Final answer:
To find the molar mass of Mn(ClO3)4, add the atomic masses of Mn, Cl, and O in the compound, resulting in 388.74 g/mol.
Step-by-step explanation:
To determine the molar mass of Mn(ClO3)4, we need to calculate the sum of the atomic masses of all the atoms in the compound. First, we find the atomic masses (in g/mol) for each type of atom from the periodic table: Mn (Manganese) has an atomic mass of approximately 54.94 g/mol, Cl (Chlorine) has an atomic mass of about 35.45 g/mol, and O (Oxygen) has an atomic mass of about 16.00 g/mol.
There is one Mn atom, four Cl atoms and twelve O atoms in one formula unit of Mn(ClO3)4. We can calculate the molar mass as follows:
- Mn: 1 × 54.94 g/mol = 54.94 g/mol
- Cl: 4 × 35.45 g/mol = 141.80 g/mol
- O: 12 × 16.00 g/mol = 192.00 g/mol
Adding these together, the molar mass of Mn(ClO3)4 is:
54.94 g/mol + 141.80 g/mol + 192.00 g/mol = 388.74 g/mol