Certainly! To find the percentage by mass of water in the compound [tex]\(\text{FeCl}_3 \cdot 6 \text{H}_2 \text{O}\)[/tex], we'll follow a systematic process step-by-step.
### Step 1: Determine the Molar Masses of Individual Elements
First, we need the molar masses of the elements involved:
- Iron (Fe): [tex]\( \text{55.845 g/mol} \)[/tex]
- Chlorine (Cl): [tex]\( \text{35.453 g/mol} \)[/tex]
- Hydrogen (H): [tex]\( \text{1.008 g/mol} \)[/tex]
- Oxygen (O): [tex]\( \text{16.00 g/mol} \)[/tex]
### Step 2: Calculate the Molar Mass of [tex]\(\text{FeCl}_3\)[/tex]
[tex]\[
\text{Molar mass of FeCl}_3 = \text{Molar mass of Fe} + 3 \times \text{Molar mass of Cl}
\][/tex]
[tex]\[
\text{Molar mass of FeCl}_3 = 55.845 + 3 \times 35.453 = 162.204 \, \text{g/mol}
\][/tex]
### Step 3: Calculate the Molar Mass of [tex]\(\text{H}_2\text{O}\)[/tex]
[tex]\[
\text{Molar mass of H}_2\text{O} = 2 \times \text{Molar mass of H} + \text{Molar mass of O}
\][/tex]
[tex]\[
\text{Molar mass of H}_2\text{O} = 2 \times 1.008 + 16.00 = 18.016 \, \text{g/mol}
\][/tex]
### Step 4: Calculate the Molar Mass of [tex]\(\text{FeCl}_3 \cdot 6 \text{H}_2\text{O}\)[/tex]
[tex]\[
\text{Molar mass of FeCl}_3 \cdot 6 \text{H}_2\text{O} = \text{Molar mass of FeCl}_3 + 6 \times \text{Molar mass of H}_2\text{O}
\][/tex]
[tex]\[
\text{Molar mass of FeCl}_3 \cdot 6 \text{H}_2\text{O} = 162.204 + 6 \times 18.016 = 270.3 \, \text{g/mol}
\][/tex]
### Step 5: Calculate the Mass of Water in the Compound
[tex]\[
\text{Mass of 6 H}_2\text{O} = 6 \times \text{Molar mass of H}_2\text{O}
\][/tex]
[tex]\[
\text{Mass of 6 H}_2\text{O} = 6 \times 18.016 = 108.096 \, \text{g/mol}
\][/tex]
### Step 6: Calculate the Percentage by Mass of Water
[tex]\[
\text{Percentage by mass of water} = \left( \frac{\text{Mass of water}}{\text{Total molar mass of compound}} \right) \times 100
\][/tex]
[tex]\[
\text{Percentage by mass of water} = \left( \frac{108.096}{270.3} \right) \times 100 \approx 39.991 \%
\][/tex]
Therefore, the percentage by mass of water in the compound [tex]\(\text{FeCl}_3 \cdot 6 \text{H}_2 \text{O}\)[/tex] is approximately [tex]\(39.99\%\)[/tex].
