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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. enter image description here

Data source: Organic Compounds: Physical and Thermochemical Data

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    $\begingroup$ What is LDF? Where did the data come from? $\endgroup$
    – Buck Thorn
    Apr 14, 2021 at 6:46
  • $\begingroup$ @BuckThorn London Dispersion Forces. $\endgroup$
    – abtoiew
    Apr 14, 2021 at 12:37
  • $\begingroup$ What are the units of the quantities in your data table? $\endgroup$
    – Karsten
    Apr 14, 2021 at 13:03
  • $\begingroup$ @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. $\endgroup$
    – abtoiew
    Apr 14, 2021 at 13:28
  • $\begingroup$ 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). $\endgroup$
    – Karsten
    Apr 14, 2021 at 14:20

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The biggest contributor to the heat of combustion is the difference in covalent bond energy of reactants and products. The intermolecular forces that have to be overcome for combustion of liquids (with products all gases) are an order of magnitude lower (adapted from https://chemistry.stackexchange.com/a/125013):

$$ \begin{array}{lrr} \hline n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline \text{Methane} & 112 & 8176\\ \text{Ethane} & 184 & 14640\\ \text{Propane} & 231 & 18832\\ n\text{-Butane} & 273 & 22390\\ n\text{-Pentane} & 309 & 26352\\ n\text{-Hexane} & 342 & 28850\\ n\text{-Heptane} & 372 & 31800\\ n\text{-Octane} & 399 & 33972\\ n\text{-Nonane} & 424 & 36910\\ n\text{-Decane} & 447 & 38750\\ \hline \end{array} $$

[OP] 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.

Some of these alkanes are gases, so there are no dispersion forces to overcome. Also, you got the sign of the contribution wrong. If a reactant has higher LDF than the products (which have none), this would make the reaction less exothermic. So you would expect lower magnitudes for the larger alkanes, opposite of what is observed.

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.

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See Table I, "Alkane Heats of Combustion". All the heats of combustion can be expressed as about 148 kcal/"mol -CH2-", barring some small details such as ring strain. But when you express them per mole of the alkane, well, the longer the alkane, the more -CH2-s, the more enthalpy, in nearly direct proportion. The enthalpy per mass burned remains about the same.

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