To determine the molar mass of MnClO₄₃, we need to calculate the sum of the atomic masses of all the elements in the compound.
Let's break down the compound into its constituent elements:
- Mn: Manganese
- Cl: Chlorine
- O: Oxygen
To find the atomic masses of these elements, we can refer to the periodic table.
The atomic mass of Mn (Manganese) is approximately 54.938045 g/mol. The atomic mass of Cl (Chlorine) is approximately 35.453 g/mol. The atomic mass of O (Oxygen) is approximately 15.999 g/mol.
Now, let's determine the molar mass of MnClO₄₃:
Molar mass of Mn = 1 × 54.938045 g/mol = 54.938045 g/mol Molar mass of Cl = 3 × 35.453 g/mol = 106.359 g/mol Molar mass of O = 12 × 15.999 g/mol = 191.988 g/mol
Adding these masses together, we get: Molar mass of MnClO₄₃ = 54.938045 g/mol + 106.359 g/mol + 191.988 g/mol = 353.285 g/mol
Therefore, the molar mass of MnClO₄₃ is approximately 353.285 g/mol.