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Understanding Percent Yield in Chemistry

Learn the concept of percent yield, its calculation, and importance in chemistry reactions with practical examples. Find out the theoretical and actual yield, and how to determine the percent yield accurately.

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Understanding Percent Yield in Chemistry

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  1. Chapter 9, section 3, part 2 Percent Yield

  2. Why percent yield? • Usually, not all the product possible is actually formed. • theoretical yield • maximum amount of product possible • is calculated using limiting reactant and stoichiometry • actual yield • the measured amount formed in lab reaction • always less than or equal to theoretical yield

  3. Example 1 • If 72.0 g of C2H2 reacts with an excess of Br2 and 729 g of the product is recovered, what is the percent yield? C2H2 + Br2 CHBr2CHBr2 First calculate theoretical yield: 72.0 g C2H2 1 mol C2H2 1mol CHBr2CHBr2 345.64 g CHBr2CHBr2 26.04 g C2H2 1 mol C2H2 1 mol CHBr2CHBr2 = 956 g CHBr2CHBr2 Then calculate percent yield: 729 g 100% = 76.3% 956 g

  4. Example 2 • If the percent yield of the reaction below is 73.8% and the reaction began with 24.3 g of CaO, how many grams of Ca(OH)2 were created? CaO + H2O  Ca(OH)2 First, calculate theoretical yield: 24.3 g CaO 1 mol CaO 1 mol Ca(OH)2 74.09 g Ca(OH)2 = 32.1 g Ca(OH)2 56.07 g CaO 1 mol CaO 1 mol Ca(OH)2 Then use given percent yield to calculate actual yield: 32.1 g Ca(OH)2 x 0.738 = 23.7 g Ca(OH)2

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