Why are enthalpy of combustion and molar mass of organic compounds proportional?

I noticed that when you graph heat of combustion against molar mass, you get a completely straight line with a correlation of 1. I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases. Though, I don't know why the curve is completely straight. You can see heat of combustion against molar mass of alkanes in the graph below.

• What is LDF? Where did the data come from? Apr 14, 2021 at 6:46
• @BuckThorn London Dispersion Forces. Apr 14, 2021 at 12:37
• What are the units of the quantities in your data table?
– Karsten
Apr 14, 2021 at 13:03
• @KarstenTheis Since I am showing a relationship between two variables which can be found even in the question's title, I do not think units or error calcuations are relevant as long as there is consistency within the data. Apr 14, 2021 at 13:28
• You don't need the units to show linearity, but you do need the units when comparing the order of magnitude of the combustion enthalpy to the enthalpy of breaking London dispersion forces. And you need the sign to by consistent (these reactions are exothermic, and breaking London forces is endothermic).
– Karsten
Apr 14, 2021 at 14:20

$$\begin{array}{lrr} \hline n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline \text{Methane} & 112 & 8176\\ \text{Ethane} & 184 & 14640\\ \text{Propane} & 231 & 18832\\ n\text{-Butane} & 273 & 22390\\ n\text{-Pentane} & 309 & 26352\\ n\text{-Hexane} & 342 & 28850\\ n\text{-Heptane} & 372 & 31800\\ n\text{-Octane} & 399 & 33972\\ n\text{-Nonane} & 424 & 36910\\ n\text{-Decane} & 447 & 38750\\ \hline \end{array}$$
Instead, the heat of combustion scales with the number of carbons (see e.g. this table and graph) because the number of $$\ce{C=O}$$ and $$\ce{O-H}$$ bonds made (and the number of $$\ce{C-C}$$ and $$\ce{C-H}$$ bonds broken) also scales with the number of carbons.