The Food Energy in a Marshmallow - ChemPRIME

The Food Energy in a Marshmallow

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We're not making this up.

Roasting marshmallow

Marshmallow was originally used medicinally to soothe sore throats. It was made from the root of the marsh mallow plant, Althaea officinalis, sometimes mixed with sugars or other ingredients and whipped to make something like the modern marshmallow.[1] Modern Campfire® marshmallows contain Corn Syrup, Sugar, Modified Food Starch (Corn), Dextrose, Water, Gelatin, Natural and Artificial Flavor, Tetrasodium Pyprophosphate and Blue 1[2], and the gelatin protein made from bones and hides makes them off limits to strict vegetarians. We'll consider the marshmallow to be 7.5 g of pure sugar (sucrose) for the calculations below.

Let's investigate the fate of a marshmallow when you eat it, and explain in part where the food energy comes from.

Contents

Aerobic Metabolism

We'll need the energy supplied by the overall reaction for the aerobic metabolism of sucrose, which occurs when plenty of oxygen is available:


C12H22O11 (s) + 12 O2(g)→12 CO2(g) + 11 H2O(l)    ΔHm1       (1)


But that reaction lumps together a lot of interesting processes. The hydrolysis (cleavage by water) of sucrose into the simple sugars glucose and fructose occurs in saliva, but not without the enzyme sucrase that catalyzes the reaction:


C12H22O11 (s) + 1 H2O(l)→C6H12O6(glucose) + C6H12O6(fructose}  ΔHm2    (2)

Anaerobic Metabolism

Lactobaccili in our mouth partially convert the simple sugars to lactic acid (which causes tooth decay), in the overall reaction for glycolysis plus fermentation:



C6H12O6 (s) → 2 C3H6O3      ΔHm3       (3)


This reaction provides energy to sustain the bacteria, but it also occurs in our bodies when sugar is metabolized anaerobically (with limited oxygen), and the lactic acid is responsible for muscle ache the day after we exert our muscles. Lactobaccilli are used in controlled recipes to create lactic acid that creates the tart or sour taste of yogurt and sauerkraut.

If bacteria don't metabolize the glucose, we do, using it to produce ATP in a process called glycolysis which involves about ten different reactions that end with the production of pyruvic acid (C3H4O3). If our muscles are well oxygenated, the pyruvic acid is converted to CO2 and H2O and we have overall reaction (1), yielding energy that we'll calculate below. During extended exercise, glycolysis comes to a halt when it runs out of oxygen to make an essential reactant, NAD+. Then anaerobic fermentation takes over, producing the NAD+ and converting the pyruvic acid to lactic acid (C3H6O3), which builds up in muscles as a result of reaction (3). This produces a lot less energy than aerobic metabolism, as we'll see below.

Lactic Acid

The process makes just 2 ATP instead of many more that would be produced if the pyruvic acid were metabolized aerobically by overall reaction (1).

How do food chemists calculate the energy produced in all these reactions?

By now you can imagine that there are innumerable reactions involved in just food metabolism, and it would be virtually impossible to list all the thermochemical equations, along with the corresponding enthalpy changes.

Fortunately Hess' law makes it possible to list just the standard enthalpy of formation ΔHf, for each compound, and use these ΔHf values to calculate the ΔHm for any reaction of interest.

Standard Enthalpy of Formation

The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).

For example, since H2O(l) appears in Equation (1), we'll need its ΔHf to calculate the energy available from a marshmallow. If we know that ΔHf [H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation


H2(g) + ½O2(g) → H2O(l)      ΔHfm = –285.8 kJ mol–1      (4)


The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed. Equation (1) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½.

Using Enthalpies of Formation to Calculate the Energy from Aerobic Metabolism of Sucrose

In addition to (4), we'll need two other ΔHf, values to calculate the energy in a marshmallow. They are the ΔHf values for the other compounds in Equation (1), CO2 and C12H22O11. All the ΔHfmvalues can be found in standard tables like the one at the end of this section, and we can write the equations (5) and (6) knowing the definition of ΔHf:


H2(g) + ½O2(g) → H2O(l)      ΔHfm = –285.8 kJ mol–1      (4)

C(s) + O2(g) → CO2(g)       ΔHfm = –393.509 kJ mol–1      (5)


12 C(s) + 11 H2(g) + 11/2 O2(g) → C12H22O11        ΔHfm = -2222.1 kJ mol–1       (6)


By Hess' Law, we may be able to combine equations 4, 5, and 6 to get Equation (1). First, we notice that (1) has sucrose on the left, but it's on the right in (6); so reversing (6) we get


C12H22O11 (s) → 12C(s) + 11 H2(g) + 11/2 O2(g)       -ΔHm = +2222.1 kJ mol–1       (6a)


To cancel the 12 C that does not appear in (1), we'll add 12 x Equation (5) (along with 6x its enthalpy change:


12 C(s) + 12 O2(g) → 12 CO2(g)       6 x ΔHm = 12 x (-393.509) kJ mol–1      (5a)


And to add the 11 H2O(l) that appears in (1), we'll add 11 x Equation (4):


11 H2(g) + 11/2 O2(g) → 11 H2O(l)      ΔHm4 = 11 x (–285.8) kJ mol–1      (4a)


If we combine Equations 6a, 5a, and 4a according to Hess' Law, we notice that 12 H2, 12 C, and 11/2 O2(g) appear on both the left and right, and cancel to give Equation (1)!

C12H22O11 (s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)    ΔHm1      (1)


We can then combine the enthalpies to get the needed ΔHm:


ΔHm = 12 ΔHm5 + 12 ΔHm4 - ΔHm6 =12 x (-393.509) + 11 x (–285.8) - (-2222.1)kJ mol–1 = -5643.8 kJ mol–1


Notice that this value appears in the Table at the end of this section. With Hess' Law, we can always calculate an enthalpy of combustion from enthalpies of formation, or vice versa! Reaction (6) corresponding to ΔHfm of sucrose does not occur, but its enthalpy can be calculated from enthalpies of reactions that do occur.

Notice that our calculation simplifies to:


ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)



The symbol Σ means “the sum of.” So we just need to add the ΔHf values for the products, and subtract the sum of ΔHf values for the reactants in Equation (1). Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient in from Equation (1) (for which ΔHm is being calculated).



Calories in a Marshmallow

Now we can calculate the food energy in the marshmallow: The molar mass of sucrose is 342.3 g/mol, so the energy per gram is -5643.8 kJ/mol / 342.3 g/mol = 16.49 kJ/g. In the 7.5 g marshmallow, remembering that 1 dietary Calorie is 4.184 kJ, we have 7.5 g x 16.49 kJ/g x (1 Cal / 4.184 kJ) = 29.6 Cal. (But who can stop at just one roasted marshmallow?)

Recap of the Calculation of a Reaction Enthalpy

Note carefully how the problem above was solved. In step 6a the reactant compound C12H22O11 (s) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH1 was opposite in sign from ΔHf. In step 5a we had the hypothetical formation of the product CO2(g) from its elements. Since 12 mol were obtained, the enthalpy change was doubled but its sign remained the same. In step 4a we had the hypothetical formation of the product H2O (l) from its elements. Since 11 mol were obtained, the enthalpy change was multiplied by 11, but its sign remained the same.

Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients.

Again, this can he summarized by the important equation

ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)

One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. That's why ΔHf for O2 doesn't appear in the calculation above; it's value is zero, corresponding to the formation of O2 from its elements. There's no change in the reaction below, so ΔHf = 0:

O2 (g) → O2 (g)      ΔHf = 0


Standard enthalpies of formation for some common compounds are given in the table below, and more are given in Table of Some Standard Enthalpies of Formation at 25°C. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.


Energy of Sucrose Hydrolysis in Saliva

EXAMPLE 1 Use standard enthalpies of formation to calculate ΔHm for the reaction

C12H22O11 (s) + 1 H2O(l) → C6H12O6 (glucose) + C6H12O6 (fructose)


Solution

ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)

From the table below, ΔHf for glucose, fructose, sucrose and water are -1271, -1266.6 (they may actually be the same, but measured by different methods), -2222.1, and -285.8 kJ mol–1 respectively. Note that we were careful to use ΔHf [H2O(l)] not ΔHf [H2O(g)] or (l). Substituting these values in the equation above gives

ΔHm = [1 mol glucose x (-1271 kJ mol–1) + 1 mol fructose x (-1266.6 kJ mol–1] - [1 mol sucrose x (-2222.1 kJ mol–1 + 1 mol water x -285.8 kJ mol–1] = -29.7 kJ mol–1.

The process is actually exothermic, releasing a small amount of heat energy. The measured energy for hydrolysis of maltose to 2 glucose units is only -4.02 kJ[3].

Energy in Glucose Metabolism and Anaerobic Formation of ATP

EXAMPLE 2 Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction below (glycolysis + fermentation), which is associated with the production of 2 mol ATP (as well as NADH) in anaerobic metabolism in your body. The ΔHf for lactic acid and glucose are -687 and -1271 kJ mol–1 respectively.

C6H12O6 (s) → 2 C3H6O3      ΔHm3      (3)


Solution


ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)

       = [2 ΔHf (C3H6O3) - [ΔHf (C6H12O6)]

       = [2 mol lactic acid x (–687) kJ mol–1] – [1 mol glucose x (–1271 kJ mol–1)

       = –1374 + 2222.1 kJ = -103 kJ.

This energy is used in part to make 2ATP molecules rather than being released entirely as heat. Note that -5643.8 kJ mol–1 resulted from the aerobic metabolism of sucrose (above) but only 2 (-103)kJ = -206 kJ would result from its anaerobic metabolism (since 1 mol of sucrose yields 2 mol of glucose).



Compound
ΔHf
kJ mol–1
ΔHf
kcal mol–1
ΔHc
kJ mol–1
H2O(g)
–241.818
–57.79
H2O(l)
–285.8
–68.3
H2O2(l)
–187.78
–44.86
CO(g)
–110.525
–26.41
CO2(g)
–393.509
–94.05
NH3(g)
–46.11
–11.02
C2H2(g)
+226.73
+54.18
C3H6O3
lactic acid
-687[4]
-164.08[5]
C3H4O3
pyruvic acid
–584.5[6]
C6H12O6
glucose
-1271[7]
+
–2803[8]
C6H12O6
galactose
–1286[9]-1286.3[10]
–2803.7[11]
C6H12O6
fructose
–1265.6[12]
–2812[13]
C12H22O11
sucrose
-2375 1[14]-2222.1[15][16]
–5645[17]–5646[18]-5644[19]
C12H22O11
maltose
–5644[20]
C6H12O6
lactose
−2236.7[21]
-
–5648[22]-5629.5[23]
C2H6O1
ethanol
-
–1367[24]
C6H14O6
sorbitol
–56441[25]
-
C18H34O2
oleic acid
–7721[26]
-
C18H30O2
linolenic acid
–6651[27]
-
C57H104O6
triolein
–23901[28]-2193.7[29]
-35224[30]-35099.6 [31]

1. Estimate, based on a theoretical calculation

The most general references are NIST, this bond energy based calculator and for QSPR calculated values, Int. J. Mol. Sci. 2007, 8, 407-432.

References

  1. http://en.wikipedia.org/wiki/Marshmallow
  2. http://www.campfiremarshmallows.com/Regular-Marshmallows.asp
  3. http://www.jbc.org/content/264/7/3966.full.pdf
  4. http://www.lactic.com/index.php/lacticacid
  5. http://www.lactic.com/index.php/lacticacid
  6. http://www.brainmass.com/homework-help/chemistry/physical-chemistry/11390
  7. http://en.wikipedia.org/wiki/Glucose
  8. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  9. http://www.brynmawr.edu/Acads/Chem/sburgmay/chem104/GW4answers07.html
  10. http://www.nist.gov/srd/PDFfiles/jpcrd719.pdf
  11. http://www.springerlink.com/content/y1143825t118916w/
  12. http://www.nist.gov/srd/PDFfiles/jpcrd719.pdf
  13. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  14. http://www.mdpi.org/ijms/papers/i8050407.pdf
  15. http://www.brynmawr.edu/Acads/Chem/sburgmay/chem104/GW4answers07.html
  16. http://webbook.nist.gov/cgi/cbook.cgi?ID=C57501&Units=SI&Mask=2#Thermo-Condensed
  17. http://home.fuse.net/clymer/rq/hoctable.html
  18. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  19. http://webbook.nist.gov/cgi/cbook.cgi?ID=C57501&Units=SI&Mask=2#Thermo-Condensed
  20. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  21. http://www.nist.gov/srd/PDFfiles/jpcrd719.pdf
  22. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  23. http://www.springerlink.com/content/y1143825t118916w/
  24. http://www.science.uwaterloo.ca/~cchieh/cact/applychem/propertyc.html
  25. http://www.mdpi.org/ijms/papers/i8050407.pdf
  26. http://www.mdpi.org/ijms/papers/i8050407.pdf
  27. http://www.mdpi.org/ijms/papers/i8050407.pdf
  28. http://www.mdpi.org/ijms/papers/i8050407.pdf
  29. http://webbook.nist.gov/cgi/cbook.cgi?ID=C122327&Units=SI&Mask=2#Thermo-Condensed
  30. http://home.fuse.net/clymer/rq/hoctable.html
  31. http://webbook.nist.gov/cgi/cbook.cgi?ID=C122327&Units=SI&Mask=2#Thermo-Condensed


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