Final answer:
A single formula unit of cerium(IV) sulfate, Ce(SO₄)₂·4 H₂O, contains 1 Ce atom, 2 S atoms, 12 O atoms, 8 H atoms. The percentage by mass of water in the compound is approximately 20.5%.
Step-by-step explanation:
In cerium(IV) sulfate, the subscript following Ce indicates there is 1 cerium atom in each formula unit. For sulfate (SO₄)²⁻, the subscript 2 indicates that there are 2 sulfate ions in one formula unit, leading to 2 sulfur atoms and 8 oxygen atoms (4 per sulfate ion). The coefficient 4 in front of H₂O indicates that there are 4 water molecules associated with each formula unit, contributing 8 hydrogen atoms and 4 oxygen atoms.
To calculate the molar mass of Ce(SO₄)₂·4 H₂O, we sum the individual atomic masses. The molar mass of water (H₂O) is approximately 18.02 g/mol, and Ce(SO₄)₂ contributes 2 × 140.12 (Ce), 2 × 32.07 (S), and 8 × 16.00 (O). Adding these gives the molar mass of the entire compound.
The percentage by mass of water is calculated by dividing the molar mass of water by the total molar mass of the compound and multiplying by 100:
Percentage by mass of water = (Molar mass of water / Total molar mass of compound) × 100
Substituting the values, we find that approximately 20.5% of the compound's mass comes from water. This percentage represents the mass ratio of water in the overall formula unit of cerium(IV) sulfate.