If equal masses of the following compounds undergo complete combustion, which will yield the greates — Stoichiometry Chemistry Question

STEPS:

1. Understand the chemistry of complete combustion: During the complete combustion of any organic compound containing carbon, hydrogen, and/or oxygen, all of the carbon atoms in the reactant are converted entirely into carbon dioxide ($\text{CO}_2$) gas.
2. Relate reactant mass to product mass: Since we are starting with equal masses of each compound, the mass of $\text{CO}_2$ produced is directly proportional to the mass percent of carbon in the starting reactant:
$$\text{Mass of }\text{CO}_2 = \text{Mass of sample} \times (\%\text{ C by mass}) \times \frac{\text{Molar Mass of }\text{CO}_2}{\text{Molar Mass of C}}$$
Therefore, the compound with the highest mass percent of carbon will yield the greatest mass of $\text{CO}_2$ upon complete combustion.
3. Determine the mass percent of carbon in each compound: Because calculators are not permitted on Section I of the AP Chemistry exam, a student can quickly estimate these percentages by comparing the mass contributed by carbon to the total molar mass of each molecule:
* Benzene ($\text{C}_6\text{H}_6$): Carbon contributes $6 \times 12 = 72\text{ g/mol}$ out of a total molar mass of $\approx 78\text{ g/mol}$.
$$\%\text{ C} = \frac{72}{78} \times 100\% \approx \mathbf{92.3\%}$$
* Cyclohexane ($\text{C}_6\text{H}_{12}$): Carbon contributes $72\text{ g/mol}$ out of a total molar mass of $\approx 84\text{ g/mol}$.
$$\%\text{ C} = \frac{72}{84} \times 100\% \approx \mathbf{85.7\%}$$
* Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$): Carbon contributes $72\text{ g/mol}$ out of a total molar mass of $\approx 180\text{ g/mol}$.
$$\%\text{ C} = \frac{72}{180} \times 100\% = \mathbf{40.0\%}$$
* Methane ($\text{CH}_4$): Carbon contributes $12\text{ g/mol}$ out of a total molar mass of $\approx 16\text{ g/mol}$.
$$\%\text{ C} = \frac{12}{16} \times 100\% = \mathbf{75.0\%}$$
4. Compare and select the correct option: Benzene ($\text{C}_6\text{H}_6$) has the highest mass percent of carbon among all the choices ($\approx 92.3\%$). Thus, burning a sample of benzene will yield the greatest mass of $\text{CO}_2$, making Option A the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect (Cyclohexane): Although cyclohexane has the same number of carbon atoms per molecule as benzene, it has twice as many hydrogen atoms. This extra hydrogen adds non-carbon mass to the molecule, reducing the carbon mass percentage to $\approx 85.7\%$ and yielding less $\text{CO}_2$ per gram.
• Option C is incorrect (Glucose): Glucose has a very low carbon mass percentage ($40.0\%$) because oxygen—which is highly electronegative and heavy—makes up more than half of the molecule's total mass. Consequently, it yields the least $\text{CO}_2$ per gram of all the options.
• Option D is incorrect (Methane): Although methane has a high ratio of carbon to total atoms (1 out of 5), hydrogen is very light. Methane's carbon mass percentage ($75.0\%$) is lower than both benzene's and cyclohexane's, meaning it yields less $\text{CO}_2$ per gram of reactant.