Chapter 13, Problem 19E

To determine

Fill in the blanks in the provided table for the element sodium, $\text{Na}$.

Answer to Problem 19E

The complete table for the element sodium, $\text{Na}$ is,

$\begin{array}{ccc}\text{Moles}& \text{Atoms}& \text{Mass}\\ 1.00& \underset{\_}{6.02\times {10}^{23}}& \underset{\_}{23.0\text{ g}}\\ \underset{\_}{3.00}& 18.06\times {10}^{23}& \underset{\_}{69.0\text{ g}}\\ \underset{\_}{2.00}& \underset{\_}{12.04\times {10}^{23}}& 46.0\text{ g}\end{array}$

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,

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

The number of atoms in $1\text{ mole}$ of sodium is thus equal to $6.02\times {10}^{23}$ atoms.

The mass of $1\text{ mole}$ of a substance is equal to the formula mass or molar mass of a substance. The formula mass of sodium is $23.0\text{ g}$. Hence, $1\text{ mole}$ of sodium has a mass equal to $23.0\text{ g}$.

Thus, the values $6.02\times {10}^{23}$ atoms and $23.0\text{ g}$ is filled in the blanks beside $1\text{ mole}$ in the table.

The number of moles in $18.06\times {10}^{23}$ atoms of sodium is calculated as,

$\begin{array}{c}6.02\times {10}^{23}\text{ atoms}=\text{1 mole}\\ \text{1 atom}=\frac{1}{6.02\times {10}^{23}}\text{mole}\\ 18.06\times {10}^{23}\text{ atoms}=\left(\frac{1}{6.02\times {10}^{23}}\right)\left(18.06\times {10}^{23}\right)\text{moles}\\ =\text{3 moles}\end{array}$

$1\text{ mole}$ of a sodium has a mass equal to $23.0\text{ g}$. Hence, $3\text{ moles}$ of sodium has a mass equal to thrice of $23.0\text{ g}$ that is $69.0\text{ g}$.

Thus, the values $\text{3 moles}$ and $69.0\text{ g}$ is filled in the blanks beside $18.06\times {10}^{23}$ atoms in the table.

The value $23.0\text{ g}$ is the mass of $1\text{ mole}$ of sodium. Hence, the number of moles in $46.0\text{ g}$ of sodium is calculated as,

$\begin{array}{c}23.0\text{ g}=\text{1 mole}\\ \text{1}\text{.0 g}=\frac{1}{23.0}\text{mole}\\ \text{46}\text{.0 g}=\left(\frac{1}{23.0}\right)\left(\text{46}\text{.0}\right)\text{moles}\\ =\text{2 moles}\end{array}$

$1\text{ mole}$ of a sodium contains $6.022\times {10}^{23}$ atoms. Hence, $2\text{ moles}$ of sodium contains twice the number of atoms that are $12.04\times {10}^{23}$ atoms.

Thus, the values $\text{2 moles}$ and $12.04\times {10}^{23}$ atoms are filled in the blanks beside $\text{46}\text{.0 g}$ in the table.

The complete table is for the element sodium is,

$\begin{array}{ccc}\text{Moles}& \text{Atoms}& \text{Mass}\\ 1.00& \underset{\_}{6.02\times {10}^{23}}& \underset{\_}{23.0\text{ g}}\\ \underset{\_}{3.00}& 18.06\times {10}^{23}& \underset{\_}{69.0\text{ g}}\\ \underset{\_}{2.00}& \underset{\_}{12.04\times {10}^{23}}& 46.0\text{ g}\end{array}$

Conclusion:

Therefore, the complete table for the element sodium, $\text{Na}$ is,

$\begin{array}{ccc}\text{Moles}& \text{Atoms}& \text{Mass}\\ 1.00& \underset{\_}{6.02\times {10}^{23}}& \underset{\_}{23.0\text{ g}}\\ \underset{\_}{3.00}& 18.06\times {10}^{23}& \underset{\_}{69.0\text{ g}}\\ \underset{\_}{2.00}& \underset{\_}{12.04\times {10}^{23}}& 46.0\text{ g}\end{array}$