To determine the density of the metal in g/cm³, we need to follow several steps involving stoichiometry and crystallography. Let's break down the process:

Step 1: Determine the Molar Mass of the Metal (M)

Given that the compound $MO_2$ is 63.19% M by mass, we can set up an equation to find the molar mass of M.

1. Molecular weight of $MO_2$: $\text{Let the molar mass of } M = M \, \text{g/mol}$ $\text{Molar mass of } O_2 = 2 \times 16 = 32 \, \text{g/mol}$ $\text{Molar mass of } MO_2 = M + 32 \, \text{g/mol}$

2. Mass percent calculation: $\frac{M}{M + 32} = 0.6319$

3. Solving for M: $M = 0.6319(M + 32)$ $M = 0.6319M + 20.2208$ $M - 0.6319M = 20.2208$ $0.3681M = 20.2208$ $M = \frac{20.2208}{0.3681} \approx 54.92 \, \text{g/mol}$

Step 2: Calculate the Volume of the Unit Cell

For a cubic closest packed (ccp) crystal structure, also known as face-centered cubic (fcc), the relation between the atomic radius and the unit cell edge length (a) is:

1. Relation between atomic radius (r) and unit cell edge length (a): $a = 2\sqrt{2}r$

2. Given the distance between centers of touching atoms: $\text{Given distance} = 341.18 \, \text{pm} = 341.18 \times 10^{-10} \, \text{cm}$ $r = \frac{341.18}{2} \, \text{pm} = 170.59 \, \text{pm} = 170.59 \times 10^{-10} \, \text{cm}$

3. Calculate the edge length (a): $a = 2\sqrt{2} \times 170.59 \times 10^{-10} \, \text{cm} = 482.07 \times 10^{-10} \, \text{cm}$ $a = 4.8207 \times 10^{-8} \, \text{cm}$

4. Volume of the unit cell (V): $V = a^3 = (4.8207 \times 10^{-8})^3 \, \text{cm}^3$ $V \approx 1.12 \times 10^{-22} \, \text{cm}^3$

Step 3: Calculate the Density of the Metal

The density ($\rho$) of the metal can be found using the formula: $\rho = \frac{\text{mass}}{\text{volume}}$

1. Mass of atoms in the unit cell: In a face-centered cubic structure, there are 4 atoms per unit cell. $\text{Mass of 4 atoms} = 4 \times \frac{54.92 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}}$ $\text{Mass of 4 atoms} = 3.64 \times 10^{-22} \, \text{g}$

2. Calculate the density: $\rho = \frac{3.64 \times 10^{-22} \, \text{g}}{1.12 \times 10^{-22} \, \text{cm}^3} \approx 3.25 \, \text{g/cm}^3$

Thus, the density of the metal is approximately $3.25 \, \text{g/cm}^3$.

Do you want more details or have any questions?

Here are some related questions:

1. How is the distance between the centers of atoms related to the unit cell edge length in different crystal structures?
2. What are the differences between face-centered cubic and body-centered cubic structures?
3. How do you calculate the atomic packing factor for different crystal structures?
4. What are the properties of materials that have a cubic closest packed structure?
5. How does the stoichiometry of a compound affect its percentage composition by mass?
6. What methods can be used to determine the molar mass of an element experimentally?
7. How does the size of an atom influence the density of the metal it forms?
8. What are some common applications of metals with cubic closest packed structures?

Tip: Always double-check the units in your calculations to ensure accuracy, especially when converting between different measurement systems.