0
$\begingroup$

I have three questions about waters of hydration, the first two of which I'm really stuck on:

  1. 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.

  2. Epsom salts is $\ce{MgSO4.xH2O}$. The hydrate was found to contain $71.4\%$ oxygen. Calculate the value of $x$.

  3. 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:

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

I couldn't get any further than this.

$\endgroup$
  • 2
    $\begingroup$ Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. Thanks! $\endgroup$ – jonsca Jun 8 '14 at 5:45
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I realize the percent of oxygen would go up by about 3.5% each time I add in an extra H2O. It is possible to do trial and error until I get to 71.4% but there has to be a more efficient way of doing this right? $\endgroup$ – Mandy Quan Jun 8 '14 at 18:16
  • $\begingroup$ sure, set up an equation like above but insert "x" times the atomic weights for the contributions from the H and O in the water $\endgroup$ – ron Jun 8 '14 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.