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Lecture 3: (Lec3A) Atomic Theory
Sections (Zumdahl 6th Edition) 3.1-3.4
Outline:
•Mass of Atoms
•The Concept of the Mole
•The mass of a mole of atoms and the mass of a mole of
molecules
• The composition of a molecule by number and weight.
Ch 3 Problems: 3.23-25, 3.27, 3.30-31, 3.33-34, 3.36, 3.37-39.
3.41, 3.43-45, 3.48, 3.52
Discussion: 3:1-2, 3.10-16, 3.14**, 3.19, 3.21
**Tough but cool problem
How do we know how much mass an atom contains?
3.1)
Determine the Average Atomic mass of Cu from the
Mass Spec analyzer
MOLE – just a number, like a dozen
• The Mole is a number like a dozen (12 items) Or a gross (144
items); it is Avogadro’s number of items.
• One mole of anything then is NA of that thing.
1 Mole of items = 6.022 ⋅1023 items
1 Gross of eggs = 1.44 ⋅102 eggs
1 Mole of eggs = 6.022 ⋅1023 eggs
1 = 6.022 ⋅1023 eggs
= 6.022 ⋅1023 eggs
Mole of eggs
Mole of eggs
N A = 6.022 ⋅1023
mole
=1
• The Mole is exactly the number of carbon-12 (i.e., 12Co) atoms
that weigh 12 grams. Therefore the Mole is that number of items,
of any type of thing. We can have a mole of stars for example or
a mole of Carbon-12 atoms, or a mole of H2 molecules.
Using NA
• Have 2 moles of water, how many molecules?
This is correct: x = 2 moles This is incorrect: x moles = 2
This is correct: N A = 6.022 ⋅1023 molecules
moles of molecules
1 mole (of molecules ) = 6.022 ⋅1023 molecules
therefore
1 = N A = 6.022 ⋅1023 molecules
mole (of molecules)
x = 2 moles ⋅ (1) = 2 moles ⋅ ( N A ) = 12. ⋅1023 ( molecules )
The value of x changes only as the units of x change.
How big (volume) is 2 moles of water?
Dalton: Atomic Mass Unit, amu
• One 12C atom weighs 12 amus
• One mole of 12C atoms weighs 12 grams
mass − mole 12C = 12 gram
mass 12C = 12 amu
Divide (Ratio the two)
mass − mole 12C 12 gram
gram
=
=1
12
mass C
12 amu
amu
Cancel Units
gram
1 mole = 1
⇒ 1gram = 1 mole − amu
amu
gram
1
= 1 mole ⋅ N A = 1 mole ⋅ 6.022 ⋅10+23
mole
amu
gram
1
= 6.022 ⋅10+23
⇒ 1gram = 6.022 ⋅10+23 amu
amu
Summary: Avogadro’s Number
N A = 6.02214 ⋅10
+23
Is a conversion factor
between items (molecules) and moles
AND
between grams and amu (or Daltons):
mol
The point of Avogadro’s number
NA is so useful because we can talk about molecules
reacting and having weights in UAMUs (or Daltons),
and then that means we can talk about moles of
molecules reacting having weights in grams.
1 g r a m = 6 .0 2 2 ⋅ 1 0 + 2 3 a m u
1 m o l e ( o f M o l e c u l e s ) = 6 .0 2 2 ⋅ 1 0 + 2 3 M o l e c u l e s
g r a m s 6 .0 2 2 ⋅ 1 0 + 2 3 a m u
m o le
y = x
⋅
⋅
m o le
gram
6 .0 2 2 ⋅ 1 0 + 2 3 M o l e c u l e
am u
= x
M o le c u le
gram s
am u
5
= 5
m o le
M o le c u le
If you have one mole of a substance that weights 5 grams, then one
molecule of the substance weights 5 Daltons (UAMU).
Compare Avogadro’s Number with the
number of stars in the universe
How many stars are there in the universe? -- There are about 10
billion (10^10) stars in an average galaxy, and there are about
10 billion galaxies that we can observe in the universe, so the
answer is 100 billion-billion or 10^20. John Hawley (Argonne)
How many universes do we need to get Avogadro's number of
stars? (6,000).
Avogadro’s number of anything is truly inconceivably large!!!
How many nucleons are there in our Sun (roughly)?
Our sun is about 2 ⋅1033 g
The Mass/Mole relationship
12 red marbles @ 7g each = 84g 55.85g Fe => 6.022 x 1023 atoms Fe
12 yellow marbles @4g each=48g 32.07g S => 6.022 x 1023 atoms S
Each element has its own atomic mass, on the periodic table.
Why is Carbon not 12.00?
Reminder: Generally the atomic mass is very close to the
mass number.
Why is Chlorine so far from an atomic mass number?
A Reminder: “Average” Atomic Mass of an Element
Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24Mg ( 78.7%);
25Mg (10.2%); 26Mg (11.1%).
24Mg
(78.7%)
25Mg (10.2%)
26Mg (11.1%)
23.98504 amu x 0.787 = 18.876226 amu
24.98584 amu x 0.102 = 2.548556 amu
25.98636 amu x 0.111 = 2.884486 amu
24.309268 amu
With Significant Digits = 24.3 amu
Mass/Mole Relationship Atoms/Elements
Atomic masses (or weights) are conversion factors to go from
number to mass (for atoms).
1 atom ( H ) = 1.008 amu
∴ AW ( H ) = 1 = 1.008 amu
= 1.008 amu
atom
= 1.008 amu
6.022 ⋅1023 atoms
NA
atom N
A
mole
atom 6.022 ⋅1023 amu
gram
= 1.008 g
AW ( H ) = 1 = 1.008 amu
atom
mol
1 atom ( Fe ) = 55.85 amu
AW ( Fe ) = 55.85 amu
1 mol ( Fe ) = 55.85 g
atom
= 55.85 g
mol
Molecular Mass vs Molar Mass ( MW )
The Molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the
compound expressed in grams , called its molar mass.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecules of water = 18.02 amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
= 2 ( 1.008 g ) + 16.00 g = 18.02 g
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
Use Atomic Masses to obtain Molecular Mass
• Use atomic masses to compute the mass of
a molecule.
• The periodic table gives us the average
mass of each atom type (“atomic weight”)
• We need to use the molecular formulas and
learn how to compute the molecular masses
from them.
M W ( H 2O ) = 2 ⋅ AW ( H ) + AW ( O )
Calculate the Molecular Mass of Glucose:
C6H12O6
M W ( C6 H12O6 ) = 6 ⋅ AW ( C ) + 12 ⋅ AW ( H ) + 6 ⋅ AW ( O )
M W ( C6 H12O6 ) = 180.16 g
• Carbon
mol
= 180.16 amu
6 x 12.01 amu =
molecule
72.06 amu
• Hydrogen 12 x 1.008 amu = 12.096 amu
• Oxygen
6 x 16.00 amu =
96.00 amu
180.16 amu
A shortcut: carbohydrate formula is (CH2O)6. What does one mole
of glucose weigh?
Masses of 1 mole of
Common substances
CaCO3 100.09 g
Oxygen 32.00 g
Copper 63.55 g
Water 18.02 g
Note: This balloon
of oxygen (O2) gas
is a bit too small.
It should be >20 L
unless highly
pressurized.
Know Atomic and Molecular
Masses
• We have reproduced Cannizzaro’s solution
to the puzzle of molecular formulae.
• If we know the masses then we can infer the
molecular formula/structure (the
stoichiometry)
• If we know the molecular formula we can
infer the atomic masses.
The Percent Composition of Molecules
• Know the masses of each type of atom
• So you know the total mass
• Can compute the fractional or percentage of each
atom type in the molecule.
• Eg: Glucose, C6H12O6, done above is 180 Da total, 72 Da
C, 12 Da H, 96 Da O. So the percent of this molecule that
is carbon, by weight (called the mass percent):
g
m ( C ) = 6 ⋅ AW ( C ) = 6 ⋅12
mol
g
= 72
mol
m (C )
72
fC =
=
= 0.40 C
Glucose
MW
72 + 12 + 96
= 40% C by mass
The RYP “molecule”
Atomic Masses given on each atom
What is the molecular Formula: R3 Y3 P2
Red, Yellow, Purple
Mass Fraction and Mass %
Total Mass = Mass Red +Mass Purple + Mass Yellow
Mass of Red Balls =
Mass Fraction Red =
Mass % Red =
Mass Fraction Purple =
Similarly, mass fraction yellow =
Check:
Mass Fraction and Mass %
Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g
Mass Fraction Red = 9.0 g / 16.0 g total = 0.56
Mass % Red = 0.56 x 100% = 56% red
Mass Fraction Purple =
2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25%
Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19
Check: 56% + 25% + 19% = 100%
Why do we need to go between moles and grams?
• Molecules go together as atoms (or moles)
• Molecules can only be weighed to
determine amount.
• We need to convert back and forth because
reactions preserve atoms (or moles of
atoms)
• Molecules represent the rearrangement and
joining together of numbers of atoms of
different types (or moles of atoms).
Extra Information
• More Information and extra problems.
• When working extra problems (or problems
in the text): Be sure to cover up the solution
before you start the problem.
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x
4 mol H x
3 mol O x
Molar mass = M =
%N =
%H =
%O =
g N x 100%
g cpd
gH
x 100%
g cpd
=
=
g O x 100% =
g cpd
Check:
100% total ?
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g/mol = 4.032 g H
3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
This is the molecular weight:
%N =
x 100%
=
%H =
x 100% =
%O =
x 100% =
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g/mol = 4.032 g H
3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
%N = 28.02g N x 100%
80.05g
= 35.00%
%H =
4.032g H x 100% =
80.05g
%O =
48.00g O x 100% = 59.96%
80.05g
99.997%
5.037%
2 H2(g) + O2(g) → 2 H2O(g)
2 dozen H2 molecules react with exactly
1 dozen O2 molecules to give exactly
2 dozen H2O molecules.
2 mole of H2 molecules react with exactly
1 mole of O2 molecules to give exactly
2 moles of H2O molecules.
Why do we do this? Because these last sizes are
in the gram range and easy to weigh.
Conventions: 1 mole of 12C atoms weighs 12 g exactly.
1 atom of 12C weighs 12 Da (amu) exactly.
(amu = atomic mass unit = ~mass of a proton or neutron)
2 H2(g) + O2(g) → 2 H2O(g)
2 dozen H2 molecules react with exactly
1 dozen O2 molecules to give exactly
2 dozen H2O molecules.
2 mole of H2 molecules
4 x 1.008 g = 4.032 g
2 x 16.00 g = 32.00 g
1 mole of O2 molecules
2 moles of H2O molecules
2 x (2x1.008 g + 16.00 g) = 36.03 g
Why do we do this? Because these last sizes are
in the 1-1000 gram range and easy to weigh.
Conventions: 1 mole of 12C atoms weighs 12 g exactly.
1 atom of 12C weighs 12 amu exactly.
(amu = atomic mass unit = ~mass of a proton or neutron)
Summary: AMU, Dalton, 12C Std.
Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom
on this scale Hydrogen has a mass of 1.008 AMU.
Dalton (D) = The new name for the Atomic Mass Unit,
one dalton = one Atomic Mass Unit
on this scale, 12C has a mass of 12.00 daltons.
Isotopic Mass = The relative mass of an Isotope relative to the
12C isotope standard = 12.00 amu.
Atomic Mass or “Atomic Weight” of an element = the average of the
masses of its naturally occurring isotopes weighted
according to their abundances.
The “Average” Atomic Mass of an Element
Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24Mg ( 78.7%);
25Mg (10.2%); 26Mg (11.1%). Data from Mass Spec.
24Mg
(78.7%) 23.98504 amu
25Mg (10.2%) 24.98584 amu
26Mg (11.1%) 25.98636 amu
Total =
With Significant Digits =
amu
.
Calculating the Moles and Number of Molecules or
Formula Units in a given Mass of a known Cpd.
Problem: Sodium Phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We determine the formula from the name, and the molecular mass
from the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na3PO4. Calculating the molar mass:
M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen =
= 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol
= 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol
Converting mass to moles:
Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol of Na3PO4)
163.94 g Na3PO4
= 0.23545 mol Na3PO4
Number of molecules (Formula units)
= 0.23545 mol Na3PO4 x 6.022 x 1023 formula units/mole
1 mol Na3PO4
= 1.46 x 1023 formula units
g = M⋅m
g
m
n
n = NA ⋅ m
.
Problem: Estimate the abundance of the two Bromine isotopes,
given that the average mass of Br is 79.904 amu. Since exact
masses of isotopes not given, estimate from: mass in amu = #p+ + #n:
79Br = 79 g/mol and 81Br = 81 g/mol (approximately).
Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0
Solution:
X(79) + Y(81) = 79.904
X + Y = 1.00 therefore X = 1.00 - Y
(1.00 - Y)(79) + Y(81) = 79.904
79 - 79 Y + 81Y = 79.904
2 Y = 0.904 or 1 with/ sig. figs.
so Y = 0.5
X = 1.00 - Y = 1.00 - 0.5 = 0.5
%X = % 79Br = 0.5 x 100% = 50% (Actual: 50.67% = 79Br)
%Y = % 81Br = 0.5 x 100% = 50% (Actual: 49.33% = 81Br)
Mole - Mass Relationships of Atoms/Elements
Element
Atomic Mass
Molar Mass Number of Atoms
1 atom of H = 1.008 Da 1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 u 1 mole of Fe = 55.85 g = 6.022 x 1023atoms
1 atom of S = 32 Da
1 atom of O = 16
u
1 mole of S =
1 mole of O =
Molecular mass:
1 molecule of O2 =
1 mole of O2 =
1 molecule of S8 =
1 mole of S8 =
g =
g=
atoms
atoms
amu
g =
molecules
amu
g =
molecules