Consider a mixture of the two solids BaCl2.2H2O (FM 244.26) and KCl (FM 74.551). When the mixture is heated to 160C for 1 hour, the water of crystallization is drive off:
BaCl2.2H2O --> BaCl2(s) + 2H2O(g)
A sample originally weighing 1.7839g weighed 1.5623g after heating. Calculate the weight percent of Ba, K, and Cl in the original sample.
I calculated the mass of the water lost which is (1.7839g-1.5623g)= .2216 g. Trying to calculate %Ba first. The answer for Ba is 47.35%, but my calculations for getting this answer is not quite there.
For Ba, convert 0.2216g H2O lost to grams Ba. The easy way to do that (and it isn't taught in class anymore that I'm aware of) is to use a chemical factor. That will be 0.2216 x (formula wt Ba/2*formula wt H2O) = 0.8446g Ba.
Then (0.8446/1.7839)*100 = 47.35%.