(i) 1 mole (44 g) of CO2 contains 12 g of carbon.
$\therefore$ 3.38 g of CO2 will contain carbon = $\frac{12 \mathrm{g}}{44 \mathrm{g}} x 3.38 \mathrm{g}$
= 0.9217 g

18 g of water contains 2 g of hydrogen.
$\therefore$ 0.690 g of water will contain hydrogen = $\frac{2 \mathrm{g}}{18 \mathrm{g}} x 0.690$
= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g
= 0.9984 g
$\therefore$ Percent of C in the compound = $\frac{0.9217 \mathrm{g}}{0.9984 \mathrm{g}} x 100$
= 92.32%

Percent of H in the compound = $\frac{0.0767\mathrm{g}}{0.9984\mathrm{g}} x 100$
= 7.68%
Moles of carbon in the compound = $\frac{92.32}{12.00}$
= 7.69
Moles of hydrogen in the compound = $\frac{7.68}{1}$
= 7.68
$\therefore$ Ratio of carbon to hydrogen in the compound = 7.69: 7.68 = 1: 1

Hence, the empirical formula of the gas is CH.

(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6 g
$\therefore$ Weight of 22.4 L of gas at STP = $\frac{11.6\mathrm{g}}{10.0\mathrm{L}} \mathrm{x} 22.4 \mathrm{L}$
= 25.984 g
≈ 26 g
Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g

$\mathrm{n} = \frac{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{gas}}{\mathrm{Empirical} \mathrm{formula} \mathrm{mass} \mathrm{of} \mathrm{gas}}$
$= \frac{26\mathrm{g}}{13\mathrm{g}}$
n = 2
$\therefore$ Molecular formula of gas = ${\left(\mathrm{CH}\right)}_{\mathrm{n}}$
= ${\mathrm{C}}_2{\mathrm{H}}_2$