Question

A salt has percentage by mass, 20.00% with respect to magnesium, 26.66% with respect to sulfur and the balance is oxygen... What is the empirical formula of the material?

Answer 1

`"Empirical formula"-=MgSO_4`...

Explanation:

As always with these types of problems, we assume an `100*g` mass of stuff....and then interrogate the molar quantities of each element.....

`"Moles of magnesium"=(20.0*g)/(24.31*g*mol^-1)=0.823*mol.`

`"Moles of sulfur"=(26.66*g)/(32.06*g*mol^-1)=0.832*mol.`

`"Moles of oxygen"=(53.33*g)/(16.00*g*mol^-1)=3.33*mol.`

We divide thru by the SMALLEST molar quantity to get an empirical formula of....

`Mg_((0.823*mol)/(0.823*mol))S_((0.832*mol)/(0.823*mol))O_((3.33*mol)/(0.823*mol))=MgS_(1.01)O_4~=MgSO_4`..

...i.e. we gots `"magnesium sulfate"` `MgSO_4`...

And note that this empirical formula is consistent with the given formula mass....we would not generally call an inorganic salt molecular.

Answer 2

Let's assume that you have a certain amount of the compound, and the mass of that amount is 100g.
This would mean that It is 20g Magnesium, 26.66g Sulfur, and 53.33g Oxygen.

Now we can convert each to moles.

20g Mg = 0.83 mol Mg
26.66g S = 0.83 mol S
53.33g O = 3.33 mol O

Now we can form a ratio.
0.83 mol of Mg equals 0.83 mol of S, meaning the two elements are present in equal amounts, or a 1:1 ratio.
`(3.33mol)/(0.83mol) ~~ 4`

This means the ratio of Oxygen to Sulfur and Oxygen to Magnesium is 4:1.

Because the question is asking for an empirical formula, we don't need to worry about the 120 part, that would only be used if the question asked for the molecular formula.

O : S = 4 : 1
Mg : S = 1 : 1

So, the formula is `MgSO_4`