# Atomic Scale Paleontology

- Page ID
- 50773

The Earth is about 4 billion years old, but what evidence is there for the earliest time when * life* appeared on Earth? There are several different

*biomarkers*or “signatures of life” that geologists use when analyzing samples for early life, but among the most important is the \({}_{\text{6}}^{\text{12}}\text{C}\) to \({}_{\text{6}}^{\text{13}}\text{C}\) ratio

^{[1]}.

## Isotopes and Isotopic Masses

In carbonate rocks, the two naturally occuring isotopes of carbon are found in the ratio 98.99% \({}_{\text{6}}^{\text{12}}\text{C}\) with isotopic weight 12.000 and 1.11% 136C of isotopic weight 13.003. Carbonate rocks are abiotic (not formed from plant or animal remains). But in samples containing microbial remains, the 126C to 136C ratio is significantly higher.

**Figure**\(\PageIndex{1}\)

*Burgess Shale with Marrella, the most abundant Burgess Shale organism, see*

^{[2]}In the environment, the element carbon is mostly \({}_{\text{6}}^{\text{12}}\text{C}\) so the Atomic weight of carbon (see below) is 12.011 amu. If all the carbon in an average person's body were 136C, however, the person would weigh about 2 lb more. But the presence of 136C would otherwise make very little difference, because isotopes are virtually identical in all their chemical properties. One property is slightly different: 136C reacts more slowly, so over time, living cells become enriched in C-12. The 136C deficiency biomarker can be preserved for billions of years in rock.

Geologists use a scale for the level of deficiency in \({}_{\text{6}}^{\text{13}}\text{C}\) : 1% deficiency of C-13 is called “-10 per mil”, 2% drop is “-20 per mil”, and so forth (1% = 0.010, and "mil" means 1/1000, so 0.010 is 10 mil). Various samples have the following deficiencies:

**Table** \(\PageIndex{1}\) *Samples with Deficiencies*

Sample |
\({}_{\text{6}}^{\text{13}}\text{C}\) deficiency |
---|---|

Mammoth bones | -21 |

3.1 billion year old microbial mats in South Afcrican Sandstones | -25 to -27 |

Fossil coal | -24 to -25 |

Burgess Shale British Columbia 540 million years old | -25 to -27 |

Photosynthetic microbes (< 2 billion years old) | -20 to -30 |

More primitive single celled organisms (in 3.8 billion year old sediments) | -50 |

Ancient microbial samples look like black splotches in limestone, shale, or other sedimentary rock. The island of Akilia near Greenland has 3.85 billion year old rocks with \({}_{\text{6}}^{\text{13}}\text{C}\) deficiencies of -37 recorded by Stephen J. Mojzsis (*Nature*, November 7, 1996). Mojzsis interpreted the intensely folded Akilia rock as metamorphized oceanic sediments. Was this the earliest life? Christopher Fedo and Martin Whitehouse found the rocks to be an ancient molten igneous mass that solidified from 1000^{o}C, so no living matter would have survived in them. The 136C deficiencies must be contamination, or result from another geologic process that also fractionates carbon isotopes. Clearly, geological interpretation must be used to determine if 136C deficiency is a reliable biomarker.

But geologists must understand how *Atomic Weights* are determined, because without atomic weights, we could not recognize \({}_{\text{6}}^{\text{13}}\text{C}\) (or other) deficiencies. Mass spectrometry is used to determine the masses and percent abundances of naturally occurring isotopes, and the atomic weight of the element is calculated as shown below. Mass spectrometry is used to determine the isotope deficiencies in biogenic samples).

## Atomic Weights

All atoms of a given element do not necessarily have identical masses. But all elements obey the law of definite proportions, so they behave *as if* they had just one kind of atom. In order to solve this dilemma, we define the * atomic weight* as the weighted average mass of all naturally occurring (occasionally radioactive) isotopes of the element.

A weighted average is defined as

Atomic Weight =

\(\left(\dfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right)~ ~ ~ +\)

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

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 =

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

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

## Atomic Weight of Carbon

Since carbon consists of two isotopes, 98.99% \({}_{\text{6}}^{\text{12}}\text{C}\) isotopic weight 12.000 and 1.11% \({}_{\text{6}}^{\text{13}}\text{C}\) of isotopic weight 13.003, the average atomic weight of carbon is

\(\dfrac{\text{98}\text{.89}}{\text{100}\text{.00}}\text{ }\times \text{ 12}\text{.000 + }\dfrac{\text{1}\text{.11}}{\text{100}\text{.00}}\text{ }\times \text{ 13}\text{.003}=\text{12}\text{.011}\)

Example \(\PageIndex{1}\): Atomic Weight of Lead

Naturally occurring lead is found to consist of four isotopes:

1.40% \({}_{\text{82}}^{\text{204}}\text{Pb}\) whose isotopic weight is 203.973.

24.10% \({}_{\text{82}}^{\text{206}}\text{Pb}\) whose isotopic weight is 205.974.

22.10% \({}_{\text{82}}^{\text{207}}\text{Pb}\) whose isotopic weight is 206.976.

52.40% \({}_{\text{82}}^{\text{208}}\text{Pb}\) whose isotopic weight is 207.977.

Calculate the atomic weight of an average naturally occurring sample of lead.

**Solution**

Atomic Weight =

\(\dfrac{\text{98.93}}{\text{100.00}} \times \text{203.973} + \dfrac{\text{24.10}}{\text{100.00}} \times \text{205.974} + \dfrac{\text{22.10}}{\text{100.00}} \times \text{206.976} +\dfrac{\text{52.40}}{\text{100.00}} \times \text{207.997} = \text{207.22}\)

Thus the atomic weight of lead is 207.2, as mentioned earlier in the discussion.

We have seen that atomic weights may vary slightly depending on whether a sample of an element is biogenic or not. This can occur for many other reasons. For example, lead derived from transmutation of uranium contains a much larger percentage of \({}_{\text{82}}^{\text{206}}\text{Pb}\) than the 24.1 percent given in the example for the average sample. Consequently the atomic weight of lead found in uranium ores is less than 207.2 and is much closer to 205.974, the isotopic weight of \({}_{\text{82}}^{\text{206}}\text{Pb}\).

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, rather than on the atomic weight of oxygen (which had been convenient for early chemists who used combining weights to determine atomic weighs of elements that reacted with oxygen). The standard that has been chosen is \({}_{\text{6}}^{\text{12}}\text{C}\) , and it is 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 \({}_{\text{6}}^{\text{12}}\text{C}\) atom.

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 \({}_{\text{6}}^{\text{12}}\text{C}\) 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 \({}_{\text{6}}^{\text{12}}\text{C}\) . 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

*N*= 6.02214179 × 10

_{A}^{23}mol

^{–1}. This is accurate to 0.00000001 percent and contains five more significant figures than 6.022 × 10

^{23}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× 10

^{23}mol

^{–1}will certainly suffice for most calculations needed.

From ChemPRIME: 4.13: Average Atomic Weights

## References

## Contributors and Attributions

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.