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?

1 Answer
Oct 24, 2015

Answer:

#"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, #"MgSO"_4 * color(blue)(n)"H"_2"O"#.

So, if you start with a sample of hydrated salt that has a mass of #"9.50 g"#, and evaporate all the water of hydration it contains, you are left with

#m_"anhydrous" = m_"hydrate" - m_"water"#

#m_"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, #"MgSO"_4#.

You now know that the hydrate contained

  • #"4.85 g"# of water
  • #"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

#"MgSO"_4 * color(blue)(7)"H"_2"O"#