Question

To determine the percent iron in an ore, a \(1.500-\mathrm{g}\) sample of the ore containing \(\mathrm{Fe}^{2+}\) is titrated to the equivalence point with \(18.6 \mathrm{~mL}\) of \(0.05012-\mathrm{M} \mathrm{KMnO}_{4} .\) The products of the titration are \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\). Calculate the weight percent of iron in the ore.

Short answer

The weight percent of iron in the ore is 17.34%.

Step-by-step solution

Write the Balanced Equation

The balanced redox equation for the reaction between ferrous ion \(\mathrm{Fe}^{2+}\) and permanganate ion \(\mathrm{MnO}_4^-\) is: \[ 5\mathrm{Fe}^{2+} + \mathrm{MnO}_4^- + 8\mathrm{H}^+ \rightarrow 5\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} \]

Calculate Moles of \(\mathrm{KMnO}_4\) Used

The titration uses 18.6 mL of 0.05012 M \(\mathrm{KMnO}_4\). Convert mL to L and calculate moles:\[ \text{Volume of } \mathrm{KMnO}_4 = \frac{18.6 \text{ mL}}{1000} = 0.0186 \text{ L} \]\[ \text{Moles of } \mathrm{KMnO}_4 = 0.0186 \text{ L} \times 0.05012 \text{ mol/L} = 0.000932232 \text{ mol} \]

Determine Moles of \(\mathrm{Fe}^{2+}\) in Sample

From the balanced equation, 1 mole of \(\mathrm{MnO}_4^-\) reacts with 5 moles of \(\mathrm{Fe}^{2+}\). Calculate moles of \(\mathrm{Fe}^{2+}\):\[ \text{Moles of } \mathrm{Fe}^{2+} = 5 \times \text{Moles of } \mathrm{KMnO}_4 = 5 \times 0.000932232 = 0.00466116 \]

Calculate Mass of Iron in the Ore

Using the molar mass of iron (\(55.85 \text{ g/mol}\)), calculate the mass of iron:\[ \text{Mass of Fe} = 0.00466116 \text{ mol} \times 55.85 \text{ g/mol} = 0.2601599 \text{ g} \]

Calculate Percent Weight of Iron in the Ore

Calculate the weight percentage by dividing the mass of iron by the mass of the ore sample and multiplying by 100:\[ \text{Percent Iron} = \left(\frac{0.2601599 \text{ g}}{1.500 \text{ g}}\right) \times 100 = 17.34399\% \approx 17.34\% \]

Key concepts

Redox Reaction

Redox reactions are chemical processes where oxidation and reduction occur simultaneously. In these reactions, one substance loses electrons (oxidation) and another gains electrons (reduction). The essence of a redox reaction is the transfer of electrons between two substances.
In the titration involving iron and potassium permanganate (KMnO_4), the redox equation is: \[5\mathrm{Fe}^{2+} + \mathrm{MnO}_4^- + 8\mathrm{H}^+ \rightarrow 5\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\] Here, \text{Fe}^{2+} ions are oxidized to \text{Fe}^{3+} by losing electrons. On the other hand, the permanganate ion \text{MnO}_4^- is reduced to \text{Mn}^{2+} by gaining electrons. Understanding redox reactions is essential as it helps to identify what happens to electrons, allowing us to use redox reactions to calculate concentrations and masses in chemical analysis.

Percent Composition

Percent composition refers to the percentage by mass of a particular element within a compound. It is a vital concept in chemistry, often used to determine the amount of an element in a given sample.
For the iron ore problem, after the titration, we calculated the mass of iron present in the sample to be approximately 0.260 g. To find out how much iron there is relative to the whole ore sample, we compute the percent composition using the formula: \[\text{Percent Iron} = \left(\frac{\text{mass of iron}}{\text{mass of sample}}\right) \times 100\] This calculation results in a percent iron of about 17.34%. This value represents how much of the ore sample's mass can be attributed to iron, offering insights into the ore's quality and suitability for use.

Moles Calculation

Moles Calculation is a fundamental concept in chemistry that involves determining the number of moles of a substance. The mole is a standard unit for measuring large quantities of very small entities, such as atoms or molecules, to bridge the gap between the atomic scale and the real-world measurement.
In the titration problem, we first calculated the moles of \text{KMnO}_4 used in the reaction by considering its volume and molarity: \[\text{Moles of } \mathrm{KMnO}_4 = \text{Volume (L)} \times \text{Molarity (mol/L)}\] Using these moles, we applied the stoichiometry from the balanced redox equation to determine the moles of \text{Fe}^{2+} ions in the sample, multiplying the moles of \text{KMnO}_4 by 5, as indicated by the mole ratio: \[\text{Moles of } \mathrm{Fe}^{2+} = 5 \times \text{Moles of } \mathrm{KMnO}_4\] This step is crucial as it lays the groundwork for calculating the mass of iron, and subsequently, its percent composition in the ore.