Relative atomic mass This is the average mass of all of the naturally occurring isotopes of an element. Example Chlorine has two common isotopes 35Cl and 37Cl. These have natural abundances of 75% and 25% respectively. The relative atomic mass is the average of all of these atoms. (75 x 35) + (25 x 37) = average atom 100 Relative atomic mass of chlorine = 35.5 Now try these 1. Boron has two common isotopes 10B and 11B with relative abundances of 20% and 80%. Calculate the relative atomic mass of boron. 2. Bromine has two isotopes 79Br and 81Br with relative abundances of 50% of each. Calculate the relative atomic mass of bromine. 3. Sulphur has three isotopes 32S and 34S and 36S with relative abundances of 80%, 16% and 4% respectively. Calculate the relative atomic mass of sulphur.
Pages to are hidden for
"Relative_atomic_mass"Please download to view full document