To calculate the molar weight of iron(II) ammonium sulfate hexahydrate, represented as Fe(NH₄)₂(SO₄)₂·6H₂O, we need to take into account the molar mass of each component in the compound.
1. **Iron (Fe)**: The molar mass is approximately 55.85 g/mol.
2. **Ammonium (NH₄)**: The molar mass of NH₄ is about 18.04 g/mol. Since there are two ammonium ions, we multiply this by 2: 2 x 18.04 g/mol = 36.08 g/mol.
3. **Sulfate (SO₄)**: The molar mass of SO₄ is approximately 96.06 g/mol. As there are two sulfate ions, we multiply this by 2: 2 x 96.06 g/mol = 192.12 g/mol.
4. **Water (H₂O)**: The molar mass of one water molecule is about 18.02 g/mol. With six water molecules in hexahydrate, we multiply this by 6: 6 x 18.02 g/mol = 108.12 g/mol.
Now, we can add all these values together to find the total molar mass:
Molar Mass = 55.85 g/mol (Fe) + 36.08 g/mol (2 NH₄) + 192.12 g/mol (2 SO₄) + 108.12 g/mol (6 H₂O) = 392.17 g/mol
Based on the options provided, the closest value to our calculated molar weight is 392.14 g/mol (option e).
Thus, the molar weight of iron(II) ammonium sulfate hexahydrate is 392.14 g/mol.