The reaction of 1.250 g k with o2 forms 2.273 g of an oxide of potassium that is used in self-contained breathing devices. determine the formula for this oxide of potassium.

KO2 First, determine how many moles of potassium you have by looking up the atomic weight. atomic weight of potassium = 39.0983 atomic weight of oxygen = 15.999 1.250 g / 39.0983 g/mol = 0.031970699 mol Now figure out the mass of oxygen by getting the difference between the product mass and the original mass of the potassium 2.273 g - 1.250 g = 1.023 g Moles of oxygen = 1.023 / 15.999 = 0.063941496 mol We now have a ratio of 0.031970699 : 0.063941496 for potassium and oxygen. We want to find a ratio of small integers that closely approximates that ratio. So divide all of the numbers by 0.031970699 (picked because it's the smallest value) and we get 1 : 2.000003 which is pretty darn close to 1 : 2. So the empirical formula for this potassium oxide is KO2

Both models are effective because they show the valence electrons, and these valence electrons indicate the reactivity of an element.