Question

On heating a \(0.200 \mathrm{~g}\) sample of a certain semimetal \(\mathrm{M}\) in air, the corresponding oxide \(\mathrm{M}_{2} \mathrm{O}_{3}\) was obtained. When the oxide was dissolved in aqueous acid and titrated with \(\mathrm{KMnO}_{4}, 10.7 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required for complete reaction. The unbalanced equation is \(\mathrm{H}_{3} \mathrm{MO}_{3}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{MO}_{4}(a q)\) \(+\mathrm{Mn}^{2+}(a q)(\) in acid \()\) (a) Balance the equation. (b) How many moles of oxide were formed, and how many moles of semimetal were in the initial \(0.200 \mathrm{~g}\) sample? (c) What is the identity of the semimetal M?

Short answer

M is Iridium (Ir).

Step-by-step solution

Balance the Redox Reaction

Begin by separating the reaction into oxidation and reduction half-reactions. The oxidation is just the conversion of \( \mathrm{H}_{3}\mathrm{MO}_{3} \) to \( \mathrm{H}_{3}\mathrm{MO}_{4} \), losing two electrons. The reduction is \( \mathrm{MnO}_{4}^{-} \) to \( \mathrm{Mn}^{2+} \), gaining five electrons.- Oxidation Half: \( \mathrm{H}_{3}\mathrm{MO}_{3} \rightarrow \mathrm{H}_{3}\mathrm{MO}_{4} + 2e^{-} \)- Reduction Half: \( \mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} \)Balance the number of electrons transferred by multiplying the oxidation half by 5 and the reduction half by 2. Then, combine the half-reactions to get the balanced equation:\[ 2 \mathrm{H}_{3}\mathrm{MO}_{3} + 2 \mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} \rightarrow 2 \mathrm{H}_{3}\mathrm{MO}_{4} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_2\mathrm{O} \]

Calculate Moles of Permanganate Used

To find the moles of \( \mathrm{MnO}_{4}^{-} \) used, use the molarity and volume of the solution: \[ \text{Moles of } \mathrm{MnO}_{4}^{-} = 0.100 \ \mathrm{M} \times 0.0107 \ \mathrm{L} = 0.00107 \ \text{moles} \]

Relate Moles of MnO4- to Moles of Oxide

Since the stoichiometry of the balanced equation indicates 2 moles of \( \mathrm{MnO}_{4}^{-} \) react with 1 mole of \( \mathrm{M}_2\mathrm{O}_3 \), we find the moles of \( \mathrm{M}_2\mathrm{O}_3 \) formed:\[ \text{Moles of } \mathrm{M}_2\mathrm{O}_3 = \frac{0.00107}{2} = 0.000535 \text{ moles} \]

Determine Moles of Semimetal M

From the formula \( \mathrm{M}_2\mathrm{O}_3 \), each mole of oxide contains 2 moles of the semimetal \( \mathrm{M} \). Using the moles of \( \mathrm{M}_2\mathrm{O}_3 \) calculated earlier:\[ \text{Moles of M} = 2 \times 0.000535 = 0.00107 \text{ moles} \]

Calculate Atomic Mass of Semimetal M

Using the initial mass of the semimetal and number of moles, calculate the atomic mass:\[ \text{Atomic mass of M} = \frac{0.200 \text{ g}}{0.00107 \text{ moles}} \approx 187 \text{ g/mol} \]The calculated atomic mass is closest to the atomic mass of Iridium (Ir).

Conclusion

The semimetal M is Iridium (Ir). Based on the balanced redox equation and sample calculations, Iridium appears in the compound \( \mathrm{M}_2\mathrm{O}_3 \) and reacts with permanganate, consistent with its reactivity and oxidation potentials.

Key concepts

Balancing Chemical Equations

Balancing chemical equations ensures the principle of conservation of mass is maintained. For any chemical reaction, the number of atoms for each element must be the same on both sides of the equation. In redox reactions involving semimetals, balancing takes on an added layer of complexity because electrons also need to be balanced.
This particular exercise involves balancing the conversion of \( \mathrm{H}_{3}\mathrm{MO}_{3} \) to \( \mathrm{H}_{3}\mathrm{MO}_{4} \), and \( \mathrm{MnO}_{4}^{-} \) to \( \mathrm{Mn}^{2+} \).
Each half-reaction is balanced for mass and charge separately before combining them into a complete balanced equation.
The technique of splitting the reaction into oxidation and reduction half-equations helps organize the balancing of both atoms and electrons effectively.
This careful balancing is crucial for correctly interpreting the stoichiometry of the reaction, leading to accurate calculations of quantities involved in the reaction.

Stoichiometry

Stoichiometry involves calculating the amounts of reactants and products in a chemical reaction. In this context, it allows us to determine the quantities of semimetal oxide and permanganate involved in the reaction.
The balanced chemical equation indicates the molar relationships between the substances reacting and what is formed.
By knowing the balanced equation, you can determine exactly how much of one reactant you need to completely react with another or how much product will be produced.

• This requires translating the coefficients of the balanced equation into mole ratios.
• For instance, we calculate that 2 moles of \( \mathrm{MnO}_{4}^{-} \) will react with each mole of \( \mathrm{M}_2\mathrm{O}_{3} \).

With stoichiometry, we understand that knowing one piece of the puzzle, like the amount of one reactant, gives us the power to calculate the rest.

Oxidation and Reduction

Understanding oxidation and reduction is essential to mastering redox reactions. These processes involve the transfer of electrons between substances.
In redox reactions, oxidation refers to the loss of electrons, while reduction involves the gain of electrons.
In the exercise reaction, the semimetal in \( \mathrm{H}_{3}\mathrm{MO}_{3} \) loses electrons, which is oxidation, while the permanganate ion \( \mathrm{MnO}_{4}^{-} \) gains electrons, signifying reduction.

• The balanced half-reactions highlight clearly how electrons move between compounds.
• One part of the reaction will always pair with its opposite (oxidation with reduction) to ensure all electrons are accounted for.

To fully balance a redox equation, we align electron transfer, which is a key step in understanding how reactant molecules come together and change.

Chemical Identity

Identifying chemical substances accurately is crucial in chemistry. This involves determining the atomic or molecular identity of elements or compounds.
In this case, identifying the semimetal \( M \) is an element of puzzle-solving in chemistry.
By balancing the chemical equation and performing stoichiometric calculations, we determine the chemical identity by comparing calculated atomic masses to known values on the periodic table.

• The exercise reveals the semimetal \( M \) to be Iridium after calculations show its atomic mass to closely match that of Ir.

This aspect of the exercise strengthens one's understanding of the elements and their typical reactivities, as each element has a unique set of properties.

Semimetals in Chemistry

Semimetals, sometimes known as metalloids, hold a unique place between metals and non-metals in the periodic table. They exhibit properties shared by both, creating fascinating opportunities for chemical reactions and materials science.
In this exercise, \( M \) is a semimetal reacting to form an oxide. Such reactions are significant in producing materials with desirable properties.

• Semimetals can conduct electricity like metals but often do not allow heat to move through them as freely.
• This dual nature ensures that semimetals like Iridium find uses in specialized applications, including electronics and catalysis.

Understanding how semimetals react and balance chemically broadens the perspective on their role in both industrial applications and academic research.