How do you write an equation for the enthalpy change of combustion of an alcohol

How would you write the equations for the enthalpy change of combustion for methanol and pentan-1-ol . I understand how to calculate the enthalpy change of combustion but how do you write an equation for the specific alcohol?

1 mol of the alcohol + as much oxygen as needed for complete combustion (ca be fractional) ==> carbon dioxide + water

Thank you, but how do I know how much oxygen was needed for the combustion? I haven't been told

All of the carbon atoms in the alcohol (or any other organic molecule) turns to carbon dioxide.
So if there are 4 carbon atoms they make four carbon dioxide molecules, which requires four oxygen molecules.

All of the hydrogen turns to water.
So if there are four hydrogen atoms in the molecule they turn to two water molecules requiring one oxygen molecule.

Don't forget that alcohols already have one oxygen atom in the structure when you are working out the number of oxygen molecules needed.

Ahh I think that makes sense now thank you !

so if it was methanol (CH3OH)

then would the formula be

CH3OH+2O2----->CO2+H2O??

the starting mass of the methanol is 188.15g which gives me around 5.9 mol but I am not sure how to include this into the equation thank you so much for your help so far

oh right

so if two water molecules are formed then would the equation be : 2 CH3OH + 3 O2 ---> 2 CO2 + 4 H2O ?

You did not read the first line of my original post!

1 mol of the alcohol + as much oxygen as needed for complete combustion (can be fractional) ==> carbon dioxide + water

Oh this is confusing! so is it CH3OH+3/2O2-->CO2+2H2O

I don't think it can be a fractional number in front though can it?

I already said that the amount of oxygen can be fractional. You are defining the enthalpy of combustion, which HAS to be 1 mol of alcohol burned.

CH₃OH + 3/2O₂ --> CO₂ + 2H₂O

Is the correct answer ...