Chapter 5, Problem 5.118P

5-118 Isooctane, which has a chemical formula C₈H₁₈ is the component of gasoline from which the term octane rating derives.

(a) Write the balanced chemical equation for the combustion of isooctane.

(b) The density of isooctane is 0.792 g/mL. How many kg of C02 are produced each year by the annual U.S. gasoline consumption of L?

(c) What is the volume in liters of this CO₂ at STP?

(d) The chemical formula for isooctane can be represented by (CH₃)₃CCH₂CH(CH₃)₂. Draw a Lewis structure of isooctane.

(e) Another molecule with the same molecular formula is octane, which can be represented by:

When comparing isooctane and octane, one structure is observed to have a boiling point of 99°C, while another is known to have a boiling point Of 125°C. Which substance, isooctane or octane, is expected to have the higher boiling point?

(f) Determine whether isooctane or octane is expected to have the greater vapor pressure.

Interpretation Introduction

Interpretation:

The balanced chemical equation for the combustion of isooctane should be determined.

Concept Introduction:

In a balanced chemical equation, all the constituents present in the reaction have equal number of atoms on both side of the reaction arrow.

Answer to Problem 5.118P

$2{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+25{\text{O}}_2\left(g\right)\to 16{\text{CO}}_2\left(g\right)+18{\text{H}}_{\text{2}}\text{O }\left(g\right)$.

Explanation of Solution

Isooctane on combustion in the presence of oxygen produces carbon dioxide gas along with water vapor. The balanced chemical equation of the reaction is taking place as depicted below:

$2{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+25{\text{O}}_{\text{2}}\left(g\right)\to 16{\text{CO}}_{\text{2}}\left(g\right)+18{\text{H}}_{\text{2}}\text{O }\left(g\right)$.

Interpretation Introduction

(b)

Interpretation:

The mass in kg of CO₂ are produced each year by the annual US gasoline consumption should be determined.

Concept Introduction:

Mass of carbon dioxide produces can be obtained from the balanced reaction i.e. $228.46\text{g}$ of gasoline produces $704.16\text{g}$ of carbon dioxide.

Density of a substance is defined as mass per unit volume. It is mathematically represented as follows:

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

Here, m is mass and v is volume of the substance.

Answer to Problem 5.118P

Mass of carbon dioxide produces yearly is $\text{11}\text{.2}{\text{×10}}^{\text{11}}\text{ kg}\text{.}$.

Explanation of Solution

Volume of gasoline consumed in 1 year is 4.6 × 10¹⁰ L.

Density of the fuel = 0.792 g/mL.

To calculate mass of gasoline, multiply volume of gasoline consumed with its density.

$\begin{array}{c}\text{Mass = Volume × Density}\\ \text{=4}{\text{.6×10}}^{\text{10}}\text{ L ×}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\text{ ×0}\text{.792 }\frac{\text{g}}{\text{mL}}\\ \text{= 3}{\text{.64 × 10}}^{13}\text{g}\end{array}$

Therefore, the mass of gasoline = 3.64 × 10¹³ g.

The balanced combustion reaction of gasoline is as follows.

$2{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+25{\text{O}}_{\text{2}}\left(g\right)\to 16{\text{CO}}_{\text{2}}\left(g\right)+18{\text{H}}_{\text{2}}\text{O }\left(g\right)$

From the balanced reaction, $228.46\text{g}$ of gasoline produces $704.16\text{g}$ of carbon dioxide.

To calculate the mass of carbon dioxide produces from the $3.64\times {10}^{13}\text{g}$ of gasoline is as follows:

$\text{3}{\text{.64×10}}^{\text{13}}\text{g Gasoline × }\frac{\text{704}{\text{.16 g CO}}_{\text{2}}}{\text{228}\text{.46 g Gasoline}}\text{ = 11}{\text{.2×10}}^{\text{13}}{\text{g CO}}_{\text{2}}$

To convert the mass of carbon dioxide from g to kg as follows:

$11.2\times {10}^{13}\text{g}\text{C}{\text{O}}_2\times \frac{\text{1}\text{kg}}{{\text{10}}^{\text{2}}\text{g}}=11.2\times {10}^{11}\text{kg}$

Therefore, mass of carbon dioxide produces yearly is $\text{11}\text{.2}{\text{×10}}^{\text{11}}\text{kg}\text{.}$.

Interpretation Introduction

(c)

Interpretation:

The volume of CO₂ in liter at STP should be determined.

Concept Introduction:

Volume CO₂ in liter at STP can be determined by calculating moles of CO₂ and then using the concept that at STP for each mode of gas occupies 22.4 L.

Answer to Problem 5.118P

The volume of carbon dioxide occupied is $\text{5}{\text{.70 x 10}}^{\text{13}}\text{L }{\text{CO}}_{\text{2}}$.

Explanation of Solution

To calculate the number of the moles of carbon dioxide, divide mass of carbon dioxide with its molar mass.

$\text{11}{\text{.2×10}}^{\text{13}}{\text{g CO}}_{\text{2}}\text{× }\frac{\text{1}{\text{mol CO}}_{\text{2}}}{\text{44}{\text{.01 g CO}}_{\text{2}}}\text{ = 0}{\text{.254×10}}^{\text{13}}{\text{mol CO}}_{\text{2}}$

At STP for each mode of gas occupies 22.4 L.

Calculate the volume of carbon dioxide occupied by $0.254\times {10}^{13}$ mol CO₂ is as follows:

$\text{0}{\text{.254×10}}^{\text{13}}{\text{mol CO}}_{\text{2}}\text{× }\frac{\text{22}{\text{.4L CO}}_{\text{2}}}{{\text{1 mol CO}}_{\text{2}}}\text{ = 5}{\text{.70×10}}^{\text{13}}{\text{ l CO}}_{\text{2}}$

Therefore, the volume of carbon dioxide occupied is $\text{5}{\text{.70 x 10}}^{\text{13}}\text{L }{\text{CO}}_{\text{2}}$.

Interpretation Introduction

(d)

Interpretation:

The Lewis structure of isooctane should be drawn.

Concept Introduction:

Lewis structure are diagrams which represents the bonding between atoms of a molecule and the lone pairs of electrons that might be present in the molecule.

Answer to Problem 5.118P

Explanation of Solution

The lewis structure of isooctane is shown in the following diagram. Since, valence electrons in carbon atom are 4 and all the carbon atoms are attached with 4 other atoms (carbon and hydrogen), they have complete octets and all the electrons are engaged in bonding. Thus, there is no lone pair present on any atom.

Interpretation Introduction

(e)

Interpretation:

If isooctane or octane is expected to have the higher boiling point should be determined.

Concept Introduction:

One of the parameter for specific boiling point of any liquid can be explained by van der waals bonds.

Liquid having more van der waals bonds will show higher boiling point.

Answer to Problem 5.118P

Isooctance has fewer cohensive interactions than octane so Isooctance has lower boiling point.

Explanation of Solution

Van der Waals attractions increase with the surface areas of the interacting electron clouds. That is, the large the interacting surfaces, the greater the magnitude of the induced dipole. Because isooctance has less surface area at which van der waals interactions with other isooctane molecules can occur, it has fewer cohensive interactions than octane, and thus, a lower boiling point.

Interpretation Introduction

(f)

Interpretation:

If isooctane or octane is expected to have the greater vapor pressure should be determined.

Concept Introduction:

Vapor pressure depends on boiling point. lower boiling point shows higher vapor pressure.

Answer to Problem 5.118P

Isooctane is expected to have higher vapour pressure at certain temperature.

Explanation of Solution

As the boiling point of isooctane is lower than Octane, isooctane is expected to have higher vapour pressure at certain temperature.