Question

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the main component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.25 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(1.00 \mathrm{gal}\) of \(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18} ?\)

Short answer

(a) \(1.25\, mol\,C_8H_{18}\times\frac{25 \,mol\, O_2}{2 \,mol\, C_8H_{18}} = 15.625\, mol\, O_2\) (b) mass of \(O_2 = \frac{10.0 \mathrm{~g}}{114.23 \mathrm{~g/mol}} \times\frac{25 \,mol\, O_2}{2 \,mol\, C_8H_{18}} \times 32.00 \mathrm{~g/mol} \approx 13.9 \mathrm{~g\, O_2}\) (c) mass of \(O_2 =1.00 \, \textrm{gal}\times\frac{3.785\mathrm{~L}}{1 \, \mathrm{gal}}\times\frac{1000\mathrm{~mL}}{1\mathrm{~L}}\times 0.692\, \textrm{g/mL} \times \frac{1\,mol\, C_8H_{18}}{114.23\,g} \times\frac{25 \,mol\, O_2}{2 \,mol\, C_8H_{18}} \times 32.00 \mathrm{~g/mol} \approx 1057.4 \mathrm{~g\, O_2}\)

Step-by-step solution

Part (a): Find moles of O₂ required for the given moles of C₈H₁₈

To find the moles of \(O_2\) required for the combustions of 1.25 moles of \(C_8H_{18}\), we can use stoichiometry from the balanced chemical reaction: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ Mole ratio: \(\frac{25 \,mol\, O_2}{2 \,mol\, C_8H_{18}}\) Now, multiply given moles of \(C_8H_{18}\) (1.25 moles) by mole ratio to find moles of \(O_2\) required: \(1.25\, mol\,C_8H_{18}\times\frac{25 \,mol\, O_2}{2 \,mol\, C_8H_{18}}\ =\ n(O_2)\)

Part (b): Find mass of O₂ required for burning given mass of C₈H₁₈

To find the mass of \(O_2\) required for burning 10.0 g of \(C_8H_{18}\), first, convert the given mass of \(C_8H_{18}\) to moles using its molar mass (114.23 g/mol): \(\textrm{Moles of}\, C_8H_{18} = \frac{10.0 \mathrm{~g}}{114.23 \mathrm{~g/mol}}\) Next, use the mole ratio from the balanced chemical reaction and the moles of \(C_8H_{18}\) to find the moles of \(O_2\) required. Finally, convert the moles of \(O_2\) to mass using the molar mass of \(O_2\) (32.00 g/mol).

Part (c): Find mass of O₂ required for burning given volume of C₈H₁₈

To find the mass of \(O_2\) required for burning 1.00 gal of \(C_8H_{18}\), first, convert the volume to mass using the given density (0.692 g/mL): \(\textrm{Mass of}\, C_8H_{18} = 1.00 \, \textrm{gal}\times\frac{3.785\mathrm{~L}}{1 \, \mathrm{gal}}\times\frac{1000\mathrm{~mL}}{1\mathrm{~L}}\times 0.692\, \textrm{g/mL}\) Next, convert the mass of \(C_8H_{18}\) to moles using its molar mass (114.23 g/mol). Use the mole ratio from the balanced chemical reaction to find the moles of \(O_2\) required, and finally, convert the moles of \(O_2\) to mass using the molar mass of \(O_2\) (32.00 g/mol).

Key concepts

Stoichiometry

Understanding stoichiometry is crucial when dealing with combustion reactions like the burning of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\). In simple terms, stoichiometry helps us determine how much of each substance is involved in a chemical reaction. It's essentially the 'recipe' for making sure all reactants are used efficiently, and all products are accounted for. In the given equation, \[2 \mathrm{C}_{8} \mathrm{H}_{18}(l) + 25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g) + 18 \mathrm{H}_{2} \mathrm{O}(g)\], stoichiometry allows us to calculate the amounts of oxygen needed and CO₂ and water produced.
The coefficients in the balanced equation act as conversion factors. For example, to find out how much \(\mathrm{O}_{2}\) is needed to burn \(1.25\, \text{moles of} \ \mathrm{C}_{8} \mathrm{H}_{18}\), we use the ratio \[\frac{25\, \text{moles of } \mathrm{O}_{2}}{2\, \text{moles of} \ \mathrm{C}_{8} \mathrm{H}_{18}}\]. Simply multiply this ratio with the moles of octane you have to find out the required moles of oxygen.
This step is a practical use of stoichiometry in figuring out how much reaction input you need to achieve the desired output, balancing the equation in actual quantities—not just theory.

Molar Mass

Molar mass plays a vital role in converting between mass and moles of a substance, especially in combustion reactions. The molar mass of a compound tells you how much one mole of that compound weighs. For instance, the molar mass of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), is \(114.23 \mathrm{\ g/mol}\), and for molecular oxygen \(\mathrm{O}_{2}\), it's \(32.00 \mathrm{\ g/mol}\).
Knowing the molar mass allows you to change grams into moles and vice versa, which is useful when you want to find out how much of a reactant or product is needed in terms of everyday measurements like mass. For example, in part (b) of the exercise, you have \(10.0\, \text{g of } \ \mathrm{C}_{8} \mathrm{H}_{18}\). To determine how many moles this is, divide the mass by the molar mass: \[\frac{10.0 \mathrm{\ g}}{114.23 \mathrm{\ g/mol}} \approx 0.0875 \mathrm{\ mol}\]This information is critical for converting to moles of \(\mathrm{O}_{2}\) later using stoichiometry. Thus, mastering molar mass calculations is an essential skill for performing any chemical equation conversion.

Balanced Chemical Equations

Balanced chemical equations are fundamental in describing chemical reactions completely. A balanced equation shows how many molecules of each reactant and product are involved, ensuring that the law of conservation of mass is upheld, meaning mass is neither created nor destroyed in a chemical reaction.
In the provided combustion reaction of octane, the equation \[2 \mathrm{C}_{8} \mathrm{H}_{18}(l) + 25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g) + 18 \mathrm{H}_{2} \mathrm{O}(g)\]is balanced because it has the same number of each type of atom on both sides of the equation.

• There are \(16\, \mathrm{C}\) atoms on both sides (from 2 \(\mathrm{C}_{8}\)).
• There are \(36\, \mathrm{H}\) atoms on both sides (from 2 \(\mathrm{C}_{8} \mathrm{H}_{18}\), resulting in \(18\times2\)).
• And \(50\, \mathrm{O}\) atoms on both sides, coming mainly from the \(\mathrm{O_{2}}\).

By ensuring the equation is balanced, it allows you to reliably determine the amounts of reactants and products involved, as required in each part of this exercise. Balancing equations also ensures that computational derivations in stoichiometry are precise and meaningful.