1. Let the saturated acyclic compound be a hydroxy compound

A saturated acyclic organic compound containing C, H, and O can be written as:

$\mathrm{C_nH_{2n+2}O_m}$

Since it is saturated and acyclic, for a monofunctional oxygen-containing compound the hydrogen count follows the saturated pattern. We will determine the exact formula from the mass percentage ratios.

2. Use the given mass percentage ratio of C and H

Given:

$\%C : \%H = 4:1$

For formula $\mathrm{C_xH_yO_z}$,

• mass of C $= 12x$
• mass of H $= y$
• mass of O $= 16z$

So,

$12x : y = 4 : 1$

$\frac{12x}{y}=4 \Rightarrow y=3x$

3. Use the given mass percentage ratio of C and O

Given:

$\%C : \%O = 3:4$

So,

$12x : 16z = 3 : 4$

$\frac{12x}{16z} = \frac{3}{4}$

Cross-multiplying:

$48x = 48z \Rightarrow x=z$

Thus,

$z=x$

and from above,

$y=3x$

Hence empirical formula is:

$\mathrm{C_xH_{3x}O_x} = (\mathrm{CH_3O})_x$

4. Use saturation condition

For a saturated acyclic compound, the general relation is:

$H = 2C + 2$

for compounds with one oxygen atom such as alcohols/ethers, because oxygen does not affect hydrogen deficiency.

So if the formula is $\mathrm{C_nH_{2n+2}O}$, compare with ratio-derived formula:

$\mathrm{C_nH_{3n}O_n}$

Equating hydrogens:

$3n = 2n + 2$

$n=2$

Therefore,

$\boxed{\mathrm{X=C_2H_6O_2}}$

Check:

• C mass $=24$
• H mass $=6$
• O mass $=32$

Then:

$C:H = 24:6 = 4:1$

$C:O = 24:32 = 3:4$

Correct.

5. Write combustion reaction

For one mole of $\mathrm{C_2H_6O_2}$:

$\mathrm{C_2H_6O_2 + aO_2 \rightarrow 2CO_2 + 3H_2O}$

Count oxygen atoms on RHS:

• from $2CO_2$: $4$ O atoms
• from $3H_2O$: $3$ O atoms

Total oxygen atoms needed on RHS $=7$

Compound already contains $2$ oxygen atoms, so oxygen atoms needed from $\mathrm{O_2}$:

$7-2=5$

Thus,

$a=\frac{5}{2}$

So, 1 mole of compound requires:

$\frac{5}{2}\text{ moles of }O_2$

For 2 moles of compound:

$2 \times \frac{5}{2} = 5$

6. Final answer

The moles of $\mathrm{O_2}$ required for complete combustion of 2 moles of $\mathrm{X}$ is:

$\boxed{5}$