At STP one litre of a gas weight 1.25 grams. The gas contains `85.71%` of carbon and `14.29%` of hydrogen. The formula of the compound is

Options

A. `CH_(4)`
B. `C_(2)H_(6)`
C. `C_(3)H_(8)`
D. `C_(2)H_(4)`

Explanation

To solve the problem, we need to find the molecular formula of the gas based on the given information. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the mass of carbon and hydrogen in 100 grams of the gas
We are given that the gas contains:
- 85.71% carbon
- 14.29% hydrogen

Assuming we have 100 grams of the gas:
- Mass of carbon = 85.71 grams
- Mass of hydrogen = 14.29 grams

### Step 2: Calculate the number of moles of carbon and hydrogen
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

- Molar mass of carbon (C) = 12 g/mol
- Molar mass of hydrogen (H) = 1 g/mol

Calculating the moles:
- Moles of carbon = \(\frac{85.71 \, \text{g}}{12 \, \text{g/mol}} \approx 7.1425 \, \text{mol}\)
- Moles of hydrogen = \(\frac{14.29 \, \text{g}}{1 \, \text{g/mol}} = 14.29 \, \text{mol}\)

### Step 3: Find the simplest mole ratio
To find the simplest ratio, we divide each number of moles by the smallest number of moles calculated:
- For carbon: \(\frac{7.1425}{7.1425} = 1\)
- For hydrogen: \(\frac{14.29}{7.1425} \approx 2\)

Thus, the ratio of carbon to hydrogen is \(1:2\).

### Step 4: Write the empirical formula
From the mole ratio, we can write the empirical formula:
- Empirical formula = \(C_1H_2\) or simply \(CH_2\).

### Step 5: Calculate the empirical formula mass
To find the empirical formula mass:
- Mass of \(C\) in \(CH_2\) = \(12 \, \text{g/mol}\)
- Mass of \(H_2\) in \(CH_2\) = \(2 \times 1 \, \text{g/mol} = 2 \, \text{g/mol}\)

Empirical formula mass = \(12 + 2 = 14 \, \text{g/mol}\).

### Step 6: Determine the molecular mass of the gas
At STP, we know that:
- 1 liter of the gas weighs 1.25 grams.
- 1 mole of any gas occupies 22.4 liters at STP.

To find the molecular mass:
\[ \text{Molecular mass} = \text{Weight of 22.4 L of gas} = 22.4 \, \text{L} \times 1.25 \, \text{g/L} = 28 \, \text{g/mol} \]

### Step 7: Calculate the value of \(n\)
Using the formula:
\[ n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{28 \, \text{g/mol}}{14 \, \text{g/mol}} = 2 \]

### Step 8: Write the molecular formula
Now, we can find the molecular formula by multiplying the subscripts in the empirical formula by \(n\):
- Molecular formula = \(C_1H_2\) becomes \(C_{1 \times 2}H_{2 \times 2} = C_2H_4\).

### Final Answer
The molecular formula of the compound is \(C_2H_4\).

---