Problem 73 - Chemistry and Chemical Reactivity

Question

To analyze an iron-containing compound, you convert all the iron to \(\mathrm{Fe}^{2+}\) in aqueous solution and then titrate the solution with standardized \(\mathrm{KMnO}_{4} .\) The balanced, net ionic equation is \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+5 \mathrm{Fe}^{2+}(\mathrm{aq})+8 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Mn}^{2+}(\mathrm{aq})+5 \mathrm{Fe}^{3+}(\mathrm{aq})+12 \mathrm{H}_{2} \mathrm{O}(\ell) $$ A \(0.598-\mathrm{g}\) sample of the iron-containing compound requires \(22.25 \mathrm{mL}\) of \(0.0123 \mathrm{M} \mathrm{KMnO}_{4}\) for titration to the equivalence point. What is the mass percent of iron in the sample?

Step-by-Step Solution

Answer

The mass percent of iron in the sample is 12.75%.

Step 1: Calculate Moles of Permanganate

Use the volume and concentration of the \( \mathrm{KMnO}_4 \) solution to find moles of \( \mathrm{MnO}_4^- \). Convert 22.25 mL to liters: \( 22.25 \ \mathrm{mL} = 0.02225 \ \mathrm{L} \). Multiply by the molarity: \[ 0.02225 \ \mathrm{L} \times 0.0123 \ \mathrm{M} = 0.000273075 \ \text{mol} \ \mathrm{MnO}_4^- \]

Step 2: Use Stoichiometry to Find Moles of \( \mathrm{Fe}^{2+} \)

From the balanced equation, one mole of \( \mathrm{MnO}_4^- \) reacts with 5 moles of \( \mathrm{Fe}^{2+} \). Use the mole ratio to calculate moles of \( \mathrm{Fe}^{2+} \): \[ 0.000273075 \ \text{mol} \ \mathrm{MnO}_4^- \times \frac{5 \ \text{mol} \ \mathrm{Fe}^{2+}}{1 \ \text{mol} \ \mathrm{MnO}_4^-} = 0.001365375 \ \text{mol} \ \mathrm{Fe}^{2+} \]

Step 3: Calculate Mass of Iron

Multiply the moles of \( \mathrm{Fe}^{2+} \) by the molar mass of iron: \( 55.85 \ \mathrm{g/mol} \). \[ 0.001365375 \ \text{mol} \ \mathrm{Fe}^{2+} \times 55.85 \ \mathrm{g/mol} = 0.07623754375 \ \mathrm{g} \ \text{Fe} \]

Step 4: Find the Mass Percent of Iron

Divide the mass of iron by the sample mass and multiply by 100 to find the mass percent: \[ \frac{0.07623754375 \ \mathrm{g}}{0.598 \ \mathrm{g}} \times 100 = 12.75\% \]

Key Concepts

Stoichiometry

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.

Net Ionic Equation

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 Calculation

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.