An organic compound contains `40%C, 6.6%H`. The empirical formula of the compound is

Options

A. `CH_3 O`
B. `C_4 HO_4`
C. `CHO`
D. `CH_2 O`

Answer

To find the empirical formula of the organic compound containing 40% carbon and 6.6% hydrogen, we will follow these steps:

### Step 1: Determine the percentage composition of each element
We have the following percentage compositions:
- Carbon (C): 40%
- Hydrogen (H): 6.6%
- Since the total percentage must equal 100%, we can find the percentage of oxygen (O):
\[ O = 100\% - (40\% + 6.6\%) = 100\% - 46.6\% = 53.4\% \]

### Step 2: Convert the percentages to grams
Assuming we have 100 grams of the compound, we can convert the percentages directly to grams:
- Carbon: 40 g
- Hydrogen: 6.6 g
- Oxygen: 53.4 g

### Step 3: Convert grams to moles
Next, we will convert the mass of each element to moles using their 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 moles:
- Moles of Carbon:
\[ \text{Moles of C} = \frac{40 \text{ g}}{12 \text{ g/mol}} = 3.33 \text{ moles} \]
- Moles of Hydrogen:
\[ \text{Moles of H} = \frac{6.6 \text{ g}}{1 \text{ g/mol}} = 6.6 \text{ moles} \]
- Moles of Oxygen:
\[ \text{Moles of O} = \frac{53.4 \text{ g}}{16 \text{ g/mol}} = 3.34 \text{ moles} \]

### Step 4: Find the simplest mole ratio
To find the simplest ratio, we will divide each mole value by the smallest number of moles calculated:
- For Carbon:
\[ \frac{3.33}{3.33} = 1 \]
- For Hydrogen:
\[ \frac{6.6}{3.33} \approx 2 \]
- For Oxygen:
\[ \frac{3.34}{3.33} \approx 1 \]

### Step 5: Write the empirical formula
From the mole ratios, we can write the empirical formula:
- C: 1
- H: 2
- O: 1

Thus, the empirical formula of the compound is:
\[ \text{Empirical Formula} = \text{C}_1\text{H}_2\text{O}_1 \text{ or simply } \text{CH}_2\text{O} \]

### Final Answer:
The empirical formula of the compound is **CH₂O**.

---