Percentage yield is defined as the actual mass of product divided by the theoretical maximum mass of product, multiplied by 100%. We can similarly express percentage yield in terms of moles.
There are essentially three types of questions:
| 1. Calculate the % yield for a reaction Here, you would divide the actual moles of product by the theoretical maximum moles of product, and multiply by 100%. e.g., The actual moles of product was 10 mol and the theoretical maximum moles of product was 20 mols. Percentage yield = 10 / 20 x 100% = 50% 2. Calculate the actual moles of product Here, you would multiply the theoretical maximum moles by the percentage yield, and divide by 100% e.g., The theoretical maximum moles of product was 20 mols and the percentage yield was 50%. Actual moles of product = 20 x 50% / 100 = 10 mols. 3. Calculate the theoretical maximum moles of product Here, you would multiply the actual moles of product and 100%, and divide by the percentage yield. e.g., The actual moles of product was 10 mol and the percentage yield was 50%. Theoretical maximum moles of product = 10 x 100% / 50% = 20 mol |
The final type of calculation is especially interesting because, when combined with a balanced symbol equation it allows us to work out how much reactant is required to make a given actual amount of product!
| Question: Buta-1,3-diene can be formed by the reaction of 2,3-dibromobutane with NaOH. Reaction: C4H8Br2 + 2 NaOH -> C4H6 + 2 NaBr + 2 H2O The percentage yield for the reaction is 80% and the chemist needs to make 10.8 g of buta-1,3-diene. Determine the masses of 2,3-dibromobutane and NaOH required. Answer A useful first step would be to work out the Mr of C4H6 (54) and C4H8Br2 (216). n C4H6 =10.8 / 54 = 0.2 mol (i.e., actual moles of product) n C4H6 =0.2 x 100% / 80% = 0.25 mol (i.e., theoretical maximum) Using the balanced symbol equation, the initial moles of C4H8Br2 and NaOH can now be calculated: n C4H8Br2 =1 x 0.25 = 0.25 mol n NaOH =2 x 0.25 = 0.50 mol Using the Mr of C4H8Br2 (216) and NaOH (40), we can now work out the required masses. m C4H8Br2 = 0.25 x 216 = 54 g n NaOH =0.50 x 40 = 20 g |