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 atoms in the paper clip. (See Appendix A for required data.)
The Correct Answer and Explanation is :
To calculate the number of unit cells and the number of atoms in a paperclip made of body-centered cubic (BCC) iron, we will need to follow these steps:
Given Data:
- Mass of paper clip = 0.59 g
- Iron (Fe) has a body-centered cubic (BCC) structure.
- Molar mass of iron (Fe) = 55.85 g/mol (from the periodic table).
- Density of BCC iron = 7.87 g/cm³ (from reference tables).
- In a BCC unit cell, there are 2 atoms per unit cell.
(a) The Number of Unit Cells:
- Convert the mass of the paperclip to moles:
[
\text{Moles of Fe} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.59 \, \text{g}}{55.85 \, \text{g/mol}} = 0.0106 \, \text{mol}
] - Convert moles to number of atoms using Avogadro’s number ( (6.022 \times 10^{23} \, \text{atoms/mol}) ):
[
\text{Number of atoms} = 0.0106 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.39 \times 10^{21} \, \text{atoms}
] - Calculate the volume of the paperclip using the density formula:
[
\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{0.59 \, \text{g}}{7.87 \, \text{g/cm}^3} = 0.075 \, \text{cm}^3
] - Calculate the number of unit cells:
In a BCC structure, there are 2 atoms per unit cell. First, calculate the volume of a single unit cell. Using the density and molar mass, the volume per unit cell can be derived:
[
\text{Volume of unit cell} = \frac{\text{molar mass}}{\text{density} \times \text{Avogadro’s number}} = \frac{55.85 \, \text{g/mol}}{7.87 \, \text{g/cm}^3 \times 6.022 \times 10^{23} \, \text{atoms/mol}} = 7.09 \times 10^{-23} \, \text{cm}^3/\text{unit cell}
] - Determine the number of unit cells in the volume of the paperclip:
[
\text{Number of unit cells} = \frac{\text{Volume of paperclip}}{\text{Volume of one unit cell}} = \frac{0.075 \, \text{cm}^3}{7.09 \times 10^{-23} \, \text{cm}^3/\text{unit cell}} = 1.06 \times 10^{21} \, \text{unit cells}
]
(b) The Number of Atoms in the Paperclip:
Since each unit cell contains 2 atoms in a BCC structure, the number of atoms in the paperclip can be calculated by multiplying the number of unit cells by 2:
[
\text{Number of atoms} = 1.06 \times 10^{21} \, \text{unit cells} \times 2 \, \text{atoms/unit cell} = 2.12 \times 10^{21} \, \text{atoms}
]
Explanation:
In this problem, we first calculate the moles of iron from the mass of the paperclip, then convert it to the total number of atoms using Avogadro’s number. The volume of the paperclip is determined from its mass and the known density of iron, which allows us to calculate the number of unit cells that fit in the volume. Given the atomic structure of BCC iron, where each unit cell contains 2 atoms, we then find the total number of atoms in the paperclip by multiplying the number of unit cells by 2. This process ties together concepts of atomic structure, density, and molecular quantities to provide an accurate estimate of the number of atoms in a typical paperclip.