# 9.50g of a sample of hydrated MgSO4 are heated and 4.85g of H20 are lost, how would you calculate the empirical formula of this hydrated salt?

Oct 24, 2015

$\text{MgSO"_4 * 7"H"_2"O}$

#### Explanation:

The idea here is that you will use the mass of evaporated water and the mass of the initial hydrated sample to find the formula of the hydrate, $\text{MgSO"_4 * color(blue)(n)"H"_2"O}$.

So, if you start with a sample of hydrated salt that has a mass of $\text{9.50 g}$, and evaporate all the water of hydration it contains, you are left with

${m}_{\text{anhydrous" = m_"hydrate" - m_"water}}$

${m}_{\text{anhydrous" = "9.50 g" - "4.85 g" = "4.65 g}}$

This is the mass of the anhydrous salt, which in your case is magnesium sulfate, ${\text{MgSO}}_{4}$.

You now know that the hydrate contained

• $\text{4.85 g}$ of water
• $\text{4.65 g}$ of anhydrous magnesium sulfate

What you need to do next is work out how many moles of each you had in the hydrate. To do that, use their respective molar masses

4.85color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.26922 moles H"_2"O"

and

4.65color(red)(cancel(color(black)("g"))) * ("1 mole MgSO"_4)/(120.37color(red)(cancel(color(black)("g")))) = "0.03863 moles MgSO"_4

Now divide both these numbers by the smallest one to get the mole ratio that exists between magnesium sulfate and water in the hydrate

"For H"_2"O: " (0.26922color(red)(cancel(color(black)("moles"))))/(0.03863color(red)(cancel(color(black)("moles")))) = 6.97 ~~7

"For MgSO"_4:" " (0.03863color(red)(cancel(color(black)("moles"))))/(0.03863color(red)(cancel(color(black)("moles")))) = 1

For every mole of mgnesium sulfate, the hydrate contained $7$ moles of water. This means that the empirical formula of the hydrate is

$\text{MgSO"_4 * color(blue)(7)"H"_2"O}$