A rock sample weighing 5.88 x 10-4 g is known to contain calcium, phosphorus, and oxygen. The amount of the first two elements in this rock is found to be 2.28 x 10-4 g and 1.18 x 10-4 g respectively. What is the empirical formula for the compound in this rock sample?

Ca₃P₂O₈.

Explanation:

Firstly, we can determine the mass of oxygen:

The mass of oxygen in the rock = the mass of the rock - (the mass of Ca + the mass of P) = (5.88 x 10⁻⁴ g) - (2.28 x 10⁻⁴ g + 1.18 x 10⁻⁴ g) = 2.42 x 10⁻⁴ g.

We need to determine the no. of moles of each element in the rock:

no. of moles = mass/atomic mass.

no. of moles of Ca = (2.28 x 10⁻⁴ g) / (40.078 g/mol) = 5.689 x 10⁻⁶ mol.

no. of moles of P = (1.18 x 10⁻⁴ g) / (30.97 g/mol) = 3.81 x 10⁻⁶ mol.

no. of moles of O = (2.42 x 10⁻⁴ g) / (16.0 g/mol) = 1.51 x 10⁻⁵ mol.

To find the mole ratio of (Ca: P: O), we divide by the lowest no. of moles that of P (3.81 x 10⁻⁶ mol).

The mole ratio of (Ca: P: O) will be ((5.689 x 10⁻⁶ mol) / (3.81 x 10⁻⁶ mol) : (3.81 x 10⁻⁶ mol) / (3.81 x 10⁻⁶ mol) : (1.51 x 10⁻⁵ mol) / (3.81 x 10⁻⁶ mol)).

∴ The mole ratio of (Ca: P: O) will be (1.5: 1.0: 4.0).

Multiplying by 2.0 to get the empirical formula of the rock:

The empirical formula of the rock is Ca₃P₂O₈.