Problem 102 - Chemistry The Central Science In Si Units

Question

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Step-by-Step Solution

Answer

(a) \( \mathrm{C}_{2}\mathrm{H}_{3} \); (b) 0.801 g of \( \mathrm{O}_2 \); (c) 1.068 g of \( \mathrm{O}_2 \).

Step 1: Determine moles of combustion products

To find the empirical formula, start by converting the masses of the combustion products to moles. For \( \mathrm{CO} \), use its molar mass of \( 28.01 \text{ g/mol} \):\[\text{Moles of } \mathrm{CO} = \frac{0.467 \text{ g}}{28.01 \text{ g/mol}} = 0.0167 \text{ moles}\]For \( \mathrm{CO}_{2} \), use its molar mass of \( 44.01 \text{ g/mol} \):\[\text{Moles of } \mathrm{CO}_{2} = \frac{0.733 \text{ g}}{44.01 \text{ g/mol}} = 0.0167 \text{ moles}\]For \( \mathrm{H}_{2} \mathrm{O} \), use its molar mass of \( 18.02 \text{ g/mol} \):\[\text{Moles of } \mathrm{H}_{2}\mathrm{O} = \frac{0.450 \text{ g}}{18.02 \text{ g/mol}} = 0.0250 \text{ moles}.\]

Step 2: Determine moles of C and H in the hydrocarbon

From \( \mathrm{CO} \) and \( \mathrm{CO}_{2} \), both provide carbon atoms. Since 1 mole of each provides 1 mole of C:\[\text{Total moles of C} = 0.0167 + 0.0167 = 0.0334 \text{ moles}\]From \( \mathrm{H}_{2}\mathrm{O} \), 1 mole provides 2 moles of hydrogen atoms:\[\text{Moles of H} = 2 \times 0.0250 = 0.0500 \text{ moles}\]

Step 3: Calculate empirical formula

Divide the moles of C and H by the smallest number of moles:\[\text{Moles of C}: \frac{0.0334}{0.0334} = 1\]\[\text{Moles of H}: \frac{0.0500}{0.0334} \approx 1.5\]To get whole numbers, multiply all by 2:\[\text{C} = 2, \quad \text{H} = 3\]Thus, the empirical formula is \( \mathrm{C}_{2}\mathrm{H}_{3} \).

Step 4: Determine grams of \( \mathrm{O}_{2} \) used

From previously calculated moles:Total carbon contributed by \( \mathrm{CO} \) and \( \mathrm{CO}_{2} \) uses:\[\text{For } \mathrm{CO}: 0.0167 \times 0.5 \times 32 \text{ g/mol} = 0.267 \text{ g}\] \[\text{For } \mathrm{CO}_{2}: 0.0167 \times 1 \times 32 \text{ g/mol} = 0.534 \text{ g}\]Total \( \mathrm{O}_{2} \) used:\[0.267 \text{ g} + 0.534 \text{ g} = 0.801 \text{ g}\]

Step 5: Determine \( \mathrm{O}_{2} \) for complete combustion

Complete combustion forms only \( \mathrm{CO}_{2} \):\[\mathrm{CO} \text{ would become } \mathrm{CO}_{2}, \text{ requiring additional } 0.0167 \text{ moles } \times 0.5 \times 32 \text{ g/mol} = 0.267 \text{ g}\]Add \(0.267 \text{ g}\) to \(0.801 \text{ g}\):\[\text{Total } \mathrm{O}_{2} = 0.801 \text{ g} + 0.267 \text{ g} = 1.068 \text{ g}\]

Step 6: Conclusion

(a) The empirical formula is \( \mathrm{C}_{2}\mathrm{H}_{3} \), (b) 0.801 g of \( \mathrm{O}_2 \) was used in the reaction, (c) 1.068 g of \( \mathrm{O}_2 \) would be required for complete combustion.

Key Concepts

Combustion Products

When hydrocarbons burn, they interact with oxygen to create combustion products. The primary components resulting from this process are carbon dioxide \(\mathrm{CO}_2\), carbon monoxide \(\mathrm{CO}\), and water \(\mathrm{H}_2\mathrm{O}\). This transformation occurs as the hydrocarbon's carbon and hydrogen atoms react with oxygen. In our exercise, the hydrocarbon undergoes combustion yielding specific masses for each product: 0.467 grams of \(\mathrm{CO}\), 0.733 grams of \(\mathrm{CO}_2\), and 0.450 grams of \(\mathrm{H}_2\mathrm{O}\). By knowing these values, we can deduce the amounts of individual atoms present initially in the hydrocarbon, aiding in the subsequent determination of the empirical formula.Understanding combustion processes is crucial because they can vary based on oxygen availability. During our scenario, the combustion isn't complete since both \(\mathrm{CO}_2\) and \(\mathrm{CO}\) are formed, showing a limited oxygen reaction. This serves as a classic example of how complex chemical processes can yield multiple products.

Mole Calculation

Mole calculations provide a fundamental way to understand matter at the atomic level. By considering the molar masses of each compound, we can translate between mass and moles, enabling us to count atoms and molecules.For example, the conversion of masses to moles for \(\mathrm{CO}\), \(\mathrm{CO}_2\), and \(\mathrm{H}_2\mathrm{O}\) in our problem is essential in understanding the proportions of elements within the hydrocarbon. Here's how it's done:

• \(\mathrm{CO}\): With a molar mass of 28.01 g/mol, 0.467 grams converts to 0.0167 moles.
• \(\mathrm{CO}_2\): With a molar mass of 44.01 g/mol, 0.733 grams converts to 0.0167 moles.
• \(\mathrm{H}_2\mathrm{O}\): With a molar mass of 18.02 g/mol, 0.450 grams converts to 0.0250 moles.

These calculations are foundational to further steps, like determining how many atoms of carbon and hydrogen there are. The mole concept simplifies chemistry by letting us use these easily measured quantities to infer the unseen atomic world. This is particularly helpful in calculating the ratio of elements to determine the empirical formula.

Complete Combustion

Complete combustion occurs when a hydrocarbon reacts entirely with oxygen to form carbon dioxide \(\mathrm{CO}_2\) and water \(\mathrm{H}_2\mathrm{O}\). This is significantly different from incomplete combustion, where you might also find carbon monoxide \(\mathrm{CO}\) as a product.In the given exercise, the combustion wasn't complete because both \(\mathrm{CO}\) and \(\mathrm{CO}_2\) were created. Complete combustion of the same hydrocarbon would consume more oxygen and produce only \(\mathrm{CO}_2\).

• In our calculation of the initial reaction, 0.801 grams of \(\mathrm{O}_2\) were used.
• For complete combustion, additional \(\mathrm{O}_2\) would be required so all carbon turns into \(\mathrm{CO}_2\), requiring us to add 0.267 grams more, leading to a total of 1.068 grams of \(\mathrm{O}_2\).

Understanding the concept of complete combustion is important not only for mastering classwork but also for grasping real-world applications, such as fuel efficiency and aiding in reducing pollution through cleaner burning fuels.