How many atoms of oxygen are there in 26.45 g of molybdium(IV) phosphate?

Answer 1

Molybdenum(IV) phosphate = Mo₃(PO₄)₄

Molar mass = 667.705 g / mol

26.45 g x 1 mol / 667.705 g = 0.03961 moles Mo₃(PO₄)₄

Each mole of Mo₃(PO₄)₄ contains 16 moles of O atoms

0.03961 mols Mo₃(PO₄)₄ x 16 mols O atoms / mol Mo₃(PO₄)₄ = 0.6338 moles of O atoms

0.6338 mols O atoms x 6.022x10²³ O atoms / mole = 3.817x10²³ O atoms