Questions: Calculate the molar mass of manganese (III) sulfite, Mn2(SO3)3. Show your work.

Solution Steps

Manganese (III) sulfite has the chemical formula $\mathrm{Mn}_{2}\left(\mathrm{SO}_{3}\right)_{3}$. This means it contains:

• 2 manganese (Mn) atoms
• 3 sulfite ($\mathrm{SO}_{3}$) groups

Each sulfite group contains:

• 1 sulfur (S) atom
• 3 oxygen (O) atoms

Using the periodic table, we find the atomic masses:

• Manganese (Mn): 54.9380 g/mol
• Sulfur (S): 32.0650 g/mol
• Oxygen (O): 15.9994 g/mol

• Manganese: $2 \times 54.9380 \, \text{g/mol} = 109.8760 \, \text{g/mol}$
• Sulfur: $3 \times 32.0650 \, \text{g/mol} = 96.1950 \, \text{g/mol}$
• Oxygen: $3 \times 3 \times 15.9994 \, \text{g/mol} = 143.9946 \, \text{g/mol}$

Add the masses of all the elements to find the molar mass of $\mathrm{Mn}_{2}\left(\mathrm{SO}_{3}\right)_{3}$: $$109.8760 \, \text{g/mol} + 96.1950 \, \text{g/mol} + 143.9946 \, \text{g/mol} = 350.0656 \, \text{g/mol}$$

Final Answer