Chapter 8, Problem 37E

Interpretation Introduction

Interpretation:

The table is to be completed.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction.

Answer to Problem 37E

The table is completed as shown below.

Gas	Molecules	Mass	STP Volume
methane, ${\text{CH}}_{\text{4}}$	$1.50\times {10}^{23}$	$4\text{g}$	$5.58\text{L}$
ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$	$1.50\times {10}^{23}$	$\text{7}\text{.50}\text{g}$	$5.58\text{L}$
propane, ${\text{C}}_3{\text{H}}_{\text{8}}$	$1.50\times {10}^{23}$	$\text{11}\text{g}$	$5.58\text{L}$

Explanation of Solution

The table to be completed is given below.

Gas	Molecules	Mass	STP Volume
methane, ${\text{CH}}_{\text{4}}$	$1.50\times {10}^{23}$
ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$		$\text{7}\text{.50}\text{g}$
propane, ${\text{C}}_3{\text{H}}_{\text{8}}$			$5.58\text{L}$

The number of molecules of methane, ${\text{CH}}_{\text{4}}$ is $1.50\times {10}^{23}$.

The formula to determine the number of moles is given below.

$\text{Number}\text{of}\text{moles}=\frac{\text{Number}\text{of}\text{molecules}}{\text{Avogadro's}\text{number}}$ …(1)

The Avogadro’s number is $6.022\times {10}^{23}\text{molecules}$.

Substitute the number of molecules of ${\text{CH}}_{\text{4}}$ and Avogadro’s number in equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{1.50\times {10}^{23}\text{molecules}}{6.022\times {10}^{23}\text{molecules}}\\ & =& 0.249\text{mol}\end{array}$

At STP, $1$ mole of ${\text{CH}}_{\text{4}}$ occupies $22.4\text{L}$.

The unit factors are given below.

$\frac{1\text{mol}{\text{CH}}_{\text{4}}}{22.4\text{L}{\text{CH}}_{\text{4}}}\text{and}\frac{22.4\text{L}{\text{CH}}_{\text{4}}}{1\text{mol}{\text{CH}}_{\text{4}}}$

Therefore, the volume occupied by $0.249\text{mol}$ is calculated as shown below.

$\begin{array}{rll}\text{Volume}\text{occpied}\text{by}0.249\text{mol}& =& \frac{0.249\text{mol}}{1\text{mol}}\times 22.4\text{L}\\ & =& 5.58\text{L}\end{array}$

Therefore, the volume occupied at STP is $5.58\text{L}$.

The molar mass of ${\text{CH}}_{\text{4}}$ is $16.05\text{g}{\text{mol}}^{-1}$.

The unit factor to calculate grams of ${\text{CH}}_{\text{4}}$ from moles of ${\text{CH}}_{\text{4}}$ is shown below.

$\frac{16.05\text{g}{\text{CH}}_{\text{4}}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}$

The formula to determine the mass of ${\text{CH}}_{\text{4}}$ from the volume of ${\text{CH}}_{\text{4}}$ is shown below.

$\text{Mass}\text{of}{\text{CH}}_{\text{4}}=\left(\begin{array}{l}\text{Volume}\text{of}{\text{CH}}_{\text{4}}\times \text{Unit factor}\text{to}\text{obtain}\text{moles}\text{of}{\text{CH}}_{\text{4}}\\ \times \text{Unit factor}\text{to}\text{obtain}\text{mass}\text{of}{\text{CH}}_{\text{4}}\end{array}\right)$ …(2)

Substitute the volume of ${\text{CH}}_{\text{4}}$ and the unit factors in equation (2).

$\begin{array}{rll}\text{Mass}\text{of}{\text{CH}}_{\text{4}}& =& 5.58\text{L}{\text{CH}}_{\text{4}}\times \frac{1\text{mol}{\text{CH}}_{\text{4}}}{22.4\text{L}{\text{CH}}_{\text{4}}}\times \frac{16.05\text{g}{\text{CH}}_{\text{4}}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\\ & =& 5.58\text{L}{\text{CH}}_{\text{4}}\times \frac{16.05\text{g}{\text{CH}}_{\text{4}}}{22.4\text{L}{\text{CH}}_{\text{4}}}\\ & =& 3.99\text{g}\\ & \simeq & 4\text{g}\end{array}$

Therefore, the mass of ${\text{CH}}_{\text{4}}$ is $4\text{g}$.

The mass of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ is $\text{7}\text{.50}\text{g}$.

The molar mass of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ is $30.08\text{g}{\text{mol}}^{-1}$.

The formula to determine the number of moles is shown below.

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ …(3)

Substitute the mass and molar mass of ${\text{C}}_2{\text{H}}_{\text{6}}$ in equation (3).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{\text{7}\text{.50}\text{g}}{30.08\text{g}{\text{mol}}^{-1}}\\ & =& 0.249\text{mol}\end{array}$

At STP, $1$ mole of ${\text{C}}_2{\text{H}}_{\text{6}}$ occupies $22.4\text{L}$.

The unit factors are given below.

$\frac{1\text{mol}{\text{C}}_2{\text{H}}_{\text{6}}}{22.4\text{L}{\text{C}}_2{\text{H}}_{\text{6}}}\text{and}\frac{22.4\text{L}{\text{C}}_2{\text{H}}_{\text{6}}}{1\text{mol}{\text{C}}_2{\text{H}}_{\text{6}}}$

Therefore, the volume occupied by $0.249\text{mol}$ is calculated as shown below.

$\begin{array}{rll}\text{Volume}\text{occpied}\text{by}0.249\text{mol}& =& \frac{0.249\text{mol}}{1\text{mol}}\times 22.4\text{L}\\ & =& 5.58\text{L}\end{array}$

Therefore, the volume occupied by ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is $5.58\text{L}$.

The number of molecules in $22.4\text{L}$ of ${\text{C}}_2{\text{H}}_{\text{6}}$ is $6.022\times {10}^{23}\text{molecules}$.

The number of molecules in $5.58\text{L}$ of ${\text{C}}_2{\text{H}}_{\text{6}}$ is calculated as shown below.

$\begin{array}{rll}\text{Number}\text{of}\text{molecules}\text{in of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}& =& \frac{5.58\text{L}}{22.4\text{L}}\times 6.022\times {10}^{23}\text{molecules}\\ & =& 1.50\times {10}^{23}\text{molecules}\end{array}$

Therefore, the number of molecules of ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is $1.50\times {10}^{23}\text{molecules}$.

The volume of propane, ${\text{C}}_3{\text{H}}_{\text{8}}$ is at STP is $5.58\text{L}$.

At STP, $1$ mole of ${\text{C}}_3{\text{H}}_{\text{8}}$ occupies $22.4\text{L}$.

The number of moles of ${\text{C}}_3{\text{H}}_{\text{8}}$ that occupy $5.58\text{L}$ is calculated as shown below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}& =& \frac{5.58\text{L}}{22.4\text{L}}\times 1\text{mol}\\ & =& \text{0}\text{.249}\text{mol}\end{array}$

At STP, $\text{1}$ mole contains $6.022\times {10}^{23}\text{molecules}$.

Therefore, the number of molecules in $\text{0}\text{.249}\text{mol}$ of ${\text{C}}_3{\text{H}}_{\text{8}}$ is calculated as shown below.

$\begin{array}{rll}\text{Number}\text{of}\text{molecules}& =& \frac{\text{0}\text{.249}\text{mol}}{\text{1}\text{mol}}\times 6.022\times {10}^{23}\text{molecules}\\ & =& 1.50\times {10}^{23}\text{molecules}\end{array}$

Therefore, the number of molecules in ${\text{C}}_3{\text{H}}_{\text{8}}$ at STP is $1.50\times {10}^{23}\text{molecules}$.

The molar mass of ${\text{C}}_3{\text{H}}_{\text{8}}$ is $44.11\text{g}{\text{mol}}^{-1}$.

Substitute the moles and molar mass of ${\text{C}}_3{\text{H}}_{\text{8}}$ in equation (3).

$\begin{array}{rll}\text{0}\text{.249}\text{mol}& =& \frac{\text{Mass}}{44.11\text{g}{\text{mol}}^{-1}}\\ \text{Mass}& =& \text{0}\text{.249}\text{mol}\times 44.11\text{g}{\text{mol}}^{-1}\\ & =& 10.98\text{g}\\ & \simeq & \text{11}\text{g}\end{array}$

Therefore, the mass of ${\text{C}}_3{\text{H}}_{\text{8}}$ is $\text{11}\text{g}$.

The table is therefore completed as shown below.

Gas	Molecules	Mass	STP Volume
methane, ${\text{CH}}_{\text{4}}$	$1.50\times {10}^{23}$	$4\text{g}$	$5.58\text{L}$
ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$	$1.50\times {10}^{23}$	$\text{7}\text{.50}\text{g}$	$5.58\text{L}$
propane, ${\text{C}}_3{\text{H}}_{\text{8}}$	$1.50\times {10}^{23}$	$\text{11}\text{g}$	$5.58\text{L}$

Conclusion

The table is completed has been rightfully completed.