Determine the molar mass of Mn(ClO3)3.
The Correct Answer and Explanation is :
To determine the molar mass of manganese(III) chlorate, Mn(ClO₃)₃, we need to calculate the mass contribution of each element present in the compound.
- Manganese (Mn): The atomic mass of manganese is approximately 54.94 g/mol. There is one manganese atom in the formula.
- Chlorate (ClO₃): The chlorate ion is made up of one chlorine (Cl) atom and three oxygen (O) atoms.
- The atomic mass of chlorine (Cl) is 35.45 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol, and since there are three oxygen atoms in the chlorate ion, the total mass for oxygen is 3 × 16.00 = 48.00 g/mol.
Now, we can calculate the molar mass of Mn(ClO₃)₃ by adding up the contributions:
- Manganese (Mn): 54.94 g/mol × 1 = 54.94 g/mol
- Chlorine (Cl): 35.45 g/mol × 3 = 106.35 g/mol
- Oxygen (O): 16.00 g/mol × 9 = 144.00 g/mol (since there are 3 oxygen atoms per chlorate ion, and there are three chlorate ions)
Total molar mass of Mn(ClO₃)₃ = 54.94 g/mol (Mn) + 106.35 g/mol (Cl) + 144.00 g/mol (O) = 305.29 g/mol
Thus, the molar mass of Mn(ClO₃)₃ is 305.29 g/mol.
This calculation involves adding the atomic masses of each element, considering the number of atoms for each element in the compound. This is a fundamental process in chemistry to determine the molar mass of a compound, which is essential for various calculations in stoichiometry, such as determining the amount of a substance required for a reaction.