Chapter 3, Problem 158AP

Interpretation Introduction

Interpretation:

The fraction of octane converted to ${\text{CO}}_2$ in the combustion of gasoline is to be calculated, with given volume and density of octane and total mass of $\text{CO,}{\text{CO}}_2{\text{, and H}}_2\text{O}$.

Concept introduction:

Atomic mass of elements is measured in terms of atomic mass unit. One atomic mass unit is defined as the number of elementary particles present in $12.0\text{g}$ of carbon atoms.

The number of moles are given by the following expression:

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}$

The percent yield of the compound is given by the following expression:

$\text{Percent}\text{yield}=\frac{\text{observed yield}}{\text{actual yield}}\times 100$

The mass of compound can be calculated as $m=n\times M$.

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

The total mass of octane is calculated as

$\text{mass = density }\times \text{ volume}$

Answer to Problem 158AP

Solution: 86.42%

Explanation of Solution

Given information: Volume of octane $=1.00\text{ gal}$

The total mass of $\text{CO,}{\text{CO}}_2{\text{ and H}}_2\text{O}$ formed is $11.53\text{ kg}$.

Density of octane $=2.65\text{ kg}/\text{gal}$

The balanced equations for combustion of octane:

$\begin{array}{l}2{\text{C}}_{\text{8}}{\text{H}}_{18}+25{\text{O}}_2 \to 16{\text{CO}}_2+18{\text{H}}_{\text{2}}\text{O}\\ {\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}+{\text{17O}}_{\text{2}}\to \text{16CO}+{\text{18H}}_{\text{2}}\text{O}\end{array}$

Since the molar ratio of octane to that of ${\text{CO}}_2$ is $2:16$, the number of moles of ${\text{CO}}_2$ is given by the following expression:

$n_{{\text{CO}}_2}=\frac{16}{2}\times n_{\text{octane}}$

Consider the mass of octane used be $x\text{ g}$.

The number of moles is given by the following expression:

$n_{\text{octane}}=\frac{m_{\text{octane}}}{M_{\text{octane}}}$ …… (1)

Substitute $x\text{ g}$ for $m_{\text{octane}}$ and $114.2\text{ g}$ for $M_{\text{octane}}$ in the equation (1)

$n_{\text{octane}}=\frac{x\text{ g}}{114.2\text{ g}}$

Therefore, the number of moles of ${\text{CO}}_2$ is calculated as

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{16}{2}\times \frac{x\text{ g}}{114.2\text{ g}}\\ =0.07005x\text{ mol}\end{array}$

The mass of ${\text{CO}}_2$ is calculated by the following expression:

$n_{{\text{CO}}_2}\times M_{{\text{CO}}_2}=m_{{\text{CO}}_2}$

Substitute $0.07005x\text{ mol}$ for $n_{{\text{CO}}_2}$ and $44.01\text{ g}/\text{mol}$ for $M_{{\text{CO}}_2}$ in the above equation

$\begin{array}{c}0.07005x\text{ mol}\times 44.01\text{ g}/\text{mol}=m_{{\text{CO}}_2}\\ 3.083x\text{ g}=m_{{\text{CO}}_2}\end{array}$

Similarly, the molar ratio of octane to that of ${\text{H}}_2\text{O}$ is $2:18$, the number of moles of ${\text{H}}_2\text{O}$ is given by the formula:

$n_{{\text{H}}_2\text{O}}=\frac{16}{2}\times n_{\text{octane}}$

Substitute $x\text{ g}$ for $m_{\text{octane}}$ and $114.2\text{ g}$ for $M_{\text{octane}}$ in the equation (1)

$n_{\text{octane}}=\frac{x\text{ g}}{114.2\text{ g}}$

Therefore, the number of moles of ${\text{H}}_2\text{O}$ is calculated as

$\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{18}{2}\times \frac{x\text{ g}}{114.2\text{ g}}\\ =0.0788x\text{ mol}\end{array}$

The mass of ${\text{H}}_2\text{O}$ is given by the following expression:

$n_{{\text{H}}_2\text{O}}\times M_{{\text{H}}_2\text{O}}=m_{{\text{H}}_2\text{O}}$

Substitute $0.0788x\text{ mol}$ for $n_{{\text{H}}_2\text{Ol}}$ and $18.016\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the above equation

$\begin{array}{c}0.0788x\text{ mol}\times 18.016\text{ g}/\text{mol}=m_{{\text{CO}}_2}\\ 1.420x\text{ g}=m_{{\text{CO}}_2}\end{array}$

Since the density of octane is $2.65\text{ kg}/\text{gal}$ and the volume of octane is $1.00\text{ gal}$, total mass of octane is calculated as

$\text{mass = density }\times \text{ volume}$

$\begin{array}{c}\text{Mass}\text{of}\text{octane}=2.65\text{ kg}/\text{gal}\times 1.00\text{ gal}\\ =\text{2}\text{.65 kg}\\ =\text{2650 g}\end{array}$

Therefore, the mass of octane used to produce $\text{CO}$ and ${\text{H}}_2\text{O}$ is $\left(2650-x\right)$.

Similarly, the mass of $\text{CO}$ and ${\text{H}}_2\text{O}$ produced for mass of octane $\left(2650-x\right)$ is given by the following expression:

$m_{\text{CO}}=\frac{16}{2}\times \frac{\left(2650-x\right)\text{ g}}{114.2\text{ g}}\times M_{\text{CO}}$

$m_{{\text{H}}_2\text{O}}=\frac{18}{2}\times \frac{\left(2650-x\right)\text{ g}}{114.2\text{ g}}\times M_{{\text{H}}_2\text{O}}$

Substitute $28.01\text{ g}/\text{mol}$ for $M_{\text{CO}}$ and $18.01\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the above equation

For carbon monoxide

$\begin{array}{c}{m}_{\text{CO}}=\frac{16}{2}\times \frac{\left(2650-x\right)\text{ g}}{114.2\text{ g}}\times 28.01\text{ g}/\text{mol}\\ =5200\text{ g}-1.962x\text{ g}\end{array}$

For water

$\begin{array}{c}{m}_{{\text{H}}_2\text{O}}=\frac{18}{2}\times \frac{\left(2650-x\right)\text{ g}}{114.2\text{ g}}\times 18.01\text{ g}/\text{mol}\\ =3763\text{ g}-1.420x\text{ g}\end{array}$

Therefore, the mass of octane is calculated as

$\begin{array}{c}3.083x\text{ g}+\cancel{1.420x\text{ g}}+5200\text{ g}-1.962x\text{ g}+3763\text{ g}-\cancel{1.420x\text{ g}}=11530\text{ g}\\ \text{8963 g}+3.083x\text{ g}-1.962x\text{ g}=11530\text{ g}\\ \text{8963 g}+1.121x\text{ g =}11530\text{ g}\\ 1.121x\text{ g}\text{}\text{ =}11530\text{ g}-\text{8963 g}\end{array}$

Solvingfurther, we get

$\begin{array}{c}1.121x\text{ g}\text{}\text{ =}2567\text{ g}\\ x=\frac{2567\text{ g}}{1.121\text{ g}}\\ x=2290\text{ g}\end{array}$

The percent efficiency of octane is calculated by the following expression:

Percent efficiency is given by = $\frac{\text{observed yield}}{\text{actual yield}}\times 100$

Substitute $2290\text{ g}$ for $\text{observed yield}$ and $\text{2650 g}$ for $\text{actual yield}$ in the above equation

$\begin{array}{c}\%\text{ yield}=\frac{2290\text{ g}}{\text{2650 g}}\times 100\\ =0.8642\times 100\\ =86.42\%\end{array}$

Conclusion

The fraction of octane converted to ${\text{CO}}_2$ in the combustion of gasoline is $86.42\%$.