A gaseous compound is composed of `85.7%` by mass carbon and `14.3%` by mass hydrogen. Its density is `2.28` g/litre at `300 K and 1.0` atm pressure. Determine the molecular formula of the compound.

Options

A. `C_(2)H_(2)`
B. `C_(2)H_(4)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Answer

To determine the molecular formula of the gaseous compound composed of 85.7% carbon and 14.3% hydrogen, we will follow these steps:

### Step 1: Calculate the Molar Mass of the Gas
We will use the Ideal Gas Law equation, which is given by:

\[ PV = nRT \]

For one mole of gas, this simplifies to:

\[ P = \frac{RT}{V} \]

We can express the volume in terms of density:

\[ V = \frac{mass}{density} \]

Substituting this into the ideal gas equation, we get:

\[ P = \frac{RT}{\frac{mass}{density}} \]

Rearranging gives us:

\[ mass = \frac{density \times R \times T}{P} \]

Given:
- Density = 2.28 g/L
- R (gas constant) = 0.0821 L·atm/(K·mol)
- Temperature (T) = 300 K
- Pressure (P) = 1 atm

Substituting the values:

\[ mass = \frac{2.28 \, \text{g/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{1 \, \text{atm}} \]

Calculating this gives:

\[ mass = \frac{2.28 \times 0.0821 \times 300}{1} \approx 56.15 \, \text{g/mol} \]

### Step 2: Determine the Empirical Formula
Next, we need to find the empirical formula using the mass percentages of carbon and hydrogen.

1. **Convert percentages to grams** (assuming 100 g of the compound):
- Carbon: 85.7 g
- Hydrogen: 14.3 g

2. **Convert grams to moles**:
- Moles of Carbon = \(\frac{85.7 \, g}{12.01 \, g/mol} \approx 7.14 \, mol\)
- Moles of Hydrogen = \(\frac{14.3 \, g}{1.008 \, g/mol} \approx 14.13 \, mol\)

3. **Find the simplest mole ratio**:
- Ratio of C to H = \( \frac{7.14}{7.14} : \frac{14.13}{7.14} \approx 1 : 2 \)

Thus, the empirical formula is \( CH_2 \).

### Step 3: Calculate the Molecular Formula
Now, we need to find the molecular formula using the molar mass we calculated earlier.

1. **Calculate the molar mass of the empirical formula \( CH_2 \)**:
- Molar mass of \( CH_2 = 12.01 \, g/mol + 2 \times 1.008 \, g/mol = 14.026 \, g/mol \)

2. **Determine the value of n**:
- \( n = \frac{Molar \, mass \, of \, compound}{Molar \, mass \, of \, empirical \, formula} = \frac{56.15 \, g/mol}{14.026 \, g/mol} \approx 4 \)

3. **Write the molecular formula**:
- The molecular formula is \( C_{1 \times 4}H_{2 \times 4} = C_4H_8 \).

### Final Answer
The molecular formula of the compound is \( C_4H_8 \).

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