A 0.72 mg sample of phosphorus reacts with bromine to form 10.01 mg of the bromide. part a what is the empirical formula of the phosphorus bromide

PBr5 You started with 0.72 mg of phosphorus and ended up with 10.01 mg of its bromide. So the amount of bromine is 10.01 - 0.72 = 9.29 mg Now you need to determine the relative number of atoms of each element used. atomic mass of phosphorus = 30.973762 atomic mass of bromine = 79.904 relative atoms of phosphorus = 0.72 / 30.973762 = 0.023245 relative atoms of bromine = 9.29 / 79.904 = 0.116265 Now you need to look for a simple ratio of integers that closely approximates 0.023245 / 0.116265. First we'll divide the larger by the smaller. 0.116265 / 0.023245 = 5.001597 Given how close the value comes to 5. The empirical formula will be PBr5. So for every atom of phosphorus, you need 5 atoms of bromine.