Problem 15 - Analytical Chemistry

Question

What weight of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) precipitate would be obtained from a \(0.4823-\mathrm{g}\) sample of iron wire that is \(99.89 \%\) pure?

Step-by-Step Solution

Answer

0.689 grams of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) precipitate is obtained.

Step 1: Calculate the Mass of Pure Iron

To start, find the mass of pure iron in the wire. The wire is 99.89% pure, so we calculate the mass of pure Fe as follows:\[\text{Mass of pure Fe} = 0.4823\, \text{g} \times \frac{99.89}{100} = 0.48177\, \text{g}\]

Step 2: Find the Moles of Iron

Next, we need to convert the mass of iron to moles. Use the molar mass of iron, which is approximately 55.85 g/mol:\[\text{Moles of Fe} = \frac{0.48177\, \text{g}}{55.85\, \text{g/mol}} = 0.008625\, \text{mol}\]

Step 3: Determine Moles of Fe in \\(\mathrm{Fe}_{2}\mathrm{O}_{3}\\)

The reaction that forms \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) is based on iron, and it involves 2 moles of Fe for every mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\):\[\text{Moles of } \mathrm{Fe}_{2}\mathrm{O}_{3} = \frac{0.008625\, \text{mol Fe}}{2} = 0.0043125\, \text{mol}\]

Step 4: Calculate Mass of \\(\mathrm{Fe}_{2}\mathrm{O}_{3}\\) Precipitate

Find the mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) formed using its molar mass (159.7 g/mol):\[\text{Mass of } \mathrm{Fe}_{2}\mathrm{O}_{3} = 0.0043125\, \text{mol} \times 159.7\, \text{g/mol} = 0.6887\, \text{g}\]

Step 5: Conclusion

The weight of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) precipitate obtained from the given iron wire sample is approximately 0.689 grams.

Key Concepts

Chemical Reactions

A chemical reaction involves the transformation of one set of chemical substances to another. In the given exercise, iron (Fe) reacts with oxygen to form iron(III) oxide, \( \mathrm{Fe}_{2}\mathrm{O}_{3} \). This is an example of a synthesis reaction, where simpler substances combine to form a complex product. During this reaction, two moles of iron typically combine with one mole of oxygen to create the iron oxide precipitate.

Chemical reactions can be represented using balanced equations, which show the exact proportion of reactants converting to products. The balanced reaction equation for the formation of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \) from iron is:

• 4 Fe + 3 O\(_2\) → 2 \( \mathrm{Fe}_{2}\mathrm{O}_{3} \)

This tells us the precise stoichiometry of the chemicals involved in this process, ensuring that the mass is conserved in accordance with the law of conservation of mass, where the mass of the reactants equals the mass of the products. Understanding the stoichiometry of a reaction is crucial for calculating quantities like the mass of precipitate, as seen in the given problem.

Molar Mass

The concept of molar mass is fundamental in stoichiometry as it provides a bridge from the mass of a substance to moles. The molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). For example, the molar mass of iron (Fe) is 55.85 g/mol.

To convert the mass of a substance to moles, divide the sample's mass by its molar mass. In our exercise, we had a pure iron mass of 0.48177 grams. Using iron's molar mass, we calculated the moles of iron:

• \( \text{Moles of Fe} = \frac{0.48177\, \text{g}}{55.85\, \text{g/mol}} = 0.008625\, \text{mol} \)

Molar mass is also integral in the final step to find the mass of a different substance from its mole value. For the formation of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \), we used its molar mass (159.7 g/mol) to determine the mass of the precipitate from its moles calculated earlier.

Precipitate Calculation

Calculating the amount of precipitate formed in a chemical reaction is an essential stoichiometric problem. In our exercise, the chemical of interest is iron(III) oxide, \( \mathrm{Fe}_{2}\mathrm{O}_{3} \), a solid precipitate formed from the reaction of iron with oxygen.

To find how much precipitate is formed, first determine the moles of the limiting reactant, in this case, iron. Then use the stoichiometry of the balanced reaction to find the moles of the precipitate. For instance, the reaction indicates that 2 moles of Fe produce 1 mole of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \). Hence, 0.008625 moles of Fe correspond to 0.0043125 moles of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \).

Finally, multiply the moles of the precipitate by its molar mass to find the mass:

• \( \text{Mass of } \mathrm{Fe}_{2}\mathrm{O}_{3} = 0.0043125\, \text{mol} \times 159.7\, \text{g/mol} = 0.6887\, \text{g} \)

This calculation confirms that 0.6887 grams of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \) are formed, providing a concrete understanding of the reaction's yield.