Ahh I see, and to explain, my only confusion in the beginning was the reaction. I must divide the final precipitate weight by the molar mass of AgCl which will give me both moles of Ag+ and Cl-. From there I can divide the moles of Cl- by 2 since MgCl2 requires 2 Cl- ions to form. So this would then give us the total moles of MgCl2, from which we can divide the weight of the moles by the weight of the original sample, which is 1.0 g. I think this is right, and I will post back whether it works or not.
Aha, so it worked. My answer came out to .475 * 100 --> 47.5% of the original sample. Thank you all for the help, I understand this much better now!