I'm lost with empirical formulas here, just need the steps here.

I got 4 in total for this online homework to get this single question right. I just need to get on the right track for one at least and I should be able to take it from there. Also one of them has something about it that is confusing me so I will list that one as well.

Here are 2, the second has something about it confusing with the format or layout:

Give the empirical formulas of the compounds formed from the following pairs of ions:

(A) Rb+ and I-

(B) Cs+ and SO42+ (The 4 is lower, the 2 is higher, and the - is in the middile?)


Reason I am listing 2:
I need at least one to get the example, but I need to understand that part about why B is listed the way it is. In case the way I listed and explained B was unclear than someone could at least show me how to do A which would show me how to work on the rest of my questions.

The charge on a compound is 0,
so the number of + charges has to equal the number of - charges.

With Rb+ and I-, they have an equal number of + and - charges so no subscripts are needed:
RbI is the formula for rubidium iodide.

If you were combining Rb+ and O^2-,
you'd have to have 2 of the Rb+ ions to cancel out the 2- charges on the O^2-.
Rb2O is the formula for rubidium oxide.

If you were combining Cs+ and O^2-, it would be the same thing,
Cs2O results in 2 + charges and 2 - charges.

The second example follows the same logic as Cs2O.
You need 2 Cs ions to cancel out the 2- charge on the sulfate (SO4^2-) ion.
So the formula for cesium sulfate is Cs2SO4.

The sulfate ion is an example of a polyatomic ion, which is just what the name implies,
a charged particle made up of many atoms.
You treat it as you would a monatomic ion, with one exception.
If you need more than one copy of a polyatomic ion, it has to be in parentheses in the formula.

The nitrate ion is NO3^1-.
Combined with Na+, the formula is NaNO3.