Which pair of species has the same percentage composition of carbon, hydrogen, and oxygen?

• Option (1): $\text{CH}_3\text{COOH}$ and $\text{C}_6\text{H}_{12}\text{O}_6$

  • For $\text{CH}_3\text{COOH}$: Molecular formula is $\text{C}_2\text{H}_4\text{O}_2$.
    • Empirical formula: Divide subscripts by 2, which gives $\text{CH}_2\text{O}$.
  • For $\text{C}_6\text{H}_{12}\text{O}_6$: Molecular formula is $\text{C}_6\text{H}_{12}\text{O}_6$.
    • Empirical formula: Divide subscripts by 6, which gives $\text{CH}_2\text{O}$.
  • Both compounds have the same empirical formula ($\text{CH}_2\text{O}$).

• Option (2): $\text{CH}_3\text{COOH}$ and $\text{C}_2\text{H}_5\text{OH}$

  • For $\text{CH}_3\text{COOH}$: Empirical formula is $\text{CH}_2\text{O}$.
  • For $\text{C}_2\text{H}_5\text{OH}$: Molecular formula is $\text{C}_2\text{H}_6\text{O}$.
    • Empirical formula: Cannot be simplified, so $\text{C}_2\text{H}_6\text{O}$.
  • The empirical formulas are different.

• Option (3): $\text{HCOOCH}_3$ and $\text{C}_{12}\text{H}_{22}\text{O}_{11}$

  • For $\text{HCOOCH}_3$: Molecular formula is $\text{C}_2\text{H}_4\text{O}_2$.
    • Empirical formula: $\text{CH}_2\text{O}$.
  • For $\text{C}_{12}\text{H}_{22}\text{O}_{11}$: Molecular formula is $\text{C}_{12}\text{H}_{22}\text{O}_{11}$.
    • Empirical formula: Cannot be simplified, so $\text{C}_{12}\text{H}_{22}\text{O}_{11}$.
  • The empirical formulas are different.

• Option (4): $\text{C}_6\text{H}_{12}\text{O}_6$ and $\text{C}_{12}\text{H}_{22}\text{O}_{11}$

  • For $\text{C}_6\text{H}_{12}\text{O}_6$: Empirical formula is $\text{CH}_2\text{O}$.
  • For $\text{C}_{12}\text{H}_{22}\text{O}_{11}$: Empirical formula is $\text{C}_{12}\text{H}_{22}\text{O}_{11}$.
  • The empirical formulas are different.

Option (1) is the only pair where both compounds have the same empirical formula ($\text{CH}_2\text{O}$). Therefore, they will have the same percentage composition of carbon, hydrogen, and oxygen.

The final answer is $\boxed{\text{Option (1)}}$.

• Percentage of iron ($\text{Fe}$) by mass = $4.6\%$
• Number of iron atoms per molecule = 1
• Atomic mass of iron ($\text{Fe}$) $\approx 55.845 \text{ g/mol}$ (we can use $56 \text{ g/mol}$ for approximation as it's a multiple choice question).

The percentage by mass of an element in a compound is given by:

Since there is one iron atom per molecule, the mass of iron in one mole of the compound is the atomic mass of iron. Let $M$ be the molar mass of the compound.

Using the atomic mass of $\text{Fe} = 56 \text{ g/mol}$:

The calculated molar mass is approximately $1217.39 \text{ g/mol}$.
The options are:
(1) $1000 \text{ g mol}^{-1}$
(2) $1217 \text{ g mol}^{-1}$
(3) $1400 \text{ g mol}^{-1}$
(4) $1600 \text{ g mol}^{-1}$

The closest value to $1217.39 \text{ g/mol}$ is $1217 \text{ g mol}^{-1}$.

The final answer is $\boxed{\text{Option (2)}}$.