Question

Octane $\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)$ is a component of gasoline. Complete combustion of octane yields ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{CO}}_2$. Incomplete combustion produces ${\mathrm{H}}_2\mathrm{O}$ and $\mathrm{CO},$ which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon (gal) of octane is burned in an engine. The total mass of $\mathrm{CO},{\mathrm{CO}}_2$, and ${\mathrm{H}}_2\mathrm{O}$ produced is $11.53\mathrm{kg}.$ Calculate the efficiency of the process; that is, calculate the fraction of octane converted to ${\mathrm{CO}}_2$. The density of octane is $2.650\mathrm{kg}/\mathrm{gal}$.

Short answer

The efficiency of the process is approximately 100% if all mass went to ${\text{CO}}_2$, requiring correction assumptions for exact division of products.

Step-by-step solution

Calculate the Mass of Octane

Determine the mass of octane from the given volume and density. The given volume of octane is 1.000 gal, and its density is 2.650 kg/gal. Use the formula:$$\text{Mass of octane}=\text{Volume}\times \text{Density}=1.000\text{ gal}\times 2.650\frac{\text{kg}}{\text{gal}}=2.650\text{ kg}$$

Write Chemical Equations

For complete combustion:$${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$$For incomplete combustion:$${\text{C}}_8{\text{H}}_{18}+x{\text{O}}_2\to y\text{CO}+z{\text{CO}}_2+9{\text{H}}_2\text{O}$$

Calculate Moles of Octane

First, determine the molar mass of octane, ${\text{C}}_8{\text{H}}_{18}$: $$8(12.01)+18(1.01)=114.23\text{ g/mol}$$Convert the mass of octane to moles:$$\text{Moles of octane}=\frac{2650\text{ g}}{114.23\text{ g/mol}}=23.19\text{ mol}$$

Determine Mass of Products If Complete Combustion Occurs

For complete combustion of 23.19 mol of octane, calculate the mass of ${\text{CO}}_2$. Each mole of octane produces 8 moles of ${\text{CO}}_2$.$$\text{Moles of }{\text{CO}}_2=23.19\times 8=185.52\text{ mol}$$Molar mass of ${\text{CO}}_2$ is 44.01 g/mol, so the mass of ${\text{CO}}_2$ is:$$185.52\times 44.01=8166.06\text{ g}=8.166\text{ kg}$$

Calculate Efficiency

Efficiency is the ratio of the actual mass of ${\text{CO}}_2$ produced to the theoretical mass if complete combustion occurred.Here, the actual mass of all products ($\text{CO},{\text{CO}}_2,{\text{H}}_2\text{O}$) is 11.53 kg. Assuming other products' mass is negligible in efficiency calculation:$$\text{Efficiency}=\frac{\text{Actual }{\text{CO}}_2\text{ produced}}{\text{Theoretical }{\text{CO}}_2\text{ produced}}\approx \frac{8.166\text{ kg}}{8.166\text{ kg}}=1$$

Correct the Efficiency Calculation

In reality, if we account for $\text{CO}$ and ${\text{H}}_2\text{O}$, the mass dedicated to ${\text{CO}}_2$ would be the difference between 11.53 kg and the inefficiency mass:If inefficiency means presence of $\text{CO}$:$$\text{CO }\text{Efficient Mass}=11.53\text{kg}-\text{CO mass}$$Re-adjust the mass balance and clear set only for ${\text{CO}}_2$:Efficiency equation becomes complex based on specific mass division and usually involves calculation solving to determine exclusive moles or mass demand.Divide remaining $11.53\text{ kg}$ with precise CO and $H_{2}O$ completion assumption. However, the exact efficiency can be numerically reevaluated in a span of variable modification inputs.

Key concepts

Octane Combustion

Octane, with the chemical formula ${\text{C}}_8{\text{H}}_{18}$, is a crucial component of gasoline used to power internal combustion engines. The energy released during the combustion of octane is what drives the engine. During complete combustion, octane reacts with oxygen to produce carbon dioxide (${\text{CO}}_2$) and water (${\text{H}}_2\text{O}$). This process yields the maximum energy output from the fuel, thus being considered efficient. On the other hand, incomplete combustion occurs when there is not enough oxygen for the octane to fully react. Instead of forming ${\text{CO}}_2$, carbon monoxide ($\text{CO}$) is produced. The presence of $\text{CO}$ indicates that the fuel is not being used optimally, leading to reduced energy efficiency and potentially hazardous emissions. Understanding these reactions is vital for achieving high combustion efficiency in engines.

Chemical Equations

Chemical equations describe the substances involved in a chemical reaction, before and after it occurs. In the context of octane combustion, there are two key equations to note: complete and incomplete combustion.

• For complete combustion: $${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$$ This equation shows that one mole of octane reacts with twelve and a half moles of oxygen to produce eight moles of ${\text{CO}}_2$ and nine moles of ${\text{H}}_2\text{O}$.
• For incomplete combustion: $${\text{C}}_8{\text{H}}_{18}+x{\text{O}}_2\to y\text{CO}+z{\text{CO}}_2+9{\text{H}}_2\text{O}$$ In this scenario, the uncertainty (expressed by $x,y,$ and $z$) reflects the variability in the amount of oxygen and the consequential formation of $\text{CO}$ and ${\text{CO}}_2$.

These equations not only help predict the products formed but also the efficiency of octane usage in generating energy.

Molar Mass Calculation

Calculating the molar mass of a compound is a fundamental step in chemical reactions. For octane, ${\text{C}}_8{\text{H}}_{18}$, the molar mass is found by adding the atomic masses of the constituent atoms. Each carbon atom has an atomic mass of approximately 12.01 g/mol, and each hydrogen atom has a mass of approximately 1.01 g/mol. Therefore, the calculation is as follows: $$8\times 12.01+18\times 1.01=114.23\text{ g/mol}$$ This means that one mole of octane weighs 114.23 grams.
The molar mass is used to convert between mass and moles, facilitating the determination of how much of a substance is participating in a reaction. This conversion is crucial for understanding how the mass of octane translates into its moles, which then relate to the quantities of products formed during combustion.

Incomplete Combustion

Incomplete combustion occurs when there is insufficient oxygen for fuel to completely react with oxygen. In the case of octane, this leads to the formation of carbon monoxide ($\text{CO}$) instead of carbon dioxide (${\text{CO}}_2$). This not only results in lower fuel efficiency because the energy output is less than what would be achieved with complete combustion, but $\text{CO}$ is also a toxic byproduct. During incomplete combustion, less energy is extracted from octane because part of it remains unoxidized or partially oxidized. The overall reaction is: $${\text{C}}_8{\text{H}}_{18}+x{\text{O}}_2\to y\text{CO}+z{\text{CO}}_2+9{\text{H}}_2\text{O}$$ Variables $x,y,$ and $z$ present in this reaction highlight the varying amounts of oxygen present. In engine performance, achieving complete combustion is ideal as it maximizes energy efficiency and minimizes harmful emissions.