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Answer:

The percent composition of Iron (II) Phosphate is

Fe = 46.866%

P = 17.330%

O = 35.806%

The percent composition of compounds is obtained form the mass of atoms in the compounds.

The formula of Iron (II) Phosphate is Fe3(PO4)2. We now have to obtain the molar mass of the compound as follows;

Molar mass = 3(56) + 2[31 + 4(16)] = 168 + 190 = 358 g/mol

Percentage of iron =  3(56)/358 × 100/1 = 46.9%

Percentage of phosphorus = 2(3)1/358 × 100/1 = 17.3%

Percentage of oxygen = 8(16)/358  × 100/1 =  35.8 %

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