Chapter 1: Chemistry
and the Atomic/Molecular
View of Matter
Chemistry: The Molecular Nature
of Matter, 6E
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1.2- Scientific Method
 Approach to gathering information &
formulating explanations.
 Scientists perform experiments in laboratories
under controlled conditions
1. Make observations/collect data
2. Law or Scientific Law
3. Hypothesis
4. Theory
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Atomic Theory
 Most significant theoretical model of nature
Atoms
Tiny submicroscopic particles
Make up all chemical substances
Make up everything in Macroscopic world
Smallest particle that has all properties of given
element
 Composed of:
 Electrons
 Neutrons
 Protons




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1.3- Matter & Its Classifications
Matter
 Anything that has mass & occupies space
Mass
 How much matter given object has
 Measure of object’s momentum, or resistance to
change in motion
Weight
 Force with which object is attracted by gravity
Ex. Mass vs. Weight
Astronaut on moon & on earth
 Weight on moon = 1/6 weight on earth
 Same mass regardless of location
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Matter
Chemical Reactions
 Transformations that alter chemical compositions
of substances
Decomposition
 Chemical reaction where 1 substance broken
down into 2 or more simpler substances
Ex.
Molten
sodium
chloride
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Electric
current
Sodium metal Na
+
chlorine gas Cl2
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Elements
 Substances that can’t be decomposed into simpler
materials by chemical reactions
 Substances composed of only 1 type of atom
H2 , O2 , Cl2 , Br2 , Na , Ca , Fe
 More complex substances composed of elements in various
combinations
diamond = carbon
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gold
Chemistry: The Molecular Nature of Matter, 6E
sulfur
6
Chemical Symbols for Elements
Chemical Symbol
 One or two letter symbol for each element name
 First letter capitalized, second letter lower case
Ex. C = carbon
Ca = calcium
Br = bromine
Cl = chlorine
S = sulfur
Ar = argon
H = hydrogen
O = oxygen
 Used to represent elements in chemical formulas
Ex. Water = H2O
Carbon dioxide = CO2
 Most based on English name
 Some based on Latin or German names
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Chemical Symbols
English Name Chemical Symbol
Sodium
Na
Potassium
K
Iron
Fe
Copper
Cu
Latin Name
Natrium
Kalium
Ferrum
Cuprum
Silver
Gold
Mercury
Antimony
Tin
Ag
Au
Hg
Sb
Sn
Argentum
Aurum
Hydrargyrum
Stibium
Stannium
Lead
Tungsten
Pb
W
Plumbum
Wolfram
(German)
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Compound
 Formed from 2 or more atoms of different
elements
 Always combined in same fixed ratios by
mass
 Can be broken down into elements by some
chemical changes
Ex. Water decomposed to elemental hydrogen
& oxygen
Mass of oxygen =
8 × mass of hydrogen
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Pure Substance vs. Mixture
Pure substances
 Elements and compounds / H2O, Na, N2, …
 Composition always same regardless of source
Mixture
 Can have variable compositions
 Made up of two or more substances
Ex. CO2 in water—varying amounts of “fizz” in soda
 2 broad categories of mixtures:
 Heterogeneous Mixtures
 Homogeneous Mixtures
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Homogeneous Mixtures
 Same properties throughout sample
 Solution
 Thoroughly stirred homogeneous mixture
Ex.
 Liquid solution
 Sugar in water
 Salt in water
 Gas solution
 Air
 Contains nitrogen, oxygen, carbon
dioxide & other gases
Other examples
-
Flour
Milk powder
Potable water
Gasoline
 Solid solution
 US 5¢ coin – Metal Alloy
 Contains copper & nickel metals
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Heterogeneous Mixtures
 2 or more regions of different properties
 Solution with multiple phases
 Separate layers
Ex.
 Salad dressing
 Oil & vinegar
 Ice & water
 Same composition
 2 different physical states
 Ether and water
 Soil
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Physical Change
 No new substances formed
 Substance may change state or the proportions
Ex. Ice melting or water freezing
 Sugar or salt dissolving
 Melting of metals
 Condensation of vapor or evaporation of water
 Stirring iron filings & sulfur together
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Chemical Change
or Chemical Reaction




Formation of new substance or compound
Involves changing chemical makeup of substances
New substance has different physical properties
Can’t be separated by physical means
Ex.
 Fool’s gold
 Compound containing sulfur & iron
 No longer has same physical properties
of free elements
 Can’t be separated using magnet
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Learning Check:
 For each of the following, determine if it
represents a Chemical or Physical Change:
Chemical Physical
Magnesium burns when heated
Magnesium metal tarnishes in air
Magnesium metal melts at 922 K
Grape Kool-aid lightens when
water is added
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Chemistry: The Molecular Nature of Matter, 6E
X
X
X
X
15
Classification of Matter
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Learning Check: Classification
Hot Ice
White Table
Cocoa (H2O) Flour Salt
(NaCl)
Pure substance
X
X
X
X
X
Element
Compound
Molecule
Heterogeneous Mixture
X
Homogeneous Mixture
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X
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Law of Conservation of Mass
 No detectable gain or loss of mass occurs in
chemical reactions.
 Mass is conserved.
Implication:
 Reactions involve reorganization of materials.
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Learning Check: Chemical Laws
 Magnesium burns in oxygen to form
magnesium oxide. If 16.88 g of Mg are
consumed and 28.00 g of MgO are produced,
what mass of oxygen was consumed?
 28.00 g – 16.88 g = 11.12g O
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Learning Check: Chemical Laws
 In a sample of MgO, there are 16.89 g Mg
and 11.11 g O. What mass of O would there
be in a sample that contains 2.00 g of Mg?
11.00 g O
X

16.89 g Mg 2.00 g Mg
11.00 g O  2.00 g Mg
X
16.89 g Mg
X = 1.30 g O
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1.4- Dalton’s Atomic Theory
John Dalton
 Developed underlying theory to explain
 Law of Conservation of Mass
 Law of Definite Proportions
 Reasoned that if atoms exist, they have
certain properties
Dalton’s Atomic Theory
1. Matter consists of tiny particles called atoms.
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Dalton’s Atomic Theory (cont)
2. Atoms are indestructible.
 In chemical reactions, atoms rearrange
but do not break apart.
3. In any sample of a pure element, all atoms
are identical in mass & other properties.
4. Atoms of different elements differ in mass &
other properties.
5. In given compound, constituent atoms
are always present in same fixed
numerical ratio.
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1.5- Atoms and Molecules
Chemical Formulas
 Atoms combine to form more complex
substances
 Discrete particles
 Each composed of 2 or more atoms
Ex.
 Molecular oxygen, O2
 Carbon dioxide, CO2
 Ammonia, NH3
 Sucrose, C12H22O11
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Chemical Formulas
 Specify composition of substance
 Chemical symbols
 Represent atoms of elements present
 Subscripts
 Given after chemical symbol
 Represents relative numbers of each type of atom
Ex.
Fe2O3 : iron & oxygen in 2:3 ratio
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Chemical Formulas
Free Elements
 Element not combined with another in compounds
 Just use chemical symbol to represent
Ex. Iron
Fe
Na
Sodium
Neon
Ne
Aluminum Al
Diatomic Molecule
 Molecules composed of 2 atoms each
 Many elements found in nature
Ex. Oxygen
Hydrogen
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O2
H2
Nitrogen
Chlorine
Chemistry: The Molecular Nature of Matter, 6E
N2
Cl2
25
Depicting Molecules
 Want to show:
 Order in which atoms are attached to each other
 3-dimensional shape of molecule
 Three ways of visualizing molecules:
1. Structural formula
2. Ball-and-Stick model
3. Space filling model
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1. Structural Formulas
 Use to show how atoms are attached
 Atoms represented by chemical symbols
 Chemical bonds attaching atoms indicated by lines
H
H
O
H
H
C
H2O
water
H
H
CH4
methane
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3-D Representations of Molecules
Hydrogen
molecule,
H2
Oxygen
molecule,
O2
Nitrogen
molecule
N2
Chlorine
molecule,
Cl2
 Use touching spheres to indicate molecules
 Different colors indicate different elements
 Relative size of spheres reflects differing sizes
of atoms
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2. “Ball-and-Stick” Model
 Spheres = atoms
 Sticks = bonds
Methane,
CH4
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Chloroform,
CHCl3
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3. “Space-Filling” Model
 Shows relative sizes of atoms
 Shows how atoms take up space in molecule
Methane
CH4
Water
H2O
Chloroform, CHCl3
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More Complicated Molecules
 Sometimes formulas contain parentheses
 How do we translate into a structure?
Ex. Urea, CO(NH2)2
 Expands to CON2H4
 Atoms in parentheses appear twice
Ball-and-stick
model
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Space-filling model
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Hydrates
 Crystals that contain water molecules
Ex. plaster: CaSO4∙2H2O calcium sulfate dihydrate
 Water is not tightly held
 Dehydration
 Removal of water by heating
 Remaining solid is anhydrous (without water)
Blue =
CuSO4 •5H2O
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Chemistry: The Molecular Nature of Matter, 6E
White =
CuSO4
32
Counting Atoms
1. Subscript following chemical symbol
indicates how many of that element are part
of the formula
 No subscript implies a subscript of 1.
2. Quantity in parentheses is repeated a
number of times equal to the subscript that
follows.
3. Raised dot in formula indicates that the
substance is a hydrate
 Number preceding H2O specifies how
many water molecules are present.
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Counting Atoms
Ex. 1 (CH3)3COH
 Subscript 3 means 3 CH3 groups
So from(CH3)3, we get 3 × 1C = 3C
3 × 3H = 9H
#C = 3C + 1C = 4 C
#H = 9H + 1H = 10 H
#O = 1 O
Total # of atoms = 15 atoms
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Counting Atoms
Ex. 2 CoCl2 · 6H2O
 The dot 6H2O means you multiple both H2 &
O by 6
 So there are:
#H
6 × 2 = 12 H
#O
6×1= 6O
#Co
1 × 1 = 1 Co
#Cl
2 × 1 = 2 Cl
Total # of atoms = 21 atoms
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Your Turn!
Count the number of each type of atom in the
chemical formula given below
a.
b.
c.
d.
e.
Na2CO3
(NH4)2SO4
Mg3(PO4)2
CuSO4∙5H2O
(C2H5)2N2H2
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a.
b.
c.
d.
e.
___Na,
___
2
1 C, ___
3 O
___N,
___H,
___S,
___O
2
8
1
4
___Mg,
___P,
___O
3
2
8
___Cu,
___S,
___O,
___H
9
10
1
1
___C,
___H,
___N
12
4
2
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Dalton’s Atomic Theory
 We now have the tools to explain this theory
& its consequences
 All molecules of compound are alike & contain
atoms in same numerical ratio.
Ex. Water, H2O
Ratio of oxygen to hydrogen is 1 : 2
1 O atom : 2 H atoms in each molecule
O weighs 16 times as much as H
1 H = 1 mass unit
1 O = 16 mass units
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Atoms in Fixed Ratios by Mass
For water in general:
 mass O = 8 mass H
 Regardless of amount of water present
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Dalton’s Atomic Theory
Successes:
 Explains Law of Conservation of Mass
 Chemical reactions correspond to rearranging
atoms.
 Explains Law of Definite Proportions
 Given compound always has atoms of same
elements in same ratios.
 Predicted Law of Multiple Proportions
 Not yet discovered
 Some elements combine to give 2 or
more compounds
Ex. SO2 & SO3
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Using Law Of Multiple Proportions
Mass S
Mass O
sulfur
dioxide
32.06 g
32.00 g
sulfur
trioxide
32.06 g
48.00 g
 Use this data to prove law of
multiple proportions
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Law of Multiple Proportions
Compound Sample Mass of Mass of
Size
Sulfur
Oxygen
Sulfur
dioxide
64.06 g
Sulfur
trioxide
80.06 g 32.06 g
Ratio of
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32.06 g 32.06 g
48.00 g
O in SO 3 48.00g 3


O in SO 2 36.00g 2
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1.6- Chemical Reactions
 When 1 or more substances
react to form 1 or more new
substances
Ex. Reaction of methane, CH4,
with oxygen, O2, to form
carbon dioxide, CO2, &
water, H2O.
Reactants = CH4 & O2
Products = CO2 & H2O
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Chemical Equations
 Use chemical symbols & formulas to represent
reactants & products.
 Reactants on left hand side
 Products on right hand side
 Arrow () means “reacts to yield”
Ex. CH4 + 2O2  CO2 + 2H2O
 Coefficients
 Numbers in front of formulas
 Indicate how many of each type of
molecule reacted or formed
 Equation reads “methane & oxygen
react to yield carbon dioxide & water”
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Conservation of Mass in Reactions
 Mass can neither be created nor destroyed
 This means that there are the same number of each
type of atom in reactants & in products of reaction
 If # of atoms same, then mass also same
CH4 + 2O2

CO2 + 2H2O
4 H + 4O + C
=
4 H + 4O + C
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Balanced Chemical Equation
Ex.
2C4H10 + 13O2  8CO2 + 10H2O
4 C & 10 H
per
molecule
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2 O per
molecule
1C&2O
per
molecule
Chemistry: The Molecular Nature of Matter, 6E
2H&1O
per
molecule
45
Balanced Chemical Equation
Ex.
2C4H10 + 13O2  8CO2 + 10H2O
2 molecules
of C4H10
13 molecules
8 molecules
of O2
of CO2
10 molecules
of C4H10
Coefficients
 Number in front of formulas
 Indicate number of molecules of each type
 Adjusted so # of each type of atom is
same on both sides of arrow
 Can change
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Balanced Chemical Equations
 How do you determine if an equation is balanced?
 Count atoms
 Same number of each type on both sides of equation?
 If yes, then balanced
 If no, then unbalanced
Ex. 2C4H10 + 13O2  8CO2 + 10H2O
Reactants
Products
2×4 = 8 C
8×1 = 8 C
2×10 = 20 H
10×2 = 20 H
13×2 = 26 O
(8×2)+(10×1)= 26 O
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Learning Check
Fe(OH)3 + 2 HNO3  Fe(NO3)3 + 2 H2O
Reactants
Fe
1
Products
1
O
3 + (2×3) = 9
(3×3) + 2 = 11
H
3+2=5
(2×2) = 4
N
2
3
 Not Balanced
 Only Fe has same number of atoms
on either side of arrow.
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Your Turn!
How many atoms of each element appear on
each side of the arrow in the following
equation?
4NH3 + 3O2 → 2N2 + 6H2O
Reactants
Products
N
(4 × 1) = 4
(2 × 2) = 4
O
(3 × 2) = 6
(6 × 1) = 6
H
(4 × 3) = 12
(6 × 2) = 12
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* All Review Questions are required
Focus On the following
1.5
1.11
1.12
1.14
1.29
1.32
1.33
1.41
1.48
1.53
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