An organic compound containing C, H and O gives following observations:
(i) It exists in two isomeric forms (A) and (B).
(ii) 0.108 g of one of the isomers on combustion gave 0.308 g of `CO_(2)` and 0.072 g of `H_(2)O`
(iii) (A) is insoluble in NaOH and `NaHCO_(3)` while (B) is soluble in NaOH.
(iv) (A) reacts with Hl to give compound (C ) and (D). (C ) can be separated from (D) by ethanolic `AgNO_(3)` solution and (D) is soluble in NaOH.
(v) (B) readily reacts with bromine water to give compound (E) of molecular formula `C_(7)H_(5)OBr_(3)`
The empirical formula of (A) and (B) is

Options

A. `C_(7)H_(8)O`
B. `C_(7)H_(10)O`
C. `C_(7)H_(6)O_(2)`
D. `C_(8)H_(7)O_(2)`

Answer

To find the empirical formula of the organic compound containing carbon (C), hydrogen (H), and oxygen (O), we will follow these steps:

### Step 1: Determine the mass of carbon and hydrogen from combustion data
We know that 0.108 g of one of the isomers was combusted, producing 0.308 g of CO₂ and 0.072 g of H₂O.

1. **Calculate the mass of carbon from CO₂:**
- The molar mass of CO₂ = 44 g/mol (12 g/mol for C and 32 g/mol for O).
- From the combustion data, we can find the mass of carbon:
\[ \text{Mass of C} = \left( \frac{12 \text{ g}}{44 \text{ g}} \right) \times 0.308 \text{ g CO₂} = 0.084 \text{ g C} \]

2. **Calculate the mass of hydrogen from H₂O:**
- The molar mass of H₂O = 18 g/mol (2 g/mol for H and 16 g/mol for O).
- From the combustion data, we can find the mass of hydrogen:
\[ \text{Mass of H} = \left( \frac{2 \text{ g}}{18 \text{ g}} \right) \times 0.072 \text{ g H₂O} = 0.008 \text{ g H} \]

### Step 2: Calculate the mass of oxygen
- The total mass of the compound is 0.108 g. Therefore, the mass of oxygen can be calculated as:
\[ \text{Mass of O} = 0.108 \text{ g} - (0.084 \text{ g C} + 0.008 \text{ g H}) = 0.016 \text{ g O} \]

### Step 3: Calculate the percentage of each element
1. **Percentage of Carbon:**
\[ \text{Percentage of C} = \left( \frac{0.084 \text{ g}}{0.108 \text{ g}} \right) \times 100 = 77.78\% \]

2. **Percentage of Hydrogen:**
\[ \text{Percentage of H} = \left( \frac{0.008 \text{ g}}{0.108 \text{ g}} \right) \times 100 = 7.407\% \]

3. **Percentage of Oxygen:**
\[ \text{Percentage of O} = 100\% - (77.78\% + 7.407\%) = 14.813\% \]

### Step 4: Calculate the number of moles of each element
1. **Moles of Carbon:**
\[ \text{Moles of C} = \frac{77.78}{12} = 6.48 \]

2. **Moles of Hydrogen:**
\[ \text{Moles of H} = \frac{7.407}{1} = 7.407 \]

3. **Moles of Oxygen:**
\[ \text{Moles of O} = \frac{14.813}{16} = 0.9258 \]

### Step 5: Determine the simplest ratio
To find the simplest ratio, divide each mole value by the smallest number of moles calculated (0.9258):

1. **Ratio of Carbon:**
\[ \frac{6.48}{0.9258} \approx 7 \]

2. **Ratio of Hydrogen:**
\[ \frac{7.407}{0.9258} \approx 8 \]

3. **Ratio of Oxygen:**
\[ \frac{0.9258}{0.9258} = 1 \]

### Step 6: Write the empirical formula
From the ratios, we can conclude that the empirical formula of the compound is:
\[ \text{Empirical Formula} = C_7H_8O \]

### Final Answer
The empirical formula of the organic compound (A and B) is **C₇H₈O**.

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