Atomic Mass Agenda Review (whiteboards) Read 4.7 & 4.8 Notes Homework Parts of Atomic Theory Atoms of the same element are identical Atoms can rearrange – reactions Atoms can mix in whole number ratios – compounds Atoms are indivisible – no subatomic particles according to Dalton How do isotopes differ? Number of neutrons in nucleus How do ions differ? Number of electrons Cations have lost electrons to gain positive charge Anions have gained electrons to have negative charge 14 C 6 ____ p+ 6 ____ e- 6 ____ no 8 9 Be+2 4 ____ p+ 4 ____ e- 2 ____ no 5 36 Cl17 ____ p+ 17 ____ e- 18 ____ no 19 READ Grab book Read section 4.7 & 4.8 w/ partner Six inch voices Average Atomic Mass date Average atomic mass A WEIGHTED average dependent on the percent abundance of each isotope. If all isotopes were in a bag, how frequently would you pull out each separate isotope. Measured in amu’s Atomic Mass Unit An amu is the mass of a single proton. 1 amu = 1.67*10-24 g Determining Average Mass Multiply Mass number by Percent abundance. Add the answers. Answer should be close to most abundant isotope. Example Cesium is 75% 133Cs, 20% 132Cs, 5% 134Cs. What is the average atomic mass? 133*.75 132*.20 134*.05 132.85 amu Closest to the most abundant isotope Example Chromium has four istopes. 4.35% 50Cr, 83.79% 52Cr, 9.5% 53Cr, 2.36% 54Cr. What is the average atomic mass of Cr? 50*.0435 52*.8379 53*.095 54*.0236 52.0552 amu Determining % abundance Create an equation equaling average atomic mass Mass1 * %X + Mass2 * %Y = average atomic mass Create an equation equaling 100% X+Y=1 Use substitution to solve for x. Mass1 * %X + Mass2 * %(1-X) = average atomic mass The average atomic weight of boron is 10.812amu. Boron has two isotopes: boron-10 with a mass of 10.013 and boron-11 with a mass of 11.009. What are the % abundances of each isotope? Let one isotope be X and the other 1-X 10.013X + 11.009(1-x) = 10.812 10.013X + 11.009 – 11.009X = 10.812 0.197 = 0.996X 0.1978 = X Boron-10 is 19.78% abundant Boron-11 is 80.22% abundant What are the % abundances of the two isotopes of rubidium? Average atomic mass 85.4678 amu Rb-85, 84.9918 amu Rb-87, 86.9092 amu Let X be for Rb-85 and 1-X for Rb-87 84.9918X + 86.9092(1-X) = 85.4678 84.9918X + 86.9092 – 86.9092X = 85.4678 1.4414 = 1.9174X 0.7517 = X Rb-85 is 75.17% abundant Rb-87 is 24.83% abundant Homework Complete isotopes and average atomic mass worksheet Review scientists and discoveries. Lab report due Rough draft Wednesday Final draft Friday