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 compound `C_xH_yO_z` is:

Options

A. `C_(3)H_(6)O_(3)`
B. `C_(2)H_(4)O`
C. `C_(3)H_(3)O_(2)`
D. `C_(2)H_(4)O_(3)`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Determine the Mass Ratio of Carbon and Hydrogen
Given that the mass percent ratio of carbon (C) to hydrogen (H) in the compound \( C_xH_yO_z \) is 6:1, we can express this as:
- Mass of C = 6 parts
- Mass of H = 1 part

### Step 2: Relate Mass to Moles
Using the atomic masses:
- Atomic mass of Carbon (C) = 12 g/mol
- Atomic mass of Hydrogen (H) = 1 g/mol

From the mass ratio:
- Mass of C = \( 12x \)
- Mass of H = \( 1y \)

Setting up the equation based on the mass ratio:
\[ 12x = 6 \cdot 1y \]
This simplifies to:
\[ 12x = 6y \quad \Rightarrow \quad y = 2x \]

### Step 3: Write the Combustion Reaction
The combustion of \( C_xH_y \) can be represented as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 4: Determine Oxygen Requirement for Combustion
For complete combustion, we need to balance the reaction. The products will be:
- \( x \) moles of \( CO_2 \) (which requires \( x \) moles of O)
- \( \frac{y}{2} \) moles of \( H_2O \) (which requires \( \frac{y}{2} \) moles of O)

Thus, the total oxygen required is:
\[ \text{Total O required} = x + \frac{y}{2} \]

### Step 5: Substitute for y
Since we have \( y = 2x \):
\[ \text{Total O required} = x + \frac{2x}{2} = x + x = 2x \]

### Step 6: Determine Oxygen in the Compound
According to the problem, the compound \( C_xH_yO_z \) contains half as much oxygen as required for complete combustion:
\[ z = \frac{1}{2} \cdot 2x = x \]

### Step 7: Find the Ratio of x, y, and z
Now we have:
- \( x \) (for Carbon)
- \( y = 2x \) (for Hydrogen)
- \( z = x \) (for Oxygen)

Thus, the ratio of \( x : y : z \) is:
\[ x : 2x : x = 1 : 2 : 1 \]

### Step 8: Write the Empirical Formula
From the ratio \( 1 : 2 : 1 \), we can write the empirical formula as:
\[ C_1H_2O_1 \quad \text{or simply} \quad CH_2O \]

### Final Answer
The empirical formula of the compound \( C_xH_yO_z \) is:
\[ \text{Empirical Formula} = CH_2O \]