If a particular ore contains 58.5% calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

Calcium Phospate formula: Ca3 (PO4) 2

Atomic weight:

Ca = 40; P = 31; O = 16

Ca3 = 40 * 3 = 120

P = 31

O4 = 16 * 4 = 64

(PO4) 2 = (31 + 64) * 2 = 95 * 2 = 190

Ca3 (PO4) 2 = 120 + 190 = 310 g/mol

31/310 = 10% P in the calcium phospate

1.00kg * 1000 g/kg = 1000 g of phosphorus.

1000 g / 10% = 10,000 g of calcium phosphate

10,000 g / 58.5% = 17,094 grams of ore.

The minimum mass of the ore should be 17,094 grams or 17.094 kg.