When a hydrocarbon burns, water is produced as well as carbon dioxide. (For this reason, clouds of condensed water droplets are often seen coming from automobile exhausts, especially on a cold day.) The density of gasoline is \(0.79 \mathrm{~g} \cdot \mathrm{mL}^{-1}\). Assume gasoline to be represented by octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), for which the combustion reaction is $$ \begin{aligned} 2 \mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}(\mathrm{l}) &+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \\ & 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{aligned} $$ Calculate the mass of water produced from the combustion of \(1.0 \mathrm{~L}\) of gasoline.

At the heart of chemistry lie chemical reactions, where substances, known as reactants, transform into new substances, called products. In the example of hydrocarbon combustion, such as octane \textbf{(C8H18)}, it reacts with oxygen (\textbf{O2}) to produce carbon dioxide (\textbf{CO2}) and water (\textbf{H2O}). This type of reaction is exothermic, releasing energy in the form of heat and light.

Understanding the process involves balancing the chemical equation, where the number of atoms for each element is the same on both the reactants and products side. The balanced equation for the combustion of octane is: \[2 \text{C8H18}(l) + 25 \text{O2}(g) \rightarrow 16 \text{CO2}(g) + 18 \text{H2O}(l)\]
This balance is crucial for stoichiometric calculations, which allow us to predict the amounts of products formed from given reactants.

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It allows us to predict how much of each substance is required or produced in a reaction. For example, the combustion equation tells us that 2 moles of octane yield 18 moles of water. This stoichiometric ratio is a critical concept as it guides the proportions in which chemicals react.

When working with stoichiometry, it is essential to start with a balanced chemical equation. Once the equation is balanced, the coefficients represent the mole ratios of the substances involved. With the knowledge of these ratios, we can perform calculations to find out, for example, how much water is produced when a certain amount of octane burns. A deeper understanding of stoichiometry can even lead to predictions about the energy released or absorbed during a reaction.

The molar mass of a substance is the weight of one mole of that substance. It is typically expressed in grams per mole (g/mol). For compounds, the molar mass is calculated by adding up the molar masses of each constituent element, multiplied by the number of atoms of that element in the compound.

For instance, the molar mass of octane (\textbf{C8H18}) is calculated by taking into consideration the molar masses of carbon (12.01 g/mol) and hydrogen (1.008 g/mol) as follows:\[ \text{Molar mass of C8H18} = (8 \times 12.01) + (18 \times 1.008) = 114.23 \text{ g/mol}\]
Knowing the molar mass of a substance is fundamental in converting between mass and moles, which is often required in stoichiometric calculations.

Converting moles to grams and vice versa is a crucial operation in stoichiometry. This conversion relies on the molar mass of the substance of interest. To convert moles to grams, you multiply the number of moles by the molar mass of the substance. Conversely, to convert grams to moles, divide the mass by the molar mass.

For our hydrocarbon combustion example, the mass of water produced is calculated using the moles of water and the molar mass of water (\textbf{H2O}), which is approximately 18.015 g/mol:\[ \text{Mass of water} = \text{Moles of water} \times \text{Molar mass of water} = 62.19 \text{ mol} \times 18.015 \text{ g/mol} \approx 1120 \text{ g}\]
This conversion is the final step in determining the measurable quantity of product from a given amount of reactant, helping to bridge the gap between the abstract world of chemical formulas and the tangible materials we can see and measure.