Isotopes and Relative Atomic Mass
Dr. Davinder Bhachu
Teacher
Introduction & Definitions
Mass Number (A): The total number of protons and neutrons in the nucleus of an atom.
Atomic Number (Z): The number of protons in an atom; it defines the element.
Neutrons: Neutral subatomic particles in the nucleus. The number of neutrons can vary between atoms of the same element, forming isotopes.
Determining the Number of Fundamental Particles
For an atom or ion, the number of subatomic particles can be found using:
Neutrons = Mass Number (A) - Atomic Number (Z)
Electrons (for neutral atoms) = Protons
Electrons in Ions = Protons - Charge
Example: Find the number of protons, neutrons, and electrons in a sodium ion (Na⁺) with a mass number of 23.
Protons = 11 (from atomic number of Na)
Neutrons = 23 - 11 = 12
Electrons = 11 - 1 = 10 (due to the +1 charge)
Isotopes
Atoms of the same element with the same number of protons (Z) but different numbers of neutrons (A)
Example: Carbon-12 (12C) and Carbon-14 (14C) are isotopes of carbon. C-12 has 6 neutrons, C-14 has 8 neutrons. Both have 6 protons.
No answer provided.
Relative Atomic Mass
Isotopes exist in different proportions. For example, approximately 99% of all carbon atoms exist as carbon-12.
The average mass of all isotopes, accounting for their relative abundances, is the relative atomic mass found in the periodic table.
Definition:
The weighted mean mass of an atom of an element, compared to 1/12th the mass of a carbon-12 atom.
1/12th the mass of an atom of carbon-12 is defined as 1 and so used as the reference for other atomic masses.
Calculating Relative Atomic Mass ()
= ∑(isotopic mass×relative abundance)/∑(total abundance).
Worked Example 1: Relative Atomic Mass of Chlorine
Chlorine has two isotopes: (75%) and (25%).
Calculate the relative atomic mass ().
Answer:
= (35×75)+(37×25)/100 = 35.5
Worked Example 2: Relative Atomic Mass of Magnesium
Magnesium has three isotopes:
(78.6%)
(10.1%)
(11.3%)
Calculate the relative atomic mass ().
Answer:
= (24×78.6)+(25×10.1)+(26×11.3)/100 = 24.3
Solving Relative Atomic Mass Problems with Algebra
An understanding that the total percentage abundance must equal 100% can allow more challenging problems to be solved.
Example
A sample of an element X consists of three isotopes: , , and .
The relative atomic mass of X is 112.9. The relative abundance of is 60%.
Calculate the relative abundances of and
112.9 = [(112 x 60) + (114 x y) + (115 x z)]/100
60 + y + z = 100
Therefore, z can be expressed as 100-60-y = 40-y
11290 = (112 x 60) + (114 x y) + (115 (40 – y))
11290 = 6720 + 114y + 4600 – 115y
11290 = 11320 – y
30 = y
Abundances = 30%, = 40-y = 10%
No answer provided.
Interpreting a Mass Spectrum
The information required to calculate relative atomic mass is obtained from mass spectrometry.
The x-axis shows mass-to-charge ratio (m/z), which often equals the relative isotopic mass for singly charged ions (z = 1).
The y-axis shows relative abundance.
Used to identify elements, the relative abundances of isotopes in a sample and Ar.
For example, the mass spectrum for neon below
= [(20 x 90.9) + (21 x 0.3) + (22 x 8.8)]/100
= 20.2
