Answer
See the answer below.
Work Step by Step
Enthalpies of formation(kJ/mol):
$C_3H_8(g): -104.7, C_4H_{10}(g): -127.1, C_8H_{18}(l):-259.3, C_2H_5OH(l): -277.0, H_2O(g):-241.83, CO_2(g): -393.509$
Propane:
$C_3H_8(g)+5\ O_2(g)\rightarrow 3\ CO_2(g) + 4\ H_2O(g)$
$(4\times -241.83+3\times -393.509 - -104.7)kJ/mol\div 44.10\ g/mol=-46.3\ kJ/g$
Butane:
$C_4H_{10}(g)+13/2\ O_2(g)\rightarrow 4\ CO_2(g) + 5\ H_2O(g)$
$(5\times -241.83+4\times -393.509 - -127.1)kJ/mol\div 58.12\ g/mol=-45.7\ kJ/g$
Isooctane:
$C_8H_{18}(l)+25/2\ O_2(g)\rightarrow 8\ CO_2(g) + 9\ H_2O(g)$ $(9\times -241.83+8\times -393.509 - -259.3)kJ/mol\div 114.23\ g/mol=-44.3\ kJ/g$
Ethanol:
$C_2H_5OH(l)+3\ O_2(g)\rightarrow 2\ CO_2(g) + 3\ H_2O(g)$
$(3\times -241.83+2\times -393.509 - -277.0)kJ/mol\div 46.07\ g/mol=-26.8\ kJ/g$
Shorter un-oxigenated chains release more energy per unit of mass in combustion.
