1. Use the given mass percent ratio of C and H

For the compound $C_XH_YO_Z$, the masses of carbon and hydrogen in one mole are:

• Carbon mass $= 12X$
• Hydrogen mass $= Y$

Given, $\text{mass percent ratio of C : H} = 6:1$ So, $\frac{12X}{Y} = \frac{6}{1}$ $12X = 6Y$ $2X = Y$

Hence, $Y = 2X$

2. Use the oxygen condition

We are told that one molecule of $C_XH_YO_Z$ contains half as much oxygen as required to burn one molecule of $C_XH_Y$ completely.

First, write the combustion of $C_XH_Y$: $C_XH_Y + \left(X + \frac{Y}{4}\right)O_2 \to XCO_2 + \frac{Y}{2}H_2O$

So, the number of oxygen atoms required for complete combustion of one molecule of $C_XH_Y$ is: $2\left(X + \frac{Y}{4}\right) = 2X + \frac{Y}{2}$

Given that the molecule $C_XH_YO_Z$ contains half of this amount of oxygen atoms: $Z = \frac{1}{2}\left(2X + \frac{Y}{2}\right)$ $Z = X + \frac{Y}{4}$

Now substitute $Y = 2X$: $Z = X + \frac{2X}{4} = X + \frac{X}{2} = \frac{3X}{2}$

Thus, $Y=2X, \qquad Z=\frac{3X}{2}$

For integer subscripts, $X$ must be even. Take the smallest even value: $X=2$ Then, $Y=4, \qquad Z=3$

So the empirical formula is: $C_2H_4O_3$

3. Check options

• A: $C_2H_4O_3$ ✅
• B: $C_3H_6O_3$ has $C:H$ mass ratio $=36:6=6:1$, but this is not empirical since it reduces to $C_1H_2O_1$ only if oxygen scaled proportionally; also from derived relation $Z=3X/2$, for $X=3$, $Z=4.5$ not possible.
• C: $C_2H_4O$ gives wrong oxygen relation.
• D: $C_3H_4O_2$ gives wrong $C:H$ mass ratio since $36:4 = 9:1$.

Therefore, the correct option is A.

4. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored answer.