A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of `CO_(2)` produced in the reaction is thrice the volume of hydrocarbon under the same conditions
What is the ratio of molecular weight to emperical formula weight of the hydrocarbon?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the ratio of the molecular weight to the empirical formula weight of the hydrocarbon based on the information provided about its combustion.

### Step-by-Step Solution:

1. **Understanding the Combustion Reaction**:
The general combustion reaction for a hydrocarbon (CₓHᵧ) can be represented as:
\[ C_xH_y + \frac{x + \frac{y}{4}}{1} O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

2. **Setting Up the Volumes**:
According to the problem:
- The volume of the hydrocarbon (let's assume it is 1 unit) is \( V_{C_xH_y} = 1 \).
- The volume of oxygen consumed is 5 times the volume of the hydrocarbon, so:
\[ V_{O_2} = 5 \times V_{C_xH_y} = 5 \text{ units} \]
- The volume of carbon dioxide produced is three times the volume of the hydrocarbon:
\[ V_{CO_2} = 3 \times V_{C_xH_y} = 3 \text{ units} \]

3. **Identifying Variables**:
From the combustion reaction:
- The volume of \( CO_2 \) produced corresponds to \( x \) (the number of carbon atoms).
- Therefore, \( x = 3 \).

4. **Using the Oxygen Volume**:
The volume of oxygen consumed can be expressed in terms of \( x \) and \( y \):
\[ \frac{x + \frac{y}{4}}{1} = 5 \]
Substituting \( x = 3 \):
\[ 3 + \frac{y}{4} = 5 \]

5. **Solving for \( y \)**:
Rearranging the equation:
\[ \frac{y}{4} = 5 - 3 = 2 \]
Multiplying both sides by 4:
\[ y = 8 \]

6. **Determining the Molecular Formula**:
The molecular formula of the hydrocarbon is:
\[ C_3H_8 \]

7. **Calculating the Molecular Weight**:
The molecular weight of \( C_3H_8 \):
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
\[ \text{Molecular Weight} = (3 \times 12) + (8 \times 1) = 36 + 8 = 44 \text{ g/mol} \]

8. **Calculating the Empirical Formula Weight**:
The empirical formula of the hydrocarbon is also \( C_3H_8 \) (since it is already in its simplest form).
\[ \text{Empirical Formula Weight} = 44 \text{ g/mol} \]

9. **Finding the Ratio**:
The ratio of molecular weight to empirical formula weight is:
\[ \text{Ratio} = \frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}} = \frac{44}{44} = 1 \]

### Final Answer:
The ratio of molecular weight to empirical formula weight of the hydrocarbon is **1**.