Iron (Fe) has an atomic radius of 126 pm. What would the density of Fe be in g/cm’ if it crystallized in the following crystal structures: Simple cubic (3 pts) Body centered cubic (3 pts)
The Correct Answer and Explanation is:
To find the density of iron (Fe) in g/cm³ under different crystal structures, we use the following formula:Density=Mass of atoms in unit cellVolume of unit cell\text{Density} = \frac{\text{Mass of atoms in unit cell}}{\text{Volume of unit cell}}Density=Volume of unit cellMass of atoms in unit cell
Given:
- Atomic radius of Fe = 126 pm = 1.26×10−81.26 \times 10^{-8}1.26×10−8 cm
- Atomic mass of Fe = 55.85 g/mol
- Avogadro’s number NA=6.022×1023N_A = 6.022 \times 10^{23}NA=6.022×1023 atoms/mol
1. Simple Cubic (SC):
- Atoms per unit cell (SC) = 1
- Edge length a=2r=2×1.26×10−8 cm=2.52×10−8 cma = 2r = 2 \times 1.26 \times 10^{-8} \text{ cm} = 2.52 \times 10^{-8} \text{ cm}a=2r=2×1.26×10−8 cm=2.52×10−8 cm
Step 1: Volume of unit cell:
V=a3=(2.52×10−8)3=1.60×10−23 cm3V = a^3 = (2.52 \times 10^{-8})^3 = 1.60 \times 10^{-23} \text{ cm}^3V=a3=(2.52×10−8)3=1.60×10−23 cm3
Step 2: Mass of one atom:
Mass=55.856.022×1023=9.28×10−23 g\text{Mass} = \frac{55.85}{6.022 \times 10^{23}} = 9.28 \times 10^{-23} \text{ g}Mass=6.022×102355.85=9.28×10−23 g
Step 3: Density:
Density=9.28×10−231.60×10−23≈5.8 g/cm3\text{Density} = \frac{9.28 \times 10^{-23}}{1.60 \times 10^{-23}} \approx 5.8 \text{ g/cm}^3Density=1.60×10−239.28×10−23≈5.8 g/cm3
2. Body-Centered Cubic (BCC):
- Atoms per unit cell (BCC) = 2
- In BCC, edge length a=4r3=4×1.26×10−83=2.91×10−8 cma = \frac{4r}{\sqrt{3}} = \frac{4 \times 1.26 \times 10^{-8}}{\sqrt{3}} = 2.91 \times 10^{-8} \text{ cm}a=34r=34×1.26×10−8=2.91×10−8 cm
Step 1: Volume of unit cell:
V=a3=(2.91×10−8)3=2.46×10−23 cm3V = a^3 = (2.91 \times 10^{-8})^3 = 2.46 \times 10^{-23} \text{ cm}^3V=a3=(2.91×10−8)3=2.46×10−23 cm3
Step 2: Mass of 2 atoms:
Mass=2×55.856.022×1023=1.856×10−22 g\text{Mass} = 2 \times \frac{55.85}{6.022 \times 10^{23}} = 1.856 \times 10^{-22} \text{ g}Mass=2×6.022×102355.85=1.856×10−22 g
Step 3: Density:
Density=1.856×10−222.46×10−23≈7.5 g/cm3\text{Density} = \frac{1.856 \times 10^{-22}}{2.46 \times 10^{-23}} \approx 7.5 \text{ g/cm}^3Density=2.46×10−231.856×10−22≈7.5 g/cm3
Final Answers:
- Simple Cubic: ≈ 5.8 g/cm³
- Body-Centered Cubic: ≈ 7.5 g/cm³
Explanation (Textbook Style – 300 Words):
To determine the density of a substance based on its crystal structure, we need to calculate both the mass of atoms in a unit cell and the volume of that unit cell. The density formula, derived from mass/volume, provides insight into how closely packed the atoms are within a specific crystal arrangement.
For iron (Fe), which has an atomic radius of 126 picometers (pm), we consider two possible crystal structures: simple cubic (SC) and body-centered cubic (BCC).
In the simple cubic structure, only one atom occupies each unit cell. The edge of the cube is twice the atomic radius, giving an edge length of 2.52 × 10⁻⁸ cm. The volume of the cube is found by cubing the edge length. The mass of one atom is calculated using the atomic mass of iron and dividing by Avogadro’s number. Dividing this atomic mass by the volume gives the density of about 5.8 g/cm³.
In contrast, the body-centered cubic structure contains 2 atoms per unit cell. Its geometry is such that the cube’s body diagonal equals 4 atomic radii, so the edge length is found using the relation a=4r3a = \frac{4r}{\sqrt{3}}a=34r, resulting in a slightly larger cube. The volume is calculated by cubing this edge length. Since there are two atoms per cell, the total mass is doubled. Calculating the mass per volume gives a higher density, about 7.5 g/cm³, reflecting the more efficient atomic packing of the BCC structure compared to the SC structure.
This analysis illustrates how the arrangement of atoms in space directly affects a metal’s density, even when atomic size and mass remain the same.
