White phosphorus, P4, spontaneously bursts into flame in oxygen. if 6.5g of white phosphorus reacts with sufficient oxygen to form 11.54g of a phosphorus oxide, what is the empirical formula of this oxide?

Answer 1

you have 0.05246 moles of P4
the 11.54g of oxide contain 5.04 g of oxygen, which is 0.315 moles of O
0.315/0.05246 = 6
That means you get 6 moles of O for every mole of P4
See what you can do with that.

Answer 2

Assuming all of the 6.5 grams P₄ is consumed in an excess of O₂ to produce 11.54 grams PₓOᵥ then …
%P per 100-wt = (6.5g/11.54g)100% = 56.3258% => 56.3258g/30.9738g/mol = 1.8185 mole P
%O per 100-wt = 100% - 56.3258% = 43.6740% => 43.6740g/15.9994g/mol = 2.7297 mole O
P:O Empirical Ratio = 1.8185/1.8185 : 2.7297/1.8185 = (1:1.5) x 2 => 2:3 Empirical Ratio
=> P₂O₃ Empirical Formula