Sure! Let's calculate the molar mass of each component of the hydrate iron (II) sulfate heptahydrate [tex]\( (FeSO_4 \cdot 7H_2O) \)[/tex] step-by-step:
1. Molar Mass of [tex]\( FeSO_4 \)[/tex]:
- Iron (Fe) has an atomic mass of approximately 55.85 g/mol.
- Sulfur (S) has an atomic mass of approximately 32.07 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
The formula for anhydrous iron (II) sulfate [tex]\( FeSO_4 \)[/tex] includes one atom of iron, one atom of sulfur, and four atoms of oxygen.
[tex]\[
\text{Molar Mass of } FeSO_4 = 55.85 \, (\text{Fe}) + 32.07 \, (\text{S}) + 4 \times 16.00 \, (\text{O}) = 151.92 \, g/mol
\][/tex]
So, the molar mass of the anhydrous salt [tex]\( FeSO_4 \)[/tex] is 151.92 g/mol.
2. Molar Mass of Water [tex]\( H_2O \)[/tex]:
- Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
Each molecule of water [tex]\( H_2O \)[/tex] has two atoms of hydrogen and one atom of oxygen.
[tex]\[
\text{Molar Mass of } H_2O = 2 \times 1.01 \, (\text{H}) + 16.00 \, (\text{O}) = 18.02 \, g/mol
\][/tex]
So, the molar mass of water in this hydrate is 18.02 g/mol.
3. Molar Mass of the Hydrate [tex]\( FeSO_4 \cdot 7H_2O \)[/tex]:
- We have already calculated the molar mass of [tex]\( FeSO_4 \)[/tex] as 151.92 g/mol.
- There are 7 molecules of water in the hydrate.
[tex]\[
\text{Molar Mass of } FeSO_4 \cdot 7H_2O = 151.92 \, (\text{FeSO}_4) + 7 \times 18.02 \, (\text{H}_2O)
\][/tex]
[tex]\[
= 151.92 + 7 \times 18.02
\][/tex]
[tex]\[
= 151.92 + 126.14
\][/tex]
[tex]\[
= 278.06 \, g/mol
\][/tex]
Accordingly, the molar mass of the hydrate [tex]\( FeSO_4 \cdot 7H_2O \)[/tex] is 278.06 g/mol.
Let's summarize the answers:
- The molar mass of the anhydrous salt [tex]\( FeSO_4 \)[/tex] = 151.92 g/mol.
- The molar mass of water [tex]\( H_2O \)[/tex] in this hydrate = 18.02 g/mol.
- The molar mass of the hydrate [tex]\( FeSO_4 \cdot 7H_2O \)[/tex] = 278.06 g/mol.
