The ratio of mass percent of C and H of an organic compound ($$C_XH_YO_Z$$) is 6 : 1. If one molecule of the above compound ($$C_XH_YO_Z$$) contains half as much oxygen as required to burn one molecule of compound $$C_XH_Y$$ completely to $$CO_2$$ and $$H_2O$$. The empirical formula of the compound $$C_XH_YO_Z$$ is:

An organic compound has the general formula $$C_{x}H_{y}O_{z}$$.

According to the statement, the mass-percent ratio of carbon to hydrogen is 6 : 1. Using atomic masses $$12\;{\rm u}$$ for carbon and $$1\;{\rm u}$$ for hydrogen, the ratio of their masses present in one empirical-formula unit is

$$\dfrac{12x}{1\cdot y}= \dfrac{6}{1}.$$

Hence,

$$12x = 6y \quad\Longrightarrow\quad y = 2x.$$

So, inside the empirical formula the number of hydrogen atoms is twice the number of carbon atoms.

Now we use the second piece of information. First state the combustion equation of the hydrocarbon $$C_{x}H_{y}:$$

$$C_{x}H_{y} + \left(x + \dfrac{y}{4}\right)O_{2}\; \longrightarrow\; x\,CO_{2} + \dfrac{y}{2}\,H_{2}O.$$

Therefore, one molecule of $$C_{x}H_{y}$$ needs $$\left(x + \dfrac{y}{4}\right)$$ molecules of $$O_{2}$$, that is

$$2\left(x + \dfrac{y}{4}\right)=2x+\dfrac{y}{2}$$ oxygen atoms in total.

The problem says that one molecule of $$C_{x}H_{y}O_{z}$$ contains half of this oxygen. So the oxygen atoms present in the organic molecule satisfy

$$z = \dfrac{1}{2}\left(2x+\dfrac{y}{2}\right) = x+\dfrac{y}{4}.$$

We already have $$y = 2x,$$ so substituting this into the above relation gives

$$z = x + \dfrac{2x}{4} = x + \dfrac{x}{2} = \dfrac{3x}{2}.$$

For $$z$$ to be an integer, $$x$$ must be even. Choosing the smallest even value, $$x = 2,$$ we obtain

$$y = 2x = 4, \qquad z = \dfrac{3(2)}{2} = 3.$$

Thus the empirical formula becomes

$$C_{2}H_{4}O_{3}.$$

Comparing with the given options, this corresponds to Option A.

Hence, the correct answer is Option A.