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A hydrate of magnesium sulfate has a mass of $\pu{12.50 g}$.This sample is heated until no water remains. The anhydrous $\ce{MgSO4}$ has a mass of $\pu{5.30 g}$. Find the formula of the hydrate.

What I tried to do was I divided 12.50 by 5.30 and got 2.3ish. Then rounded it down to 2. After that I times the $\ce{MgSO4}$ by two to get $\ce{Mg2S2O8}$.

I know this isn't right but this is my initial thought process. Can anyone tell me the steps to I need to solve this? This is on a review guide so I just need to know the process.

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closed as off-topic by Tyberius, Todd Minehardt, airhuff, Jan, ron Nov 20 '17 at 17:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

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    $\begingroup$ Do not repost questions, you should have edited original. $\endgroup$ – Mithoron Nov 20 '17 at 0:58
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You must've confused a hydrate with something else. Just like many other hydrates, magnesium sulfate hydrate dehydrates as follows:

$$\ce{MgSO4 * xH2O ->[\Delta] MgSO4 + x H2O}$$

To determine the formula, you need to find $x$, which can be done using molecular weight of the hydrate:

$$x = \frac{M(\ce{MgSO4 * x H2O}) - M(\ce{MgSO4})}{M(\ce{H2O})} \tag{1} \label{eq:1}$$

Molecular weight of the unknown hydrate, on the other hand, can be found from the amount of anhydrous salt as those two are in $1 : 1$ ratio (see the reaction).

$$M(\ce{MgSO4 * x H2O}) = \frac{m(\ce{MgSO4 * x H2O})}{n(\ce{MgSO4 * x H2O})} = \frac{m(\ce{MgSO4 * x H2O})}{n(\ce{MgSO4})} = \frac{m(\ce{MgSO4 * x H2O}) \cdot M(\ce{MgSO4})}{m(\ce{MgSO4})} = \frac{\pu{12.50 g} \cdot \pu{120.37 g mol-1}}{\pu{5.30 g}} = \pu{283.89 g mol-1} \tag{2}$$

Plugging the found molecular weight to \eqref{eq:1} will allow you to determine the formula.

$$x = \frac{\pu{283.89 g mol-1} - \pu{120.37 g mol-1}}{\pu{18.02 g mol-1}} = 9.07 \approx 9 \tag{3}$$ This is magnesium sulfate nonahydrate $\ce{MgSO4 * 9 H2O}$.

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  • $\begingroup$ I think it would be better to just describe how OP should do it, not provide full answer. $\endgroup$ – Mithoron Nov 20 '17 at 1:01
  • $\begingroup$ @Mithoron I edited the question to address your concern and put the exact answer as well as the key calculations under the spoilers. $\endgroup$ – andselisk Nov 20 '17 at 1:15

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