# Is there a way to estimate the nutritional calorific value of a chemical

The chemical in question is L Carvone. It is an ingredient of some flavors. Tastes and smells like spearmint.

What's a good way to estimate its calorific value? If any. I cannot see it as a carb / protein / fat etc.

Does it have a calorific value at all?

• Calorimetry . Or thermochemistry calculations based on tabulated data. As for fat sugars fuels etc. However from a nutritional view point one should know how the metabolism of the specific molecule work. At extreme, a compound can enter the body and either be excreted intact or accumulated intact. – Alchimista Oct 27 '17 at 12:00
• @curious_cat I agree with Alchimista. It probably isn't too hard to obtain an estimate of the energy of a compound (Nando shows some nice ways below), but that's quite different from the amount of energy you would get from eating it, which would likely require a more detailed understanding of human metabolism. – Tyberius Oct 28 '17 at 13:20

If carvone ($\ce{C10H14O1}$) is digested by the body using oxygen to generate carbon dioxide and water, you can estimate the heat of this chemical reaction:

$$\ce{C10H14O1 + aO2 -> xCO2 + yH2O}$$

After balancing the equation:

$$\ce{C10H14O1 + 13O2 -> 10CO2 + 7H2O}$$

The required enthalpy can be know from the formation heats of each molecule multiplied by the found factors.

### A quick estimation

I just did this exercise using computational chemistry (B3LYP/6-31G(d) in Gaussian09) and found the following energies (in kcal/mol):

• carvone: -291456.0217
• oxygen: -94322.73333
• carbon dioxide: -118326.709
• water: -47947.30569

The change in energy of the reaction is then:

$$\begin{multline}10\times (\pu{-118327kcal/mol})+7\times (\pu{-47947kcal/mol})\\-1\times (\pu{-291456kcal/mol})-13\times (\pu{-94323kcal/mol}) = \pu{-1137 kcal/mol}\end{multline}$$

this is $\pu{-1137 kcal/mol}$ (exothermic, as expected).

Since molecular weight of carvone is $\pu{150.2 g/mol}$, we can convert to $\mathrm{kcal/g}$, the negative value indicates it releases heat:

$$\frac{\pu{-1137 kcal/mol}}{\pu{150.2 g/mol}} = \pu{-7.6 kcal/g}$$

• Oh god … I would have assumed that someone who has reached the level of calculating with Gaussian had learnt how to use units correctly … To allow me to upvote in peace with myself I have edited your post to do so. – Jan Oct 29 '17 at 2:31
• Also re units: SI units would of course be preferred ;p – Jan Oct 29 '17 at 2:31
• Sorry, this is just my second answer. I haven't learned much about your strict rules yet. Also, I thought calories would be better since its the common units for food energy. Thanks for edits! – Nando Oct 29 '17 at 2:53
• Well yes, the calorie just won’t die any time soon from nutrition tables; that is unfortunately true. No worries, the answer itself was good but you should really make sure each equation including units adds up … Also, you can ping me by adding @Jan if I commented on your question. While SE does implement some auto-pinging mechanism, they may fail. You as the post owner are always pinged. – Jan Oct 29 '17 at 2:55
• @Nando you can include an image from a book so long as you cite where it came from. – Tyberius Oct 29 '17 at 14:38