"ifoods": Eat Isotopes to Live Longer - ChemPRIME

# "ifoods": Eat Isotopes to Live Longer

back to Average Atomic Weights

You may have seen or heard the hype about "ifoods", or nonradioactive, "isotopic foods", that are purported to have the potential to combat ageing.

Is there any truth to this claim?

All naturally occurring atoms of a given element do not necessarily have identical masses. For example, hydrogen is about 99.98% 11H ("proteum") and 0.02% 21H ("deuterium", symbolized "D"). There are also traces of naturally occurring tritium, 31H. 11H and 21H are stable (non-radioactive), but 31H is radioactive. The percentages are somewhat variable due to a small difference in the physical and chemical properties of the isotopes, caused by the "kinetic isotope effect". For example, the uptake of 21H compounds may be enhanced in some plants, while 11H is concentrated in gaseous species. The kinetic isotope effect is especially important for hydrogen because 21H is 100% heavier than 11H; other elements have small kinetic isotope effects, which can usually be ignored (so we can say that all atoms of an element are chemically identical).

Elements 1 to 29 in the Table of the Nuclides, a plot of # neutrons vs. # protons for all known nuclides. Stable isotopes are in blue.

The fact that bonds to deuterium are broken a little more slowly than bonds to hydrogen led M.S. Shchepinov[1] to suggest that increasing the deuterium content in our bodies might lead to resistence to the free radical attack that causes aging. A company ("Retrotope"[2]) was formed to investigate the production of "ifoods" or isotopic foods containing the heavy stable isotopes 21H, 136C, or 157N. One source of deuterium would be "heavy water", deuterium oxide (D2O). Some sources claim that deuterium oxide is toxic, but it only becomes toxic at levels where it replaces over about 50% of the water in our bodies (which are 70% water, so we'd have to drink many pounds of D2O)[3].

## What's "Normal" Hydrogen, and What's "Deuterium-enriched"?

If hydrogen exists as a mixture of isotopes, how can we assign just one "atomic weight" to "normal" hydrogen? That's possible because the isotopic concentration of ocean water is close to constant[4], and because most hydrogen compounds (excepting gas, and artificially enriched samples) have very close to this composition, we can use the weighted average mass of the isotopes as a single "atomic weight":

A weighted average is defined as

Atomic Weight =
$\left(\tfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right)~ ~ ~ +$

$\left(\tfrac{\%\text{ abundance isotope 2}}{100}\right)\times \left(\text{mass of isotope 2}\right)~ ~ ~ + ~ ~ ...$

Similar terms would be added for all the isotopes. The calculation is analogous to the method used to calculate grade point averages in most colleges:

GPA =
$\left(\tfrac{\text{Credit Hours Course 1}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 1}\right)~ ~ ~ +$

$\left(\tfrac{\text{Credit Hours Course 2}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 2}\right)~ ~ ~ + ~ ~ ...$

## Example: The Atomic Weight of Hydrogen

Calculate the atomic weight of hydrogen given the following average compostion of the element:

Isotope Mass (amu) Abundance (%)

11H

1.007 825 0320 7(10)[5] 99.988 5(70)

21H

2.014 101 777 8(4)[6] 0.011 5(70)

31H

3.016 049 277 7(25)[7] negligible
(1 in 1016 atoms

Solution Atomic Weight =

$\left(\tfrac{\text{99.9885}\%}{100}\right)\times \left(\text{1.00782503207 amu}\right)~ ~ ~ +$

$\left(\tfrac{\text{0.0115}\%}{100}\right)\times \left(\text{2.0141017778 amu}\right)~ ~ ~ + ~ ~ ...$

$\left(\tfrac{\text{0}\%}{100}\right)\times \left(\text{3.0160492777 amu}\right)~$

= 1.007 9 amu

This is the atomic weight given on most periodic tables. It is much closer to the mass of 11H than 21H because 11H has a much larger natural abundance.

The hydrogen atoms in gaseous hydrogen tend to be enriched in 11H, and some biological species enrich hydrogen in one isotope or the other.

## "Heavy Water", D2O

Pure D2O is prepared by an expensive and energy-intensive process of repeated distillation of normal water or by chemical reactions influenced by the kinetic isotope effect[8]. It's boiling point is 101.4 °C or 214.56 °F, and a density of 1.1056 g/mL at O°C compared to 0.9982 g/ml for normal water. Normal water contains 2 x 1.0079 amu of hydrogen per average oxygen atom (15.9994 amu), but deuterium oxide contains 2 x 2.01410 amu of hydrogen per atom of oxygen. Deuterium oxide weighs about 11% more than normal water. The cost in 2010 for deuterium oxide is about \$1/g [9].

## ifoods

The relatively high cost of D2O makes it a fairly expensive everyday food. But the main problem with ifoods may be that it is difficult to target the replacement of H with D in biomolecules at the bonds which are subject to attack by ageing agents like free radicals. Replacing a small percentage of the H with D may not be very effective. Since the kinetic isotope effect is much smaller for 136C and 157N, these would be even less effective in ifoods. Furthermore, ageing is the result of a multitude of processes in addition to free radical attack.

## Standard Atomic Weights

After the possibility of variations in the isotopic composition of the elements was recognized, it was suggested that the scale of relative masses of the atoms (the atomic weights) should use as a reference the mass of an atom of a particular isotope of one of the elements. The standard that was eventually chosen was 126C, and it was assigned an atomic-weight value of exactly 12.000 000. Thus the atomic weights given in the Table of Atomic Weights are the ratios of weighted averages (calculated as in the Example) of the masses of atoms of all isotopes of each naturally occurring element to the mass of a single 126C atom. Since carbon consists of two isotopes, 98.99% 126C isotopic weight 12.000 and 1.11% 136C of isotopic weight 13.003, the average atomic weight of carbon is

$\frac{\text{98}\text{.89}}{\text{100}\text{.00}}\text{ }\times \text{ 12}\text{.000 + }\frac{\text{1}\text{.11}}{\text{100}\text{.00}}\text{ }\times \text{ 13}\text{.003}=\text{12}\text{.011}$

for example. Deviations from average isotopic composition are usually not large, and so the average atomic weights serve quite well for nearly all chemical calculations. In the study of nuclear reactions, however, one must be concerned about isotopic weights. This is discussed further in the section on Nuclear Chemistry.

The SI definition of the mole also depends on the isotope 126C and can now be stated. One mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in exactly 0.012 kg of 126C. The elementary entities may be atoms, molecules, ions, electrons, or other microscopic particles. This official definition of the mole makes possible a more accurate determination of the Avogadro constant than was reported earlier. The currently accepted value is NA = 6.02214179 × 1023 mol–1. This is accurate to 0.00000001 percent and contains five more significant figures than 6.022 × 1023 mol–1, the number used to define the mole previously. It is very seldom, however, that more than four significant digits are needed in the Avogadro constant. The value 6.022× 1023 mol–1 will certainly suffice for most calculations needed.