The pair of species having different percentage (mass) of carbon is :

Options

A. `CH_(3)COOH and C_(6)H_(12)O_(6)`
B. `CH_(3)COOH and C_(2)H_(5)OH`
C. `HCOOCH_(3) and HCOOH`
D. `C_(2)H_(5)OH and CH_(3)OCH_(3)`

Answer

To find the pair of species having different percentage (mass) of carbon, we will calculate the percentage of carbon in each compound listed in the options. The percentage of carbon can be calculated using the formula:

\[ \text{Percentage of Carbon} = \left( \frac{\text{Total mass of Carbon}}{\text{Molecular weight of the compound}} \right) \times 100 \]

### Step-by-Step Solution:

1. **Calculate the percentage of carbon in CH₃COOH (Acetic Acid)**:
- Molecular formula: CH₃COOH
- Number of carbon atoms: 2
- Molecular weight calculation:
- Carbon: 2 × 12 = 24
- Hydrogen: 4 × 1 = 4
- Oxygen: 2 × 16 = 32
- Total molecular weight = 24 + 4 + 32 = 60
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{24}{60} \right) \times 100 = 40\% \]

2. **Calculate the percentage of carbon in C₆H₁₂O₆ (Glucose)**:
- Molecular formula: C₆H₁₂O₆
- Number of carbon atoms: 6
- Molecular weight calculation:
- Carbon: 6 × 12 = 72
- Hydrogen: 12 × 1 = 12
- Oxygen: 6 × 16 = 96
- Total molecular weight = 72 + 12 + 96 = 180
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{72}{180} \right) \times 100 = 40\% \]

3. **Calculate the percentage of carbon in C₂H₅OH (Ethanol)**:
- Molecular formula: C₂H₅OH
- Number of carbon atoms: 2
- Molecular weight calculation:
- Carbon: 2 × 12 = 24
- Hydrogen: 6 × 1 = 6
- Oxygen: 1 × 16 = 16
- Total molecular weight = 24 + 6 + 16 = 46
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{24}{46} \right) \times 100 \approx 52.17\% \]

4. **Calculate the percentage of carbon in HCOOCH₃ (Methyl Formate)**:
- Molecular formula: HCOOCH₃
- Number of carbon atoms: 2
- Molecular weight calculation:
- Carbon: 2 × 12 = 24
- Hydrogen: 3 × 1 = 3
- Oxygen: 2 × 16 = 32
- Total molecular weight = 24 + 3 + 32 = 59
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{24}{59} \right) \times 100 \approx 40.68\% \]

5. **Calculate the percentage of carbon in HCOOH (Formic Acid)**:
- Molecular formula: HCOOH
- Number of carbon atoms: 1
- Molecular weight calculation:
- Carbon: 1 × 12 = 12
- Hydrogen: 2 × 1 = 2
- Oxygen: 2 × 16 = 32
- Total molecular weight = 12 + 2 + 32 = 46
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{12}{46} \right) \times 100 \approx 26.08\% \]

6. **Calculate the percentage of carbon in CH₃OCH₃ (Dimethyl Ether)**:
- Molecular formula: CH₃OCH₃
- Number of carbon atoms: 2
- Molecular weight calculation:
- Carbon: 2 × 12 = 24
- Hydrogen: 6 × 1 = 6
- Oxygen: 1 × 16 = 16
- Total molecular weight = 24 + 6 + 16 = 46
- Percentage of Carbon:
\[ \text{Percentage of Carbon} = \left( \frac{24}{46} \right) \times 100 \approx 52.17\% \]

### Conclusion:
- **CH₃COOH** and **C₆H₁₂O₆** both have 40% carbon.
- **CH₃COOH** and **C₂H₅OH** have 40% and 52.17% carbon respectively.
- **HCOOCH₃** and **HCOOH** have approximately 40.68% and 26.08% carbon respectively.
- **C₂H₅OH** and **CH₃OCH₃** both have 52.17% carbon.

Thus, the pairs with different percentages of carbon are:
- **CH₃COOH and C₂H₅OH**
- **HCOOCH₃ and HCOOH**