A compound has the empirical formula `C_(10)H_(8)Fe`. A solution of 0.26 g of the compound in 11.2 g of benzene (`C_(6)H_(6)`)boils at `80.26^(@)C`. The boiling point of benzene is `80.10^(@)C`, the` K_(b)` is `2.53^(@)C`/molal. What is the molecules formula of the compound?

Options

A. `C_(30)H_(24)Fe_(3)`
B. `C_(10)H_(8)Fe`
C. `C_(5)H_(4)Fe`
D. `C_(20)H_(16)Fe_(2)`

Answer

To find the molecular formula of the compound with the empirical formula \( C_{10}H_8Fe \), we will follow these steps:

### Step 1: Calculate the elevation in boiling point
The elevation in boiling point (\( \Delta T_b \)) can be calculated as:
\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) = 80.26^\circ C - 80.10^\circ C = 0.16^\circ C \]

### Step 2: Use the boiling point elevation formula
The boiling point elevation can also be expressed using the formula:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the ebullioscopic constant and \( m \) is the molality of the solution. Rearranging gives:
\[ m = \frac{\Delta T_b}{K_b} = \frac{0.16^\circ C}{2.53^\circ C/\text{molal}} \approx 0.0632 \text{ molal} \]

### Step 3: Calculate the number of moles of solute
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent. We have 11.2 g of benzene, which is 0.0112 kg. Therefore, we can calculate the number of moles of solute:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \implies \text{moles of solute} = m \cdot \text{mass of solvent in kg} \]
\[ \text{moles of solute} = 0.0632 \text{ molal} \cdot 0.0112 \text{ kg} \approx 0.000707 \text{ moles} \]

### Step 4: Calculate the molar mass of the solute
Now, we can find the molar mass of the solute using the mass of the solute (0.26 g):
\[ \text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{0.26 \text{ g}}{0.000707 \text{ moles}} \approx 367 \text{ g/mol} \]

### Step 5: Determine the molecular formula
The empirical formula \( C_{10}H_8Fe \) has a molar mass calculated as follows:
- Carbon (C): 12 g/mol × 10 = 120 g/mol
- Hydrogen (H): 1 g/mol × 8 = 8 g/mol
- Iron (Fe): 55 g/mol × 1 = 55 g/mol
- Total = 120 + 8 + 55 = 183 g/mol

Now, we find the ratio of the molar mass of the compound to the molar mass of the empirical formula:
\[ \text{Ratio} = \frac{367 \text{ g/mol}}{183 \text{ g/mol}} \approx 2 \]

### Step 6: Write the molecular formula
Since the ratio is approximately 2, we multiply the subscripts in the empirical formula by 2:
\[ \text{Molecular formula} = C_{20}H_{16}Fe_2 \]

### Final Answer
The molecular formula of the compound is \( C_{20}H_{16}Fe_2 \).

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