Given molar mass (M) = 74 g⁻¹ and percentage of oxygen by mass = 43.24\%.

Given molar mass ($M$) = $74 \text{ g}\cdot\text{mol}^{-1}$ and percentage of oxygen by mass = $43.24\%$.
Mass of oxygen = $0.4324 \times 74 \text{ g}\cdot\text{mol}^{-1} = 31.9976 \text{ g}\cdot\text{mol}^{-1}$.

The molar mass of oxygen is $16 \text{ g}\cdot\text{mol}^{-1}$.
Number of oxygen atoms ($z$) = $\frac{31.9976 \text{ g}\cdot\text{mol}^{-1}}{16 \text{ g}\cdot\text{mol}^{-1}} \approx 2$.
So, there are 2 oxygen atoms in the compound. The formula is $C_xH_yO_2$.

Mass of 2 oxygen atoms = $2 \times 16 \text{ g}\cdot\text{mol}^{-1} = 32 \text{ g}\cdot\text{mol}^{-1}$.
Remaining mass for carbon and hydrogen = $74 \text{ g}\cdot\text{mol}^{-1} - 32 \text{ g}\cdot\text{mol}^{-1} = 42 \text{ g}\cdot\text{mol}^{-1}$.

Let the formula for the carbon and hydrogen part be $C_xH_y$.
We know $12x + 1y = 42$.
By trial and error, or by considering common organic structures:
If $x=1$, $y=30$ (too many hydrogens).
If $x=2$, $y=18$ (too many hydrogens).
If $x=3$, $y=42 - (12 \times 3) = 42 - 36 = 6$. This gives $C_3H_6$.
This ratio ($C_3H_6$) is plausible for an organic compound.

The molecular formula is $\boxed{C_3H_6O_2}$.

2.5.2
The molecular formula $C_3H_6O_2$ can represent several functional isomers. Two common functional groups for this formula are carboxylic acids and esters.

1. Propanoic acid (a carboxylic acid):

   $$CH_3CH_2COOH$$

   This is a 3-carbon chain with a carboxylic acid group.

2. Methyl ethanoate (an ester):

   $$CH_3COOCH_3$$

   This is an ester formed from ethanoic acid and methanol.