Calculating moles, ratio, empirical formula

[wellY--3]

when you have to work out the amount of moles in 3 grams of magnesiun oxide (2 grams are magnesium) would you go 0.3 divided by 40.31?
then the answer is 7.44x10^-3 so that means that's the nuymber of moles of magnesium and also the number of moles of oxygen. So if u had to find the ratio of number of moles of magneium to oxygen it would just be 1:1?

but I'm confused about how you would find the empirical formula of that. Can you help?
P.S the grams are wrong because these were found in a disaster of an experiment

 

[symbolipoint]

What are the atomic weights of Magnesium and of Oxygen? 24.312, and 15.999, respectively. The compound of magnesium oxide is MgO. So what is the formula weight of Magnesium oxide? It is 40.311 grams per mole. That is the ratio to use.

You want to know how many moles in 3 grams of MgO?
3 grams MgO * (1 mole MgO)/(40.311 grams MgO) = 7.44*10^(-2) moles.

You must have know most of what you were doing; you were off by one order of magnitude. Are you still confused about any of this?

 

[wellY--3]

oh yer 7.44*10^(-2) sorry
so if i had to work out the number of moles of Mg and O seperately it is just 7.44*10^(-2) mol?
so the ratio n(Mg0:n(O) would be 1:1?
so then that makes the empirical formula just MgO?

 

[symbolipoint]

wellY--3 still asks:

so the ratio n(Mg0:n(O) would be 1:1?
so then that makes the empirical formula just MgO?

Look at the charge of a Magnesium cation: +2
... and look at the "charge" of Oxygen anion: -2;
They can combine as MgO. Note that as you guessed, 1 formula unit MgO has one unit Oxygen and one unit Magnesium. Yes, the empirical formula is MgO.