Question

The common Eastern mole, a mammal, typically has a mass of ${}^{\mathbf{75}\mathbf{}\mathbf{g}}$, which corresponds to about ${}^{\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{moles}}$of atoms. (A mole of atoms is${}^{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}}$.) In atomic mass units (u), what is the average mass of the atoms in the common Eastern mole?

Short answer

The average mass of the atoms in the common Eastern mole is${}^{\mathbf{10}\mathbf{}\mathbf{u}}$.

Step-by-step solution

Given data

A mole of atoms corresponds to ${}^{\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}}$.

${}^{\mathbf{75}\mathbf{}\mathbf{g}}$of common Eastern mole corresponds to ${}^{\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{atoms}}$.

Understanding the conversion factor

Use the given conversion factor to calculate the number of moles in a common easter mole. Using this number and the number of atoms in mole, calculate the average mass of one atom in the common eastern mole.

Determination of the mass of one mole of atom in a common eastern mole

The mass of one mole of atoms in the common Eastern mole ${}^{{\mathbf{m}}_{\mathbf{c}}}$is,

$$\begin{gathered} {\mathit{m}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{75}}{\mathbf{7}\mathbf{.}\mathbf{5}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{}\mathit{g} \end{gathered}$$

Thus, the mass of one mole of atom in the common eastern mole is ${}^{\mathbf{10}\mathbf{}\mathbf{g}}$.

Determination of the mass of one atom in a common eastern mole

The average mass of one atom in the common Eastern mole is ${}^{{\mathbf{m}}_{\mathbf{c}}}$divided by Avogadro’s Number $N_A$.

$$\begin{gathered} \frac{{\mathbf{m}}_{\mathbf{c}}}{{\mathbf{N}}_{\mathbf{A}}}\mathbf{=}\frac{\mathbf{10}\mathbf{}\mathbf{g}}{\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{66}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{}\mathit{u} \end{gathered}$$

Thus, the average mass of one atom in the common eastern mode is${}^{\mathbf{10}\mathbf{}\mathbf{u}}$ .