A solid contains CaCl
2, CaI
2, and sugar and weighs X grams. It is dissolved in water to give sugar (sucrose)
molecules, Ca
2+, Cl
-, and I
-. The total number of moles of calcium in this solid sample is Y. After this the solution which contains the original amounts of chloride and iodide, is treated with sufficient AgNO
3 soln to precipitate the Cl
- and I
- as AgCl
(s) and AgI
(s). The precipitate is isolated, dried, and weighed and found to weigh Z grams. The sugar remains in soln and is not affected.
a) Call the weight of CaCl
2 in the original solid x and the weight of CaI
2 y. Using the notation AgCl for the formula weights, set up equations defining the stoichiometry of the problem, which when solved will permit you to find the unknowns x and y from the experimental data Y and Z.
b) Suppose X=41.2683 g, Y=0.140524 mol, and Z=48.5083 g. Find the percentage by weight of each of the three compounds in the original solid mixture.
I've written equations and such but I can't figure out how to make it so that you can solve for x and y. I've tried writing separate equations but that doesn't work because you don't know the individual masses of AgCl and AgI. I think I can figure out the rest of it once I get the equation right. Is it safe to assume that 2Y is the number of moles of I and Cl since for every Ag there should be 2 I's or Cl's.
4AgNO
3 + CaCl
2 + CaI
2 2AgCl
(s) + 2AgI
(s) +4NO
3- + 2Ca
2+So far I've come up with : 4AgNO
3 + x + y = Z + 40.078Y + 4NO
3- But this won't work because you have two unknowns. And I don't even know if that's right. It seems like you would need 2 equations but you can't because you don't know individual masses of AgCl
(s) and AgI
(s). If you can I need an answer to this by morning, I would have done it yesterday but I didn't have time.