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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.

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  1. 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%.

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