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?
(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)`

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 determine the molecular formula of the compound with the empirical formula \( C_{10}H_{8}Fe \), we will follow these steps:

### Step 1: Calculate the change in boiling point (\( \Delta T \))
The change in boiling point can be calculated using the formula:
\[ \Delta T = T_2 - T_1 \]
Where:
- \( T_2 = 80.26^\circ C \) (boiling point of the solution)
- \( T_1 = 80.10^\circ C \) (boiling point of pure benzene)

Calculating \( \Delta T \):
\[ \Delta T = 80.26 - 80.10 = 0.16^\circ C \]

### Step 2: Use the boiling point elevation formula
The boiling point elevation can be expressed as:
\[ \Delta T = K_b \times m \]
Where:
- \( K_b = 2.53^\circ C/molal \)
- \( m \) is the molality of the solution.

Rearranging the formula to find molality:
\[ m = \frac{\Delta T}{K_b} = \frac{0.16}{2.53} \approx 0.0632 \, mol/kg \]

### Step 3: Calculate the number of moles of solute
Molality (\( m \)) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
Given that the mass of benzene (solvent) is \( 11.2 \, g \) or \( 0.0112 \, kg \):
\[ 0.0632 = \frac{\text{number of moles}}{0.0112} \]
Calculating the number of moles:
\[ \text{number of moles} = 0.0632 \times 0.0112 \approx 0.000707 \, moles \]

### Step 4: Calculate the molar mass of the compound
The molar mass can be calculated using the formula:
\[ \text{Molar mass} = \frac{\text{mass of solute}}{\text{number of moles}} \]
Given the mass of the compound is \( 0.26 \, g \):
\[ \text{Molar mass} = \frac{0.26}{0.000707} \approx 367.5 \, g/mol \]

### Step 5: Calculate the molar mass of the empirical formula
The empirical formula is \( C_{10}H_{8}Fe \). Now, we calculate its molar mass:
- Carbon (C): \( 12 \times 10 = 120 \, g/mol \)
- Hydrogen (H): \( 1 \times 8 = 8 \, g/mol \)
- Iron (Fe): \( 56 \, g/mol \)

Adding these together:
\[ \text{Molar mass of } C_{10}H_{8}Fe = 120 + 8 + 56 = 184 \, g/mol \]

### Step 6: Determine the factor \( x \)
The relationship between the molecular formula and empirical formula is given by:
\[ \text{Molar mass} = x \times \text{Molar mass of empirical formula} \]
Substituting the values we have:
\[ 367.5 = x \times 184 \]
Solving for \( x \):
\[ x = \frac{367.5}{184} \approx 2 \]

### Step 7: Determine the molecular formula
The molecular formula is given by:
\[ \text{Molecular formula} = x \times \text{Empirical formula} \]
Thus:
\[ \text{Molecular formula} = 2 \times C_{10}H_{8}Fe = C_{20}H_{16}Fe_{2} \]

### Conclusion
The molecular formula of the compound is \( C_{20}H_{16}Fe_{2} \).

### Final Answer
**(d) \( C_{20}H_{16}Fe_{2} \)**

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