Which one of the following samples contains the smallest number of atoms ?

Options

A. 1 g of `CO_(2)`
B. 1 g of `C_(8) H_(18)`
C. 1 g of `C_(2) H_(6)`
D. 1 g of LIF

Answer

To determine which sample contains the smallest number of atoms, we will calculate the number of atoms in each sample based on the given mass and the molar mass of each compound.

### Step-by-Step Solution:

1. **Identify the Samples and Their Molar Masses**:
- Sample 1: CO₂ (Carbon Dioxide)
- Sample 2: C₈H₈ (Octene)
- Sample 3: C₂H₆ (Ethane)
- Sample 4: LiF (Lithium Fluoride)

2. **Calculate the Molar Mass of Each Compound**:
- **CO₂**:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
- Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol

- **C₈H₈**:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol
- Molar mass of C₈H₈ = (8 × 12) + (8 × 1) = 96 + 8 = 104 g/mol

- **C₂H₆**:
- Molar mass of C₂H₆ = (2 × 12) + (6 × 1) = 24 + 6 = 30 g/mol

- **LiF**:
- Lithium (Li) = 7 g/mol
- Fluorine (F) = 19 g/mol
- Molar mass of LiF = 7 + 19 = 26 g/mol

3. **Calculate the Number of Moles for Each Sample**:
- For 1 gram of each sample, the number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- **CO₂**:
\[ n = \frac{1 \text{ g}}{44 \text{ g/mol}} = 0.02273 \text{ moles} \]

- **C₈H₈**:
\[ n = \frac{1 \text{ g}}{104 \text{ g/mol}} = 0.009615 \text{ moles} \]

- **C₂H₆**:
\[ n = \frac{1 \text{ g}}{30 \text{ g/mol}} = 0.03333 \text{ moles} \]

- **LiF**:
\[ n = \frac{1 \text{ g}}{26 \text{ g/mol}} = 0.03846 \text{ moles} \]

4. **Calculate the Number of Atoms in Each Sample**:
- The total number of atoms in a compound can be calculated by multiplying the number of moles by Avogadro's number (N = \(6.022 \times 10^{23}\) atoms/mol) and the number of atoms in one molecule of the compound.

- **CO₂**:
- Number of atoms = \(0.02273 \times 6.022 \times 10^{23} \times 3 = 0.02273 \times 6.022 \times 10^{23} \times 3 = 4.09 \times 10^{22}\) atoms

- **C₈H₈**:
- Number of atoms = \(0.009615 \times 6.022 \times 10^{23} \times 10 = 0.009615 \times 6.022 \times 10^{23} \times 10 = 5.79 \times 10^{21}\) atoms

- **C₂H₆**:
- Number of atoms = \(0.03333 \times 6.022 \times 10^{23} \times 8 = 0.03333 \times 6.022 \times 10^{23} \times 8 = 1.60 \times 10^{22}\) atoms

- **LiF**:
- Number of atoms = \(0.03846 \times 6.022 \times 10^{23} \times 2 = 0.03846 \times 6.022 \times 10^{23} \times 2 = 4.63 \times 10^{22}\) atoms

5. **Compare the Number of Atoms**:
- CO₂: \(4.09 \times 10^{22}\) atoms
- C₈H₈: \(5.79 \times 10^{21}\) atoms
- C₂H₆: \(1.60 \times 10^{22}\) atoms
- LiF: \(4.63 \times 10^{22}\) atoms

6. **Conclusion**:
- The sample with the smallest number of atoms is **C₈H₈** (Octene) with \(5.79 \times 10^{21}\) atoms.