0.2 g off an organic compound contains C, H and O. On combustion, it yields 0.15 g `CO_(2) and 0.12" g "H_(2)O`. The percentage of C, H and O respectively is

Options

A. `C=15%,H=20%,O=65%`
B. `C=10%,H=8.2%,O=81.8%`
C. `C=12.2%,H=8.8%,O=79%`
D. `C=20%,H=6.66%,O=73.34%`

Explanation

To find the percentage of carbon (C), hydrogen (H), and oxygen (O) in the given organic compound, we can follow these steps:

### Step 1: Write down the given data
- Weight of organic compound = 0.2 g
- Weight of carbon dioxide (CO₂) produced = 0.15 g
- Weight of water (H₂O) produced = 0.12 g

### Step 2: Calculate the molecular weights
- Molecular weight of CO₂ = 12 (C) + 16×2 (O) = 44 g/mol
- Molecular weight of H₂O = 1×2 (H) + 16 (O) = 18 g/mol

### Step 3: Calculate the moles of CO₂ and H₂O produced
- Moles of CO₂ = weight of CO₂ / molecular weight of CO₂
\[ \text{Moles of CO₂} = \frac{0.15 \, \text{g}}{44 \, \text{g/mol}} \approx 0.00341 \, \text{mol} \]

- Moles of H₂O = weight of H₂O / molecular weight of H₂O
\[ \text{Moles of H₂O} = \frac{0.12 \, \text{g}}{18 \, \text{g/mol}} \approx 0.00667 \, \text{mol} \]

### Step 4: Calculate the mass of carbon and hydrogen in the compound
- Each mole of CO₂ contains 1 mole of C, so:
\[ \text{Mass of C} = \text{Moles of CO₂} \times \text{Atomic weight of C} = 0.00341 \, \text{mol} \times 12 \, \text{g/mol} \approx 0.041 \, \text{g} \]

- Each mole of H₂O contains 2 moles of H, so:
\[ \text{Mass of H} = \text{Moles of H₂O} \times \text{Atomic weight of H} \times 2 = 0.00667 \, \text{mol} \times 1 \, \text{g/mol} \times 2 \approx 0.01334 \, \text{g} \]

### Step 5: Calculate the mass of oxygen in the compound
- Total mass of the organic compound = mass of C + mass of H + mass of O
- Rearranging gives us:
\[ \text{Mass of O} = \text{Weight of organic compound} - (\text{Mass of C} + \text{Mass of H}) \]
\[ \text{Mass of O} = 0.2 \, \text{g} - (0.041 \, \text{g} + 0.01334 \, \text{g}) \approx 0.14566 \, \text{g} \]

### Step 6: Calculate the percentages of C, H, and O
- Percentage of C:
\[ \text{Percentage of C} = \left(\frac{\text{Mass of C}}{\text{Weight of organic compound}}\right) \times 100 = \left(\frac{0.041}{0.2}\right) \times 100 \approx 20.5\% \]

- Percentage of H:
\[ \text{Percentage of H} = \left(\frac{\text{Mass of H}}{\text{Weight of organic compound}}\right) \times 100 = \left(\frac{0.01334}{0.2}\right) \times 100 \approx 6.67\% \]

- Percentage of O:
\[ \text{Percentage of O} = 100 - (\text{Percentage of C} + \text{Percentage of H}) = 100 - (20.5 + 6.67) \approx 72.83\% \]

### Final Result
The percentages of C, H, and O in the organic compound are approximately:
- C: 20.5%
- H: 6.67%
- O: 72.83%