Problem 66 - Chemistry

Question

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to Fe(II) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

Step-by-Step Solution

Answer

The percent by mass of iron in the ore is 45.16%.

Step 1: Write the balanced chemical equation

The reaction between Fe(II) ions and KMnO_4 can be written as: \[ 5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O \] This tells us that 1 mole of KMnO_4 reacts with 5 moles of Fe(II).

Step 2: Calculate moles of KMnO_4

Using the volume and concentration of KMnO_4, calculate the moles:\[ \text{Moles of } KMnO_4 = 0.0194 \; \text{M} \times 0.02330 \; \text{L} = 4.52 \times 10^{-4} \; \text{mol} \]

Step 3: Determine moles of Fe(II)

Since 1 mole of KMnO_4 reacts with 5 moles of Fe(II), we can determine:\[ 5 \times 4.52 \times 10^{-4} \; \text{mol} = 2.26 \times 10^{-3} \; \text{mol Fe} \]

Step 4: Calculate mass of iron

Using the molar mass of iron (approximately 55.85 g/mol), find the mass:\[ \text{Mass of Fe} = 2.26 \times 10^{-3} \; \text{mol} \times 55.85 \; \text{g/mol} = 0.1261 \; \text{g} \]

Step 5: Calculate percent by mass of iron in the ore

The percent by mass is calculated using the formula:\[ \frac{0.1261 \; \text{g}}{0.2792 \; \text{g}} \times 100\% = 45.16\% \]

Key Concepts

Iron Oxidation

Iron oxidation is a chemical reaction involving the transformation of iron(II) ions to iron(III) ions. In the given problem, iron initially in the form of Fe(II) is transformed into Fe(III) by reacting with potassium permanganate, a strong oxidizing agent. This transformation involves the loss of electrons by iron, which is the essence of oxidation. Iron(II) ions lose an electron to become iron(III) ions, represented as Fe²⁺ to Fe³⁺. Understanding this shift is vital in stoichiometry as it implies the balance of electrons and atoms involved in reactions. This balanced flow allows us to predict the ratio in which substances react and are produced.

Molar Mass

Molar mass is a crucial concept in stoichiometry that refers to the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). In this context, molar mass helps to convert moles to grams, facilitating the solution of stoichiometry problems. For iron, the molar mass is approximately 55.85 g/mol. This value is derived from the average mass of iron atoms, considering isotopic abundances. Whenever you calculate the mass of a substance from moles, you multiply the moles by its molar mass. This essential conversion factor ensures every mole of a substance is accurately represented by its weight.

Percent Composition

Percent composition measures how much of a given element or compound is present in a mixture. It is expressed as a percentage of the total mass. In the exercise, the percent by mass of iron in the ore is calculated by dividing the mass of iron obtained during the experiment by the initial sample mass, then multiplying by 100. This tells us what fraction of the original ore is composed of iron. The formula used for calculating percent composition is: \[ \frac{\text{{mass of element}}}{\text{{total mass of compound}}} \times 100 \% \]. This calculation is vital in determining the composition of alloys, compounds, or any sample where specific elemental content information is desired.

Chemical Equations

Chemical equations represent reactions where substances transform and rearrange, providing a snapshot of the reactants and products. The balanced equation for iron's oxidation to Fe(III) demonstrates this: \[ 5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O \]. Here, it illustrates how one mole of potassium permanganate reacts with five moles of iron(II) ions, showing precisely how moles interact. Balancing chemical equations is foundational for stoichiometry as it ensures atom conservation and accurate stoichiometric calculations. It dictates proper proportions, enabling precise determination of reactant quantities and prediction of product amounts.