- Determine the moles of permanganate used: \(n_{MnO4} = 0.00063 \text{ moles}\).
- Calculate the moles of iron: \(n_{Fe} = 5 \times n_{MnO4} = 0.00315 \text{ moles}\).
- Find the mass percent of iron: \(\%Fe = 35.18235\%\).
- Calculate the mass percent as \(Fe_2O_3\): \(\%Fe_2O_3 = 50.301405\%\).
- Final Answer: \(\boxed{35.18235\%, 50.301405\%}\)
Explanation
- Problem Analysis
The objective is to determine the mass percent of iron (Fe) in a \(0.5000 \text{ g}\) ore sample and the corresponding mass percent if the iron were present as hematite (\(Fe_2O_3\)). We are provided with the sample mass (\(m_{sample} = 0.5000 \text{ g}\)), the concentration of the permanganate titrant (\(C_{KMnO4} = 0.0200 \text{ M}\)), and the volume of titrant used (\(V_{KMnO4} = 31.50 \text{ mL} = 0.03150 \text{ L}\)). The reaction is \(MnO_4^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\), indicating that \(1 \text{ mole}\) of \(MnO_4^{-}\) reacts with \(5 \text{ moles}\) of \(Fe^{2+}\). The molar masses involved are approximately \(55.845 \text{ g/mol}\) for iron and \(159.687 \text{ g/mol}\) for \(Fe_2O_3\).
- Calculating Mass Percent of Fe
First, we calculate the number of moles of permanganate (\(MnO_4^{-}\)) consumed in the titration:
\[n_{MnO4} = C_{KMnO4} \times V_{KMnO4} = 0.0200 \text{ M} \times 0.03150 \text{ L} = 0.00063 \text{ moles}.\]
Next, using the stoichiometry of the reaction, we find the moles of iron present in the sample:
\[n_{Fe} = 5 \times n_{MnO4} = 5 \times 0.00063 = 0.00315 \text{ moles}.\]
Now, calculate the mass of iron in the sample:
\[m_{Fe} = n_{Fe} \times M_{Fe} = 0.00315 \text{ moles} \times 55.845 \text{ g/mol} = 0.17591175 \text{ g}.\]
The mass percent of iron in the sample is:
\[\%Fe = \left(\frac{m_{Fe}}{m_{sample}}\right) \times 100\% = \left(\frac{0.17591175 \text{ g}}{0.5000 \text{ g}}\right) \times 100\% = 35.18235\%.\]
- Calculating Mass Percent of Fe2O3 and Conclusion
If the iron in the sample were present as hematite (\(Fe_2O_3\)), we would use the relationship between the moles of \(Fe\) and \(Fe_2O_3\). Since each mole of \(Fe_2O_3\) contains two moles of \(Fe\), we have:
\[n_{Fe2O3} = \frac{n_{Fe}}{2} = \frac{0.00315}{2} = 0.001575 \text{ moles}.\]
The total mass of \(Fe_2O_3\) is:
\[m_{Fe2O3} = n_{Fe2O3} \times M_{Fe2O3} = 0.001575 \text{ moles} \times 159.687 \text{ g/mol} = 0.251507025 \text{ g}.\]
The mass percent of \(Fe_2O_3\) in the sample is:
\[\%Fe_2O_3 = \left(\frac{m_{Fe2O3}}{m_{sample}}\right) \times 100\% = \left(\frac{0.251507025 \text{ g}}{0.5000 \text{ g}}\right) \times 100\% = 50.301405\%.\]
Thus, the mass percent of \(Fe\) is \(35.18235\%\) and the mass percent if present as \(Fe_2O_3\) is \(50.301405\%\). Final results are: \(\boxed{35.18235\%, 50.301405\%}\).
Examples
In the mining and steel industries, precise chemical analysis of ore samples is crucial for assessing economic viability and determining process parameters. For example, by titrating dissolved ore samples with potassium permanganate, engineers can accurately measure the iron content. This helps in optimizing the furnace operations for metal extraction and ensuring quality control in the production of steel alloys. Such analytical techniques form the backbone of industrial chemistry and material quality management.