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

Hint: 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 \[MgS{O_4} \cdot 7{H_2}O\].