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Chapter 3: Problem 48

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: \(\mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{C}_{10} \mathrm{H}_{8}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.

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