Atomic Structure Powerpoint #2

The Mass Spectrometer The mass spectrometer can be used to measure the mass of individual atoms. The mass of an individual hydrogen atom is 1.67 x 10-24g, and the mass of a carbon atom is 1.99 x 10-25g. The masses of all the elements are in the range of 10-24 to 10-22g. Since these numbers are so small, it makes sense to use relative masses instead. The relative atomic mass of an element, Ar, is the average mass of an atom of the element taking into account all its isotopes and their relative abundance, compared to one atom of carbon-12

The Mass Spectrometer The results of the analysis of the mass spectrometer are presented in the form of a mass spectrum. The horizontal axis shows the mass/charge ratio of the different ions, and the relative abundance of the ions is shown on the vertical axis. Ex. The mass spectrum of gallium shows that in a sample of 100 atoms, 60 have a mass of 69, and 40 have a mass of 71. Calculate the relative atomic mass of the element. Total mass= (60 x 69) + (40 x 71) = 6980 Average mass= total mass/ number of atoms 6980/100= 69.80

The Mass Spectrometer Example: Deduce the relative atomic mass of the element rubidium from the data given of 100 atoms. 77 706050403020100 23 85 86 87 88

The Mass Spectrometer Total mass = (85 x 77) + ( 87 x 23) = 8546 Relative atomic mass = average mass of atoms Total mass/ number of atoms 8546/100 = 85.46

Boron exists in two isotopic forms. 10B and 11B Boron exists in two isotopic forms. 10B and 11B. 10B is used as a control for nuclear reactors. Use your Periodic Table to find the abundances of the two isotopes. Solution: Consider a sample of 100 atoms. Let x be 10B atoms. The remaining atoms are 11B. Number of 11B atoms = 100 – x Total mass = 10x + (100 – x)11 = 10x + 1100 – 11x = 1100 – x Average mass= total mass/ # of atoms = 1100 – x/100 From the periodic table: the relative atomic mass of boron = 10.81 10.81 = 1100-x/100 1081 = 1100 – x X = 1100 – 1081 X = 19.00 The abundance for 10B= 19.00% and 11B= 81.00%

Questions to be answered Which ion would be deflected most in a mass spectrometer? A) 35Cl+ B) 37Cl+ C) 37Cl+2 D) (35Cl37Cl)+ 2) What is the same for an atom of phosphorus-26 and phosphorus-27? A) atomic # and mass # B) # of protons and electrons C) # of neutrons and electrons D) # of protons and neutrons 3) Use the periodic table to find the percentage abundance of neon-20, assuming that neon has only one other isotope, neon-22.

Electron Arrangement Electromagnetic radiation comes in different forms of differing energy. All electromagnetic waves travel at the same speed (c) but can be distinguished by their different wavelengths (λ). The distance between two successive crests (or troughs) is called the wavelength. The frequency (f) is the number of waves which pass a point in one second. The wavelength and frequency are related by the equation c=f λ where c= speed of light. c= 3 x 108m/s

Electron Arrangement