bwf8398 3 Report post Posted April 30, 2016 I was recently doing some practice problems for my IB chemistry SL test and ran across this problem. "A 2.450 g sample of a mixture of NaCl and CaCl_{2} was dissolved in distilled water. The chloride solution was treated with excess AgNO_{3} (aq) solution. The precipitated AgCl (s) was collected, washed, and dried. The mass of the dried AgCl was 6.127 g. Calculate percent by mass of the NaCl and CaCl_{2} in the original mixture." (2 marks) AgCl has a molar mass of 143.45 g/mol, so I calculated that the amount of Cl in the sample was 6.127*(35.45/143.35) = 1.515 g Cl This means that together, there has to be (2.450 - 1.515) = .935 g of Na and Ca combined. At this point, I made a system of equations. Let x represent the number of moles of Na and let y represent the number of moles of Ca. .935 = 23x + 40y 1.515 = 35.45x + 70.90y The first equation expresses the mass of the Na and Ca, and the second expresses the mass of Cl (35.45 is doubled since there are 2 Cl for every Ca.) Ultimately, solving the equation gave me that x = .027 which led to there being 1.58g NaCl. This made the percent composition 64.5% NaCl and 35.5% CaCl_{2} My question is this: is there some really simple way to do this problem that I overlooked? It's only worth 2 marks, and it took me about 10 minutes to figure out how to solve it (and I might have made some mistakes in my solution). In case it matters, this is from an IB Chem SL Pearson book, not a previous IB chem paper. Share this post Link to post Share on other sites

kw0573 1,241 Report post Posted April 30, 2016 You do not want to solve systems of equations on the chem exam! I don't really have an easy way, but I do have a tip on naming variables, which simplifies some work. You should define only one variable such that either it's what you are finding, or you can get to there in one step. A good variable would be % of NaCl by mass. Here I am going to use mass of NaCl to be m. So that 2.45 - m = mass of CaCl_{2.}So that once we know m, we can just do m/2.45 * 100% to get the answer. Start with some relationship: mol (NaCl) + 2*mol(CaCl_{2}) = mol (Cl^{-}). Plug values in: m / (22.99 + 35.45) + 2*(2.45-m) / (40.08 + 2*35.45) = 6.127 / (143.32) = 0.04275. I used slightly different values but the final answer didn't change much. ^ At this point don't find common denominator just multiply everything by (40.08 + 2* 35.45). Here you should note that having unknowns in the numerator is so much better than the denominator. I got a similar answer in the end. In your solution, you used mole ratio implicitly but in mine I just substituted everything into this simple relationship. So ideally you want to use a minimal number of relationships that ties everything together. Mole ratio is often a better one than conservation of mass. Share this post Link to post Share on other sites

kw0573 1,241 Report post Posted April 30, 2016 You are able to use a graphing calculator on chemistry paper 2 and 3, so if you know how to solve systems of linear equations on a graphing calculator then what you have done is a fine approach. Share this post Link to post Share on other sites