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Iron(II) Phosphate Formula Mass: Calculation and Applications

Understanding Iron(II) Phosphate (Fe₃(PO₄)₂)

Iron(II) phosphate, also known as ferrous phosphate, is an inorganic compound with the chemical formula Fe₃(PO₄)₂. It is commonly used in fertilizers, lithium-ion batteries, and as a corrosion inhibitor. Calculating its formula mass (molar mass) is essential for laboratory and industrial applications.

Calculating the Formula Mass of Iron(II) Phosphate

The formula mass is the sum of the atomic masses of all atoms in the compound.

Step-by-Step Calculation:

1. Identify the elements and their quantities:
– Iron (Fe) – 3 atoms
– Phosphorus (P) – 2 atoms
– Oxygen (O) – 8 atoms (from 2 phosphate groups, PO₄²⁻)

2. Use atomic masses from the periodic table:
– Fe: 55.85 g/mol
– P: 30.97 g/mol
– O: 16.00 g/mol

3. Calculate the total mass contribution of each element:
– Fe: 3 × 55.85 = 167.55 g/mol
– P: 2 × 30.97 = 61.94 g/mol
– O: 8 × 16.00 = 128.00 g/mol

4. Add them together:
– Total formula mass = 167.55 + 61.94 + 128.00 = 357.49 g/mol

Practical Applications

– Fertilizers: Provides essential iron and phosphorus for plant growth.
– Batteries: Used in lithium iron phosphate (LiFePO₄) cathodes for energy storage.
– Corrosion Prevention: Acts as a protective coating for metals.

Conclusion

The formula mass of Fe₃(PO₄)₂ is 357.49 g/mol, crucial for precise chemical measurements. Its industrial uses make it an important compound in agriculture and technology.