Question

12) What is the mass percent of phosphorus in MgKPO4 6H2O? (15 pts)

Answer 1

11.62%

Step-by-step explanation:

We are given;

• A compound MgKPO₄.6H₂O

We are required to calculate the mass percent of phosphorus in the above compound.

• Mass % of an element =(Mass of the element in the compound ÷ Formula mass of the compound) × 100

In this case;

Atomic mass of Mg = 24.305 g

Atomic mass of K = 39.098 g

Atomic mass of P = 30.974 g

Atomic mass of O = 16.0 g

Molar mass of water is 18.02 g/mol

• Therefore; the Formula mass of MgKPO₄.6H₂O will be;

= (1 × 24.305 g) + (1 × 39.098 g) + (1 × 30.974 g) + (4 × 16.0 g) + (6 ×18.02 g)

= 266.497 g

But, mass of P in the compound is 30.974 g

Therefore;

Mass % of P = (30.974 g ÷ 266.497 g) × 100

= 11.62%

Thus, the mass percent of phosphorus in the compound is 11.62%