Energy release depends on heat of combustion per gram, favoring hydrogen-rich fuels

Steps:

• Compute molar masses: benzene (C6H6) 78 g/mol, heptane (C7H16) 100 g/mol, octane (C8H18) 114 g/mol, propane (C3H8) 44 g/mol.
• Note heats of combustion (ΔHc) per mole: benzene ~3260 kJ/mol, heptane ~4820 kJ/mol, octane ~5510 kJ/mol, propane ~2220 kJ/mol.
• Calculate ΔHc per gram: divide molar ΔHc by molar mass (benzene ~41.8 kJ/g, heptane ~48.2 kJ/g, octane ~48.3 kJ/g, propane ~50.5 kJ/g).
• Compare values: benzene has the lowest energy release per gram.

Why A is correct:

• Benzene's aromatic structure and low hydrogen content (H/C ratio of 1) yield lower ΔHc per gram than alkanes, per standard combustion enthalpy data.

Why the others are wrong:

• B: Heptane's higher hydrogen content (H/C ≈2.3) gives more energy from C-H bonds oxidized to water.
• C: Octane, like heptane, has high hydrogen per gram, maximizing combustion energy.
• D: Propane's small size and high H/C ratio (≈2.7) release the most energy per gram.

Final answer: A