Problem 43 - Chemistry: Principles and Reactions

Question

A \(0.500-g\) sample of steel is analyzed for manganese. The sample is dissolved in acid and the manganese is oxidized to permanganate ion. A measured excess of \(\mathrm{Fe}^{2+}\) is added to reduce \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\). The excess \(\mathrm{Fe}^{2+}\) is determined by titration with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{27}\). If \(75.00 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{FeSO}_{4}\) is added and the excess requires \(13.50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to oxidize \(\mathrm{Fe}^{2+}\), calculate the percent by mass of \(\mathrm{Mn}\) in the sample.

Step-by-Step Solution

Answer

Answer: The mass percent of manganese in the steel sample is 2.802%.

Step 1: Calculate the moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) consumed in the reaction

Using the volume and concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), we can calculate the moles consumed in the titration. The given volume and concentration are \(13.50 \mathrm{~mL}\) and \(0.100 \mathrm{M}\) respectively. Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} = \text{Volume} \times \text{Concentration} = (13.50 \times 10^{-3}) \times 0.100\) Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} = 1.35 \times 10^{-3}\) moles

Step 2: Calculate the moles of excess \(\mathrm{Fe}^{2+}\) used in the reaction

The reaction between \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and \(\mathrm{Fe}^{2+}\) can be represented by the balanced equation: \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}\) From the balanced equation, 1 mole of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) reacts with 6 moles of \(\mathrm{Fe}^{2+}\), therefore: Moles of excess \(\mathrm{Fe}^{2+} = 1.35 \times 10^{-3} \times 6 = 8.1 \times 10^{-3}\) moles

Step 3: Calculate the moles of \(\mathrm{Fe}^{2+}\) initially added

The given volume and concentration of \(\mathrm{FeSO}_{4}\) solution are \(75.00 \mathrm{~mL}\) and \(0.125 \mathrm{M}\) respectively. Moles of \(\mathrm{Fe}^{2+}\) initially added = \(\mathrm{Volume} \times \mathrm{Concentration} = (75.00 \times 10^{-3}) \times 0.125\) Moles of \(\mathrm{Fe}^{2+}\) initially added = 9.375 \times 10^{-3}$ moles

Step 4: Calculate the moles of \(\mathrm{Fe}^{2+}\) used to reduce \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\)

The moles of \(\mathrm{Fe}^{2+}\) used to reduce \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\) can be calculated by subtracting the moles of excess \(\mathrm{Fe}^{2+}\) from the total moles of \(\mathrm{Fe}^{2+}\) initially added: Moles of \(\mathrm{Fe}^{2+}\) used = 9.375 \times 10^{-3} - 8.1 \times 10^{-3} = 1.275 \times 10^{-3}$ moles

Step 5: Calculate the moles of \(\mathrm{Mn}\) in the sample

The balanced equation for reduction of \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\) by \(\mathrm{Fe}^{2+}\) is: \(\mathrm{MnO}_{4}^{-} + 5 \mathrm{Fe}^{2+} + 8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H}_{2} \mathrm{O}\) 1 mole of \(\mathrm{MnO}_{4}^{-}\) react with 5 moles of \(\mathrm{Fe}^{2+}\), and we have moles of \(\mathrm{Fe}^{2+}\) used to reduce \(\mathrm{MnO}_{4}^{-}\) from the previous step. Hence: Moles of \(\mathrm{Mn} = \frac{1}{5} \times 1.275 \times 10^{-3} = 2.55 \times 10^{-4}\) moles

Step 6: Calculate the mass of \(\mathrm{Mn}\) in the sample

Using the molar mass of \(\mathrm{Mn}\), which is \(54.94 \mathrm{g/mol}\): Mass of \(\mathrm{Mn} = (2.55 \times 10^{-4}) \times 54.94 = 1.401 \times 10^{-2}\) g

Step 7: Calculate the mass percent of \(\mathrm{Mn}\) in the sample

Finally, we can calculate the mass percent of \(\mathrm{Mn}\) in the sample using the mass of the sample and the mass of manganese: Mass percent of \(\mathrm{Mn} = \frac{1.401 \times 10^{-2}}{0.500} \times 100 \% = 2.802 \%\) The mass percent of manganese in the sample is \(2.802 \%\).

Key Concepts

Titration Analysis

Titration is an essential technique used in chemistry to determine the concentration of an unknown solution by allowing it to react with a standard solution of known concentration, known as the titrant. This process involves adding the titrant to the unknown solution from a burette, a process known as titration.

During the process, a color change or an indicator is often used to determine the end point, which signifies the completion of the reaction. In the context of the provided exercise, titration is used to determine the excess \text{\(\text{Fe}^{2+}\)} after it has reduced the \text{\(\text{MnO}_{4}^{-}\)} to \text{\(\text{Mn}^{2+}\)}. The excess \text{\(\text{Fe}^{2+}\)} is titrated with \text{\(\text{K}_{2} \text{Cr}_{2} \text{O}_{7}\)}, a redox titration where the end point can be identified by the change in color of the solution, without the need of an external indicator, as the reactants themselves act as self-indicators.

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar mass, we sum the atomic masses of all atoms in the molecule as provided in the periodic table of elements.

For instance, in our exercise, the molar mass of \text{\(\text{Mn}\)} (manganese) is given as 54.94 g/mol. Knowing the molar mass allows us to convert between mass and number of moles, a crucial step in stoichiometric calculations. In the problem, the mass of manganese present in the steel sample was calculated by first determining the moles of manganese and then using its molar mass to find the corresponding mass.

Redox Reactions

Redox reactions are chemical processes where the oxidation states of atoms are changed due to the transfer of electrons. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. These reactions are key in many analytical techniques, including titration.

In the provided exercise, the titration involves a redox reaction between \text{\(\text{Fe}^{2+}\)} and \text{\(\text{MnO}_{4}^{-}\)}. The \text{\(\text{Fe}^{2+}\)} ions are oxidized to \text{\(\text{Fe}^{3+}\)} ions, while the \text{\(\text{MnO}_{4}^{-}\)} ions are reduced to \text{\(\text{Mn}^{2+}\)}. Understanding the stoichiometry of redox reactions is crucial as it determines the amount of titrant needed to reach the end point, which in turn gives us information about the unknown concentration or the quantity of a substance within a sample.