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8/8/2013 Isotopes • The number of protons determine the element and rarely changes – If the number of protons changes the atom changes identity • The number of electrons can change in relation to the number of protons and determines charges – If the number of electrons changes the atom becomes either positive (lost electrons) or negative (gained) • When the number of neutrons change • Only the mass of the atom changes…. So it is just a heavier or lighter version of the same atom… we call them isotopes Mass Number - vs - Average Atomic Mass Isotopes • So why is atomic mass an average? – We have different isotopes of the same atom • They are the same type of atom but they have a different mass because they have a different number of neutrons • Isotopes - Have the same number of protons and a different number of neutrons Isotope Notations Mass Number: 1. # of Protons + # of Neutrons = Mass # 2. Is a whole number because you can not have parts of a proton or neutron 3. Used in hyphen notation and nuclear notation for isotopes Average Atomic Mass: 1. Average mass of all the isotopes a. Based on relative abundance (the more abundant an isotope is the more it contributes to the average) 2. Has numbers after a decimal because it is calculated 3. The atomic masses on the periodic table are average atomic masses Relative Abundance • Relative abundance – refers to the abundance of naturally occurring isotopes Hyphen Notation Mass Number Helium - 3 Name or Abbreviation Nuclear Notation Mass Number 3 He 2 Abbreviation Atomic Number Sample problems 1. Calculate the average atomic mass for Boron if 19.9% of Boron atoms are B-10 and 80.1% are B-11. – An example is Chlorine (Cl), which has 2 isotopes: 1) Chlorine – 35 2) Chlorine – 37 relative abundance = 75.7% relative abundance = 24.3% • Calculating the weighted Average atomic mass: – Multiply the mass # by the relative abundance for each isotopes, then add them all together 35 x 0.757 = 26.5 Multiply % by mass # 37 x 0.243 = 9.00 Add them all together Weighted Avg : 26.5 + 9.00 = 35.5 amu 1 8/8/2013 Sample problems • An unknown element has two common isotopes. The first isotope has a mass of 85 amu and a relative abundance of 72.2%. The mass of the second isotope is 87 and has a relative abundance of 27.8%. Calculate the average atomic mass of the unknown element. 2
