An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound?

Ca3(PO4)2

Explanation:

We assume that the sample is 100.0 g.The sample contains 38.7 g of Ca, 19.9 g of P, and 41.2 g of O as the proportions show.Then we can calculate the number of moles of each component (n = m/atomic mass).The number of moles of Ca = 38.7/40.078 = 0.96 mol.The number of moles of P = 19.9/30.97 = 0.64 mol.The number of moles of O = 41.2/15.99 = 2.75 mol.Now, we can get the molar ratios of different components (Ca, P, and O) in the sample by dividing the number of moles of each component by the lower number of moles, that we should divide the number of moles by (0.64).Ca: P: O = (0.96/0.64) : (0.64/0.64) : (2.75/0.64) = 1.5 : 1 : 4.To avoid the fraction of the ratios, we can multiply all ratios by 2.0.Now, the ratio of Ca : P : O will be 3 : 2 : 8.That main the empirical formula of the compound is Ca3P2O8 which can be expressed as calcium phosphate (Ca3(PO4)2).

The total number of electrons is 14