Answer:
1. Percent by mass of H₂O = 45.3%
2. Percent by mass of anhydrous Salt (FeSO₄) = 54.7 %
Explanation:
Data Given
Formula of the Molecule = FeSO₄ • 7H₂O
% by mass water (H₂O) = ?
% by mass FeSO₄ = ?
________________________________________
> First of all find the atomic masses of each component in a molecule
For H₂O atomic masses are given below
H = 1 g/mol
O = 16 g/mol
> Then find the total mass of H₂O in haydrated salt
7H₂O = 7 (2x1 +1x16) g/mol
7H₂O = 7 (2+16) g/mol
7H₂O = 7 (18) g/mol
7H₂O = 126 g/mol
> find total Molar Mass of Molecule:
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 16x4] + 7(2+16)
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126
Molar Mass of FeSO₄ • 7H₂O = 278
Now to find the mass by percent of H₂O
Formula used to find the mass by percent of a component
Percent composition of H₂O = mass of H₂O in Molecule / molar mass of FeSO₄ • 7H₂O x 100%
Put the values
Percent by mass of H₂O = 126 (g/mol) / 278 (g/mol) x 100%
Percent by mass of H₂O = 0.4532 x 100%
Percent by mass of H₂O = 45.3%
_______________________________________________
> First of all find the atomic masses of each component in a molecule
For anhydrus salt (FeSO₄) atomic masses are given below
Fe = 56 g/mol
S= 32 g/mol
O = 16 g/mol
> Then find the total mass of FeSO₄ in haydrated salt
FeSO₄ = (56x1 + 32 + 4x16) g/mol
FeSO₄ = ( 56 + 32 + 64) g/mol
FeSO₄ = 152 g/mol
> find total Molar Mass of Molecule:
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 16x4] + 7(2+16)
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126
Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126
Molar Mass of FeSO₄ • 7H₂O = 278
Now to find the mass by percent of FeSO₄
Formula used to find the mass by percent of a component
Percent composition of FeSO₄ = mass of FeSO₄ in Molecule / molar mass of FeSO₄ • 7H₂O x 100%
Put the values
Percent by mass of FeSO₄ = 152 (g/mol) / 278 (g/mol) x 100%
Percent by mass of FeSO₄ = 0.547 x 100%
Percent by mass of FeSO₄ = 54.7 %
