Question

For $9.50g$ of a sample of hydrated $MgS{O_4}$ are heated and $4.85g$ of ${H_2}O$ are lost, how would you calculate the empirical formula of this hydrated salt?

Answer 1

Hydrates are compounds that comprise water with a certain mass in the form of ${H_2}0$ in their molecular formula. These compounds are frequently derived in the form of a crystal which can formerly be heated in direction to eliminate the water in the form of steam. An anhydrate is the substance that vestiges after the water from a hydrate has been removed.
Unknown hydrates remain written with their base form; formerly an $'n'$ is placed before the ${H_2}O$. The $'n'$ before the ${H_2}O$ resources there is a number there, then we don't know what it is however, such as in $MgS{O_4}n{H_2}O$ for a magnesium sulfate hydrate.

Complete step by step answer:
The hint here is that you drive the mass of evaporated water and the mass of the primary hydrated example near find the formula of the hydrate, $MgS{O_4}n{H_2}O$ .
Thus, if you twitch with a example of hydrated salt that has a mass of $9.50g$, and evaporate completely the water of hydration it holds, you are leftward with
${m_{anhydrous}} = {\text{ }}{m_{hydrate}} - {\text{ }}{m_{water}}$
${m_{anhydrous}} = {\text{ }}9.50{\text{ }}g{\text{ }} - {\text{ }}4.85{\text{ }}g{\text{ }} = {\text{ }}4.65{\text{ }}g$
This is the mass of the anhydrous salt, which in your situation is magnesium sulfate, $MgS{O_4}$.
You at this time know that the hydrate contained
$4.85g$ of water
$4.65{\text{ }}g$ of anhydrous magnesium sulfate
What you need to do next is work available how many moles of each you had in the hydrate. To do that, use their individual molar masses
$4.85{\text{g}} \times \dfrac{{1{\text{mole}}{{\text{H}}_2}{\text{O}}}}{{18.015{\text{g}}}} = 0.26922{\text{moles}}{{\text{H}}_2}{\text{O}}$
And
$4.65{\text{g}} \times \dfrac{{1{\text{moleMgS}}{{\text{O}}_4}}}{{120.37{\text{g}}}} = 0.03863{\text{molesMgS}}{{\text{O}}_4}$
At present divide equally these numbers by the smallest one to convert the mole ratio that occurs between magnesium sulfate plus water in the hydrate
For ${H_2}O$ : $\dfrac{{0.26922{\text{moles}}}}{{0.03863{\text{moles}}}}{\text{}} = 6.97{\text{}} \approx {\text{}}7$
For $MgS{O_4}$: $\dfrac{{0.03863{\text{moles}}}}{{0.03863{\text{moles}}}} = 1$

Note:
For every single mole of magnesium sulfate, the hydrate confined $7$ moles of water. This yields the empirical formula of the hydrate is $MgS{O_4} \cdot 7{H_2}O$.