Data ysis?

i did a lab like this last year for chem cl, the m of the mg ribbon is the difference between your second and first number, .34 grams. to find the moles used, you divide .34 grams by the molar m of mg, which is 24.305 g. you get .0133 mols. the m of magnesium oxide formed would be the difference of the third and first number, .56 grams. so then .22 g of oxygen was combined with the mg. divide that by the molar m of oxygen to find moles of oxygen used and you get .0137 moles. given these numbers, you have a one to one ratio of moles mg and moles O used. so the empirical formula is MgO. good job with the lab unless you made up your numbers cause my cl hecka screwed up when we did the lab, we made mg2o3 or something that wasn't what we were spose'd to get. good luck with learning chem