Answer:
PI₃
Explanation:
The empirical formula is found using the molar ratio between phosphorus and iodine.
The phosphorus composes 7.523% by weight of the substance, or 7.523g of phosphorus per gram of the substance. The moles of phosphorus per gram of substance is calculated using the atomic weight of phosphorus (30.97 g/mol):
(7.523g P) / (1 g substance) x (mol/30.97g P) = (0.2429 mol P) / (1 g substance)
Similarly, the moles of iodine (atomic weight 126.90 g/mol) per gram of substance is found:
(92.48 g I) / (1 g substance) x (mol/126.90 g I) = (0.7288 mol I) / (1 g substance)
The molar ratio between P and I is found by dividing the two values calculated above:
(0.7288 mol I) / (1 g substance) ÷ (0.2429 mol P) / (1 g substance) = (3.00 mol I) / (mol P)
Thus, the ratio of iodine to phosphorus is 3:1, so the empirical formula (the simplest formula) is PI₃
