A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?

We will determine the moles of each element first

moles = mass / atomic mass

Atomic weight of Al : 27 g / mol

Atomic weight of P : 3 1g / mol

Atomic weight of O : 16 g / mol

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g

Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40 / 31 = 0.819

Moles of O = mass / atomic mass = 52.50 / 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

moles of Al = 0.819 / 0.819 = 1

moles of P = 0.819 / 0.819 = 1

moles of O = 3.28 / 0.819 = 4

So the empirical formula will be : AlPO4