91Ó°ÊÓ

Stoichiometry deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to determine how much of a substance is consumed or produced when a certain amount of another substance is involved.
In a redox titration, stoichiometry plays a key role because it guides us through the calculations necessary to find the amounts of reactants involved in the reaction sequence. In our example involving \(\mathrm{Fe}^{2+}\) and \(\mathrm{KMnO}_4\), the balanced chemical equation provides a clear stoichiometric ratio: \(1 \, \text{mol} \, \mathrm{MnO}_4^-\) reacts with \(5 \, \text{mol} \, \mathrm{Fe}^{2+}\).
This 1:5 ratio is crucial when calculating how much \(\mathrm{Fe}^{2+}\) is present based on the amount of \(\mathrm{MnO}_4^-\) used in the titration. Using this stoichiometry, we can derive the necessary quantities needed for further calculations, such as determining the mass of iron.

A net ionic equation is a simplified representation of a chemical reaction. It shows only the species that are directly involved in the chemical change, omitting spectator ions.
In the exercise's titration of iron with \(\mathrm{KMnO}_4\), the net ionic equation shows us the essentials of the reaction: \(\mathrm{MnO}_4^-(\mathrm{aq}) + 5\mathrm{Fe}^{2+}(\mathrm{aq}) + 8\mathrm{H}_3\mathrm{O}^+(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + 5\mathrm{Fe}^{3+}(\mathrm{aq}) + 12\mathrm{H}_2\mathrm{O}(\ell)\).
This highlights the core interplay: the permanganate ion (\(\mathrm{MnO}_4^-\)) acts as the oxidizing agent, while the iron(II) ion (\(\mathrm{Fe}^{2+}\)) is oxidized to iron(III) ion (\(\mathrm{Fe}^{3+}\)). This approach simplifies the analysis, focusing only on the species that undergo oxidation and reduction during the reaction, thereby allowing a clear understanding of the electron transfer and reaction stoichiometry.

Mass percent is a way of expressing a concentration or component of a mixture or compound. It helps us understand how much of a particular substance is present in a sample, relative to the total mass of that sample.
To find the mass percent of iron in an iron-containing compound, we first determine the mass of iron from the titration, then compare it to the total mass of the sample. As shown in the step-by-step solution, the mass of iron was computed as \(0.076\, \mathrm{g}\) for the \(0.598\, \mathrm{g}\) sample.
Finally, to calculate the mass percent, we use the formula:

• Divide the mass of iron by the total sample mass.
• Multiply the result by 100 to convert it to a percentage.

Using this, we find that \(\frac{0.076}{0.598} \times 100 = 12.75\%\). This result communicates the proportion of the sample's mass attributable to iron, a helpful measure when analyzing the purity and composition of the sample.