Question #1d260

2 Answers
Oct 15, 2015

Because that is the average atomic mass of its two stable isotopes.


The average atomic mass of an element is determined using the atomic masses and the abundances of its stable natural isotopes.

As you know, the identity of a chemical element is given by its atomic number, i.e. by the number of protons it has in its nucleus.

For a given element, the number of neutrons it has in its nucleus can vary quite a lot. Atoms that have the same number of protons but different numbers of neutrons in their nucleus are called isotopes.

Each isotope will contribute to the average atomic mass of the element proportionally to its abundance.

Helium only has two stable isotopes, helium-3 and helium-4.

The atomic masses and abundances for these two isotopes are

  • #""^3"He" -> "3.0160293 u"" "# and #" "0.000137%# abundance
  • #""^4"He" -> "4.002602 u"" "# and #" "99.999863%# abundance

So, what will the average atomic mass of helium be? Notice that helium-3 has a very, very small abundance, which means that its atomic mass will contribute very, very little to the average.

#"avg. atom. mass" = sum_i ("isotope"_i xx "abundance"_i)#

In helium's case, you have - I'll use decimal abundances, which are percent abundances divided by #100#

#m_a = "3.0160293 u" xx 0.00000137 + "4.002602 u" xx 0.99999863#

#m_a = "0.000004132 u" + "4.0025965 u"#

#m_a = "4.002601 u" ~~ "4.00 u"#

The average atomic mass of helium can thus be approximated to be #"4.00 u"#.

Oct 15, 2015

The atomic weight (mean relative mass) is not #"4.00 u"#. It is #"4.002602 u"# as reported on the periodic table published by the National Institute of Standards and Technology (NIST), the Webelements Periodic Table with full significant figures, and Vertex42.


When calculations are carried out to determine the mean relative mass of helium, the result is #"4.002602 u"#. To report the average atomic mass of helium to only 3 significant figures is misleading and could cause a person to infer that helium has only one isotope.