Chapter 13, Problem 20E

To determine

Fill in the blanks in the provided table for the compound carbon dioxide, ${\text{CO}}_2$.

Answer to Problem 20E

The complete table for compound carbon dioxide, ${\text{CO}}_2$ is,

$\begin{array}{ccc}\text{Moles}& \text{Molecules}& \text{Mass}\\ 3.00& \underset{\_}{18.06\times {10}^{23}}& \underset{\_}{132.0\text{ g}}\\ \underset{\_}{1.00}& \underset{\_}{6.02\times {10}^{23}}& 44.0\text{ g}\\ \underset{\_}{0.5}& 3.01\times {10}^{23}& \underset{\_}{22.0\text{ g}}\end{array}$

Explanation of Solution

The number of particles in a mole of a substance is equal to $6.02\times {10}^{23}$. This number is known by the term Avogadro’s number. It is based on the number of carbon atoms that are present in $12\text{ g}$ of carbon-twelve.

The Avogadro’s number is a calculated value and indicates the number of particles that are present in $1\text{ mole}$ of a substance. The number of particles can be atoms, ions or molecules. Hence, a mole of a substance contains $6.02\times {10}^{23}\text{ units}$. The relationship can be expressed as,

$1\text{ mole}=6.02\times {10}^{23}\text{ units}$

$1\text{ mole}$ of carbon dioxide contains $6.02\times {10}^{23}$ molecules. The $3\text{ moles}$ of carbon dioxide contains thrice the number of atoms that are $18.06\times {10}^{23}$ molecules.

The mass of $1\text{ mole}$ of a substance is equal to the formula mass or molar mass of a substance. The molar mass of $1\text{ mole}$ of carbon dioxide is $44.0\text{ g}$. Hence, $3\text{ moles}$ of carbon dioxide has a mass equal to thrice of $44.0\text{ g}$ that is $132.0\text{ g}$.

Thus, the value $18.06\times {10}^{23}$ molecules and $132.0\text{ g}$ is filled in the blanks beside $3\text{ moles}$ in the table.

The molar mass of a substance is the mass of $1\text{ mole}$ of that substance. The molar mass of carbon dioxide is $44.0\text{ g}$. Hence, $44.0\text{ g}$ of carbon dioxide is the mass of $1\text{ mole}$ of carbon dioxide.

$1\text{ mole}$ of carbon dioxide contains $6.02\times {10}^{23}$ molecules.

Thus, the value $1\text{ mole}$ and $6.02\times {10}^{23}$ molecules are filled in the blank beside $44.0\text{ g}$ in the table.

The number of moles in $3.01\times {10}^{23}$ molecules of carbon dioxide is calculated as,

$\begin{array}{c}6.02\times {10}^{23}\text{ molecules}=\text{1 mole}\\ \text{1 molecule}=\frac{1}{6.02\times {10}^{23}}\text{mole}\\ 3.01\times {10}^{23}\text{ molecules}=\left(\frac{1}{6.02\times {10}^{23}}\right)\left(3.01\times {10}^{23}\right)\text{moles}\\ =\text{0}\text{.5 mole}\end{array}$

$1\text{ mole}$ of carbon dioxide has a mass of $44.0\text{ g}$. Hence, $0.5\text{ mole}$ of carbon dioxide has a mass which is half of $44.0\text{ g}$ that is $22.0\text{ g}$.

Thus, the value $0.5\text{ mole}$ and $22.0\text{ g}$ is filled in the blanks beside $3.01\times {10}^{23}$ molecules in the table.

The complete table is for the compound carbon dioxide, ${\text{CO}}_2$,

$\begin{array}{ccc}\text{Moles}& \text{Molecules}& \text{Mass}\\ 3.00& \underset{\_}{18.06\times {10}^{23}}& \underset{\_}{132.0\text{ g}}\\ \underset{\_}{1.00}& \underset{\_}{6.02\times {10}^{23}}& 44.0\text{ g}\\ \underset{\_}{0.5}& 3.01\times {10}^{23}& \underset{\_}{22.0\text{ g}}\end{array}$

Conclusion:

Therefore, the complete table for compound carbon dioxide, ${\text{CO}}_2$,

$\begin{array}{ccc}\text{Moles}& \text{Molecules}& \text{Mass}\\ 3.00& \underset{\_}{18.06\times {10}^{23}}& \underset{\_}{132.0\text{ g}}\\ \underset{\_}{1.00}& \underset{\_}{6.02\times {10}^{23}}& 44.0\text{ g}\\ \underset{\_}{0.5}& 3.01\times {10}^{23}& \underset{\_}{22.0\text{ g}}\end{array}$