[canadiankarma]

If the enthalpy of combustion of say methanol is 725 kJ/mol, and you want to determine the heat per gram of methanol, do you need to divide 725 by 2 to get the enthalpy of combustion of 1 mole of methanol(because of the balanced equation) and then divide the enthalpy of combustion by the molar mass of methanol.

THank you for your help, I'm really confused..
   

[AWK]

The data in tables are exactly for 1 mole.
It is convenient to rescale eqation for 1 mole of methanol
CH4O + 1.5O2 = CO2 + 2H2O - 725kJ

and then divide the enthalpy of combustion by the molar mass of methanol.

[Donaldson Tan]

do not double-post in future. i have deleted your other post.

the enthalpy of combustion is with respect to one mole of that substance. 1 mole of methanol contains 32g of mass. in another words, 32g of methanol liberates 725kJ of energy during combustion.

so the mass basis for enthalpy of combustion is 725/32 = 22.7kJ/g of methanol