Q157CP - Chemistry: Molecular Nature Of Matter And Change

Question

In a sample of any metal, spherical atoms pack closely together, but the space between them means that the density of the sample is less than that of the atoms themselves. Iridium (Ir) is one of the densest elements: 22.56 g/cm³. The atomic mass of Ir is 192.22 amu, and the mass of the nucleus is 192.18 amu. Determine the density (in g/cm³) of (a) an Ir atom and (b) an Ir nucleus. (c) How many Ir atoms placed in a row would extend 1.00 cm [radius of Ir atom = 1.36 Å; the radius of Ir nucleus = 1.5 femtometers (fm); V of a sphere =4.3πr³ ]?

Step-by-Step Solution

Answer

1. The density of an Ir atom is $\mathbf{30}\mathbf{.}\mathbf{40}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$
2. The density of an Ir nucleus is$\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$
3. The number of Ir atoms is $\mathbf{3}\mathbf{.}\mathbf{68}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{lr}\mathbf{ }\mathbf{atoms}$

Step 1: Calculating the density of an Ir atom

The volume of the atom is calculated using the given formula,

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{36}\dot{\dot{\mathbf{A}}\mathbf{\times }\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{10}}\mathbf{m}}{\mathbf{1}\dot{\mathbf{A}}}}\mathbf{\times }\frac{\mathbf{100}\mathbf{c}\mathbf{m}}{\mathbf{1}\mathbf{m}}\mathbf{)}}^{\mathbf{3}} \\ \mathbf{V}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{23}}{\mathbf{cm}}^{\mathbf{3}} \end{gathered}$$

On solving for density,

$$\begin{gathered} \mathbf{d}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}\mathbf{=}\frac{\mathbf{192}\mathbf{.}\mathbf{22}\mathbf{amu}}{\mathbf{1}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{23}}{\mathbf{cm}}^{\mathbf{3}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{g}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{amu}} \\ \mathbf{d}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{40}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}} \end{gathered}$$

Step 2: Calculating the density of an Ir nucleus

On solving the volume of the atom,

$$\begin{gathered} \mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi r}}^{\mathbf{3}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{\pi }{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{fm}\mathbf{\times }\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{15}}\mathbf{m}}{\mathbf{1}\mathbf{fm}}\mathbf{\times }\frac{\mathbf{100}\mathbf{cm}}{\mathbf{1}\mathbf{m}}\mathbf{)}}^{\mathbf{3}} \\ \mathbf{V}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{38}}{\mathbf{cm}}^{\mathbf{3}} \end{gathered}$$

On solving for density,

$$\begin{gathered} \mathbf{d}\mathbf{=}\frac{\mathbf{192}\mathbf{.}\mathbf{18}\mathbf{amu}}{\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{38}}{\mathbf{cm}}^{\mathbf{3}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{g}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{amu}} \\ \mathbf{d}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}} \end{gathered}$$

Step 3: Calculating the number of Ir atoms

For finding the diameter,

$$\begin{gathered} \mathbf{diameter}\mathbf{=}\mathbf{2}\mathbf{r} \\ \mathbf{=}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{36}\dot{\mathbf{A}\mathbf{)}} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{72}\dot{\mathbf{A}} \\ \mathbf{2}\mathbf{.}\mathbf{72}\dot{\mathbf{A}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{72}\dot{\mathbf{A}}\mathbf{\times }\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}}\mathbf{m}}{\mathbf{1}\dot{\mathbf{A}}}\mathbf{\times }\frac{\mathbf{100}\mathbf{cm}}{\mathbf{1}\mathbf{m}} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{72}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{8}}\mathbf{cm} \end{gathered}$$

Ir atoms in 1 cm would be 

$\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{8}}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{68}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{lr}\mathbf{ }\mathbf{atoms}$