1. The problem statement, all variables and given/known data A 16.00g sample of a mixture of KClO3 and KCl is heated until all the oxgen has been removed from the sample. The product entirely KCl (no O), has a total mass of 9.00g. What is the percentage of KClO3 present in the original mixture? 16.00g mixture 9.00g mixture without oxygen 3.00g oxygen without mixture 2. Relevant equations Percent Composition (percent = mass of element/mass of compound) Various mole equations (eg. moles to mass) 3. The attempt at a solution I received this question on a test awhile back and, while I have received the answer and the steps to reach it, I cannot understand why it is that it is done that way. From what I can see, we do not need to worry about percent composition of the mixture until the end. At the moment, we do not have KCl by itself, as O3 is bonded to it. I think I need to find the mass of KCl in the original mixture, and so to do that, I chose to calculate the moles in KCl and O. 9.00gKCl/(75g/mol) = 0.12molKCl 3.00gO/(16g/mol)= 0.1875mol O By this point, I had already figured that this was likely going nowhere, as I still did not know the amount of KCl in KClO3 by itself, but rather just the moles of KCLO3+KCL without the oxygen. ------------------------------ Next, is my teacher's solution. gKClO3 = 3.00gO x (1.22gKClO3/48.9gO) = 7.6625 % = (7.6625/12.00) x 100% = 63.9% I understand the percent part, but what I do not understand is why 1.22gKClO3 is divided by 48.0gO. If I could have insight onto that, then I would be most appreciative. Thanks for reading my wall of text!