0.9 gm of a volatile solid organic compound (molecular weight =90) containing carbon, hydrogen and oxygen was heated with 224 ml of oxygen at 1 atm and `0^(@)` c. After combustion, the total volume of gases was 560 ml at same T and P. On treatment with KOH, the volume decreased to 112ml. Determine the value of x+y+z if molecular formula of organic compound is `C_(x)H_(y)O_(z).`

Answer

To solve the problem step by step, we will analyze the combustion of the organic compound and the subsequent reactions with KOH.

### Step 1: Determine the moles of the organic compound
Given:
- Mass of the organic compound = 0.9 g
- Molecular weight = 90 g/mol

We can calculate the number of moles (n) of the organic compound using the formula:
\[ n = \frac{\text{mass}}{\text{molecular weight}} \]
\[ n = \frac{0.9 \, \text{g}}{90 \, \text{g/mol}} = 0.01 \, \text{mol} \]

### Step 2: Analyze the combustion reaction
The combustion of the organic compound (CₓHᵧOᵧ) in the presence of oxygen (O₂) produces carbon dioxide (CO₂) and water (H₂O):
\[ C_xH_yO_z + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Calculate the volume of CO₂ produced
Given:
- Initial volume of O₂ = 224 mL
- Total volume of gases after combustion = 560 mL
- Volume of gases after treatment with KOH = 112 mL

The volume of CO₂ produced can be calculated as:
\[ \text{Volume of CO₂} = \text{Total volume} - \text{Volume after KOH treatment} \]
\[ \text{Volume of CO₂} = 560 \, \text{mL} - 112 \, \text{mL} = 448 \, \text{mL} \]

### Step 4: Convert volumes to moles
Using the ideal gas law, we can convert the volumes to moles at standard conditions (1 atm, 0°C):
- Molar volume of gas at STP = 22400 mL/mol

Moles of CO₂ produced:
\[ n_{CO_2} = \frac{448 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.02 \, \text{mol} \]

Moles of O₂ remaining after combustion:
\[ n_{O_2 \, \text{remaining}} = \frac{112 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.005 \, \text{mol} \]

### Step 5: Write the stoichiometric equation
From the combustion reaction, we can write:
\[ C_xH_yO_z + O_2 \rightarrow x \, CO_2 + \frac{y}{2} \, H_2O \]

The stoichiometric relationship gives us:
- Moles of O₂ used = Moles of O₂ initially - Moles of O₂ remaining
- Moles of O₂ used = \( 0.01 - 0.005 = 0.005 \, \text{mol} \)

### Step 6: Set up the stoichiometric equation
From the combustion reaction, we know:
\[ \text{Moles of O₂ used} = \frac{x + \frac{y}{4} - \frac{z}{4}}{2} \]

Setting the equation:
\[ 0.005 = \frac{x + \frac{y}{4} - \frac{z}{4}}{2} \]

### Step 7: Solve for x, y, and z
We know:
1. Moles of CO₂ produced = 0.02, hence \( x = 0.02 \)
2. Moles of H₂O produced = \( \frac{y}{2} \) (we will find y)
3. Moles of O₂ used = \( 0.005 \)

From the equations:
- For CO₂: \( x = 0.02 \)
- For H₂O: \( y = 0.02 \times 2 = 0.04 \)
- For O₂: Rearranging gives \( z = 0.04 \)

### Step 8: Calculate x + y + z
Now, we can find:
\[ x + y + z = 2 + 2 + 4 = 8 \]

### Final Answer
The value of \( x + y + z = 8 \).

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