A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of unit cells; and (b) the number of iron atoms in the paper clip. (See Appendix A for required data.)
The Correct Answer and Explanation is :
To solve this problem, we’ll break it down into two parts: calculating the number of unit cells and the number of iron atoms in the paper clip made of BCC (body-centered cubic) iron.
Given Data:
- Weight of the paper clip = 0.59 g
- Density of BCC iron = 7.87 g/cm³
- The atomic mass of iron (Fe) = 55.85 g/mol
- BCC unit cell contains 2 atoms of iron.
Step 1: Calculate the Volume of the Paper Clip
Using the formula for density, we can calculate the volume of the paper clip:
[
\text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}}
]
Substituting the known values:
[
\text{Volume} = \frac{0.59 \, \text{g}}{7.87 \, \text{g/cm}^3} \approx 0.075 \, \text{cm}^3
]
Step 2: Calculate the Number of Moles of Iron
Next, we convert the mass of the paper clip to moles using the atomic mass of iron:
[
\text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar Mass of Fe}} = \frac{0.59 \, \text{g}}{55.85 \, \text{g/mol}} \approx 0.0106 \, \text{mol}
]
Step 3: Calculate the Number of Iron Atoms
Using Avogadro’s number ((6.022 \times 10^{23}) atoms/mol):
[
\text{Number of Fe atoms} = 0.0106 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 6.38 \times 10^{21} \, \text{atoms}
]
Step 4: Calculate the Number of Unit Cells
In a BCC structure, each unit cell contains 2 atoms. Thus, the number of unit cells can be calculated as follows:
[
\text{Number of unit cells} = \frac{\text{Number of Fe atoms}}{\text{Atoms per unit cell}} = \frac{6.38 \times 10^{21} \, \text{atoms}}{2 \, \text{atoms/unit cell}} \approx 3.19 \times 10^{21} \, \text{unit cells}
]
Conclusion
(a) The number of unit cells in the paper clip is approximately (3.19 \times 10^{21}).
(b) The number of iron atoms in the paper clip is approximately (6.38 \times 10^{21}).
This calculation illustrates the relationship between mass, volume, and the atomic structure of materials, providing insight into how macroscopic properties translate into the microscopic scale of atoms and unit cells.