How to calculate the number of waters of hydration? I have three questions about waters of hydration, the first two of which I'm really stuck on:


A sample of cobalt(II) nitrate hydrate was heated to remove all the water of hydration. The hydrate was found to be $65.96\%$ oxygen. Calculate the number of water molecules associated with each formula unit of cobalt(II) nitrate.
Epsom salts is $\ce{MgSO4.xH2O}$. The hydrate was found to contain $71.4\%$ oxygen. Calculate the  value of $x$.
Zinc nitrate $\ce{Zn(NO3)2.xH2O}$ contains $21.98\%$ zinc by mass. What is the value of $x$?


I have absolutely no idea for the first two questions.

For the third:


Assume $\pu{100 g}$ of material
$\pu{21.98 g}$ of Zn = $\pu{0.3284 mol}$ of $\ce{Zn}$


I couldn't get any further than this.
All of the questions involve the same reasoning, so let's look at question #2.

What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?