An element has a body-centered cubic(bcc) structure with a cell edge of 288pm the density of the element is 7.2g/cm3. How many atoms are present in 208g of element
I would convert 288 pm to cm, then cube that to obtain the volume in cc.
Next, mass = volume x density = ?
Next,
(atoms/unit cell)x molar mass/6.02E23 = mass from previous step.
Solve for molar mass.
Then mols = 208g/molar mass
1 mol contains 6.02E23 atoms; how many atoms in the mols of 208 grams.
To determine the number of atoms present in the given mass of an element, you need to follow these steps:
Step 1: Calculate the volume of the unit cell
In a body-centered cubic (bcc) structure, the volume of the unit cell can be calculated using the formula:
V = a^3 * 2^(1/2)
Where V is the volume of the unit cell and a is the edge length of the unit cell.
Given that the cell edge (a) is 288 pm (picometers), you need to convert it to cm by dividing by 10^10:
a = 288 pm / 10^10 cm/pm
Now, substitute the value of a in the volume formula:
V = (288 pm / 10^10 cm/pm)^3 * 2^(1/2)
V ≈ 0.068 cm^3
Step 2: Calculate the number of unit cells in 208 g of the element
To do this, you need to know the density (ρ) of the element. The density (ρ) is given as 7.2 g/cm^3.
The mass (m) of the element is given as 208 g.
The number of unit cells (N_uc) can be calculated using the formula:
N_uc = m / (ρ * V)
Substitute the values:
N_uc = 208 g / (7.2 g/cm^3 * 0.068 cm^3)
N_uc ≈ 421.05
Step 3: Calculate the total number of atoms
In a body-centered cubic (bcc) structure, there are two atoms per unit cell. Therefore, the total number of atoms (N_atoms) can be calculated by multiplying the number of unit cells (N_uc) by 2:
N_atoms = N_uc * 2
N_atoms ≈ 421.05 * 2
N_atoms ≈ 842.10
Rounding to the nearest whole number, the total number of atoms present in 208 g of the element is approximately 842 atoms.