An ionic compound is composed of the following elements:

Nitrogen - 31.57%

Hydrogen - 9.10%

Phosphorus - 23.27%

Oxygen - 36.06%

What is the empirical formula for this compound?

Answer: N3 H12 P O3

Explanation:

From the question:

N = 31.57% H = 9.10% P = 23.27%

O = 36.06%

Divide each of the element by their respective relative atomic masses.

N = 31.57 / 14 = 2.26

H = 9.10 / 1 = 9.10

P = 23.27 / 31 = 0.750

O = 36.06 / 16 = 2.25

Divide each answer by the lowest of them all, we then have:

N = 2.26 / 0.750 = Approx = 3

H = 9.10 / 0.750 = Approx = 12

P = 0.750 / 0.750 = 1

O = 2.25 / 0.750 = Approx = 3

The empiral formula is

N3 H12 P O3

The empirical formula for the compound is (NH₄) ₃PO₃

Explanation:

Here we have

Nitrogen Molar mass = 14.01 g/mol, Percentage = 31.57%

Hydrogen Molar mass = 1.00784 g/mol, Percentage = 9.10%

Phosphorus Molar mass = 30.973762 g/mol, Percentage = 23.27%

Oxygen Molar mass = 15.999 g/mol, Percentage = 36.06%

Therefore the number of moles of each element per 100 g of the ionic compound is as follows;

Number of moles, n = Mass / (Molar Mass)

Mass of nitrogen, N in 100 g of ionic compound = 31.57 g n = 2.253 moles

Mass of Hydrogen, H in 100 g of ionic compound = 9.10 g, n = 9.029 moles

Mass of Phosphorus, P in 100 g of ionic compound = 23.27 g, n = 0.7513 moles

Mass of Oxygen, P in 100 g of ionic compound = 36.06 g, n = 2.2539 moles

Dividing by the smallest mole ratio, of 0.7513 (that of phosphorus) we have;

N = 3 moles

H = 12.0184 ≈ 12 moles

P = 1 moles

O = 3 moles

Therefore, we have the empirical formula as follows;

N₃H₁₂PO₃ or (NH₄) ₃PO₃.