A pure gas that is `14.3%` hydrogen and `85.7%` carbon by mass has a density of `2.5g L^(-1)` at `0^(@)C` and 1 atm pressure. What is the molecular formula of the gas :

Options

A. `CH_(2)`
B. `C_(2)H_(4)`
C. `C_(4)H_(8)`
D. `C_(6)H_(12)`

Explanation

To find the molecular formula of the gas that is 14.3% hydrogen and 85.7% carbon by mass, we can follow these steps:

### Step 1: Calculate the moles of each element
- **Moles of Hydrogen (H)**:
\[ \text{Moles of H} = \frac{\text{mass percentage of H}}{\text{molar mass of H}} = \frac{14.3}{1} = 14.3 \text{ moles} \]

- **Moles of Carbon (C)**:
\[ \text{Moles of C} = \frac{\text{mass percentage of C}}{\text{molar mass of C}} = \frac{85.7}{12} \approx 7.15 \text{ moles} \]

### Step 2: Determine the simplest mole ratio
To find the simplest ratio, we divide both values by the smaller number of moles (7.15):
- **Ratio of H**:
\[ \frac{14.3}{7.15} \approx 2 \]
- **Ratio of C**:
\[ \frac{7.15}{7.15} = 1 \]

This gives us a ratio of approximately 2:1 for H:C.

### Step 3: Write the empirical formula
From the ratio, we can write the empirical formula:
\[ \text{Empirical Formula} = \text{C}_1\text{H}_2 = \text{CH}_2 \]

### Step 4: Calculate the molar mass of the empirical formula
Now, we calculate the molar mass of the empirical formula (CH₂):
\[ \text{Molar mass of CH}_2 = 12 \text{ (C)} + 2 \times 1 \text{ (H)} = 14 \text{ g/mol} \]

### Step 5: Use the density to find the molar mass of the gas
Using the ideal gas law, we can find the molar mass (M) using the formula:
\[ PM = dRT \]
Where:
- \( P = 1 \text{ atm} \)
- \( d = 2.5 \text{ g/L} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 273 \text{ K} \)

Rearranging the formula for M:
\[ M = \frac{dRT}{P} \]
Substituting the values:
\[ M = \frac{2.5 \times 0.0821 \times 273}{1} \approx 56 \text{ g/mol} \]

### Step 6: Determine the molecular formula
Now, we can find the number of empirical units in the molecular formula:
\[ \text{Number of empirical units} = \frac{\text{Molar mass of gas}}{\text{Molar mass of empirical formula}} = \frac{56}{14} = 4 \]

Thus, the molecular formula is:
\[ \text{Molecular Formula} = 4 \times \text{CH}_2 = \text{C}_4\text{H}_8 \]

### Final Answer:
The molecular formula of the gas is **C₄H₈**.

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