Tom is correct in his method but he adds a couple of unnecessary steps. You only need consider atomic O and atomic S (not the compounds), so to determine the number of moles you need only divide by the atomic weight (16 for O and 32 for S) and not the molecular weights. When you take the mole ratio, you will find that there are 1.5 O for every 1 S and you easily get the S2O3 empirical formula. You cannot get the molecular formula from this problem (and there is no need to use the molecular forms of O and S in his final step). The answer is S2O3
MgS. Mg is the lightest of all of the 2A elements you listed. In MgS, it is ~57% mass S. In BaS, the heaviest 2A element you listed, it is ~19% S.
The way to determine this is by dividing the molar mass of sulfur by the molar mass of the compound and multiplying by 100.
For MgS, it would be [32.0/(24.3+32.0)]*100=~56.8% mass sulfur.
Look at a periodic table
all of these have one S
so the one with the lightest 'other part'
will have the highest percent of S
and Mg is the lightest of the choices
so it's MgS
When you get a good response,
please consider giving a best answer.
This is the only reward we get.