Verified answer

All the (SO4)2- from the cerium sulfate is precipitated out in the BaSO4

So work out the moles of BaSO4 produced, and from this the moles of (SO4)2- that was in the cerium sulfate

moles = mass / molar mass

moles BaSO4 = 0.0787 g / 233.43 g/mol

moles BaSO4 = 3.371x10^-4 moles

Each BaSO4 has 1 (SO4)2-

Therefore 3.371x10^-4 moles BaSO4 contains 3.371x10^-4 moles of (SO4)2-

Work out the mass of (SO4)2-

mass = molar mass x moles

mass (SO4)2- = 96.07 g/mol x 3.371x10^-4 mol

mass (SO4)2- = 0.03239 g

There were 0.03239 g of SO4^2- in the 0.0639 g sample of cerium sulfate...

Now you can work out the mass of Ce

mass Ce = total mass - (SO4)2-

mass Ce = 0.0639 g - 0.03239 g

mass Ce = 0.0315 g

Now, moles Ce

moles = mass / molar mass

moles Ce = 0.0315 g / 140.12 g/mol

moles Ce = 2.248x10^-4 moles

Now you know the moles of Ce and the moles of SO4^2- you can work out the empirical formula of the compound.

The ratio of moles Ce : moles SO4^2-

= 2.248x10^-4 moles : 3.371x10^-4 moles

Two get a whole number ratio divide both numbers by the smallest number

Ce : SO4

(2.248x10^-4 / 2.248x10^-4) : (3.371x10^-4 / 2.248x10^-4)

1 : 1.5

This in not a whole number ratio yet , multipy both sides x 2

Ce : SO4

2 : 3

Ce2(SO4)3