Question

A metal burns in air at under high pressure to form an oxide with formula MO₂. This compound is 23.72 % oxygen by mass. The distance between the centers of touching atoms in a cubic closest packed crystal of this metal is269.0 pm. What is this metal? What is its density?

Short answer

The metal is rhodium, and the density of the metal is $12.44g/c{m}^3$.

Step-by-step solution

Determination of the metal

Let the taken compound be 100gms. So, the mass of oxygen:

$100g\times 0.2372=23.72g$

Then the number of moles in the oxygen atom:

$$\begin{gathered} n=\frac{\text{mass}}{\text{atomic mass}} \\ =\frac{23.72g}{16.9g/mol} \\ =1.4825mol \end{gathered}$$

It is given that the formula is of the form MO₂. So the number of in the ratio 1:2

Therefore, $n\left(M\right)=0.7413$

The mass of metal $=76.28g(\because 100g-23.72g)$

Now, theatomic massof $M=\frac{\text{Mass}}{\text{number of moles}}$

$$\begin{gathered} =\frac{76.28g}{0.7413mol} \\ =102.9g/mol \end{gathered}$$

So, the atomic mass of the metal is similar to rhodium.

Determine the density of the metal.

Distance between centers of touching atoms $=134.5pm\left(\because \frac{269pm}{2}\right)$

The arrangement is a cubic closed packing crystal.

So, $a^2+a^2={\left(4r\right)}^2$

By solving, we get $a=3.8\times {10}^{-8}cm$

The volume of metal $V=a$

$$\begin{gathered} V={\left(3.8\times {10}^{-8}cm\right)}^3 \\ V=5.49\times {10}^{-23}c{m}^3 \end{gathered}$$

The mass of rhodium per unit cell:

$$\begin{gathered} m=\frac{4}{6.022\times {10}^{23}mo{l}^{-1}}\times 102.9g/mol \\ m=6.83\times {10}^{-22}g \end{gathered}$$

Now, thedensityof metal:

$$\begin{gathered} \rho =\frac{m}{V} \\ =\frac{6.83\times {10}^{-23}g}{5.49\times {10}^{-23}c{m}^3} \\ =12.44g/c{m}^3 \end{gathered}$$

Hence, the density of the metal is 12.44g/cm³.