Problem 48 - General Chemistry: Principles and Modern Applications

Question

Without doing detailed calculations, explain which of these compounds produces the greatest mass of \(\mathrm{H}_{2} \mathrm{O}\) when \(1.00 \mathrm{g}\) of the compound is burned in an excess of oxygen: \(\mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{C}_{10} \mathrm{H}_{8}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

Step-by-Step Solution

Answer

The compound \(C_{10}H_{8}\) with the most hydrogen atoms is predicted to produce the greatest mass of \(H_{2}O\) when \(1.00 g\) of the compound is burned in excess oxygen.

Step 1: Identifying hydrogen atoms in each compound

Identify the number of hydrogen atoms in each given compound. This can be done by looking at the subscript following H in the molecular formula. For \(CH_{4}\), it is 4, for \(C_{2}H_{5}OH\), it is 6, for \(C_{10}H_{8}\), it is 8, and for \(C_{6}H_{5}OH\), it is 6.

Step 2: Comparing hydrogen atoms in the compounds

Compare the number of hydrogen atoms in each compound. It is expected that a compound with more hydrogen atoms will yield more water upon combustion. Looking at our results from step 1, \(C_{10}H_{8}\) has the most hydrogen atoms.

Step 3: Considering the mass of the compounds

While a greater amount of hydrogen atoms can generally determine a higher production of \(H_{2}O\), the initial mass of the compounds also plays a crucial role. Remember that we're starting with \(1.00 g\) of each compound. We must consider the molar mass of each compound, as it can influence the actual amount of \(H_{2}O\) produced per gram of compound combusted. This aspect however, is not required in this exercise as it is stated to solve without detailed calculations.

Key Concepts

Hydrogen Atoms

Hydrogen atoms are crucial in determining the amount of water produced during a combustion reaction. When a compound containing hydrogen is burned, its hydrogen atoms combine with oxygen to form water.
The molecular formula of a compound tells us how many hydrogen atoms are present. For example, in methane (\( \text{CH}_4 \)), the subscript '4' indicates there are four hydrogen atoms in each molecule. It is essential to carefully read the molecular formula to count these atoms accurately.
More hydrogen atoms generally mean more potential water production. This is because each hydrogen atom has the potential to produce a molecule of water when oxidized. However, it's not just about counting the atoms; other factors can impact water production, such as the compound's molecular structure and combustion efficiency.

• \( \text{CH}_4 \): 4 hydrogen atoms
• \( \text{C}_2\text{H}_5\text{OH} \): 6 hydrogen atoms
• \( \text{C}_{10}\text{H}_8 \): 8 hydrogen atoms
• \( \text{C}_6\text{H}_5\text{OH} \): 6 hydrogen atoms

Water Production

Water production in combustion reactions is linked to the number of hydrogen atoms present in the reactant. When burned, the hydrogen atoms react with oxygen to form water, a process that is predictable and follows chemical principles.
Calculating water production isn't just about the total number of hydrogen atoms, but also about the way they interact during the combustion. For each hydrogen atom, one molecule of water ( \( \text{H}_2\text{O} \)) can form, given that enough oxygen is available. Therefore, compounds with more hydrogen atoms can potentially produce more water.
Let's look deeper into our compounds:
\( \text{C}_{10}\text{H}_8 \), with its eight hydrogen atoms, is expected to generate more water than \( \text{CH}_4 \), which only has four. This means \( \text{C}_{10}\text{H}_8 \) has the highest potential for water production based on the hydrogen count alone.

• Hydrogen atoms bond with oxygen during combustion.
• More hydrogen generally means more water production.

Molecular Formula Analysis

Molecular formula analysis helps us understand the composition of each compound involved in a chemical reaction. By examining the molecular formula, we identify how many atoms of each element are present. This information is vital when predicting the outcomes of reactions such as combustion.
Each of our compounds— \( \text{CH}_4 \), \( \text{C}_2\text{H}_5\text{OH} \), \( \text{C}_{10}\text{H}_8 \), and \( \text{C}_6\text{H}_5\text{OH} \)— has a unique molecular composition that determines its behavior in chemical reactions, particularly in how much water is formed upon burning.
Understanding this begins with counting hydrogen atoms, as we did, but also requires considering the total molecular mass. Although this exercise does not require detailed calculations, knowing the molar masses can provide deeper insights. By looking at these formulas:

• \( \text{CH}_4 \): simple molecular structure, fewer atoms, predictable outcomes.
• \( \text{C}_2\text{H}_5\text{OH} \): ethanol, with its OH group, impacts how it combusts.
• \( \text{C}_{10}\text{H}_8 \): complex structure, potentially impacts reaction completeness.

Molecular formulas are foundational tools in chemistry that help predict reaction outcomes.