Determining % Yield of a Chemical Reaction

• 1). Write a Balanced Chemical Equation for the Reaction.
  
  Suppose you burn 20 grams of powdered aluminum (Al) in an airtight vessel containing 30 grams of oxygen and recover 58.25 grams of purified aluminum oxide (Al2O3). The balanced equation for this reaction is:
  
  2Al + 3O2 ------> 2Al2O3
  
  
• 2). Determine the Molecular Weights of Each of the Reactants and the Product of Interest. The molecular weight of each of the reactants is the number of each of the atoms in each compound times the atomic weight of the atom. Atomic weights can be found on most periodic tables. For example, for Al2O3, the atomic weights are aluminum (Al) 26.98 g/mole; and oxygen (O)16.00 g/mole. The molecular weight of aluminum oxide is 2 x 26.98 + 3 x 16 = 101.96 g/mole. Similarly, the weight of molecular oxygen (O2) is 32.00 g / mole.
  
  
• 3). Determine the of Limiting Reagent. Determine the number of moles of each reactant present. Divide the number of grams added by the molecular weight of the reactant. For elemental aluminum, you have 20 g / 26.98 g/mole = 0.7413 moles. For molecular oxygen you have 30g / 32g/mole = 0.9375 moles.
  
  
• 4). Determine the Stoichiometrically Limiting Reagent. Stoichiometrically limiting means which material limits the amount of product that can be produced on a molar basis. At first glance, there are fewer moles of aluminum than molecular oxygen, so one could wrongly conclude that aluminum is the limiting reagent. However, one must look at the coefficients in the balanced equation. For every two moles of aluminum, three moles of molecular oxygen are required to produce two moles of silver chloride. Determine the mole ratio of the materials in the equation. Divide the coefficients in the balanced equation by the smallest coefficient (in this example 2).
  
  This yields:
  
  1 Al + 1.5 O2 1 Al2SO3, or a mole ratio of 1:1.5:1
  
  This means for every mole of aluminum provided, 1.5 moles of oxygen are required to convert it to all to aluminum oxide. In this example, you have 0.7413 moles of aluminum and need 1.5 x 0.7412 = 1.1120 moles of molecular oxygen for 100% conversion. You only have 0.9375 moles of molecular oxygen. Thus, molecular oxygen is the stoichiometrically limiting reagent.
  
  
• 5). Determine Theoretical Yield. Use the mole ratio of the stoichiometrically limiting reagent to the product to determine the theoretical number of moles of product that can be produced. In this example, the mole ratio of oxygen to aluminum oxide is 1.5:1. For every 1.5 mole of molecular oxygen, only 1 mole of aluminum oxide is produced. Since there are 0.9375 moles of molecular oxygen, at most, only 0.9375 / 1.5 = 0.6250 moles of aluminum oxide can be produced.
  
  
• 6). Convert the Theoretical Yield in Moles to Grams. Multiply the maximum number of moles of product by the product molecular weight to get the theoretical yield in grams. In this example, 0.6250 moles x 101.96 g/mole = 63.73 grams of Al2O3.
  
  
• 7). Calculate the Percent Yield. Divide the amount of product recovered in grams by the theoretical yield in grams and multiply by 100 to get the per yield. In this case 58.25g/63.75g x 100 = 91.41% yield.