Question

An oxide of phosphorus is 56.34% phosphorus and the rest is oxygen, what is the empirical formula for this compound?

Answer 1

56.34g x (1 mole P/30.97g) = 1.82 mol
43.66g x (1 mole O/16g) = 2.73 mol
P 1.82/1.82 = 1
O 2.73/1.82 = 1.5 -> x2
ANSWER: P2O3

Answer 2

P2O3

Step-by-step explanation:

The empirical formula shows the minimal ratio between the elements in a compound, so if you calculate the number of moles in the molecule, you can determine the ratio and so the empirical formula.

If you suppose you have 100 grams of the oxide of phosphorus, then you have 56.34 g of P and 43.66 g of O.

Then you calculate the number of moles of each element and finally divide each number by the smallest one, it would allowed you to know the atomic ratio between P and O.

P= 56.34 g/ 31 g per mole= 1.81 moles

O= 43.66 g/16 g per mole 2.73

Divide by 1.81

P= 1 and O=1.5, multiply by 2 because you can´t use non integer numbers. P2O3