# Question #533df

Oct 8, 2015

$M g S {O}_{4} .7 {H}_{2} O$

#### Explanation:

The hydrated magnesium sulfate contains water of crystallisation which is driven off when heated:

$M g S {O}_{4} . x {H}_{2} {O}_{\left(s\right)} \rightarrow M g S {O}_{4 \left(s\right)} + x {H}_{2} {O}_{\left(g\right)}$

By recording the loss in mass we can find the mass of water and then convert masses into moles to find $x$.

Mass $M g S {O}_{4} . x {H}_{2} O = 4.312 \text{g}$

Mass $M g S {O}_{4} = 2.107 \text{g}$

So mass water lost = $4.312 - 2.107 = 2.205 \text{g}$

Now we need to convert grams into moles using:

Number of moles = mass in grams/mass of 1 mole

Or:

$n = \frac{m}{M} _ r \text{ } \textcolor{red}{\left(1\right)}$

${M}_{r}$ is the relative molecular mass for a molecular compound or the relative formula mass for an ionic compound.

We get this by adding all the relative atomic masses (${A}_{r}$) together as they occur in the formula. You should always be provided with these in the question or on a data sheet.

Using accurate ${A}_{r}$ values we get:

${M}_{r} \left[M g S {O}_{4}\right] = 120.37$

${M}_{r} \left[{H}_{2} O\right] = 18.01$

You should use the ${A}_{r} ' s$ you are given - they will be close.

Now using $\textcolor{red}{\left(1\right)} \Rightarrow$

$n M g S {O}_{4} = \frac{2.107}{120.37} = 0.0175$

and:

$n {H}_{2} O = \frac{2.205}{18.01} = 0.12243$

So we now know that the ratio in moles of:

$M g S {O}_{4} : {H}_{2} O$ is:

$0.0175 : 0.12243$

We like to express this ratio in whole numbers according to how the atoms/molecules combine so we can divide both numbers by $0.0175 \Rightarrow$

$\frac{0.0175}{0.0175} : \frac{0.12243}{0.0175} \Rightarrow$

$1 : 6.996$

Which is good enough (in my book) as:

$1 : 7$

So $x = 7$ which gives the formula for hydrated magnesium sulfate to be:

$M g S {O}_{4} .7 {H}_{2} O$

Here is a full lab on hydrate and how to calculate the formula of a hydrate, I hope this helps more: