Question

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

Answer 1

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.