An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave`C,38.71%` and `H,9.67%`. The empirical formula of the compound would be `:`

Options

A. `CH_(3)O`
B. `CH_(2)O`
C. `CHO`
D. `CH_(4)O`

Explanation

To determine the empirical formula of the organic compound containing carbon, hydrogen, and oxygen based on the given percentages, we will follow these steps:

### Step 1: Determine the percentage of oxygen
Given:
- Percentage of carbon (C) = 38.71%
- Percentage of hydrogen (H) = 9.67%

To find the percentage of oxygen (O), we subtract the sum of the percentages of carbon and hydrogen from 100%:

\[ \text{Percentage of oxygen} = 100\% - (\text{Percentage of carbon} + \text{Percentage of hydrogen}) \]

Calculating:

\[ \text{Percentage of oxygen} = 100\% - (38.71\% + 9.67\%) = 100\% - 48.38\% = 51.62\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound, the masses of each element will be equal to their percentages:

- Mass of carbon = 38.71 g
- Mass of hydrogen = 9.67 g
- Mass of oxygen = 51.62 g

### Step 3: Convert grams to moles
Using 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

Now we convert the masses to moles:

\[ \text{Moles of carbon} = \frac{38.71 \text{ g}}{12 \text{ g/mol}} \approx 3.2258 \text{ mol} \]

\[ \text{Moles of hydrogen} = \frac{9.67 \text{ g}}{1 \text{ g/mol}} = 9.67 \text{ mol} \]

\[ \text{Moles of oxygen} = \frac{51.62 \text{ g}}{16 \text{ g/mol}} \approx 3.2263 \text{ mol} \]

### Step 4: Determine the mole ratio
Now we find the mole ratio by dividing each mole value by the smallest number of moles calculated:

- Smallest number of moles = 3.2258 (for C and O)

\[ \text{Ratio of carbon} = \frac{3.2258}{3.2258} = 1 \]

\[ \text{Ratio of hydrogen} = \frac{9.67}{3.2258} \approx 3 \]

\[ \text{Ratio of oxygen} = \frac{3.2263}{3.2258} \approx 1 \]

### Step 5: Write the empirical formula
From the ratios, we can conclude:

- Carbon (C) = 1
- Hydrogen (H) = 3
- Oxygen (O) = 1

Thus, the empirical formula of the compound is:

\[ \text{Empirical formula} = \text{C}_1\text{H}_3\text{O}_1 \text{ or simply } \text{C}_1\text{H}_3\text{O} \]

### Final Answer:
The empirical formula of the compound is **C₁H₃O**.

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