A compound contains `C=40%,O=53.5%`, and `H=6.5%` the empirical formula formula of the compound is:

Options

A. `CH_2O`
B. `C_2H_4O`
C. `C_6H_(12)O_6`
D. `C_2H_4O_2`

Explanation

To find the empirical formula of the compound containing carbon (C), hydrogen (H), and oxygen (O) with the given percentages, we can follow these steps:

### Step 1: Convert percentages to grams
Assume we have 100 grams of the compound. This means:
- Carbon (C) = 40 grams
- Hydrogen (H) = 6.5 grams
- Oxygen (O) = 53.5 grams

### Step 2: Determine the number of moles of each element
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- 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 moles for each element:
- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{40 \text{ g}}{12 \text{ g/mol}} \approx 3.33 \text{ moles} \]

- Moles of Hydrogen (H):
\[ \text{Moles of H} = \frac{6.5 \text{ g}}{1 \text{ g/mol}} = 6.5 \text{ moles} \]

- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{53.5 \text{ g}}{16 \text{ g/mol}} \approx 3.34 \text{ moles} \]

### Step 3: Find the simplest mole ratio
Now, we need to divide each mole value by the smallest number of moles calculated:
- Smallest number of moles = 3.33 (for Carbon)

Calculating the ratios:
- Ratio of Carbon (C):
\[ \frac{3.33}{3.33} = 1 \]

- Ratio of Hydrogen (H):
\[ \frac{6.5}{3.33} \approx 1.95 \approx 2 \]

- Ratio of Oxygen (O):
\[ \frac{3.34}{3.33} \approx 1 \]

### Step 4: Write the empirical formula
From the ratios, we can see:
- C: 1
- H: 2
- O: 1

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

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

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