Please Help- Experimental %?

Question

If Cobalt (II) Sulfate is heated too strongly this reaction occurs:
CoSO4 (s) ---> CoO(s) + SO3 (g)
If a sample of CoSO4*6H2O is heated & this reaction along with dehydration occurs,what will happen to the experimental % water?
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I thought the % would go down b/c there will be a loss of water from cobalt sulfate (in the form of water vapor) but that is incorrect. Can someone please explain what would happen to me. I would really appreciate it! Thanks so much.

Answer 1

you miss the point. the REAL % H2O is

wt H2O x 6 /(wt CoSO4*6H2O)

this is determined by driving off the water (I suppose)
to get "wt CoSO4"; (wt CoSO4 + Wt H2O) = total wt

But if the wt CoSO4 is reduced because of loss of SO3
the apparent weight of the water lost will include the lost SO3 and be a bigger number (increased %).
In other words the weight loss will include both water and some SO3 when it should only include water so it will be too large - a larger percentage of the total.
Think about this: if you really really heated it up you would vaporize everything and the weight loss would be 100% and you'd "assume" that it was all water (unless you had half a brain and realized that using a high powered laser to heat a sample might be over-kill). But the point is that the inadvertent loss of SO3 increases the experimental water loss (assuming only water "should" have been lost)

Answer 2

If you overheat the material and drive off some of the sulfur trioxide along with the water, the weight loss will be larger than if just the water was lost. So, when you subtract final weight (after heating) from the starting weight to get weight of water lost, the difference is larger than it should be. This will make it seem as if there was more water in the sample than there really was! It would appear as if % water was larger! I hope this explains it for you adequately.