Question

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

A. Ca2PO4

B. Ca3PO6

C. Ca4P2O4

D. Ca3P2O8 (or Ca3(PO4)2)

E. CaPO4

Answer 1

The percentages of Ca, P and O are given; therefore, 100 g of the compound contains 38.7 g Ca, 19.9 g P and 41.2 g O.

The atomic masses of the elements are known.

Ca: 40.078 g/mol

P: 30.9738 g/mol

O: 15.9994 g/mol

Find the number of moles of the elements by diving the masses of each element by the atomic masses.

Moles Ca = (38.7 g)/(40.078 g/mol) = 0.9656 mole.

Moles P = (19.9 g)/(30.97380 g/mol) = 0.6425 mole.

Moles O = (41.2 g)/(15.9994 g/mol) = 2.5751 mole.

Find the simplest ratio of the number of moles of Ca, P and O by dividing by the smallest number of moles.

Ratio of number of moles of Ca:P:O = 0.9656:0.6425:2.5751 = (0.9656/0.6425):(0.6425/0.6425):(2.5751/0.6425) = 1.5029:1.000:4.0079 ≈ 3.0:2.0:8.0 (multiply each by 2 to get the simplest ratio).

The simplest ratio of the moles of Ca, P and O is 3:2:8 and hence, the empirical formula is Ca₃P₂O₈ or Ca₃(PO₄)₂; therefore, the correct answer is (D).