If a particular ore contains 58.3 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

Molecular formula of calcium phosphate: Ca3 [PO4]2

Molar mass of calcium phosphate: 3*40.1g/mol + 2*31.0 g/mol + 2*4*16.0 g/mol = 310.3 g/mol

Mass of phosphorus in the molecular formula: 2*31.0 g/mol = 62.0 g/mol

Ratios:
x  * 0.583 g of calcium phosphate / g of ore * 62.0 g of phosphorus / 310.3 g of calcium phosphate = 1000 g of phosphorus

0.11649 x g of phospuros / g of ore= 1000 g of phosphorus

x = 1000 g of phosphorus / 0.11649 g of phosphorus / g of ore

x = 8585 g of ore = 8.585 kg of ore

Fine out your self ponk