The density of an unknown metal is 8.94 g/cm^3 and its atomic radius is 0.126 nm . It has a face-centered cubic lattice. Find the atomic weight of the unknown metal.
60.9
4*radius = sqrt(2)*a
solve for a. I recommend converting radius to cm (from nm).
a is the length of a side of the crystal; therefore, a^3 will be the volume of the cube in cc.
mass of the unit cell = volume x density.
Substitute mass into
mass = 4unitcells*molar mass/6.022 x 10^23. and solve for molar mass.
Well, since you have given me the density and atomic radius of the metal, I could really use some help putting the pieces together. Or maybe I can just glue them and call it an art project? Anyway, let's get to the calculation.
The density of an unknown metal is given as 8.94 g/cm³. Now, to find the atomic weight, we need to determine the mass of a single atom. And for that, we need the volume of the unit cell.
Since the metal has a face-centered cubic (FCC) lattice, each corner atom contributes only (1/8)th of its volume to the unit cell, while each face-centered atom contributes the whole volume. So, considering this convoluted party, the volume of the unit cell can be calculated as:
Volume of the unit cell = [(1/8) x atomic radius³] + atomic radius³
Now, we need to solve for the atomic weight using the equation:
Density = (Atomic weight x Avogadro's number) / Volume of the unit cell
Plugging in the values we know:
8.94 g/cm³ = (Atomic weight x 6.022 x 10²³) / [((1/8) x 0.126 nm)³ + (0.126 nm)³]
By doing some number crunching and converting nanometers into centimeters, you should be able to determine the atomic weight! Good luck, and may your calculations be as stable as a unicycle-riding clown!
To find the atomic weight of the unknown metal, we can use the equation:
Density = (atomic weight * Avogadro's number) / (volume per atom * 1 g/cm^3),
where volume per atom = (4/3) * π * (atomic radius)^3.
First, convert the given atomic radius from nm to cm:
0.126 nm = 0.126 x 10^-7 cm = 1.26 x 10^-8 cm.
Next, calculate the volume per atom:
volume per atom = (4/3) * π * (1.26 x 10^-8 cm)^3 = 1.24 x 10^-23 cm^3.
Now, substitute the values into the equation and solve for the atomic weight:
8.94 g/cm^3 = (atomic weight * 6.022 x 10^23) / (1.24 x 10^-23 cm^3).
Cross multiply and solve for atomic weight:
atomic weight = (8.94 g/cm^3 * 1.24 x 10^-23 cm^3) / (6.022 x 10^23).
atomic weight ≈ 1.825 x 10^-22 g.
Therefore, the atomic weight of the unknown metal is approximately 1.825 x 10^-22 g.
To find the atomic weight of the unknown metal, we need to relate the given information about its density, lattice structure, and atomic radius.
Here are the steps to solve this problem:
Step 1: Recognize the lattice structure.
The face-centered cubic (FCC) lattice has atoms in each corner of a cube and additional atoms at the center of each face.
Step 2: Determine the number of atoms per unit cell.
In an FCC lattice, there are 4 atoms per unit cell. This can be derived by considering that each corner atom is shared by 8 adjacent unit cells, and each face atom is shared by 2 adjacent unit cells.
Step 3: Calculate the volume of the unit cell.
The volume of a unit cell can be determined by considering its lattice structure. For an FCC lattice, the volume can be calculated as follows:
Volume of the unit cell = (4 * r^3) / 3
Given the atomic radius r = 0.126 nm, we can substitute this value into the equation to find the volume of the unit cell.
Step 4: Calculate the mass of the unit cell.
The mass of the unit cell can be calculated by multiplying the number of atoms in the unit cell by the atomic weight of each atom.
Step 5: Relate density and mass to find the atomic weight.
The density is given as 8.94 g/cm^3. To relate this to the mass of the unit cell, we can use the following equation:
Density = mass / volume
We now have all the information needed to solve the problem. Let's apply these steps:
Step 1: The FCC lattice has 4 atoms per unit cell.
Step 2: The atomic radius is given as 0.126 nm.
Step 3: Calculating the volume of the unit cell:
Volume of the unit cell = (4 * r^3) / 3
= (4 * (0.126 nm)^3) / 3
≈ 0.003986 nm^3
Step 4: Calculating the mass of the unit cell:
Let's assume the atomic weight of the unknown metal is W amu.
Mass of the unit cell = (4 * W amu) / Avogadro's Number
Step 5: Relating density and mass to find the atomic weight:
Density = mass / volume
8.94 g/cm^3 = [(4 * W amu) / Avogadro's Number] / (0.003986 nm^3)
To simplify the equation, we need to convert the units:
1 g = 1 amu (approximately)
1 cm^3 = 1 nm^3
8.94 g/cm^3 = [(4 * W) / Avogadro's Number] / (0.003986)
8.94 amu/nm^3 = (4 * W) / Avogadro's Number
We can rearrange the equation to solve for W:
W = (8.94 amu/nm^3 * Avogadro's Number) / 4
Using the Avogadro's Number (approximately 6.022 x 10^23 mol^-1), we can calculate W:
W ≈ (8.94 amu/nm^3 * 6.022 x 10^23 mol^-1) / 4
By evaluating the right side of the equation, we can find the atomic weight of the unknown metal.