Chapter 4, Problem 90QP

(a)

Interpretation Introduction

Interpretation:

The percent composition of iron in $\text{FeO}$ is to be determined.

Explanation of Solution

The percent composition of any substance is calculated by dividing the mass of substance in a sample by total mass of the sample. This ratio is multiplied by $100\text{ \%}$ .

To determine the percent composition let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound $\text{FeO}$ as follows:

The mass of Fe from periodic table is $55.85\text{ g/mole}$ . For $1\text{ mole}$ , the mass is,

$\begin{array}{c}m=1\text{ mole}\times \text{55}\text{.85 g/mole}\\ =\text{55}\text{.85 g}\end{array}$

The mass of O from periodic table is $16.00\text{ g/mole}$ . For $1\text{ mole}$ , the mass is,

$\begin{array}{c}m=1\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{16}\text{.00 g}\end{array}$

Therefore, the total mass for $1\text{ mole}$ of $\text{FeO}$ is,

$\begin{array}{c}\text{total mass}=55.85\text{ g}+16.00\text{ g}\\ =71.85\text{ g}\end{array}$

Now, the percent composition of Fe is,

$\begin{array}{c}\%\text{ Fe}=\frac{55.85\cancel{\text{g}}}{71.85\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{77}\text{.73 \%}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percent iron in ${\text{Fe}}_2{\text{O}}_3$ is to analyzed.

Explanation of Solution

To determine the percent composition let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound ${\text{Fe}}_2{\text{O}}_3$ as follows:

The mass of Fe from periodic table is $55.85\text{ g/mole}$ . For $2\text{ mole}$ , the mass is,

$\begin{array}{c}m=2\text{ mole}\times \text{55}\text{.85 g/mole}\\ =\text{111}\text{.7 g}\end{array}$

The mass of O from periodic table is $16.00\text{ g/mole}$ . For $3\text{ mole}$ , the mass is,

$\begin{array}{c}m=3\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{48}\text{.00 g}\end{array}$

Therefore, the total mass for $1\text{ mole}$ of ${\text{Fe}}_2{\text{O}}_3$ is,

$\begin{array}{c}\text{total mass}=111.7\text{ g}+48.00\text{ g}\\ =159.7\text{ g}\end{array}$

Now, the percent composition of Fe is,

$\begin{array}{c}\%\text{ Fe}=\frac{111.7\cancel{\text{g}}}{159.7\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{69}\text{.94 \%}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The percent iron in ${\text{Fe}}_3{\text{O}}_4$ is to analyzed.

Explanation of Solution

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound ${\text{Fe}}_3{\text{O}}_4$ as follows:

The mass of Fe from periodic table is $55.85\text{ g/mole}$ . For $3\text{ mole}$ , the mass is,

$\begin{array}{c}m=3\text{ mole}\times \text{55}\text{.85 g/mole}\\ =\text{167}\text{.55 g}\end{array}$

The mass of O from periodic table is $16.00\text{ g/mole}$ . For $4\text{ mole}$ , the mass is,

$\begin{array}{c}m=4\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{64}\text{.00 g}\end{array}$

Therefore, the total mass for $1\text{ mole}$ of ${\text{Fe}}_3{\text{O}}_4$ is,

$\begin{array}{c}\text{total mass}=167.55\text{ g}+64.00\text{ g}\\ =231.55\text{ g}\end{array}$

Now, the percent composition of Fe is,

$\begin{array}{c}\%\text{ Fe}=\frac{167.55\cancel{\text{g}}}{231.55\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{72}\text{.36 \%}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The percent iron in ${\text{FeCO}}_3$ is to analyzed.

Explanation of Solution

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound ${\text{FeCO}}_3$ as follows:

The mass of Fe from periodic table is $55.85\text{ g/mole}$ . For $1\text{ mole}$ , the mass is,

$\begin{array}{c}m=1\text{ mole}\times \text{55}\text{.85 g/mole}\\ =\text{55}\text{.85 g}\end{array}$

The mass of C from periodic table is $12.01\text{ g/mole}$ . For $1\text{ mole}$ , the mass is,

$\begin{array}{c}m=1\text{ mole}\times \text{12}\text{.01 g/mole}\\ =\text{12}\text{.01 g}\end{array}$

The mass of O from periodic table is $16.00\text{ g/mole}$ . For $3\text{ mole}$ , the mass is,

$\begin{array}{c}m=3\text{ mole}\times \text{16}\text{.00 g/mole}\\ =48.\text{00 g}\end{array}$

Therefore, the total mass for $1\text{ mole}$ of ${\text{FeCO}}_3$ is,

$\begin{array}{c}\text{total mass}=55.85\text{ g}+12.01\text{ g}+48.00\text{ g}\\ =115.86\text{ g}\end{array}$

Now, the percent composition of Fe is,

$\begin{array}{c}\%\text{ Fe}=\frac{55.85\cancel{\text{g}}}{115.86\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{48}\text{.20 \%}\end{array}$

Out of all the iron minerals, wustite $\left(\text{FeO}\right)$ has the highest percentage of iron, which is $77.73\text{ \%}$ .