Download Mass to Atoms - River Dell Regional School District

The Atom
Chapter 4
http://www.youtube.com/watch?v
=Uy0m7jnyv6U
I.
II.
III.
IV.
History of the Atomic Theory
A. Democritus
B. Aristotle
C. Lavoisier
D. Proust
E. Dalton
F. Modern Atomic Theory
History of Atomic Structure
A. Thomson
B. Milikan
C. Rutherford
D. Bohr
E. Chadwick
F. Quantum Atom
Subatomic Particles
A. Atomic Number
B. Mass Number and Isotopes
C. Electrons and Ions
D. Nuclear and Hyphenation Notation
E. Average Atomic Mass
Weighing and Counting Atoms
A. Mole
Atoms
B. Mole
Mass
C. Mass Atoms
I. History of the Atomic Theory
Remember: a scientific theory explains
behaviors and the ‘nature’ of things
 Theories can be revised when new
discoveries are made
 The theory describing the composition of
matter has been revised many times

I. History of the Atomic Theory
Democritus (460-370 BC)
1.Matter is made up of “atoms”
that are solid, indivisible and
indestructible
2.Atoms constantly move in space
3.Different atoms have different
size and shape
4.Changes in matter result from
changes in the grouping of
atoms
5. Properties of matter result from
size, shape and movement
A.
I. History of the Atomic Theory
B. Aristotle (384-322 BC ) & Others
1. Four kinds of matter
a. Fire – Earth – Water – Air
2. One kind of matter can transform
into another
3. Rejected idea of the “atom” (idea then
ignored for almost 2000 years
4. This theory was more popular and
it was easier to accept
Aristotle’s Theory of Matter
I. History of the Atomic Theory
C. Antoine Lavoisier (1770s)
1. Experiment:
2 Sn + O2

2 SnO
tin
oxygen
tin (II) oxide
mass before reaction = mass after reaction
2. Law of Conservation of Mass
a. Matter cannot be created or destroyed
(in a chemical or physical change)
I. History of the Atomic Theory
D.
Joseph Proust (1779)
1. Develops Law of Definite Composition- all
samples of a specific substance contain the
same mass ratio of the same elements
a. ex: all samples of CO2 contains 27.3%
carbon and 72.7% oxygen
b. therefore ‘elements’ are combining
in a whole number ratio – WHY????
I. History of the Atomic Theory - Dalton
E.
John Dalton (1803)
1. Develops Law of Multiple Proportions
a. describes the ratio of elements by mass in
two different compounds composed of the
same elements
2. Example: carbon monoxide carbon dioxide
1 part oxygen : 2 parts oxygen
*when compared to the same amount of
carbon in each compound
I. History of the Atomic Theory- Dalton
3. Dalton collects data and develops his
atomic theory in 1803
4. Dalton’s Background
a.
b.
c.
d.
Dalton became a school teacher at the
age of 12 (he left school at age 11)
loved meteorology - pioneer in this field
studied works of Democritus, Boyle and
Proust
Wrote New System of Chemical
Philosophy in 1808
5. Dalton’s Atomic Theory
1. Matter is made of small particles-atoms
2. Atoms of a given element are identical in
size, mass, but differ from those of other
elements*.
3. Atoms cannot be subdivided or destroyed*.
( supports law of conservation of mass)
4.Atoms combine in small whole number ratios
to form compounds. (def comp,Mult prop)
5. Atoms combine, separate, or rearrange in
chemical reactions.
* Modified in Modern Atomic Theory
JUST A THEORY…….

But it lead to the
Modern
Atomic theory
F. Modern Atomic Theory
1. All matter is made up of small particles
called atoms.
2. Atoms of the same element have the same
chemical properties while atoms of different
elements have different properties
3. Not all atoms of an element have the same
mass, but they all have a definite average
mass which is characteristic. (isotopes)
F. Modern Atomic Theory
4.
5.
Atoms of different elements combine to
form compounds and each element in
the compound loses its characteristic
properties.
Atoms cannot be subdivided by chemical
or physical changes – only by nuclear
changes
I. History of the Atomic Theory
1803
1897
1909
1913
1935
Today
solid
particle
electron
proton
e- orbit
nucleus
neutron
Quantum
Atom
theory
Dalton
Thomson
Rutherford
Bohr
Chadwick
Schrodinger
and others
II.
History of the Atomic Structure
A. J.J. Thomson (1856-1940)
J.J. Thomson (1887)
1.
Experiments with cathode ray tubes
a. atoms have (-) charged particles
which are smaller than atoms
b. determined charge/mass ratio of the
“electron”
Voltage source
-
+
Vacuum tube
Metal Disks
Voltage source
-
+
Voltage source
-
+
Voltage source
-
+
Voltage source

+
Passing an electric current makes a beam
appear to move from the negative to the
positive end
Voltage source

+
Passing an electric current makes a beam
appear to move from the negative to the
positive end
Voltage source

+
Passing an electric current makes a beam
appear to move from the negative to the
positive end
Voltage source

+
Passing an electric current makes a beam
appear to move from the negative to the
positive end
Voltage source

By adding an electric field
Voltage source
+
 By adding an electric field
Voltage source
+
 By adding an electric field
Voltage source
+
 By adding an electric field
Voltage source
+
 By adding an electric field
Voltage source
+
 By adding an electric field
Voltage source
+
 By adding an electric field he found that the
moving pieces were negative

Demonstration of the cathode ray
experiment.
2. Thomson’s Model
The Pudding Model
a. electrons present
 b. atom is like plum
pudding - bunch of
positive stuff
(pudding), with the
electrons suspended
(plums)

II.
History of the Atomic Structure
B. Robert Milikan (1868-1953)
1. Oil Drop Experiment (1909)
a. Discovered mass and actual charge of
electron (-1)
b. Mass is 1/1840 the mass of a hydrogen
atom
1) e – has a mass of 9.11 x 10-28 g
Oil Drop
II.

History of the Atomic Structure –
Summary thus far
So, at this point we know:
- Atoms are divisible into smaller particles
– Electrons are negatively charged
– The mass of an electron is very small
HOWEVER
– Atoms should have a (+) portion to balance
the negative part
- Electrons are so small that some other
particles must account for mass
II. History of the Atomic Structure
Ernest Rutherford (1871-1937)
C. Ernest Rutherford (1909)
1. Discovered the proton p+
2. Received Nobel Prize in Chemistry
3. Gold Foil Experiment (Expectations)
a. Shot alpha particles at atoms of gold
b. expected them to pass straight
through
Lead
block
Uranium
Florescent
Screen
Gold Foil
He thought this would happen:
According to Thomson Model
He thought the mass of the positive charge was
evenly distributed in the atom
Here is what he observed:
4. Gold Foil Experiment Results
a. Most positive alpha particles pass right
through
b. However, a few were deflected
c. Rutherford reasoned that the positive
alpha particle was deflected or repelled
by a concentration of positive charge
The positive region accounts for deflection
5. Gold Foil Experiment Conclusions
a. the atom is mostly empty space
b. the atom has a small, dense positive center
surrounded by electrons
Rutherford Model of the Atom
II. History of the Atomic Structure

At this point in 1909, we know:
– p+ = 1.67 x 10-24 g
– e- = 9.11 x 10-28 g
– The charges are balance!

But,
– How are the electrons arranged?
– There is still mass that is unaccounted for
II. History of the Atomic Structure
D.
Niels Bohr (1913)
1. Electrons orbit nucleus in
predictable paths
II.
E.
History of the Atomic Structure
E. Chadwick (1891 – 1974)
Chadwick (1935)
1. Discovers neutron in
nucleus
2. Neutron is neutral - does
not have a charge n0
3. Mass is 1.67 x 10-24 g
a. slightly greater than
the mass of a proton
II. History of the Atomic Structure
F. The Quantum Atom Theory
1. The atom is mostly empty
space
2. Two regions:
a. Nucleus- protons and neutrons
b. Electron cloud- region where
you have a 90% chance of
finding an electron
II. History of the Atomic Structure
Charges balanced
 Mass accounted for
 However –
what about the
behavior of the
electrons?

III.
Subatomic Particles
A. Comparing Particles
Relative Actual
mass (g)
Name Symbol Charge mass
Electron
e-
-1
Proton
p+
+1
1amu 1.67 x 10-24
Neutron
n0
0
1amu
0
9.11 x 10-28
1.67 x 10-24
III.
Subatomic Particles
B. Atomic Number and Mass Number
1. Atomic number
1. the number of protons in the nucleus
of an atom
a. identifies the element
b. no two elements have the same
atomic number
2. Ex. C is 6, N is 7 and O is 8
carbon
nitrogen
oxygen
III.
Subatomic Particles
B. Atomic Number and Mass Number
2. Mass number
a. the number of protons plus neutrons in the
nucleus of an atom
b. mass number is very close to the mass of an
atom in amu (atomic mass units)
c. two atoms with the same atomic number but
different mass number are called isotopes
1) (mass #) – (atomic #) = #n 0
III.
Subatomic Particles
C. Ions
1. Electrons and Ions
a. For neutral atoms, #e- = #p+
b. If there are more electrons, a negative ion
forms (anion)
c. If there are less electrons, a positive ion
forms (cation)
For now, we will work only with neutral atoms
ions
Subatomic Particles
C. Formation of Ions
Examples of Ions
Atom loses electrons Atom gain electrons
and form cations
and forms anions
Cations (+ ions)
Anions (- ions)
K+
BrCa2+
O2Al3+
N3-
C. Formation of Ions From Atoms
Na loses an electron
and forms a cation
Na – e- --> Na+
Cl gains an electron
and forms an anion
Cl + e- --> Cl-
III.
Subatomic Particles
D. Changing Number of Particles
1. You can never change the number of
protons and have the same element
2. If you change the number of neutrons in
an atom, you get an isotope
3. If you change the number of electrons in
an atom, you get an ion

III.
Subatomic Particles
E. Nuclear Notation
1. Nuclear Notation is one method for
depicting isotopes of an element
2. contains the symbol of the element,
the mass number, and the atomic
number
Mass
number
Atomic
number
X
III.
Subatomic Particles
E. Nuclear Notation
23
Na
11
 How many protons?
 How many neutrons?
 How many electrons?
III.
Subatomic Particles
F. Hyphen Notation
1. Element symbol or name – mass #
2. EXAMPLES
a. Fluorine-19
b. C-14
c. U-238
IV.
Mass of Atoms
A. Atomic Mass
1. Mass of an atom
a. too small to measure in grams
b. use relative mass (amu)
1) atomic mass unit
2) 1 amu is defined as 1/12 the mass
of one C-12 atom
IV. Mass of Atoms
B. Average Atomic Mass
1. weighted average mass of all known
isotopes
a. weighted means that the frequency of
an isotope is considered
b. mass of each isotope is multiplied by
its percent occurrence in nature – then
masses of all isotopes is added to get
the average atomic mass
IV. Mass of Atoms
C. The Mole and Molar Mass
1. measures the amount of substance
a. 1 mole = 6.02x1023 (Avogdro’s #) of
particles (atoms, molecules, ions, electrons)
b. standard – 1mole is the number of atoms in
12g of C-12 isotope
2. Molar mass – mass in grams of one mole
(mol) of any substance
a. numerically equal to atomic mass in amu
b. unit is grams/mol
Average Atomic Mass
What is average atomic mass?
Average atomic mass is the weighted
average of the atomic masses of the
naturally occurring isotopes of an
element
Calculating a Weighted Average
Example
A box contains two size of marbles. If
25.0% have masses of 2.00 g and 75.0%
have masses of 3.00 g what is the weighted
average?
(.250) (2.00) + (.750) (3.00) = .500 + 2.25
=
2.75g
Calculating Average Atomic Mass
EXAMPLE
Boron has two isotopes:
B-10 (mass 10.013 amu) 19.8% abundance
B-11 (mass 11.009 amu) 80.2% abundance
Calculate the average atomic mass.
(.198) (10.013) + (.802) ( 11.009) =
1.98 amu + 8.83 amu = 10.81
amu
Calculating Average Atomic Mass
Calculate the average atomic mass of Mg.
Isotope 1 - 23.985 amu (78.99%)
Isotope 2 - 24.986 amu (10.00%)
Isotope 3 – 25.982 amu (11.01%)
(23.985)(.7899)+(24.986)(.1000)+(25.982)(.1101)
18.95 amu + 2.498 amu + 2.861 amu = 24.31
amu
Average Atomic Mass
Helium has two naturally occurring isotopes,
He-3 and He-4. The atomic mass of
helium is 4.003 amu. Which isotope is
more abundant in nature?
He-4 is more abundant in nature because
the atomic mass is closer to the mass of
He-4 than to the mass of He-3.
Calculating Average Atomic Mass
Process
1. Multiply %occurrence x mass of isotope
2. Add products for each isotope
isotope occurrence
isotope occurrence
Ex. X- 40 (30.0% )
X-30 (70.0%)
(40 x .300) + (30 x .700) =
12 + 21 = 33 amu
Isotopic Pennies – number of pre and post 1982
a. Let X be the number of pre-1982 pennies
b. Let 10-X be the number of post-1982 pennies
c. (X)(3.1g) + (10-X)(2.5g) = mass of 10 pennies
pre-82
post-82
EXAMPLE (Mass of a sample of pennies is 31.0g)
(X)(3.1g) + [(10-X)(2.5g)] = 31.0 g
3.1X + 25 - 2.5X = 31.0g
.6X + 25 = 31.0g
.6X = 6.0g
X = 6.0g/.6
X = 10 pre-82 pennies
10-X = 0 post-82 pennies
Isotopic Penny Lab- Average Atomic Mass
Calculate percent of pre-82 and post-82 pennies
# of pre-82 pennies x 100%
10
# post-82 pennies x 100%
10
Calculate the average atomic mass of coinium
(% pre-82)(3.1g) + (% post-82)(2.5) = average atomic
mass
V.
Radioactive Decay
A. The Process
1. What is radioactive decay?
a. spontaneous nuclear change in which
unstable nuclei emit radiation and lose
energy
1) radiation – rays and particles
emitted by radioactive materials
2. Why do atoms undergo decay?
a. produce a nucleus that is more stable
V.
Radioactive Decay of Elements
B. Comparison of alpha, beta, gamma
Alpha
Form particle
Beta
particle
Gamma ____
electromagnetic
radiation
Symbol
Mass
Charge
Notation
4 amu
+2
no mass
-1
no mass
none
Alpha Particle
V.
1.
2.
Radioactive Decay
C. Types of Decay
Beta Decay (neutronproton + electron)
a. beta particle (electron) is given off
b. atomic number increases by one
c. mass number stays the same
Alpha Decay
a. alpha particle (2p++2n0) is given off
b. atomic number decreases by 2
c. mass number decreases by 4
V.
Radioactive Decay
D. Examples of Decay
Beta Decay (n0 p+ + e-) e- released
Parent Nuclei
Daughter Nuclei
Co-60 --------------> Ni-60 + e(z = 27)
(z = 28)
C-14 ---------------> N-14 + e(z = 6)
(z = 7)
Beta Decay
Beta Decay
Beta Decay
+ beta particle
+ beta particle
+ beta particle
+ beta particle
+ beta particle
Radioactive Decay
D. Examples of Decay
Alpha Decay (alpha particle (2n0 + 2p+)
released)
Parent Nuclei
Daughter Nuclei
Th-232 ----------------> Ra-228 + alpha
(z = 90)
(z = 88)
Ra-226 ---------------> Rn-222 + alpha
(z = 88)
(z = 86)
Alpha Decay
Alpha Decay
Alpha Decay
Alpha Decay
+ alpha particle
+ alpha particle
+ alpha particle
+ alpha particle
+ alpha particle
Radioactive Decay of Uranium
Radioactive Decay of Uranium
Alpha Decay
+ alpha particle
+ alpha particle
+ alpha particle
+ alpha particle
+ alpha particle
Radiation
Radiation
Radiation
The Mole
What
is a mole in chemistry?
What conversion factors are
associated with the mole?
Types of conversions involving
mole equalities
What is a Mole?
What are mole equalities
A mole is equal to 6.02 x 1023 particles
 Particles can be atoms, molecules or ions
 6.02 x 1023 is Avogadro's Number


Mole Equalities
- 1 mole = molar mass
- 1 mole = 6.02 x 1023 particles
Mole Conversions [mass-mole-atoms]
Type
Equality Used
1. MOLES  MASS
2. MASS  MOLES
1 mole= molar mass (g)
3. MOLES  ATOMS
4. ATOMS  MOLES 1 mole = 6.02 x 1023 atoms
5. MASS  ATOMS
6. ATOMS MASS
1 mole = 6.02 x 1023 atoms
1 mole = molar mass (g)
Mole Calculations
Using Conversion Factors
1 mole
6.02 x 1023
molar mass
1 mole
PARTICLES <----> MOLES <----> MASS
6.02 x 1023
1 mole
1 mole
molar mass
Solving Mole Problems
EXAMPLES
1.
2.
1.00 mole of He = 4.00 g.
2.00 mole of He = _____g
2.00 mol He X 4.00g He =
1
1 mole He
He
8.00 g
EXAMPLES
3.
1.00 mole He = 6.02 X 1023 atoms
2.00 mole He = ________atoms He
2.00 mole He x 6.02 x 1023 atoms He = 12.04 x 1023
1
1 mole He
1.20 x 1024
atoms He
5.
16.00g He = _____ moles He
16.0 g He x 1 mole He =
1
4.00 g He
4.00 moles He
4.
EXAMPLES
6.
3.01 X 1023 atoms He = _____ moles
3.01 x 1023 atoms He x
1
1 mole He
=
6.02 x 1023 atoms He
.500 mol He
7.
8.00g He =______atoms He
8.00 g He x 1 mole He x 6.02 x 1023 atoms He =
1
4.00g He
1 mole He
12.04 x 1023 atoms He = 1.20 x 1024 atoms He
Sample Problems
1.
Moles to mass.
Find the mass of 3.50 moles of carbon.
2.
Mass to Moles
How many moles of carbon are contained in
60.0 g of carbon?
3.
Moles to Atoms
How many atoms of carbon are found in 4.00
moles of carbon?
Sample Problems
4.
Atoms to Moles
How many moles of carbon are represented
by
1.806 x 1024 atoms of carbon?
5.
Mass to Atoms
How many carbon atoms are found in 36.0g of
carbon?
6.
Atoms to Mass
What is the mass of 1.204 x 1024 atoms of
carbon?
More Sample Problems
2.00 moles of Cu =
60.0 grams of C =
3.00 x 1023 atoms He =
2.50 moles Al =
28.0 grams N =
1.80 x 1023 atoms Mg =
atoms of Cu
moles of C
moles of He
grams of Al
atoms of N
grams of Mg
Mole Calculations
(Mass of Helium is 4.00 )
1. 1.00 mole of helium = 4.00g
2. 2.00 mole of helium = 8.00g
3. 1.00 mole of helium = 6.02 x 1023 atoms
4. 2.00 mole of helium = 1.20 x 1024 atoms
5. 16.0g of helium = 4.00 mol
6. 3.0l x 1023 atoms of helium = .500 moles
7. 8.00g of helium = 1.20 x 1024 atoms.