The vapour density of an organic compound is 23.0. It contains 52.17% C and 13% H. It gives iodoform test. The compound is :

Options

A. Ethanol
B. dimethyl ether
C. Acetone
D. Methanol

Answer

To solve the problem step by step, we need to determine the molecular formula of the organic compound based on the given data and then identify the compound.

### Step 1: Calculate the Molecular Mass
The vapor density (VD) of the compound is given as 23.0. The molecular mass (M) can be calculated using the formula:
\[ M = 2 \times \text{Vapor Density} \]
\[ M = 2 \times 23.0 = 46 \, \text{g/mol} \]

### Step 2: Determine the Percentage Composition
The percentage composition of the compound is given as:
- Carbon (C): 52.17%
- Hydrogen (H): 13%
- The remaining percentage is assumed to be Oxygen (O):
\[ \text{Oxygen} = 100\% - (52.17\% + 13\%) = 34.83\% \]

### Step 3: Convert Percentage to Moles
Next, we convert the percentage composition to moles by dividing by the atomic masses:
- Atomic mass of Carbon (C) = 12 g/mol
- Atomic mass of Hydrogen (H) = 1 g/mol
- Atomic mass of Oxygen (O) = 16 g/mol

Calculating the number of moles:
- Moles of Carbon:
\[ \text{Moles of C} = \frac{52.17}{12} \approx 4.35 \]
- Moles of Hydrogen:
\[ \text{Moles of H} = \frac{13}{1} = 13 \]
- Moles of Oxygen:
\[ \text{Moles of O} = \frac{34.83}{16} \approx 2.18 \]

### Step 4: Find the Simplest Whole Number Ratio
To find the simplest whole number ratio, we divide each value by the smallest number of moles calculated:
- Smallest number of moles is approximately 2.18 (for Oxygen).

Calculating the ratio:
- Ratio of Carbon:
\[ \frac{4.35}{2.18} \approx 2 \]
- Ratio of Hydrogen:
\[ \frac{13}{2.18} \approx 6 \]
- Ratio of Oxygen:
\[ \frac{2.18}{2.18} = 1 \]

Thus, the simplest whole number ratio is:
- C: 2
- H: 6
- O: 1

### Step 5: Write the Empirical Formula
From the whole number ratio, the empirical formula of the compound is:
\[ \text{C}_2\text{H}_6\text{O} \]

### Step 6: Confirm the Molecular Formula
Since the molecular mass calculated is 46 g/mol and the empirical formula mass of C2H6O is also:
\[ (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 \, \text{g/mol} \]
This confirms that the empirical formula is also the molecular formula.

### Step 7: Identify the Compound
The compound gives an iodoform test, which is characteristic of compounds that have a methyl ketone group or a secondary alcohol with a methyl group adjacent to the hydroxyl group. The molecular formula C2H6O corresponds to ethanol (C2H5OH), which can give a positive iodoform test.

### Conclusion
The organic compound is **Ethanol (C2H6O)**.

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