Posted by Roy on Saturday, October 8, 2011 at 9:01am.
You can do this the long way or a short way or one that's in between. The short way isn't taught anymore I don't think. The long way is to write an equation, balance it, and solve as if it were a stoichiometry problem. Here is a worked example that will solve all of your stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
The short way is
2.763g x (molar mass 0.5H2O/molar mass CaSO4.0.5H2O) = ? g H2O.
The in between way is to calculate the %H2O in CaSO4.1/2H2O.
That is (1/2 molar mass H2O/molar mass CaSO4.1/2H2O) x 100 = %H2O
Then (%H2O/100)*2.763 = g H2O (note the short way and the in between way are really the same thing but the approach is different)
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