To determine the molecular formula of the compound, we'll follow these steps:
1. Calculate the moles of carbon (C) and hydrogen (H):
- Given:
- Mass of Carbon (C) = 32.0 grams
- Mass of Hydrogen (H) = 8.0 grams
- Molar mass of Carbon (C) = 12.0 g/mol
- Molar mass of Hydrogen (H) = 1.0 g/mol
- Moles of Carbon:
[tex]\[
\text{moles of C} = \frac{\text{grams of C}}{\text{molar mass of C}} = \frac{32.0}{12.0} \approx 2.67 \text{ moles}
\][/tex]
- Moles of Hydrogen:
[tex]\[
\text{moles of H} = \frac{\text{grams of H}}{\text{molar mass of H}} = \frac{8.0}{1.0} = 8.0 \text{ moles}
\][/tex]
2. Determine the simplest whole number ratio of the moles:
- To find the ratio, divide the moles of each element by the smallest number of moles calculated:
The smallest number of moles among the two elements is [tex]\( \approx 2.67 \)[/tex].
- Ratio of Carbon:
[tex]\[
\text{ratio of C} = \frac{2.67}{2.67} = 1
\][/tex]
- Ratio of Hydrogen:
[tex]\[
\text{ratio of H} = \frac{8.0}{2.67} \approx 3
\][/tex]
- After rounding to the nearest whole number:
[tex]\[
\text{ratio of C} = 1, \quad \text{ratio of H} = 3
\][/tex]
3. Form the empirical formula:
- Based on the ratios determined, the empirical formula is [tex]\( CH_3 \)[/tex].
4. Confirm the empirical formula matches the given molar mass:
- The given molar mass of the compound = 30.0 g/mol.
- Calculate the molar mass of the empirical formula [tex]\( CH_3 \)[/tex]:
- Carbon (C): [tex]\( 1 \times 12.0 = 12.0 \)[/tex]
- Hydrogen (H): [tex]\( 3 \times 1.0 = 3.0 \)[/tex]
Total molar mass of [tex]\( CH_3 \)[/tex] = [tex]\( 12.0 + 3.0 = 15.0 \)[/tex] g/mol
- Since the empirical formula mass (15.0 g/mol) does not match the given molar mass (30.0 g/mol), multiply the empirical formula by an integer that scales the empirical formula mass to the molar mass:
[tex]\[
\frac{30.0}{15.0} = 2
\][/tex]
- Therefore, the molecular formula is:
[tex]\[
(CH_3)_2 = C_2H_6
\][/tex]
Hence, the molecular formula of the compound is [tex]\( \mathbf{C_2H_6} \)[/tex].
Thus, the correct answer is:
[tex]\[
C_2H_6
\][/tex]
