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?
2 atoms/cell x 22.9898/6.02E23 = mass of unit cell = ?
Then mass = volume x density and
volume = mass/density = ?
a = V^1/34.3E-8 cm but you should go through all of the calculations. This is an estimate as are all of the calculations that follow. For the body centered unit cell
4r = a(3)^1/2 and solve for r. I get approx 186 pm.
B. What do you mean it would float if it didn't react so vigorously. It DOES float, vigorous reaction or not. I can't answer B part: I don't know to what information they refer.
A. Ah, let's get to the pointy end of science! To estimate the atomic radius of sodium, we can start by calculating the volume of the unit cell. Since sodium adopts a body-centered cubic (bcc) structure, the volume of the unit cell can be expressed as the cube of the edge length (a): V = a^3.
Now, the density of sodium is given as 0.97 g/cm^3. We know that the density (ρ) is equal to the mass (m) divided by the volume (V): ρ = m/V.
Since the molar mass of sodium (Na) is approximately 23 g/mol, we can use Avogadro's number (6.022 x 10^23) to calculate the mass of one sodium atom (m):
m = (atomic mass of Na) / (Avogadro's number) = 23 g/mol / (6.022 x 10^23 /mol) = 3.82 x 10^-23 g.
Now we can equate the two expressions for density, knowing that the unit cell contains two sodium atoms:
0.97 g/cm^3 = (2 x 3.82 x 10^-23 g) / (a^3).
Solving for a^3, we get:
(a^3) = (2 x 3.82 x 10^-23 g) / 0.97 g/cm^3 = 7.88 x 10^-23 cm^3.
Finally, taking the cube root of both sides, we find:
a ≈ 1.42 x 10^-7 cm.
This estimated value represents the body diagonal of the bcc structure, so to obtain the atomic radius (r), we divide by the square root of 3:
r ≈ (1.42 x 10^-7 cm) / √3 ≈ 8.20 x 10^-8 cm.
B. Now, onto the floating sodium situation! Cubic close-packed (ccp) structures have a packing efficiency of approximately 74%. This means that around 74% of the total volume is occupied by atoms. Since the density of bcc sodium was given (0.97 g/cm^3), we can calculate the density of ccp sodium using its packing efficiency:
Density of ccp sodium = Density of bcc sodium / Packing efficiency = 0.97 g/cm^3 / 0.74 ≈ 1.31 g/cm^3.
Considering that the density of water is approximately 1.0 g/cm^3, it appears that the density of ccp sodium is greater than that of water. Hence, if sodium were a cubic close-packed metal, it would not float on water. It seems sodium's floating dreams will remain a salty fantasy!
To estimate the atomic radius of sodium (Na), we'll use the information given: the body-centered cubic (BCC) structure and the density of 0.97 g/cm^3.
A. Estimating the Atomic Radius:
First, we need to calculate the molar mass of sodium (Na).
From the periodic table, the molar mass of sodium is approximately 22.99 g/mol.
Using Avogadro's number (6.022 × 10^23 atoms/mol), we can calculate the volume occupied by one sodium atom in the BCC structure.
1. Calculate the volume of one sodium atom:
Density = Mass / Volume
Rearranging the formula: Volume = Mass / Density
Since we have the density in g/cm^3 and Avogadro's number represents 1 mol, we'll convert the density to kg/m^3:
Density = 0.97 g/cm^3 * 1000 kg/m^3 = 970 kg/m^3
Using the molar mass of sodium:
Volume = (Molar Mass / Density) * Avogadro's number
Volume = (22.99 g/mol / 970 kg/m^3) * (6.022 × 10^23 atoms/mol)
2. Calculate the length of the edge of the body-centered cubic structure (a):
In the body-centered cubic structure, there are two sodium atoms per unit cell. Hence, the volume occupied by these two atoms is:
Volume occupied by two atoms = 2 * Volume
The volume of a cube with an edge length (a) is:
Volume = a^3
Therefore, we have:
a^3 = Volume occupied by two atoms
Now, substituting the calculated value of Volume occupied by two atoms:
a^3 = 2 * Volume
3. Calculate the atomic radius (r):
The atomic radius (r) is related to the edge length (a) by the formula:
r = √(3/4) * a
Substituting the calculated value of "a" into this formula will give us the estimated atomic radius of sodium.
B. Estimating the Density for Cubic Close-Packed (CCP) Structure:
Now, let's estimate the density of sodium if its structure were changed from BCC to a cubic close-packed (CCP) structure. Assuming the atomic radius derived from part A is applicable, we can calculate the density using the following equation:
Density = (4 * Molar Mass) / (3 * π * Atomic Radius^3)
Comparing the estimated density with the density of water (1 g/cm^3) will help determine if sodium would still float on water.
Note: The estimation assumes that the change in structure from BCC to CCP does not alter the atomic radius significantly. It is an approximation for the purpose of estimation.
Please let me know which calculations you need assistance with, A or B.
A. To estimate the atomic radius of sodium, we can use the density and Avogadro's number.
1. First, we need to find the mass of one sodium atom. Since the density is given in grams per cubic centimeter, we can assume that the volume occupied by one atom is the same as the volume of the unit cell in the body-centered cubic structure.
Density (ρ) = mass (m) / volume (V)
Rearranging the equation, we have:
mass (m) = density (ρ) * volume (V)
2. The volume of a body-centered cubic structure can be calculated using the formula:
V = (4 * r^3) / 3, where r is the atomic radius.
3. We can substitute this value of volume into the equation from step 1 to find the mass of one sodium atom.
4. Next, we know that Avogadro's number (NA) represents the number of atoms in one mole, which is 6.022 x 10^23 atoms/mol. We can use Avogadro's number to find the number of moles of sodium atoms present.
5. Finally, we can use the mass of one sodium atom and the number of moles to calculate the atomic mass, which will help us estimate the atomic radius.
B. To estimate the density of sodium if its structure were that of a cubic close-packed (CCP) metal, we can calculate the volume occupied by one sodium atom in the CCP structure using the estimated atomic radius from part A.
1. The volume occupied by one sodium atom in a CCP structure is given by the formula:
V = (4 * r^3) / 3, where r is the atomic radius.
2. The CCP structure has a packing efficiency of 74%, meaning that 74% of the total volume is occupied by atoms. We can use this packing efficiency to estimate the volume of the unit cell in the CCP structure.
3. We can then divide the total volume of the unit cell by the number of sodium atoms present to find the volume occupied by one sodium atom in the CCP structure.
4. Using the estimated atomic radius and the volume occupied by one sodium atom, we can calculate the density of sodium in the CCP structure using the equation:
density (ρ) = mass (m) / volume (V)
We can now proceed to calculate the estimates for A and B.