Ammonia react with phosphorus acid to form a compound that contains 28.2% of nitrogen 20.8%, phosphorus 8.1% of hydrogen 42.9%oxygen. Calculate the empirical formula of this compound

Empirical formula will be (NH₄) ₃PO₄, which matches the molecular formula

Explanation:

This is the reaction:

NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O

In 100 g of compound we have:

28.2 g N

20.8 g of P

8.1 g of H

42.9 g of O

Now we divide each between the molar mass:

28.2 g / 14 g/mol = 2.01 mol

20.8 g / 30.97 g/mol = 0.671 mol

8.1 g / 1 g/mol = 8.1 mol

42.9 g / 16 g/mol = 2.68 mol

And we divide again between the lowest value of moles

2.01 mol / 0.671 mol → 3

0.671 mol / 0.671 mol → 1

8.1 mol / 0.671 mol → 12

2.68 mol / 0.671 mol → 4

Molecular formula will be: N₃PH₁₂O₄ → (NH₄) ₃PO₄

Empirical formula will be (NH₄) ₃PO₄, which matches the molecular formula