Isotopes

Isotopes are variants of a particular chemical element such that, while all isotopes of a given element share the same number of in each atom, they differ in neutron numbers. Thus, different isotopes of a single element occupy the same position on the periodic table.

The number of protons within the atom's nucleus is called atomic number (Z) and equals also to the number of electrons per atom. It identifies uniquely an element, but an atom of a given element may in principle have any number of  neutrons.

The number of nucleons (protons and neutrons) in the nucleus is the mass number (A), and each isotope of a given element has a different mass number.

The atomic number (Z) and mass number of an atom / element (A) can be displayed in standard notation.

     

Average Atomic Mass

Each element has at least two natural occuring isotopes, each with their own mass. To determine the atomic mass of an element you have to calculate the average atomic mass. Since the difference isotopes do not occur in the same abundances, you have to take into this into account. Therefore, the weigthed average is calculated.

Mass spectrometers are instruments that measure the charge-to-mass ratio of charged particles. A gas sample at very low pressure is bombarded with high-energy electrons. This causes electrons to be ejected from some of the gas molecules, creating positive ions. The positive ions are then focused into a very narrow beam and accelerated by an electric field toward a magnetic field. The magnetic field deflects the ions from their straight-line path. The extent to which the beam of ions is deflected depends on four factors:

The mass spectrometer is used to measure masses of isotopes as well as isotopic abundances, that is, the relative amounts of the isotopes

The distribution of isotopic masses, although nearly constant, does vary somewhat depending on the source of the element.

Determine at Atomic Mass

The atomic mass of an atom is a calculate weighted average. This take into account the mass of each isotope and also the abundance of each isotope. The following formula can be used to calculate the average atomic mass:

              Atomic Mass = Σ (Isotope Mass)(Isotope Abundance)
                                   = (mass of isotope 1)(abundance of isotope 1) + (mass of isotope 2)(abundance of isotope 2) + .......

Example: Calculate the average atomic mass of an element that contains 50.54% of an isotope with a mass of 79.0 u and 49% of an isotope with a mass of 81.0 u.              

 Atomic Mass = Σ (Isotope Mass)(Isotope Abundance)
                        = (mass of isotope 1)(abundance of isotope 1) + (mass of isotope 2)(abundance of isotope 2) + .......
                        = 79.00 u)(0.5054) + (81.00 u)(0.49)
                        = 39.9266 + 40.062
                        = 79.9892 u
                        = 79.99 u

The atomic mass of the atom is 79.99 amu
A sample of element X contains 75.53% of an isotope with a mass of 34.969 u, and 24.47% of an isotope with a mass of 36.966 u.
              
  • what is the average isotopic mass of element X?
  • Which element is most likely to be the element X, explain your answer

    .

    •  Atomic Mass = Σ (Isotope Mass)(Isotope Abundance)
                            = (mass of isotope 1)(abundance of isotope 1) + (mass of isotope 2)(abundance of isotope 2) + .......
                            = (34.969 u)(0.7553) + (36.966 u)(0.2447)
                            = 26.412 + 9.043
                            = 35.455 u

    The element is Chlorine

 

Chlorine has two nature isotopes: Cl - 35 (34.9689 a) and Cl - 37 (36.9659 u). If chlorine has an average atomic mass of 35.454 u, what is the abundance of each isoptope?

let x = abundance of Cl - 35 and y = abundance of Cl - 37

x + y = 1 therefore y = 1 - x

(Isotope Mass)(Isotope Abundance)
                        = (mass of isotope 1)(abundance of isotope 1) + (mass of isotope 2)(abundance of isotope 2) + .......
       35.454 u   =   (34.9689 u)(x) + (36.9659 u)(1 - x)
       35.454 u   =   34.9689x + (36.9659 - 36.9659x
          - 1.5119 = - 1.997x
                      x = 0.75709

y = 0.24291
therefore, Cl-35 makes up 75.709 % of the sample and Cl-37 makes up 24.291 % of the sample