Chapter 2, Problem 2.43AP

Interpretation Introduction

Interpretation:

The carbon footprint (number of pounds of ${\text{CO}}_{\text{2}}$ released into the atmosphere) of the Oxhide family car over one year is to be calculated.

Concept introduction:

The combustion is the process of heating the compounds in the presence of oxygen. The combustion of alkanes yields carbon dioxide and water. In the process of combustion of alkanes a high amount of heat is evolved. These products are given by all kinds of alkanes generally.

Answer to Problem 2.43AP

The carbon footprint (number of pounds of ${\text{CO}}_{\text{2}}$ released into the atmosphere) of the Oxhide family car over one year is $8563.78\text{lb}/\text{year}$.

Explanation of Solution

The general balanced reaction for the complete combustion of an alkane is given below as,

${\text{C}}_n{\text{H}}_{2n+2}+\frac{\left(3n+1\right)}{2}{\text{O}}_{\text{2}}\to n{\text{CO}}_{\text{2}}+\left(n+1\right){\text{H}}_{\text{2}}\text{O}$

The chemical formula for gasoline is ${\text{C}}_8{\text{H}}_{18}$. Thus, the reaction for the complete combustion of gasoline is written as,

${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_{\text{2}}\to 8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}$ … (1)

Thus, one mole of gasoline gives $8$ moles of ${\text{CO}}_{\text{2}}$. The average of car is about $12,000\text{miles}$ per year with an average mileage of about $\text{25}\text{miles}$ per gallon of gasoline. Therefore, the volume of gasoline over one year is calculated as,

$\frac{\text{12000}\text{miles}\text{per}\text{year}}{\text{25}\text{miles}\text{per}\text{gallon}}=480\text{gallons}\text{per}\text{year}$

This is converted into $\text{mL}$ using the conversion as,

$\begin{array}{rll}\text{1}\text{gallon}& =& \text{3}\text{.785}\text{L}\\ \text{1}\text{L}& =& \text{1000}\text{mL}\end{array}$

Therefore,

$\begin{array}{rll}1\text{gallon}& =& 1\times \text{3}\text{.785}\text{L}\times \frac{\text{1000}\text{mL}}{1\text{L}}\\ 480\text{gallons}/\text{year}& =& 480\times \text{3}\text{.785}\text{L}\times \frac{\text{1000}\text{mL}}{1\text{L}}\\ & =& 1816800\text{mL}/\text{year}\end{array}$

The mass of $1816800\text{mL}/\text{year}$ gasoline can be calculated using the formula as,

$d=\frac{m}{V}$

Rearrange the above equation for mass as,

$\begin{array}{rll}d& =& \frac{m}{V}\\ m& =& d\times V\end{array}$

The density of density of gasoline is $0.692\text{g}{\text{mL}}^{-1}$.

Substitute the values in the above equation as,

$\begin{array}{rll}m& =& d\times V\\ & =& 0.692\text{g}{\text{mL}}^{-1}\times 1816800\text{mL}/\text{year}\\ & =& 1257225.6\text{g}/\text{year}\end{array}$

The number of moles can be calculated using the formula as,

$n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$ … (2)

The molar mass of gasoline is $114\text{g}{\text{mol}}^{-1}$.

Substitute the values in the above equation as,

$\begin{array}{rll}n& =& \frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ & =& \frac{1257225.6\text{g}/\text{year}}{114\text{g}{\text{mol}}^{-1}}\\ & =& 11028.29\text{mol}/\text{year}\end{array}$

From equation (1), it is observed that one mole of gasoline gives $8 \text{moles}$ of ${\text{CO}}_{\text{2}}$. Therefore, $11028.29\text{mol}/\text{year}$ of gasoline will give $8\times 11028.29\text{mol}/\text{year}$ of ${\text{CO}}_{\text{2}}$ which is equal to $88226.35\text{mol}/\text{year}$ of ${\text{CO}}_{\text{2}}$. Thus, the mass of ${\text{CO}}_{\text{2}}$ obtained using (2) is as follows.

$\begin{array}{rll}n& =& \frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ \text{Given}\text{mass}& =& n\times \text{Molar}\text{mass}\end{array}$

The molar mass of ${\text{CO}}_{\text{2}}$ is $\text{44}\text{g}{\text{mol}}^{-\text{1}}$.

Substitute the values in the above equation as,

$\begin{array}{rll}\text{Given}\text{mass}& =& n\times \text{Molar}\text{mass}\\ & =& 88226.35\text{mol}/\text{year}\times \text{44}\text{g}{\text{mol}}^{-\text{1}}\\ & =& 3881959.74\text{g}/\text{year}\end{array}$

The number of pounds of ${\text{CO}}_{\text{2}}$ can be calculated using the conversion as,

$1\text{lb}=453.3\text{g}$

Therefore,

$\begin{array}{rll}3881959.74\text{g}/\text{year}& =& 3881959.74\text{g}/\text{year}\times \frac{1\text{lb}}{453.3\text{g}}\\ & =& 8563.78\text{lb}/\text{year}\end{array}$

Conclusion

The carbon footprint (number of pounds of ${\text{CO}}_{\text{2}}$ released into the atmosphere) of the Oxhide family car over one year is $8563.78\text{lb}/\text{year}$.