Sodium metal adopts a body-centered cubic structure with a density of 0.97g/cm3. A. Use this information and Avogadro's number to estimate the atomic radius of sodium. B. If it didn't react so vigorously, sodium could float on water. Use the answer from part a to estimate the density of Na if its structure were that of a cubic close-packed metal. Would it still float on water?
A. The atomic radius of sodium can be estimated using the equation for the volume of a cube, V = a3, where a is the length of one side of the cube. Since the density of sodium is 0.97 g/cm3, we can calculate the volume of one mole of sodium atoms as 0.97 cm3/mol. Using Avogadro's number (6.022 x 1023 atoms/mol), we can calculate the length of one side of the cube as 0.282 nm. The atomic radius of sodium is therefore 0.141 nm.
B. The density of sodium if its structure were that of a cubic close-packed metal would be 2.27 g/cm3. This is much higher than the density of water (1 g/cm3), so sodium would not float on water.
A. To estimate the atomic radius of sodium, we can use the density and Avogadro's number.
The density of sodium (Na) is given as 0.97 g/cm³. We know that the density formula is mass divided by volume (ρ = m/V).
Since density is given as 0.97 g/cm³, we can rewrite this as:
0.97 g/cm³ = mass / volume
We also know that Avogadro's number (NA) is equal to 6.022 x 10^23 atoms/mol.
The molar mass of sodium is approximately 22.99 g/mol.
First, we need to find the volume occupied by each Na atom.
Volume occupied by each Na atom = Volume of unit cell / Number of atoms per unit cell
For a body-centered cubic (BCC) structure, the volume of the unit cell can be calculated as:
Volume of unit cell = (4 * π * r³) / 3
(Where r is the atomic radius)
Since there is one atom per unit cell in a BCC structure, the number of atoms per unit cell is 1.
Now, we can rewrite the density equation and solve for the atomic radius (r):
0.97 g/cm³ = (22.99 g/mol / NA) / [(4 * π * r³) / 3]
Simplifying the equation, we get:
r = [(3 * 22.99 g/mol) / (4 * π * 0.97 g/cm³ * NA)]^(1/3)
Substituting the values into the equation, we can calculate the atomic radius:
r = [(3 * 22.99 g/mol) / (4 * π * 0.97 g/cm³ * 6.022 x 10^23 atoms/mol)]^(1/3)
Using a calculator, we find that the estimated atomic radius of sodium is approximately 1.86 Å (angstroms).
B. To estimate the density of Na if its structure were cubic close-packed (CCP), we need to consider that in a CCP structure, there are 4 atoms per unit cell.
The volume of a CCP unit cell can be calculated as:
Volume of unit cell = (4 * π * r³) / 3
Since there are now 4 atoms per unit cell, we need to multiply the atomic radius by the factor (4/3)^(1/3):
New atomic radius (r') = r * (4/3)^(1/3)
Now, we can calculate the new density (ρ') using the formula:
ρ' = (4 * atomic mass of Na) / (NA * volume of unit cell)
Substituting the values into the equation, we can calculate the new density:
ρ' = (4 * 22.99 g/mol) / (6.022 x 10^23 atoms/mol * [(4 * π * r'^3) / 3])
Using the atomic radius calculated in part A, we can find the new density:
ρ' = (4 * 22.99 g/mol) / (6.022 x 10^23 atoms/mol * [(4 * π * (r * (4/3)^(1/3))^3) / 3])
Using a calculator, we find that the estimated density of Na in a cubic close-packed structure is approximately 1.555 g/cm³.
Since the density of water is approximately 1 g/cm³, the density of Na in a cubic close-packed structure is higher than that of water. Therefore, it would not float on water in this structure.
To estimate the atomic radius of sodium (Na), we can use the following steps:
A. Estimating Atomic Radius:
1. Start with the density of sodium, which is given as 0.97 g/cm3.
2. Convert the density to kg/m3 by dividing it by 1000: 0.97 g/cm3 ÷ 1000 = 0.97 kg/m3.
3. Determine the molar mass of sodium using the periodic table. Sodium has an atomic mass of approximately 22.99 g/mol.
4. Use Avogadro's number (6.022 x 10^23) to calculate the number of sodium atoms per unit volume.
- Divide the density by the molar mass to get the molar volume: 0.97 kg/m3 ÷ 22.99 g/mol = 0.0422 mol/m3.
- Multiply the molar volume by Avogadro's number to get the number of sodium atoms per cubic meter: 0.0422 mol/m3 × (6.022 x 10^23 atoms/mol) = 2.541 x 10^22 atoms/m3.
5. Calculate the side length of the unit cell for the body-centered cubic (bcc) structure using the formula:
- For bcc, the number of atoms per unit cell is 2, and the volume is equivalent to the length cubed (a^3).
- Set up an equation: 2.541 x 10^22 atoms/m3 = 2/a^3, where 'a' is the side length of the unit cell.
- Solve for 'a': a^3 = 2/(2.541 x 10^22 atoms/m3), then take the cube root of both sides to find 'a'.
6. Once you have the side length, divide it by 2 to get the atomic radius (r). The bcc structure has an atom located at each corner, and one in the center, so the radius is half the side length.
- Atomic radius (r) = a/2.
B. Estimating Density for Cubic Close-Packed (ccp):
1. Calculate the volume of a unit cell for a cubic close-packed structure.
- For ccp, the number of atoms per unit cell is 4, and the volume is equivalent to the length cubed (a^3).
2. Use the atomic radius determined in part A to find the side length of the unit cell for ccp.
- Side Length (a) = 2r.
3. Calculate the volume of the unit cell using the side length.
4. Divide the molar mass of sodium by the volume of the unit cell to get the molar volume for ccp.
5. Estimate the density of sodium in the ccp structure by dividing the molar mass by the ccp molar volume.
Regarding floating on water:
If the density of sodium in the ccp structure is less than the density of water (approximately 1 g/cm3), then sodium would still be able to float on water. However, if the density of sodium in the ccp structure exceeds the density of water, it would not float.