# Average Atomic Mass

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```					Atomic Mass
 Agenda
 Review (whiteboards)
 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
6C
 ____ p+   6
 ____ e-   6
 ____ no   8
9       Be+2
4
 ____ p+      4
 ____ e-      2
 ____ no      5
36        Cl -
17
 ____ p+         17
 ____ e-         18
 ____ no         19
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.
 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

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 views: 56 posted: 10/26/2011 language: English pages: 18