Calculate the percentage of water in each of the following hydrates.

Aluminum bromide hexahydrate

I just need help setting this up. I can work it from there

Would it be:
AlBr * 6H2O

Did I write it right?

3 answers

AlBr3 * 6H20

we will be happy to critique your thinking or work.
Al = 26.981
3Br = 239.727
12H = 12.096
6O = 95.994
Total = 374.798
Total H2O = 108.090

108.090/374.798 x 100% = 28.840%

Correct? are my sig figs messed up in anyway?
It is fine. The sig figures are ok, IF those figs in the individual elements are correct.