Understanding the Concept
To solve this problem, we need to bridge the gap between the number of molecules and the mass of the constituent elements. The key concept here is the Mole. One mole of any substance contains exactly $6.022 \times 10^{23}$ particles (Avogadro's number).
Step-by-Step Solution
1. Initial State
- Initial mass of $\text{CH}_4$: $32\text{ g}$
- Molar mass of $\text{CH}_4$: $\text{C} (12) + 4 \times \text{H} (1) = 16\text{ g/mol}$
- Initial moles of $\text{CH}_4$: $\frac{32\text{ g}}{16\text{ g/mol}} = 2\text{ moles}$
2. Amount Removed
- Molecules removed: $6.023 \times 10^{23}$
- Since $6.023 \times 10^{23}$ molecules represent exactly $1\text{ mole}$, we have removed $1\text{ mole}$ of $\text{CH}_4$.
3. Final State
- Moles remaining: $2\text{ moles (initial)} - 1\text{ mole (removed)} = 1\text{ mole}$ of $\text{CH}_4$ remaining.
4. Calculating Mass of Carbon and Hydrogen
Each molecule of $\text{CH}_4$ contains 1 atom of Carbon and 4 atoms of Hydrogen. Therefore, 1 mole of $\text{CH}_4$ contains:
- Moles of Carbon: $1\text{ mole} \times 1 = 1\text{ mole C}$
- Moles of Hydrogen: $1\text{ mole} \times 4 = 4\text{ moles H}$
Now, convert moles to grams:
- Mass of Carbon: $1\text{ mole} \times 12\text{ g/mol} = 12\text{ grams}$
- Mass of Hydrogen: $4\text{ moles} \times 1\text{ g/mol} = 4\text{ grams}$
Conclusion
After the partial evacuation, there are 12 grams of carbon and 4 grams of hydrogen remaining in the vessel.