(I have now prefaced the slides for my talk on Sept. 5 with the solution
to the atomic mass problem from Sept. 3. The lecture on moles
follows those slides.)
Problem: Given the exact masses, 35Cl = 34.969mu and
37Cl = 36.966 u, and the average atomic mass of chlorine
= 35.453 u, find the abundances of the two isotopes.
Problem: Given the exact masses, 35Cl = 34.969 u and
37Cl = 36.966 u, and the average atomic mass of chlorine
= 35.453 u, find the abundances of the two isotopes.
Ave. atomic mass = a1m1 + a2m2 + . . .
Problem: Given the exact masses, 35Cl = 34.969 u and
37Cl = 36.966 u, and the average atomic mass of chlorine
= 35.453 u, find the abundances of the two isotopes.
Ave. atomic mass = a1m1 + a2m2 + . . .
35.453 u = a1(34.969 u) + a2(36.966 u)
Problem: Given the exact masses, 35Cl = 34.969 u and
37Cl = 36.966 u, and the average atomic mass of chlorine
= 35.453 u, find the abundances of the two isotopes.
Ave. atomic mass = a1m1 + a2m2 + . . .
35.453 u = a1(34.969 u) + a2(36.966 u)
Also: a1 + a2 = 1
Problem: Given the exact masses, 35Cl = 34.969 u and
37Cl = 36.966 u, and the average atomic mass of chlorine
= 35.453 u, find the abundances of the two isotopes.
Ave. atomic mass = a1m1 + a2m2 + . . .
35.453 u = a1(34.969 u) + a2(36.966 u)
Also: a1 + a2 = 1
So 35.453 u = (1 – a2)(34.969 u) + a2(36.966 u)
So 35.453 u = (1 – a2)(34.969 u) + a2(36.966 u)
35.453 u = 34.969 u – 34.969 u(a2) + 36.966 u(a2)
35.453 u = 34.969 u – 34.969 u(a2) + 36.966 u(a2)
35.453 u = 34.969 u + 1.997 u(a2)
a2
0.2424
and a1 = 1 – 0.2424 = .7576
How to Count Atoms
(when they are really really small)
Review:
1 12C atom = 12.0000 u = 1.9926x 10-23 g
(from mass spectroscopy experiments)
So
19926
.
 10 23 g
 16605
.
 10 24 g / u
12.0000 u
Goal: a number that is capable of expressing
numbers of atoms in convenient terms.
H and O react in simple numbers of atoms to form
water (H2O):
2 H : 1 O There is not a simple relation between the
masses that react.
Yet – masses are what we can readily measure
in a laboratory.
To get a quantity capable of expressing numbers of
atoms,
Define: 1 mole = number of atoms in 12.0000 g of
12C.
We now have enough information to calculate the
number of atoms in one mole of carbon:
 1 atom 12 C 
23
12
12
.
00
g
C

6
.
022

10
atoms
C


 23 
 19926
.
 10 g 
But this is also the number of atoms in one atomic mass,
expressed in grams, of any element.
This important number is known as Avogadro’s Number,
NA.
Both a mole and a dozen express quantities of things
by a collective number.
A dozen eggs weighs more than a dozen ping-pong
balls because each individual unit is heavier.
16.0 amu
12.0 amu
O
C
192.0 amu in carton
144.0 amu in carton
(A dozen oxygen atoms
weigh more than a dozen
carbon atoms)
Take enough cartons to have
6.02 x 1023 atoms
The total mass will be 12.0 gram
Take the same number of atoms
as at the left.
The total mass will be 16.0 grams
Moral: a mole of oxygen weighs more than a mole of carbon because
each individual oxygen atom weighs more.
So now, atomic masses have two interpretations:
1 atom of C weighs 12.0000 u
1 mole of C weighs 12.0000 g
Some conversion factors:
1 mole of C = 12.00 g
1 mole of C = 6.02 x 1023 atoms C
Now we can calculate the mass of any atom:
1 atom of U = 238 u


1mole U
24
1 atom U

 1.66x10 moles U
23
 6.02x10 atoms U
1.66x10
 24
 238 g U 
22
moles U 
  3.95  10 g U
 1mole U

Now we can calculate the mass of any atom:
1 atom of U = 238 u

  238 g U 
1 mole U
22

3.95

10
g
1 atom U



23
 6.02  10 atoms U  1 mole U
Find the number of carbon atoms in 3.0 g C.
 1mole C   6.02  10 23 atoms C 
  1.5  10 23 atoms C
3.0 g C 12.0 g C  
1mole C



A mole road-map
Atomic
mass
NA
atoms
Moles
grams
This can be applied to compounds, which have a
molecular mass (in amu) or a molar mass (in grams).
For instance:
In one molecule of H2O, there are
2 H atoms = 2 x 1 amu = 2 u
1 O atom = 1 x 16 amu = 16 u
18 u
1 mole of H2O has 1 mole of O atoms and
2 moles of H atoms.
1 mole of water weighs 18 g and has 6.02 x 1023 molecules
For Ca(OH)2:
1 Ca = 40.1 u
2 O = 32.0 u
2 H = 2.0 u
74.1 u
1 mole of Ca(OH)2 has a mass of 74.1 g
Ca(OH)2 is an ionic compound so we can’t talk about
molecules. Sometimes the simplest formula – which
indicates only the ratio of the ions – is referred to as a
formula unit.
How many atoms of O are there in 5.0 g of CO2?
CO 2 =
1 C = 12 u
2 O = 2x16 = 32 u
= 44 u
 1mole CO 2   2 moles O   6.02  10 23 atoms O 

 5.0 g CO 2  44 g CO   1mole CO  
1mole O


2 
2 
 1.4  10 23 atoms O