What is the molar mass of an equivalent of MnO2 in the transformation MnO2-MnSO4?
The correct answer and explanation is:
The molar mass of MnO₂ (manganese dioxide) is calculated by summing the atomic masses of its constituent atoms:
- Manganese (Mn): approximately 54.94 g/mol
- Oxygen (O): approximately 16.00 g/mol × 2 = 32.00 g/mol
Total molar mass of MnO₂ = 54.94 + 32.00 = 86.94 g/mol
To find the molar mass of an equivalent of MnO₂ in the transformation to MnSO₄ (manganese sulfate), it is necessary to understand the concept of equivalents in redox reactions.
In the transformation from MnO₂ to MnSO₄, Mn changes its oxidation state from +4 in MnO₂ to +2 in MnSO₄. This means each Mn atom gains 2 electrons during the reduction process:
MnO₂ (Mn⁴⁺) → MnSO₄ (Mn²⁺)
The change in oxidation number is 4 – 2 = 2 electrons per Mn atom.
Equivalent mass is calculated as:
Equivalent mass = Molar mass / n
Where n is the number of electrons gained or lost per formula unit in the redox reaction.
For MnO₂ reducing to MnSO₄, n = 2.
Thus,
Equivalent mass of MnO₂ = 86.94 g/mol ÷ 2 = 43.47 g/equiv
This means one equivalent of MnO₂ corresponds to 43.47 grams, which is the mass of MnO₂ that gains or loses 1 mole of electrons (in this case 2 moles of electrons per mole of MnO₂).
The concept of equivalents is useful in titration and stoichiometric calculations, where reactions involve electron transfer. By using the equivalent mass, chemists can easily relate amounts of substances involved in redox reactions based on the number of electrons transferred rather than the total molecular weight.
In summary, the molar mass of MnO₂ is about 86.94 g/mol, and because the Mn oxidation state changes by 2 in the transformation to MnSO₄, the equivalent mass of MnO₂ is approximately 43.47 g/equiv.