Need MASSIVE help for the Determination of Water in a Hydrate???

Question

Calculating the Determination of Water in a Hydrate????HELP!!!?

Here are the weights:

Weight of crucible and cover = 12.85grams
Weight of crucible, cover and sample = 16.21grams
Weight of crucible, cover and sample after first heating = 14.82grams
Weight of crucible, cover and sample after second heating = 14.74grams
Weight of crucible, cover and sample after third heating = 14.72 grams

My chemistry teacher wants me to find the calculations of these samples above. I have no F'in idea if I am right or wrong when I am working out this problem.

I think the formula is Percentage of water=(mass lost divided by mass of sample) and then you multiply it by 100.

Here are the directions for the calculations:

Calculate weight of original sample
Numerical setup:


Calculate weight lost from sample by heating
Numerical setup:


Calculate percentage of water in sample
Numerical setup:


For this one, it also asks:

sample No._____
Percentage of water_____

Answer 1

Note the crucible and cover are just added weight to everything, and subtracting their weight from the total gives the weight of the material in the crucible. So you start with about 3.4 grams of some hydrate. After 1 heating, about 2 grams of some hydrate minus some water is left. Usually, cooking off hydrate water completely requires your material in the crucible to reach some constant weight. Some more water cooks off in the second heating, but by the third, you have pretty much cooked off all water of hydration.
So the percentage of water is:
100*[weight of original sample-wt of final sample]
DIVIDED by
weight of orignial sample.