Chapter 10, Problem 75A

Interpretation Introduction

Interpretation: The compound with the highest percentage of iron by mass needs to be determined.

Concept introduction: The percentage of an element by mass can be calculated by taking a ratio of the atomic mass of that element to the molar mass of the compound multiplied by 100%. For example, if the given compound is AB, the percentage of A by mass in the compound can be calculated by taking the atomic mass of A and the molar mass of AB multiplied by 100%.

Explanation of Solution

Consider option (a), given compound is ${\text{FeCl}}_{\text{2}}$ . Atomic mass of Fe is 55.85 g/mol and that of Cl is 35.45 g/mol. The percentage of Fe by mass in ${\text{FeCl}}_{\text{2}}$ can be calculated as follows:

  $\begin{array}{c}\%\text{Fe}=\frac{55.85\text{ g/mol}}{55.85\text{ g/mol+2}\left(35.45\text{ g/mol}\right)}\times 100\%\\ =\frac{55.85\text{ g/mol}}{126.75\text{ g/mol}}\times 100\%\\ =44.06\%\end{array}$

The percentage of Fe by mass is $44.06\%$ .

Consider option (b), given compound is $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{3}}$ . Atomic mass of Fe, C, H, and O is 55.85 g/mol, 12 g/mol, 1 g/mol, and 16 g/mol respectively. The percentage of Fe by mass can be calculated as follows:

  $\begin{array}{c}\%\text{Fe}=\frac{55.85\text{ g/mol}}{55.85\text{ g/mol}+3\left[2\left(12\text{ g/mol}\right)+3\left(1\text{ g/mol}+\text{2}\left(16\text{ g/mol}\right)\right)\right]}\times 100\%\\ =\frac{55.85\text{ g/mol}}{55.85\text{ g/mol}+369\text{ g/mol}}\times 100\%\\ =\frac{55.85\text{ g/mol}}{424.85\text{ g/mol}}\times 100\%\\ =13.14\%\end{array}$

Therefore, the mass percentage of Fe in the compound is $13.14\%$ .

Consider option (c), $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ . Here, the atomic mass of Fe, O, and H is 55.85 g/mol, 16 g/mol, and 1 g/mol respectively.

The percentage by mass of Fe can be calculated as follows:

  $\begin{array}{c}\%\text{Fe}=\frac{55.85\text{ g/mol}}{55.85\text{ g/mol}+2\left(16\text{ g/mol}\right)+2\left(1\text{ g/mol}\right)}\times 100\%\\ =\frac{55.85\text{ g/mol}}{55.85\text{ g/mol}+32\text{ g/mol}+2\text{ g/mol}}\times 100\%\\ =\frac{55.85\text{ g/mol}}{89.85\text{ g/mol}}\times 100\%\\ =62.16\%\end{array}$

Therefore, the mass percentage of Fe in $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $62.16\%$ .

Consider option (d), FeO. Here, the atomic mass of Fe and O is 55.85 g/mol and 16 g/mol respectively.

The percentage by mass of Fe can be calculated as follows:

  $\begin{array}{c}\%\text{Fe}=\frac{55.85\text{ g/mol}}{55.85\text{ g/mol}+16\text{ g/mol}}\times 100\%\\ =\frac{55.85\text{ g/mol}}{\text{71}\text{.85 g/mol}}\times 100\%\\ =77.73\%\end{array}$

Therefore, the mass percentage of Fe in FeO is $77.73\%$ .

Thus, the compound with the highest mass of Fe is FeO which is 77.73% and the correct option is (d).