A compound (60 gm) on analysis gave `C=24gm` H=4gm` and `O=32gm`, its empirical formula is

Options

A. `C_2H_2O`
B. `C_2H_4O_2`
C. `CH_2O`
D. `CH_2O_2`

Explanation

To find the empirical formula of the compound based on the given masses of carbon, hydrogen, and oxygen, we can follow these steps:

### Step 1: Determine the number of moles of each element.
To find the number of moles, we will use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- For Carbon (C):
- Mass = 24 g
- Molar mass of C = 12 g/mol
- Moles of C = \(\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles}\)

- For Hydrogen (H):
- Mass = 4 g
- Molar mass of H = 1 g/mol
- Moles of H = \(\frac{4 \text{ g}}{1 \text{ g/mol}} = 4 \text{ moles}\)

- For Oxygen (O):
- Mass = 32 g
- Molar mass of O = 16 g/mol
- Moles of O = \(\frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles}\)

### Step 2: Write the mole ratio.
Now we have the number of moles:
- C: 2 moles
- H: 4 moles
- O: 2 moles

The ratio of moles of each element is:
- C : H : O = 2 : 4 : 2

### Step 3: Simplify the ratio.
To simplify the ratio, we can divide each number by the smallest number of moles, which is 2:

\[ \text{C : H : O} = \frac{2}{2} : \frac{4}{2} : \frac{2}{2} = 1 : 2 : 1 \]

### Step 4: Write the empirical formula.
From the simplified ratio, we can write the empirical formula as:

\[ \text{Empirical formula} = \text{C}_1\text{H}_2\text{O}_1 = \text{CH}_2\text{O} \]

### Step 5: Verify the molecular mass.
To ensure that the empirical formula is consistent with the total mass of the compound (60 g), we can calculate the molar mass of the empirical formula:

\[ \text{Molar mass of CH}_2\text{O} = (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \text{ g/mol} \]

Since the total mass of the compound is 60 g, we can find the number of empirical units in the compound:

\[ \text{Number of empirical units} = \frac{60 \text{ g}}{30 \text{ g/mol}} = 2 \]

Thus, the molecular formula is:

\[ \text{C}_2\text{H}_4\text{O}_2 \]

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

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