Chapter 4.8, Problem 57SEP

To determine

What is the theoretical $\text{atom \%}$ of each element in the compound $\text{FeO}$, what is the corresponding $\text{wt\%}$ of each element in the compound and write your conclusion.

Answer to Problem 57SEP

The theoretical $\text{atom \%}$ of $\text{Fe}\text{and}\text{O}$ element in the compound $\text{FeO}$ is $50\%$ respectively, and $\text{wt\%}$ of $\text{Fe}\text{and}\text{O}$ element is $\text{77}\text{.7\%}\text{and}\text{23}\text{.3\%}$ respectively.

Explanation of Solution

Express mass of one $\text{Fe}$ atom.

 $m_{\text{Fe}}=\frac{M_{\text{Fe}}}{N}$                                                                                          (I)

Here, atomic or molar mass of $\text{Fe}$ is $M_{\text{Fe}}$ and number of atom per mole is $N$.

Express mass of one $\text{O}$ atom.

 $m_{\text{O}}=\frac{M_{\text{O}}}{N}$                                                                                         (II)

Here, atomic or molar mass of $\text{O}$ is $M_{\text{O}}$.

Express $\text{wt\%}$ of $\text{Fe}$.

 $\text{wt\%Fe}=\frac{m_{\text{Fe}}}{m_{\text{Fe}}+m_{\text{O}}}\left(100\%\right)$                                                              (III)

Express $\text{wt\%}$ of $\text{O}$.

 $\text{wt\%O}=100\%-\text{wt\%Fe}$                                                                    (IV)

Conclusion:

Write the reaction for $\text{FeO}$.

 $\text{FeO}\to \text{Fe}+\text{O}$

Here, iron oxide is $\text{FeO}$, iron is $\text{Fe}$ and oxygen is $\text{O}$.

The atom percentage of $\text{Fe}\text{and}\text{O}$ in $\text{FeO}$ is 50% each, since there is one $\text{O}$ atom for each $\text{Fe}$ atom.

Hence, the theoretical $\text{atom \%}$ of $\text{Fe}\text{and}\text{O}$ element in the compound $\text{FeO}$ is $50\%$ respectively.

Write the atomic mass of iron and oxygen.

 $\begin{array}{l}{M}_{\text{Fe}}=55.85\text{g}\\ M_{\text{O}}=16\text{g}\end{array}$

Write the number of atom per mole or Avogadro number.

 $N=6.02\times {10}^{23}\text{atoms}$

Substitute $55.85\text{g}$ for $M_{\text{Fe}}$ and $6.02\times {10}^{23}\text{atoms}$ for $N$ in Equation (I).

 $\begin{array}{c}{m}_{\text{Fe}}=\frac{55.85\text{g}}{6.02\times {10}^{23}\text{atoms}}\\ =9.27\times {10}^{-23}\text{g}\end{array}$

Substitute $16\text{g}$ for $M_{\text{O}}$ and $6.02\times {10}^{23}\text{atoms}$ for $N$ in Equation (II).

 $\begin{array}{c}{m}_{\text{O}}=\frac{16\text{g}}{6.02\times {10}^{23}\text{atoms}}\\ =2.66\times {10}^{-23}\text{g}\end{array}$

Substitute $9.27\times {10}^{-23}\text{g}$ for $m_{\text{Fe}}$ and $2.66\times {10}^{-23}\text{g}$ for $m_{\text{O}}$ in Equation (III).

 $\begin{array}{c}\text{wt\%Fe}=\frac{9.27\times {10}^{-23}\text{g}}{9.27\times {10}^{-23}\text{g}+2.66\times {10}^{-23}\text{g}}\left(100\%\right)\\ =77.7\%\end{array}$

Substitute $77.7\%$ for $\text{wt\%Fe}$ in Equation (IV).

 $\begin{array}{c}\text{wt\%O}=100\%-77.7\%\\ =23.3\%\end{array}$

Hence, the $\text{wt\%}$ of $\text{Fe}\text{and}\text{O}$ element in the compound $\text{FeO}$ is $\text{77}\text{.7\%}\text{and}\text{23}\text{.3\%}$ respectively.

From above calculations of $\text{wt\%}\text{and}\text{atom\%}$, it is clear that both are different from each other. The weight percentage is given as weight of solute divided by weight of solvent whereas the percentage of each atom in the given alloy is atom percentage.