Mass percent of Douglasite????

Question

Douglasite is a mineral with the formula 2 KCl·FeCl2·2 H2O. Calculate the mass percent of douglasite in a 468.4 mg sample if it took 37.20 mL of a 0.1000 M AgNO3 solution to precipitate all of the Cl - as AgCl. Assume the douglasite is the only source of chloride ion.

Answer 1

Molarity is the number of moles of a substance in a liter of solution. So 37.2 mL of 0.1 M solution contains (0.0372 L)(0.1 mol/L) = 0.00372 moles of AgNO3. So there must also be 0.00372 moles of Cl atoms in the sample. Each particle of douglasite contains four Cl atoms, so there are only 0.00372 / 4 = 0.00093 moles of douglasite. Noting that the molecular masses of K, Cl, Fe, H, and O are 39, 35, 56, 1, and 16, respectively, the weight of one mole of douglasite is 2*(39 + 35) + 56 + 35*2 + 2*(1*2 + 16) = 310 g. (If you want a more precise molecular mass, look up the atomic weights of the elements yourself and use more precise values.) So the weight of douglasite in the sample is (0.00093 mol)(310 g/mol) = 0.2883 g = 288.3 mg. If the weight of the original sample was 468.4 mg, by weight it was 288.3 / 468.4 = 0.615 or 61.5% douglasite.