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

Options

A. `CH_(2)O`
B. `CHO`
C. `CH_(4)O_(2)`
D. `C_(2)H_(2)O`

Explanation

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

### Step 1: Convert the percentages to grams
Assuming we have 100 grams of the compound, the percentages can be directly converted to grams:
- Carbon (C): 40 g
- Oxygen (O): 53.5 g
- Hydrogen (H): 6.5 g

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

Now, we calculate the moles of each element:
- Moles of Carbon = \( \frac{40 \text{ g}}{12 \text{ g/mol}} = 3.33 \text{ moles} \)
- Moles of Oxygen = \( \frac{53.5 \text{ g}}{16 \text{ g/mol}} = 3.34 \text{ moles} \)
- Moles of Hydrogen = \( \frac{6.5 \text{ g}}{1 \text{ g/mol}} = 6.5 \text{ moles} \)

### Step 3: Determine the simplest mole ratio
Next, we find the smallest number of moles among the three:
- The smallest value is approximately 3.33 (for both Carbon and Oxygen).

Now, we divide the number of moles of each element by the smallest number of moles:
- Ratio of Carbon = \( \frac{3.33}{3.33} = 1 \)
- Ratio of Oxygen = \( \frac{3.34}{3.33} \approx 1 \)
- Ratio of Hydrogen = \( \frac{6.5}{3.33} \approx 1.95 \)

### Step 4: Convert to whole numbers
To convert the ratios to whole numbers, we can multiply each ratio by 2 (to eliminate the decimal):
- Carbon = \( 1 \times 2 = 2 \)
- Oxygen = \( 1 \times 2 = 2 \)
- Hydrogen = \( 1.95 \times 2 \approx 4 \)

### Step 5: Write the empirical formula
Now, we can write the empirical formula based on the whole number ratios:
- Empirical formula = \( C_2H_4O_2 \)

### Final Answer
The empirical formula of the compound is \( C_2H_4O_2 \).

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