Question

4.72 grams of pure phosphorus are allowed to react with oxygen in air to form an oxide. If the product weights 10.82 grams, find the empirical formula of the compound. Show all steps.

Answer 1

Answer:- P2O5 (Phosphorus pentoxide)

Explanation:

To find empirical formula,

First you want to know how many moles of each element you have in your sample.

So, you are told that a 4.72 g sample of pure phosphorus was allowed to react with oxygen to form an oxide. The oxide formed by the reaction has a mass of 10.82 g.

m(Oxide) = m(O) + m(P)

m(O) = m(Oxide) - m(P)

This means that your sample will contain

m(O) = 10.82 g−4.72 g = 6.10 g O

Now that you know how many grams of each element you have in the sample, use their respective molar masses to find how many moles of each you have

1) 4.72 g of P ×(1 mole P / 30.974 g) = 0.1524 moles P

2) 6.10 g of O × (1 mole O / 15.999 g) = 0.3813 moles O

Now, to get the mole ratio that exists between the two elements in the compound, divide both values by the smallest one.

P = (0.1524moles / 0.1524moles) = 1

O = (0.3813moles / 0.1524moles) = 2.5

The empirical formula for this compound must contain the smallest whole number ratio that exists between the elements. Since you have a 1:2.5 mole ratio, the smallest whole number ratio that matches this one is 2:5.

This means that the compound's empirical formula will be

(P1O2.5)2 = P2O5