Question

Calculate the percentage of iron in iron (III) sulphate

Answer 1

28.01 percent

Step-by-step explanation:

Iron (III) sulphate,
`Fe_2(SO_4)_3`, has 2 atoms of Fe.

One atom of Fe has a molar weight of 56. Hence, 2 atoms of Fe would have a total molar weight of 56 x 2 = 112 g/mol

Molar weight of
`Fe_2(SO_4)_3` = 399.88 g/mol

Percentage Fe in
`Fe_2(SO_4)_3` = 112/399.88 x 100%

= 28.01%

The percentage of iron is iron (III) sulphate is, therefore, 28.01 percent.