An organic compound with 68.9 % of C and 4.92 % of H, is aromatic and gives `CO_(2)` with `NaHCO_(3)`. The organic compound is

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Answer

To solve the problem of identifying the organic compound with the given percentage composition and properties, we can follow these steps:

### Step 1: Determine the percentage of oxygen
Given:
- Percentage of Carbon (C) = 68.9%
- Percentage of Hydrogen (H) = 4.92%

To find the percentage of Oxygen (O), we can use the formula:
\[ \text{Percentage of O} = 100 - (\text{Percentage of C} + \text{Percentage of H}) \]

Calculating:
\[ \text{Percentage of O} = 100 - (68.9 + 4.92) = 100 - 73.82 = 26.18\% \]

### Step 2: Calculate the number of moles of each element
Using the molar masses:
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

Calculating the number of moles:
- Moles of C = \( \frac{68.9}{12} = 5.74 \)
- Moles of H = \( \frac{4.92}{1} = 4.92 \)
- Moles of O = \( \frac{26.18}{16} = 1.63 \)

### Step 3: Find the simplest whole number ratio
To find the simplest ratio, we divide each mole value by the smallest number of moles calculated (which is 1.63):
- Ratio of C = \( \frac{5.74}{1.63} \approx 3.5 \)
- Ratio of H = \( \frac{4.92}{1.63} \approx 3 \)
- Ratio of O = \( \frac{1.63}{1.63} = 1 \)

### Step 4: Convert to whole numbers
Since we have a fractional value (3.5), we can multiply all ratios by 2 to convert them to whole numbers:
- C: \( 3.5 \times 2 = 7 \)
- H: \( 3 \times 2 = 6 \)
- O: \( 1 \times 2 = 2 \)

Thus, the empirical formula is \( C_7H_6O_2 \).

### Step 5: Identify the compound
The compound is aromatic and gives CO2 with NaHCO3, indicating it contains a carboxylic acid group. The structure that fits \( C_7H_6O_2 \) is benzoic acid (C6H5COOH).

### Conclusion
The organic compound is **benzoic acid (C7H6O2)**.

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