To calculate binding energy of an atom we find difference between the mass of whole nucleons that constitute that atom and the experimental mass of the atom from tables.

But, to calculate the binding energy of this reaction:

$$\rm C + O_2 \to CO_2$$

I think we must first find the mass defect:

$$\Delta = M(\mathrm{CO_2}) - [M(\mathrm{^{12}C}) + 2M(\mathrm{^{16}O})]$$

Then the binding energy of $\mathrm{CO_2}$ molecule or the energy that will be released from this reaction is:

$$\mathrm{BE} = \Delta C^2$$

If this procedure is true, we have $M(\mathrm{^{12}C}) = 12\,\rm u$ and $M(\mathrm{^{16}O}) = 15.994915\,\rm u$ from tables. But what is $M(\mathrm{CO_2})$, we need it to calculate $\Delta$ and I can't find its value in any table, website, or book.

If this procedure isn't true, for finding the binding energy of $\mathrm{CO_2}$ molecule or the energy that is released from that reaction what should I do, please help.

  • 3
    $\begingroup$ Chemical binding energies are usually too small to produce a measurable mass deficit. The binding energy is measured by measuring how much energy we have to put in to dissociate the molecule. $\endgroup$ Oct 17, 2017 at 4:58

1 Answer 1


The standard free energy of formation of carbon dioxide from graphite and oxygen is $394\, \rm kJ\, mol^{-1}$ and so in the formation of one molecule of carbon dioxide $6.5 \times 10^{-19} \,\rm J$ of energy are liberated.

$6.5 \times 10^{-19} \,\rm J$ of energy is $4.1 \, \rm eV$ or has the mass equivalent of $4.4 \times 10^{-9}\,\rm u$.

So you will see that even your very accurate and precise value for the mass of an oxygen atom $M(^{16}\rm O)=15.994915 \, u$ is not good enough to find the energy of formation of carbon dioxide.

This is a good example of the difference between Chemistry and Nuclear Physics in that Chemistry deals with energy changes of order $\rm eV$ whereas Nuclear Physics deals with energies of order $\rm MeV$.

  • $\begingroup$ Thanks, your answer was clearly written. Can we say what fraction of mass is transferred to energy in the reaction I have been written in my question? $\endgroup$
    – Masa
    Oct 17, 2017 at 10:37
  • $\begingroup$ Use the masses of oxygen and carbon atoms in atomic mass unit that you have quoted and work out the "molecular mass" and then find the percentage that the mass defect $4.4 \times 10^{-9}\,\rm u$ is to the "molecular mass". $\endgroup$
    – Farcher
    Oct 17, 2017 at 11:03

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