Chapter 17, Problem 8PEB

Magnetite and hematite are iron ore minerals. On the basis of their chemical formulas, which mineral has a higher percentage of iron? (Magnetite is Fe₂O₃, hematite is Fe₂O₃.)

To determine

The mineral that contains higher percentage of iron from Magnetite and hematite.

Answer to Problem 8PEB

Solution:

Magnetite

Explanation of Solution

Given data:

Magnetite is ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$.

Hematite is ${\text{Fe}}_2{\text{O}}_3$.

Formula used:

Write the mass percent of an element in a compound, which can be found from:

$\frac{\left(\begin{array}{l}\text{atomic weight of}\\ \text{ element}\end{array}\right)\left(\begin{array}{l}\text{number of atoms }\\ \text{of elements}\end{array}\right)}{\text{formula}\text{weight of compound}}\times 100\%\text{of compound}=\%\text{ of element}$

Explanation:

For Magnetite:

The formula for ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$.

The formula weight of Magnetite:

$\begin{array}{llll}Atoms\hfill & Atomicweight\hfill & \hfill & Total\hfill \\ 3\text{ of Fe}\hfill & 3\times 28.09\text{ u}\hfill & =\hfill & 167.55\text{ u}\hfill \\ 4\text{ of O}\hfill & 4\times 16\text{ u}\hfill & =\hfill & 64.00\text{ u}\hfill \\ \hfill & \text{Formula weight}\hfill & =\hfill & 231.55\text{ u}\hfill \end{array}$

Recall the mass percent of an element in a compound, which can be found from:

$\frac{\left(\begin{array}{l}\text{atomic weight of}\\ \text{ Fe}\end{array}\right)\left(\begin{array}{l}\text{number of atoms }\\ \text{of Fe}\end{array}\right)}{\text{formula}{\text{weight of Fe}}_{\text{3}}{\text{O}}_{\text{4}}}\times 100\%{\text{of Fe}}_{\text{3}}{\text{O}}_{\text{4}}=\%\text{ of Fe}$

Substitute $\left(55.85\text{ u Fe}\right)$ for atomic weight of element, $\left(3\right)$ for number of atoms and $231.55{\text{ u Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ for formula weight of compound.

$\begin{array}{c}\frac{\left(55.85\text{ u Fe}\right)\left(3\right)}{231.55{\text{ u Fe}}_{\text{3}}{\text{O}}_{\text{4}}}\times 100\%{\text{ Fe}}_{\text{3}}{\text{O}}_{\text{4}}=\%\text{ of Fe}\\ \%\text{ of Fe}=\frac{\left(55.85\text{ Fe}\right)\left(3\right)}{231.55}\times 100\%\\ =72.4\text{\% Fe}\end{array}$

For Hematite:

The formula for: ${\text{Fe}}_2{\text{O}}_3$.

The formula weight of Hematite:

$\begin{array}{llll}Atoms\hfill & Atomicweight\hfill & \hfill & Total\hfill \\ \text{2 of Fe}\hfill & 2\times 28.09\text{ u}\hfill & =\hfill & 111.70\text{ u}\hfill \\ \text{3 of O}\hfill & 3\times 16\text{ u}\hfill & =\hfill & 48.00\text{ u}\hfill \\ \hfill & \text{Formula weight}\hfill & =\hfill & 159.70\text{ u}\hfill \end{array}$

Recall the mass percent of an element in a compound, which can be found from:

$\frac{\left(\begin{array}{l}\text{atomic weight of}\\ \text{ Fe}\end{array}\right)\left(\begin{array}{l}\text{number of atoms }\\ \text{of Fe}\end{array}\right)}{\text{formula}{\text{weight of Fe}}_{\text{3}}{\text{O}}_{\text{4}}}\times 100\%{\text{of Fe}}_{\text{3}}{\text{O}}_{\text{4}}=\%\text{ of Fe}$

Substitute $\left(55.85\text{ u Fe}\right)$ for atomic weight of element, $\left(2\right)$ for number of atoms and $159.70{\text{ u Fe}}_2{\text{O}}_3$ for formula weight of compound.

$\begin{array}{c}\frac{\left(55.85\text{ u Fe}\right)\left(2\right)}{159.70{\text{ u Fe}}_2{\text{O}}_3}\times 100\%{\text{ Fe}}_2{\text{O}}_3=\%\text{ of Fe}\\ \%\text{ of Fe}=\frac{\left(55.85\text{ Fe}\right)\left(2\right)}{159.70}\times 100\%\\ =69.9\text{\% Fe}\end{array}$

Conclusion:

Magnetite has the higher percent of iron.