This problem is from my chemistry book and I am having a little trouble with the intermediary steps:
Barium metal crystallizes in a body- centered cubic lattice (atoms at lattice points only)
The unit cell edge length is 502 pm, and the density of the metal if 3.5g/cm^3. (molar mass Ba= 137.327)
Calculate Avogadros Number.
(Hint: first calc the vol. occupied by 1 mol of Barium atoms in the unit cell, then calc the volume occupied by one Ba atom in the unit cell. Assume that 68% of unit cell is occupied by Ba atoms)
I know that V=edge length cubed (must convert from pm to cm), that there are two total spheres in a bcc unit cell and that Volume= mass/density. Im assuming the volume of 1 mole of Barium is 68% of the total volume, but how do I calc the volume occupied by 1 atom if i am trying to solve for Avogadros number?
The edge length is 502 pm (convert to cm), then a*(3)1/2 = 4r where r is the radius of the atom. The volume of the atom is V = (4/3)*pi*r3
To solve this problem, you need to follow the steps provided in the hint. Here's how you can calculate Avogadro's number:
Step 1: Convert the unit cell edge length from picometers (pm) to centimeters (cm).
Given: Unit cell edge length = 502 pm
1 cm = 10^10 pm (conversion factor)
502 pm * (1 cm / 10^10 pm) = 5.02 * 10^(-8) cm
Step 2: Calculate the volume occupied by one mole of barium atoms in the unit cell.
Given: Density of barium (Ba) = 3.5 g/cm^3
Given: Molar mass of barium (Ba) = 137.327 g/mol
Density = Mass / Volume
Volume = Mass / Density
1 mol of Ba = 137.327 g (molar mass in grams)
Volume of 1 mol Ba = (Mass of 1 mol Ba) / (Density of Ba)
Volume of 1 mol Ba = (137.327 g) / (3.5 g/cm^3)
Volume of 1 mol Ba = 39.23 cm^3
Step 3: Calculate the volume occupied by one Ba atom in the unit cell.
In a body-centered cubic (bcc) lattice, 68% of the unit cell is occupied by Ba atoms, which means each Ba atom occupies 0.68 times the volume of the unit cell.
Volume occupied by one Ba atom = Volume of 1 mol Ba * 0.68
Volume occupied by one Ba atom = 39.23 cm^3 * 0.68
Volume occupied by one Ba atom = 26.71 cm^3
Step 4: Calculate the volume occupied by one Ba atom in a unit cell using the unit cell edge length.
In a bcc lattice, there are 2 total spheres in the unit cell.
Volume occupied by one Ba atom = (Volume of sphere) * (Number of spheres)
Volume occupied by one Ba atom = (4/3) * pi * r^3 * 2
Assuming Ba atoms touch each other, the radius of the Ba atom is half the edge length of the unit cell.
Edge length = 5.02 * 10^(-8) cm
Radius (r) = 0.5 * Edge length = 2.51 * 10^(-8) cm
Volume occupied by one Ba atom = (4/3) * pi * (2.51 * 10^(-8) cm)^3 * 2
Step 5: Determine Avogadro's number.
Avogadro's number (N) is defined as the number of atoms in 1 mole of a substance.
Volume occupied by one Ba atom = (Volume of one mole of Ba) / (Avogadro's number)
(4/3) * pi * (2.51 * 10^(-8) cm)^3 * 2 = (39.23 cm^3) / N
Now, you can solve for Avogadro's number (N) by rearranging the equation:
N = (39.23 cm^3) / [(4/3) * pi * (2.51 * 10^(-8) cm)^3 * 2]
Calculating the value will give you Avogadro's number.
To solve this problem and calculate Avogadro's number, we need to follow the steps outlined in the problem statement.
Step 1: Calculate the volume occupied by 1 mole of Barium atoms in the unit cell.
Since it is mentioned in the problem that 68% of the unit cell is occupied by Barium atoms, we can assume that 68% of the total volume is occupied by Barium atoms.
Given that the unit cell edge length is 502 pm (1 pm = 1 × 10^-12 m), we need to convert it to cm.
Edge length = 502 pm × (1 × 10^-12 m / 1 pm) × (1 cm / 0.01 m) = 5.02 cm
Now, we can calculate the volume of the unit cell occupied by one mole of Barium atoms using the formula:
Volume of 1 mole of Barium atoms = 0.68 × Edge length^3
Step 2: Calculate the volume occupied by one Barium atom in the unit cell.
In a body-centered cubic (bcc) structure, there are two total spheres (atoms) in the unit cell, and each sphere (atom) has a volume associated with it. We need to calculate the volume of one Barium atom in the unit cell.
Volume of one Barium atom = Volume of 1 mole of Barium atoms / Avogadro's number
Step 3: Calculate Avogadro's number.
Avogadro's number (N) can be calculated by rearranging the equation from step 2:
Avogadro's number = Volume of 1 mole of Barium atoms / Volume of one Barium atom
Now, let's perform the calculations.
Step 1:
Using the formula:
Volume of 1 mole of Barium atoms = 0.68 × (5.02 cm)^3
Calculate this value.
Step 2:
Using the given molar mass of Barium (137.327 g/mol) and the density of the metal (3.5 g/cm^3), we can calculate the volume of one Barium atom in the unit cell.
Mass of 1 mole of Barium atoms = Molar mass of Barium = 137.327 g/mol
Volume of one Barium atom = Mass of 1 mole of Barium atoms / Density of the metal
Step 3:
Using the values obtained in steps 1 and 2, calculate Avogadro's number using the formula:
Avogadro's number = Volume of 1 mole of Barium atoms / Volume of one Barium atom
Perform the necessary calculations, and you will have the value of Avogadro's number.