Consider the hydrate iron (II) sulfate heptahydrate [tex] FeSO _4 \cdot 7 H _2 O [/tex] when answering the questions below. Type in the molar masses using two decimal places.

1. What is the molar mass of the anhydrous salt [tex] FeSO _4 [/tex]? [tex]$\qquad$[/tex] ______ [tex]$g / mol$[/tex]

2. What is the molar mass of water in this hydrate? [tex]$\qquad$[/tex] ______ [tex]$g / mol$[/tex]

3. What is the molar mass of this hydrate? [tex]$\qquad$[/tex] ______ [tex]$g / mol$[/tex]



Answer :

Let's solve each part of the question step-by-step, considering the given molar masses.

### 1. Molar Mass of Anhydrous Salt [tex]\( FeSO_4 \)[/tex]

To calculate the molar mass of anhydrous iron(II) sulfate [tex]\( FeSO_4 \)[/tex], we need to sum the molar masses of each constituent atom:
- Iron (Fe): [tex]\( 55.85 \, g/mol \)[/tex]
- Sulfur (S): [tex]\( 32.07 \, g/mol \)[/tex]
- Oxygen (O): [tex]\( 16.00 \, g/mol \)[/tex] and there are 4 oxygen atoms.

Thus, the molar mass of [tex]\( FeSO_4 \)[/tex]:
[tex]\[ 55.85 + 32.07 + (4 \times 16.00) = 55.85 + 32.07 + 64.00 = 151.92 \, g/mol \][/tex]

So, the molar mass of [tex]\( FeSO_4 \)[/tex] is 151.92 g/mol.

### 2. Molar Mass of Water [tex]\( H_2O \)[/tex]

Next, we calculate the molar mass of water [tex]\( H_2O \)[/tex]. We need the molar masses of hydrogen and oxygen:
- Hydrogen (H): [tex]\( 1.01 \, g/mol \)[/tex] and there are 2 hydrogen atoms.
- Oxygen (O): [tex]\( 16.00 \, g/mol \)[/tex].

Thus, the molar mass of [tex]\( H_2O \)[/tex]:
[tex]\[ (2 \times 1.01) + 16.00 = 2.02 + 16.00 = 18.02 \, g/mol \][/tex]

So, the molar mass of [tex]\( H_2O \)[/tex] is 18.02 g/mol.

### 3. Molar Mass of the Hydrate [tex]\( FeSO_4 \cdot 7H_2O \)[/tex]

Finally, to find the molar mass of [tex]\( FeSO_4 \cdot 7H_2O \)[/tex], we add the molar mass of [tex]\( FeSO_4 \)[/tex] from part 1 to the total molar mass of 7 water molecules:
- Molar mass of [tex]\( FeSO_4 \)[/tex]: [tex]\( 151.92 \, g/mol \)[/tex]
- Molar mass of [tex]\( 7H_2O \)[/tex] (since [tex]\( H_2O = 18.02 \, g/mol \)[/tex]):
[tex]\[ 7 \times 18.02 = 126.14 \, g/mol \][/tex]

Thus, the molar mass of the hydrate:
[tex]\[ 151.92 + 126.14 = 278.06 \, g/mol \][/tex]

So, the molar mass of [tex]\( FeSO_4 \cdot 7H_2O \)[/tex] is 278.06 g/mol.

In summary:
- The molar mass of anhydrous [tex]\( FeSO_4 \)[/tex] is 151.92 g/mol.
- The molar mass of water [tex]\( H_2O \)[/tex] is 18.02 g/mol.
- The molar mass of [tex]\( FeSO_4 \cdot 7H_2O \)[/tex] is 278.06 g/mol.