Percent Yield Notes

New Vocabulary

  • Theoretical yield: Maximum product from reactant amount.
  • Actual yield: Product obtained in an experiment.
  • Percent yield: Ratio of actual to theoretical yield, expressed as a percentage.

How Much Product?

  • Reactions rarely produce the predicted product amount due to:
    • Adherence to container surfaces or evaporation.
    • Competing reactions.
    • Loss during purification or on filter paper.

Theoretical vs. Actual Yield

  • Theoretical yield: Maximum product from reactant amount.
  • Actual yield: Product obtained experimentally.
  • Percent yield: ActualYieldTheoreticalYield×100\frac{Actual Yield}{Theoretical Yield} × 100%

Percent Yield Example

  • Reaction: 2AgNO<em>3(aq)+K</em>2CrO<em>4(aq)Ag</em>2CrO<em>4(s)+2KNO</em>3(aq)2AgNO<em>3(aq) + K</em>2CrO<em>4(aq) \rightarrow Ag</em>2CrO<em>4(s) + 2KNO</em>3(aq)
  • Problem: 0.500 g AgNO<em>3AgNO<em>3 yields 0.455 g Ag</em>2CrO4Ag</em>2CrO_4.
  • Convert grams of AgNO3AgNO_3 to moles: 0.500g×1mol169.9g=2.94×103mol0.500 g \times \frac{1 mol}{169.9 g} = 2.94 × 10^{-3} mol
  • Convert moles of AgNO<em>3AgNO<em>3 to moles of Ag</em>2CrO<em>4Ag</em>2CrO<em>4: 2.94×103mol×1molAg</em>2CrO<em>42molAgNO</em>3=1.47×103mol2.94 × 10^{-3} mol \times \frac{1 mol Ag</em>2CrO<em>4}{2 mol AgNO</em>3} = 1.47 × 10^{-3} mol
  • Calculate theoretical yield of Ag<em>2CrO</em>4Ag<em>2CrO</em>4: 1.47×103mol×331.7g1mol=0.488g1.47 × 10^{-3} mol \times \frac{331.7 g}{1 mol} = 0.488 g
  • Calculate percent yield: 0.455g0.488g=93.2\frac{0.455 g}{0.488 g} = 93.2%

Marketplace Importance

  • Percent yield impacts cost-effectiveness in industrial processes.
  • Industry aims for high yield, speed, and low cost.