To determine the empirical formula of the compound given the masses of [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex] produced during combustion, we'll follow several steps:
1. Determine the moles of [tex]\( CO_2 \)[/tex] produced:
- Given: mass of [tex]\( CO_2 \)[/tex] = 11.0 g
- Molar mass of [tex]\( CO_2 \)[/tex] = 44.01 g/mol
[tex]\[
\text{Moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} = \frac{11.0 \, \text{g}}{44.01 \, \text{g/mol}} = 0.25 \, \text{mol}
\][/tex]
2. Determine the moles of [tex]\( H_2O \)[/tex] produced:
- Given: mass of [tex]\( H_2O \)[/tex] = 4.5 g
- Molar mass of [tex]\( H_2O \)[/tex] = 18.02 g/mol
[tex]\[
\text{Moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O} = \frac{4.5 \, \text{g}}{18.02 \, \text{g/mol}} = 0.25 \, \text{mol}
\][/tex]
3. Determine the moles of carbon (C) and hydrogen (H):
- Each mole of [tex]\( CO_2 \)[/tex] corresponds to 1 mole of carbon atoms.
[tex]\[
\text{Moles of C} = 0.25 \, \text{mol}
\][/tex]
- Each mole of [tex]\( H_2O \)[/tex] corresponds to 2 moles of hydrogen atoms.
[tex]\[
\text{Moles of H} = 0.25 \, \text{mol} \times 2 = 0.50 \, \text{mol}
\][/tex]
4. Find the mole ratio of C to H to obtain the empirical formula:
- Ratio of moles of C to moles of C:
[tex]\[
\text{Ratio of C} = \frac{\text{moles of C}}{\text{moles of C}} = \frac{0.25 \, \text{mol}}{0.25 \, \text{mol}} = 1.0
\][/tex]
- Ratio of moles of H to moles of C:
[tex]\[
\text{Ratio of H} = \frac{\text{moles of H}}{\text{moles of C}} = \frac{0.50 \, \text{mol}}{0.25 \, \text{mol}} = 2.0
\][/tex]
5. Determine the simplest whole number ratio:
- The ratio [tex]\(C : H\)[/tex] is [tex]\( 1.0 : 2.0 \)[/tex]. This represents the simplest ratio.
6. Empirical formula:
- Based on the mole ratio, the empirical formula of the compound is [tex]\( CH_2 \)[/tex].
So, the empirical formula of the compound is [tex]\( \boxed{CH_2} \)[/tex]. Therefore, the correct answer is (B) [tex]\(CH_2\)[/tex].
