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.
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
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 :)
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.
1 answer