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 water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate: (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

The correct option is: (i) CH, (ii) 26g, (iii) C₂H₂

Below is the step-by-step derivation and educational explanation:

(i) Determining the Empirical Formula

A welding fuel gas contains only carbon (C) and hydrogen (H). When burned completely in oxygen, all carbon is converted to carbon dioxide (CO₂) and all hydrogen is converted to water (H₂O).

First, we calculate the mass of carbon present in 3.38 g of CO₂:

$\text{Mass of C}=\frac{12.0\text{ g}}{44.0\text{ g}}\times 3.38\text{ g}=0.922\text{ g}$

Next, we calculate the mass of hydrogen present in 0.690 g of H₂O:

$\text{Mass of H}=\frac{2.0\text{ g}}{18.0\text{ g}}\times 0.690\text{ g}=0.0767\text{ g}$

Now, we find the number of moles of each element:

$\text{Moles of C}=\frac{0.922\text{ g}}{12.0\text{ g/mol}}=0.0768\text{ mol}$

$\text{Moles of H}=\frac{0.0767\text{ g}}{1.0\text{ g/mol}}=0.0767\text{ mol}$

Dividing by the smaller value to find the simplest whole-number ratio:

$\text{C : H ratio}=\frac{0.0768}{0.0767}:\frac{0.0767}{0.0767}\approx 1:1$

Therefore, the empirical formula is CH.

(ii) Calculating the Molar Mass of the Gas

At standard temperature and pressure (STP), 1 mole of any ideal gas occupies exactly 22.4 L.
We are given that 10.0 L of the gas weighs 11.6 g at STP. We can find the mass of 22.4 L to determine the molar mass:

$\text{Molar Mass}=\frac{11.6\text{ g}}{10.0\text{ L}}\times 22.4\text{ L/mol}=25.98\text{ g/mol}\approx 26\text{ g/mol}$

Therefore, the molar mass of the gas is 26 g.

(iii) Determining the Molecular Formula

First, calculate the empirical formula mass for CH:

$\text{Empirical Mass}=12.0+1.0=13.0\text{ g/mol}$

Next, find the multiplier ($n$) by dividing the molar mass by the empirical mass:

$n=\frac{\text{Molar Mass}}{\text{Empirical Mass}}=\frac{26}{13}=2$

Multiplying the empirical formula by this factor yields the molecular formula:

$\text{Molecular Formula}={\left(\text{CH}\right)}_2={\text{C}}_2{\text{H}}_2$

Thus, the molecular formula of the gas is C₂H₂.