You measured out 0.101g of Mg. You heat the sample and measure out 0.234 of the magnesium oxide compound. Given your data, what us the empirical formula for the Mg(x)O(y)? x and y are ratio x:y

If I'm doing this correctly, I divided 0.101g of Mg/24.31g of Mg = .00415 mol Mg
But I'm stuck on what to do next or even if my first step is correct or not. Help please?

5 answers

mass of O2: .234-.101 grams=.131g
moles O2: .131/32=.004
Moles Mg= .101/24=.004
divide both by the lower
Moles O2= 1
moles Mg=1

MgO2 is what you got.
I understand your math, but where did Oxygen started with 2 of them to be O2? Is it because of the charges?
If you had used mols O
you would have had .13/.16 = .008
or two mols of O for every one of Mg
so it would have been MgO2 anyway :)
I mean .131/16 = .008 mols O
IT STATE THAT IF A CHEMICAL IS IN EQUILIBRIUM AND ARE OF THE FACTORS INVOLVED IN THE EQUILIBRIUM WILL SHIFT.