g of O 2 , </li></ul><ul><li>143 g of ferric oxide, Fe 2 O 3 , is produced. What is the percent yield of ferric oxide? </li></ul>
  8. 8. References: <ul><li>I just modified slides found online at http://www.chalkbored.com/lessons/chemistry-11/percent-yield.ppt#2 . All credit goes to the teacher who created and uploaded this lesson to the web. </li></ul>

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