Question

How to calculate the mass of iron in .275 grams of iron iii sulfate?

Answer 1

Mass of iron =
`0.275 grams`

The chemical formula of iron (III) sulfate =
`Fe_(2)(SO_(4))_(3)`

First, calculate the number of moles of
`Fe_(2)(SO_(4))_(3)`

Number of moles of iron (III) sulfate =
`(given mass in g)/(molar mass)`

Molar mass of iron (III) sulfate = 399.88 g/mol

Number of moles of iron (III) sulfate =
`(0.275 g)/(399.88 g/mol)`

=
`6.87* 10^(-4) mol`

In one mole of iron (III) sulfate there are two moles of iron according to the chemical formula.

So, in
`6.87* 10^(-4) mol` of iron (III) sulfate there are =
`2* 6.87 * 10^(-4) mol` of iron.

Number of moles of iron =
`13.74* 10^(-4) mol`

Molar mass of iron =
`55.8 g/mol`

Mass of iron =
`number of moles * molar mass`

=
`13.74* 10^(-4) mol * 55.8 g/mol`

=
`766.692 * 10^(-4)  g`

=
`7.66 * 10^(-2)  g`

Hence, mass of iron =
`7.66 * 10^(-2)  g`

Answer 2

Molar mass of iron(III)sulfate that is Fe₂(SO₄)₃ =399.88 g mol⁻¹

Mass given of iron(III)sulfate that is Fe₂(SO₄)₃= 0.275 g

Number of moles of iron(III)sulfate that is Fe₂(SO₄)₃=
`(0.275)/(399.88)`

=0.00068 mol

there are two moles of Fe in 1 mole of Fe₂(SO₄)₃

So, in 0.00068 mol of Fe₂(SO₄)₃=2 ×0.00068 mol of Fe

= 0.00136 mol of Fe

molar mass of Fe=55.8 g mol⁻¹

So the mass of iron in 0.00136 mol of Fe= 55.8 g mol⁻¹ × 0.00136 mol

=0.075 g