What is the empirical formula for a compound that contains 88.8 percent copper and 11.2 percent oxygen?

Haven't got a calculator on me, but I'll get you started. 88.8% O = .888 g O 11.2% H = .112 g H .888 g O * (1 mol O / 15.999 g O) = aboutish .0555 mol O .112 g H * (1 mol H / 1.0079 g H) = about .112 mol H That's the ratio of moles of O to H, divide both of them by the smaller of the two, .0555, and it looks like you'll wind up with almost whole numbers, so just round them and those are the coefficients for that element.