The molecular mass of a compound is 88 amu having empirical formula of `C_(2)H_(4)O`. Its molecular formula will be

Options

A. `C_(2)H_(4)O_(2)`
B. `C_(4)H_(8)O_(2)`
C. `C_(8)H_(4)O_(2)`
D. `C_(2)H_(8)O_(2)`

Explanation

To find the molecular formula of the compound with a molecular mass of 88 amu and an empirical formula of C₂H₄O, follow these steps:

### Step 1: Calculate the Empirical Formula Mass
First, we need to calculate the mass of the empirical formula (C₂H₄O).

- **Carbon (C)**: There are 2 carbon atoms. The atomic mass of carbon is approximately 12 amu.
\[ 2 \times 12 = 24 \text{ amu} \]

- **Hydrogen (H)**: There are 4 hydrogen atoms. The atomic mass of hydrogen is approximately 1 amu.
\[ 4 \times 1 = 4 \text{ amu} \]

- **Oxygen (O)**: There is 1 oxygen atom. The atomic mass of oxygen is approximately 16 amu.
\[ 1 \times 16 = 16 \text{ amu} \]

Now, add these values together to find the empirical formula mass:
\[ \text{Empirical formula mass} = 24 + 4 + 16 = 44 \text{ amu} \]

### Step 2: Calculate the Value of N
Next, we need to find the value of \( N \), which is the factor by which we multiply the empirical formula to get the molecular formula. The formula to find \( N \) is:
\[ N = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} \]
Substituting the known values:
\[ N = \frac{88 \text{ amu}}{44 \text{ amu}} = 2 \]

### Step 3: Determine the Molecular Formula
Now, we multiply the empirical formula by \( N \) to find the molecular formula:
\[ \text{Molecular formula} = N \times \text{Empirical formula} = 2 \times C₂H₄O \]
This gives us:
\[ \text{Molecular formula} = C_{2 \times 2}H_{4 \times 2}O_{1 \times 2} = C₄H₈O₂ \]

### Final Answer
The molecular formula of the compound is \( C₄H₈O₂ \).

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