A powder contains FeSO4·7H2O (molar mass=278.01 g/mol), among other components. A 3.125 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3·xH2O, which was subsequently ignited to produce 0.469 g Fe2O3.

What was the mass of FeSO4·7H2O in the 3.125 g sample?

mass of FeSO4·7H2O:

The correct answer and explanation is :

Let’s work through this carefully:

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First, given data:

• Mass of sample = 3.125 g
• Final mass of Fe₂O₃ = 0.469 g
• Molar mass of FeSO₄·7H₂O = 278.01 g/mol

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Step 1: Relate Fe₂O₃ to Fe

We know Fe₂O₃ contains iron (Fe). The molar mass of Fe₂O₃ is:

[
M(\text{Fe}_2\text{O}_3) = 2(55.845) + 3(16.00) = 159.69\ \text{g/mol}
]

Now, find moles of Fe₂O₃:

[
\text{Moles of Fe}_2\text{O}_3 = \frac{0.469\ \text{g}}{159.69\ \text{g/mol}} = 0.002936\ \text{mol}
]

Each mole of Fe₂O₃ contains 2 moles of Fe:

[
\text{Moles of Fe} = 2 \times 0.002936 = 0.005872\ \text{mol}
]

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Step 2: Relate Fe to FeSO₄·7H₂O

In FeSO₄·7H₂O, there is 1 mole of Fe per mole of FeSO₄·7H₂O.

Thus:

[
\text{Moles of FeSO}_4\cdot 7\text{H}_2\text{O} = \text{Moles of Fe} = 0.005872\ \text{mol}
]

Now, find the mass of FeSO₄·7H₂O:

[
\text{Mass of FeSO}_4\cdot 7\text{H}_2\text{O} = 0.005872\ \text{mol} \times 278.01\ \text{g/mol} = 1.632\ \text{g}
]

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✅ Answer:
[
\boxed{1.632\ \text{g}}
]

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Explanation (around 300 words):

This problem involves stoichiometry and the chemical transformation of iron from one compound to another. The powder initially contains FeSO₄·7H₂O among other substances. When treated with nitric acid (HNO₃), all iron is oxidized to Fe³⁺. Adding ammonia (NH₃) precipitates hydrated iron(III) oxide, Fe₂O₃·xH₂O. Heating removes the water, leaving behind pure Fe₂O₃.

We first use the mass of Fe₂O₃ obtained after ignition to find the number of moles of Fe₂O₃. Knowing the molar mass of Fe₂O₃ (159.69 g/mol), the moles are calculated by dividing the mass (0.469 g) by the molar mass. Each mole of Fe₂O₃ contains exactly 2 moles of iron atoms because of the chemical formula (Fe₂O₃). Therefore, multiplying the moles of Fe₂O₃ by 2 gives the moles of Fe present in the original sample.

Since FeSO₄·7H₂O contains one iron atom per formula unit, the number of moles of Fe corresponds directly to the number of moles of FeSO₄·7H₂O originally present. Thus, we multiply the moles of FeSO₄·7H₂O by its molar mass (278.01 g/mol) to find the mass.

Finally, we find that 1.632 g of the 3.125 g sample was FeSO₄·7H₂O. The rest of the sample consisted of other, unspecified components. This method accurately traces the quantity of the original iron-containing compound based on how much Fe₂O₃ was ultimately recovered, an essential technique in quantitative chemical analysis.

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