A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

Explanation

To solve the problem step by step, we will calculate the empirical formula, molar mass, and molecular formula of the welding fuel gas based on the given data.

### Step 1: Calculate the mass of carbon in the gas

From the combustion of the gas, we have:
- Mass of carbon dioxide (CO₂) produced = 3.38 g

Using the molar mass of CO₂ (44 g/mol), we can find the mass of carbon in the CO₂:
\[ \text{Mass of carbon} = \left( \frac{12 \text{ g/mol}}{44 \text{ g/mol}} \right) \times 3.38 \text{ g} = 0.92 \text{ g} \]

### Step 2: Calculate the mass of hydrogen in the gas

From the combustion, we have:
- Mass of water (H₂O) produced = 0.690 g

Using the molar mass of H₂O (18 g/mol), we can find the mass of hydrogen in the H₂O:
\[ \text{Mass of hydrogen} = \left( \frac{2 \text{ g/mol}}{18 \text{ g/mol}} \right) \times 0.690 \text{ g} = 0.077 \text{ g} \]

### Step 3: Calculate the total mass of the gas

Total mass of the gas is the sum of the mass of carbon and hydrogen:
\[ \text{Total mass} = 0.92 \text{ g} + 0.077 \text{ g} = 0.997 \text{ g} \]

### Step 4: Calculate the percentage composition of carbon and hydrogen

Percentage of carbon:
\[ \text{Percentage of carbon} = \left( \frac{0.92 \text{ g}}{0.997 \text{ g}} \right) \times 100 = 92.3\% \]

Percentage of hydrogen:
\[ \text{Percentage of hydrogen} = \left( \frac{0.077 \text{ g}}{0.997 \text{ g}} \right) \times 100 = 7.7\% \]

### Step 5: Calculate the number of moles of carbon and hydrogen

Number of moles of carbon:
\[ \text{Moles of carbon} = \frac{0.92 \text{ g}}{12 \text{ g/mol}} = 0.0767 \text{ mol} \]

Number of moles of hydrogen:
\[ \text{Moles of hydrogen} = \frac{0.077 \text{ g}}{1 \text{ g/mol}} = 0.077 \text{ mol} \]

### Step 6: Determine the simplest mole ratio

The mole ratio of carbon to hydrogen is:
\[ \text{Ratio} = \frac{0.0767}{0.0767} : \frac{0.077}{0.0767} \approx 1 : 1 \]

### Step 7: Write the empirical formula

The empirical formula is:
\[ \text{Empirical formula} = \text{CH} \]

### Step 8: Calculate the molar mass of the gas

Given:
- Volume of gas = 10.0 L
- Mass of gas = 11.6 g

At STP, 1 mole of gas occupies 22.4 L. Therefore, the number of moles of the gas is:
\[ \text{Moles of gas} = \frac{10.0 \text{ L}}{22.4 \text{ L/mol}} = 0.446 \text{ mol} \]

Now, we can calculate the molar mass:
\[ \text{Molar mass} = \frac{11.6 \text{ g}}{0.446 \text{ mol}} \approx 26 \text{ g/mol} \]

### Step 9: Determine the molecular formula

Empirical formula mass of CH:
\[ \text{Empirical formula mass} = 12 \text{ g/mol (C)} + 1 \text{ g/mol (H)} = 13 \text{ g/mol} \]

Now, we find the ratio of the molar mass to the empirical formula mass:
\[ \text{Ratio} = \frac{26 \text{ g/mol}}{13 \text{ g/mol}} = 2 \]

Thus, the molecular formula is:
\[ \text{Molecular formula} = \text{C}_2\text{H}_2 \]

### Summary of Results
1. **Empirical Formula**: CH
2. **Molar Mass**: 26 g/mol
3. **Molecular Formula**: C₂H₂