In a car engine, gasoline (represented by \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) does not burn completely, and some CO, a toxic pollutant, forms along with \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). If \(5.0 \%\) of the gasoline forms \(\mathrm{CO}\) : (a) What is the ratio of \(\mathrm{CO}_{2}\) to CO molecules in the exhaust? (b) What is the mass ratio of \(\mathrm{CO}_{2}\) to CO? (c) What percentage of the gasoline must form CO for the mass ratio of \(\mathrm{CO}_{2}\) to \(\mathrm{CO}\) to be exactly \(1 / 1 ?\)

The first step in understanding the incomplete combustion of gasoline involves writing the chemical equation. Gasoline, represented by \(\text{C}_{8}\text{H}_{18}\), burns in oxygen to produce multiple products. In complete combustion, gasoline would form carbon dioxide \(\text{CO}_{2}\) and water \(\text{H}_{2}\text{O}\). However, in incomplete combustion, carbon monoxide \(\text{CO}\) is also formed because the oxygen supply is insufficient. Accordingly, the balanced chemical equation looks like this:
\[ \text{C}_{8}\text{H}_{18} + O_{2} \rightarrow a \times \text{CO}_{2} + b \times \text{CO} + c \times \text{H}_{2}\text{O} \]
The exact values of \(\text{a}\), \(\text{b}\), and \(\text{c}\) depend on the level of oxygen available and the specifics of the combustion process.

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Using stoichiometry, we can calculate the amounts of substances involved. In this exercise, we start by considering that 5.0% of gasoline forms \(\text{CO}\), and 95.0% forms \(\text{CO}_{2}\). Assuming 100 molecules of gasoline combust, we will have 1144 carbon atoms to start with (since each gasoline molecule \(\text{C}_{8}\text{H}_{18}\) contains 8 carbon atoms, and 100 molecules yield \[ 100 \times 8 = 800 \] carbon atoms).
This proportion helps us set up our equations and calculations to determine the molar and mass ratios of the gases produced in incomplete combustion.

Molar ratios help us understand the relationship between the amounts of reactants and products. With the given 5.0% of gasoline forming \(\text{CO}\), we calculate that 5.0% of carbon atoms form \(\text{CO}\), and the remaining 95.0% form \(\text{CO}_{2}\). Thus, we get:

• For \(\text{CO}\): \[ 0.05 \times 1144 = 57.2 \] molecules of \(\text{CO}\)
• For \(\text{CO}_{2}\): \[ 0.95 \times 1144 = 1086.8 \] molecules of \(\text{CO}_{2}\)

Using these values, we find the molar ratio as follows:
\(\text{CO}_{2}: \text{CO} = 1086.8: 57.2\). Simplifying further, we divide both terms by 57.2 to get the narrower molar ratio as:
\[ \frac{1086.8}{57.2} \rightarrow 19:1 \] So, for every molecule of \(\text{CO}\) produced, there are approximately 19 molecules of \(\text{CO}_{2}\).

Mass ratios help us compare the masses of different chemicals formed during the reaction. The molar mass of \(\text{CO}_{2}\) is 44 g/mol, and for \(\text{CO}\) it is 28 g/mol. By using these molar masses, we can calculate the total mass of each substance produced:

• For \(\text{CO}_{2}\): \[ 1086.8 \times 44 = 47819.2 \text{g} \]
• For \(\text{CO}\): \[ 57.2 \times 28 = 1601.6 \text{g} \]

Now, we find the mass ratio by dividing the total mass of \(\text{CO}_{2}\) by the total mass of \(\text{CO}\):
\[ \frac{47819.2}{1601.6} \rightarrow \frac{47819.2}{1601.6} \times \frac{1}{1} = 29.85:1 \] This means that the mass of \(\text{CO}_{2}\) is approximately 29.85 times the mass of \(\text{CO}\). If we need the mass ratio to be 1:1, we use the equation to solve for the percentage of gasoline that transforms into \(\text{CO}\):
\[ 44x = 28(1 - x) \]
Solving, \[ x \rightarrow \frac{28}{72} \times 100 = 38.9\% \] Thus, for a 1:1 mass ratio, about 38.9% of the gasoline must form \(\text{CO}\).