2
$\begingroup$

I have a test review question which i really need help on. It states :

A student heated a sample of the hydrate $\ce{CaSO4 * xH2O}$ in an evaporating dish over a bunsen burner to remove all the water and recorded the following data:

Mass of empty evaporating dish: $\pu{19.73g}$
Mass of evaporating dish + hydrate: $\pu{21.50g}$
Mass of evaporating dish + anhydrous $\ce{CaSO4}$: $\pu{20.80g}$

Determine A) the value of $x$ and B) the most likely name/formula of the hydrate.

I already figured out $x$ by getting the mass of water then simplifying the ratio of anhydrous to water.
But the second question which asks to name the formula of the hydrate really confused me. I would appreciate if anyone can guide me how to solve B.

$\endgroup$
  • 2
    $\begingroup$ For the hydrate part, you just add [di, tri, quadra, penta, hexa, etc.] to hydrate. So if x were 2 (I don't know, I didn't calculate it) it would be calcium sulfate dihydrate. $\endgroup$ – airhuff Mar 22 '17 at 1:59
  • $\begingroup$ It would indeed be straight forward to answer, if you had included your finding of $x$. $\endgroup$ – Martin - マーチン Jun 27 '17 at 5:51
1
$\begingroup$

From my calculations

\begin{align} x &= \frac{M(\ce{CaSO4}) \times \left[m(\ce{CaSO4 * xH2O}) - m(\ce{CaSO4})\right]}{M(\ce{H2O}) \times m(\ce{CaSO4})} =\\ &= \frac{\pu{136.14 g mol-1} \times (\pu{1.77 g} - \pu{1.07 g})}{\pu{18.02 g mol-1} \times \pu{1.07 g}} =\\ &= 4.94, \end{align}

which gives $\ce{CaSO4* 5 H2O}$, calcium(II) sulfate pentahydrate.

$\endgroup$
0
$\begingroup$

If the value of x turns out to be 2 than the given salt is gypsum or if it turns out to be half then the given salt is plaster of Paris. These are the common names of hydrates of calcium sulphate Otherwise you may use the prefixes.

$\endgroup$

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.