Acompound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?

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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

i do not get it   at all