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

The amount per 100 g is:

38.7 % calcium = 38.7g Ca / 100g compound = 38.7g

19.9 % phosphorus = 19.9g P / 100g compound = 19.9g

41.2 % oxygen = 41.2g O / 100g compound = 41.2g

The molar amounts of calcium, phosphorus and oxygen in 100g sample are calculated by dividing each element’s mass by its molar mass:

Ca = 38.7/40.078 = 0.96

P = 19.9/30.97 = 0.64

O = 41.2/15.99 = 2.57

C0efficients for the tentative empirical formula are derived by dividing each molar amount by the lesser value that is 0.64 and in this case, after that multiply wih 2.

Ca = 0.96 / 0.64 = 1.5=1.5 x 2 = 3

P = 0.64 / 0.64 = 1 = 1x2= 2

O = 2.57 / 0.64 = 4= 4x2= 8

Since, the resulting ratio is calcium 3, phosphorus 2 and oxygen 8

So, the empirical formula of the compound is Ca₃(PO₄)₂

the answer is b. dilution