1. The problem statement, all variables and given/known data Hello everyone, I have completed a lab and I’m working on my report. I’m not sure if my work is correct but I would most definitely appreciate it if anyone could check it for me? I’m particularly having trouble finishing question #3. It’s worth a lot of my grade so I want to make sure I get it right! Thanks a lot in advance for all the time and help :) Basically, the lab is a set of 2 redox titrations. For the first part, we standardized a KMnO₄ solution and in the second we titrated K₃[Fe(C₂O₄)₃]∙3H₂O 2. Relevant equations 2MnO₄²¯(aq) + 5C₂O₄²¯(aq) + 16H⁺(aq) à 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l) 3. The attempt at a solution 1. Given a mass of 0.2349 g of Na₂C₂O₄ in 100 mL, what is the concentration of Na₂C₂O₄? A:// 0.2349 g Na₂C₂O₄ = 1 mol Na₂C₂O₄ / 0.1 L = 0.0175 M 2. You used 6.9933 mL of KMnO₄, what is the concentration of the KMnO₄ solution? A:// 0.00175 mol Na₂C₂O₄ x 2 mol KMnO₄ / 5 mol Na₂C₂O₄ = 0.0007 mol KMnO₄ / 0.0069933 L = 0.1001 M 3. You dissolved 0.2196 g of K₃[Fe(C₂O₄)₃]∙3H₂O in 100 mL. Given the concentration of KMnO₄ you just calculated and assuming that you used 7.1667 mL of it: determine the Mass Percent Composition of C₂O₄²¯ in K₃[Fe(C₂O₄)₃]∙3H₂O. A:// 7.1667 mL x 1L / 1000 mL = 0.0071667 L x 0.1001mol / 1L = 0.000717386 mol KMnO₄ 0.000717386 mol KMnO₄ x 5 mol Na₂C₂O₄ / 2 mol KMnO₄ = 0.0017935 mol Na₂C₂O₄ x 491.243 g/mol = 0.08810 g Na₂C₂O₄ And then I’m stuck… 4. Compare your experimentally determined % oxalate in your product to theoretically mass percentage of C₂O₄²¯ in K₃[Fe(C₂O₄)₃]∙3H₂O. Use the following equation to aid in your discussion: Experimentally determined % oxalate / Theoretically determined % oxalate x 100% This seems fairly straight forward if I could correctly figure out #3…however, what would my theoretical mass percent composition come from? Is it from the 0.2349 g of Na₂C₂O₄? I know I'm not suppose to post more than one question per post but they are all related and if I made a mistake in the previous one then I carry it forward to the next; so pardon my breach of protocol.