EXPERIMENT 3: PREPARATION OF ALUM: KAl(SO4)2.12H2O In the course lab manual & lab demonstration there is no specification as to what the exact 'theoretical yield' should be, though the equation ' %yield = (mass of alum/theoretical yield) x 100% ' is given. I am assuming that theoretical yield is the weight/mass of initial aluminum foil that was used to begin the experiment (1.1116g), and I also assume the 'mass of alum' represents the mass of aluminum within the final dry weight/mass of the alum crystal (2 KAl(SO4)212 H2O(s)) which is 18.2114g. Though I am not sure if my assumptions are correct. Let me know if you would like more info from the lab assignment. Thanks! Same question answered, I just changed numbers, please show all calculation steps to see where im getting confused