CHAPTER ONE
The Foundations of Chemistry
Why is Chemistry Important?
Materials for our homes
Cooking
Components for computers
and other electronic devices
Fuel
Body functions
2
Some definitions / Vocabulary

Chemistry


Matter


Anything that has mass and occupies space.
(In other words: anything that has mass and volume)
Energy


Science that describes matter – its properties, the
changes it undergoes, and the energy changes that
accompany those processes
The capacity to do work or transfer heat.
Types of energy


Kinetic and potential energy
Heat energy, light energy,
chemical energy, mechanical energy
3
Natural Laws

The Law of Conservation of Mass


The Law of Conservation of Energy


During a chemical or physical change the mass of
the system remains constant
Energy cannot be created or destroyed in a
chemical reaction or in a physical change. It can
only be converted from one form to another.
The Law of Conservation
of Matter and Energy

Read at home
4
States of Matter

Solid

Liquid

Gas
5
States of Matter
Change States


heating
cooling
Ice
Steam
Water
6
Substances

Substance


Examples



matter all samples of which have
identical composition and properties
water
sulfuric acid
Properties


physical properties – physical changes
chemical properties – chemical changes
7
Physical Properties

Physical properties




changes of state
density, color, solubility
always involve only one substance
A substance cannot be broken down
or purified by physical means!
8
Mixtures

Mixture



Homogeneous mixtures



a combination of two or more substances
can be separated by physical means
have uniform properties throughout
examples: salt water; air
Heterogeneous mixtures


do not exhibit uniform properties throughout
examples: iron+sulfur; water+sand
9
Chemical Properties

Chemical properties




chemical reactions
always involve changes in composition
always involve more than one substance
Examples



burning of methane
rusting of iron
oxidation of sugar
10
Decomposition of Water
hydrogen
Element
Element
oxygen
water
Compound
11
Compounds and Elements

Compounds


If a substance can be decomposed into
simpler substances through chemical
changes, it is called a compound
Elements

If a substance cannot be decomposed
into simpler substances by chemical
means, it is called an element
12
Compounds and Elements

Important to remember



Law of Definite Proportions


both compounds and elements are substances
a compound consists of 2 or more elements
different samples of any pure compound contain
the same elements in the same proportion by mass
Symbols of elements
 found on the periodic chart (learn Table 1-2)
 www.webelements.com
13
Scientific Notation

Use it when dealing with very large
or very small numbers:

42,800,000. =

0.00000005117 =
14
Measurements in Chemistry
Quantity
 length
 mass
 time
 current
 temperature
 amt. substance
Unit
meter
kilogram
second
ampere
Kelvin
mole
Symbol
m
kg
s
A
K
mol
15
Metric Prefixes
Name
 mega kilo deci centi milli micro nano pico-
Symbol
M
k
d
c
m

n
p
Multiplier
106
103
10-1
10-2
10-3
10-6
10-9
10-12
16
Metric Prefixes: Examples
1000 m =
0.008 s =
30,000,000 g =
0.07 L =
17
Use of Numbers

Exact numbers



obtained from counting or by definition
1 dozen = 12 things for example
Measured numbers


numbers obtained from measurements
are not exact
every measurement involves an estimate
18
Significant Figures

Significant figures

digits believed to be correct by the person
making the measurement
19
Significant Figures

Side B:
13.6 mm
>13.5 mm but <13.7 mm
in my judgement!
13.6 mm
certain figures
estimated figure
20
Significant Figures
13.6 mm
certain figures + estimated figure
significant figures


we always report only 1 estimated figure
the estimated figure is always the last
one of the significant figures
21
Significant Figures - Rules
1) Exact numbers (defined quantities) have
an unlimited number of significant figures.
We do not apply the rules of significant
figures to them.
2) Leading zeroes are never significant:

0.000357 has three significant figures
3) Zeros between nonzero digits are always
significant:

20.034
1509
1.0000005
22
Significant Figures - Rules
Trailing zeros
4) Zeroes at the end of a number that contains
a decimal point are always significant:

35.7000
0.07200
40.0
41.0
5) Zeroes at the end of a number that does
not contain a decimal point may or may not
be significant (use scientific notation to
remove doubt):

173,700 may have 4, 5, or 6 significant figures
23
Significant Figures - Rules
Addition/Subtraction Rule
6) The position of the first doubtful digit
dictates the last digit retained in the
sum or difference.
Multiplication/Division Rule
7) In multiplication or division, an answer
contains no more significant figures than
the least number of significant figures
used in the operation.

Study examples 1-1 & 1-2 in the book
24
The Unit Factor Method

The basic idea of the method:


Principles:



multiplication by unity (by 1) does not
change the value of the expression
construct unit factors from any two
terms that describe identical quantity
the reciprocal of a unit factor is also a
unit factor
Study examples 1-3 through 1-9 in the book
25
The Unit Factor Method


1 ft = 12 in
Unit factors:
1 ft
12 in
12 in
1 ft
Example: Express 77.5 inches in feet
1 ft
77.5 in = 77.5 in x
= 6.46 ft
12 in

See Table 1-7 for various conversion factors
26
More examples

9.32 yrd = ? mm
1. We use the following knowledge to build
unit factors:
1 yrd = 3 ft
1 in = 2.54 cm
1 ft = 12 in
1 cm = 10 mm
2. Multiply 9.32 yrd by unit factors to get
the value expressed in mm:
3 ft
12 in
2.54 cm 10 mm
9.32 yrd x
x
x
x
= 8.52·103 mm
1 yrd
1 ft
1 in
1 cm
27
Density




mass
density =
volume
tells us how heavy a unit volume of matter is
usually expressed as “g/ml” for liquids and
solids and as “g/L” for gases
Table 1-8 lists densities of some common
substances
28
Density: Example

Example: Calculate the density of a substance
if 742 grams of it occupies 97.3 cm3.
1 cm3  1 mL  97.3 cm3  97.3 mL
742 g
d 
 7.63 g/mL
97.3 mL

Learn examples 1-11 through 1-13 in the book
29
Specific Gravity


d (substance)
Sp. Gr. =
d (water)
tells us how much heavier or lighter
a substance is compared to water:
Sp. Gr. < 1 – lighter than water
Sp. Gr. > 1 – heavier than water


specific gravity has no units – it is a
dimensionless quantity
See example 1-14 in the book
30
Specific Gravity: Example


Example 1-15: Battery acid is 40% sulfuric acid,
H2SO4, and 60% water by mass. Its specific
gravity is 1.31. Calculate the mass of pure H2SO4
in 100.0 mL of battery acid.
What do we know?
1. The mass percentage of H2SO4 and H2O in the
sample of battery acid.
2. Specific gravity of battery acid.
3. Density of water (1.00 g/mL).

To find the mass of H2SO4, we need to know
the mass of 100.0 mL of battery acid.
31
Specific Gravity: Example
m
d 
V
 m  d V
d (bat.acid) d (bat.acid)
Sp.Gr.(bat.acid) 

 1.31
d (H2O)
1.00g/mL
Therefore,
d (bat.acid)  1.31  1.00 g/mL  1.31 g/mL
m (bat.acid)  1.31 g/mL  100.0 mL  131 g
40%
m (H2SO4 )  131 g 
 52.4 g
100%
32
Heat and Temperature

Heat and Temperature are not the same thing:
 Heat is a form of energy
 T is a measure of the intensity of heat in a body

Heat always flows spontaneously from a hotter body
to a colder body – never in the reverse direction
Body 1
T1
hotter
Heat
T1 > T 2
Body 2
T2
colder
33
Temperature Scales

3 common temperature scales
 Fahrenheit
0ºF – freezing (salt+H2O)
30ºF – freezing H2O
90ºF – human body
 Celcius
 Kelvin
0ºC – freezing H2O
100ºC – boiling H2O
0 K – absolute zero
273.15 K – freezing H2O
http://home.comcast.net/~igpl/Temperature.html
34
Temperature Scales & Water
Melting (MP) and boiling (MP)
points of water on different
temperature scales



Fahrenheit
Celsius
Kelvin
MP
BP
32 oF 212 oF
0.0 oC 100 cC
273 K 373 K
35
Temperature Conversion
degrees Kelvin
degrees Celcius
? K = ?ºC + 273
?ºC = ? K - 273
degrees Fahrenheit
degrees Celcius
?ºF = (?ºC)·1.8 + 32
?ºC = (?ºF – 32)/1.8


Examples 1-16 & 1-17 in the book
http://www.lenntech.com/unit-conversioncalculator/temperature.htm
36
Heat

Chemical and Physical changes:
 evolution of heat (exothermic processes)
 absorption of heat (endothermic processes)

Units of measurement:
 joule (J) – SI units
 calorie (cal) – conventional units
 1 cal = 4.184 J

A “large calorie” (1 large cal = 1000 cal = 1 kcal)
is used to express the energy content of foods
37
Specific Heat

The specific heat (Cp) of a substance:
 the amount of heat (Q) required to raise the
temperature of 1 g of the substance 1ºC (or 1 K)
Q
Cp 
m  ΔT

Units of measurement:

J
J
or
g  C
g K
38
Specific Heat: Example 1

Knowing specific heat, we can determine how
much energy we need in order to raise the
temperature of a substance by T = T2 – T1:
 Calculate the amount of heat necessary to
raise the temperature of 250 mL of water
from 25 to 95ºC given the specific heat of
water is 4.18 J·g-1 ·ºC-1.

What do we know?




the temperature change
the specific heat of water
the volume of water
the density of water
39
Specific Heat: Example 1
heat
specific heat 
m  ΔT
heat  (specific heat)  m T
T  95C - 25C  70C
m  d V  1.00 g/mL  250 mL  250 g
J
heat  4.18
 250 g  70C  73150 J  7.3  10 4 J
g  C

Examples 1-18 through 1-20 in the book
40
Specific Heat: Example 2

Given specific heats of two different substances, we
can also calculate the heat transfer between them:
 0.350 L of water at 74.0ºC is poured into an
aluminum pot at room temperature (25.0ºC). The
mass of the pot is 200 g. What will be the
equilibrium temperature of water after it
transfers part of its heat energy to the pot? The
specific heats of aluminum and water are 0.900
and 4.18 J·g-1 ·ºC-1, respectively.
You might encounter this kind
of problem at your first exam
41
Specific Heat: Example 2

What do we know?
 the pot and water come to equilibrium, that is
eventually they have the same temperature
 the specific heat of aluminum and water
 the mass of aluminum
 the volume of water
 the density of water
 finally, the Law of conservation of energy
which tells us that the amount of heat lost by
water is the same as the amount of heat
gained by the aluminum pot
42
Specific Heat: Example 2
heat
specific heat 
m  ΔT
heat  (specific heat)  m T
Let’s denote the final temperature as Tf. Then the
changes in temperature for water and aluminum are:
T (H2O)  74.0C -Tf and T (Al) Tf  25.0C
1000 mL
m (H2O)  d V  1.00 g/mL  0.350 L 
 350 g
1L
Note that we used the unit factor
method to convert L to mL
43
Specific Heat: Example 2
heat  (specific heat)  m T
heat loss (H2O)  heat gain (Al)
J
J
4.18
 350 g  (74.0C -Tf )  0.900
 200 g  (Tf - 25.0C)
g  C
g  C
Solving this equation with respect to Tf,
we obtain
Tf = 68.6ºC

Try to solve the equation yourself and analyze
why the answer is given with 3 significant figures
44
Reading Assignment

Read Chapter 1

Learn Key Terms (pp. 40-41)


Go through Chapter 2 notes
available on the class web site
If you have time, read Chapter 2
45
Homework Assignment
Textbook problems (optional, Chp. 1):

11, 13, 15, 18, 27, 29, 30, 32, 36, 41, 43, 47,
49, 57, 62, 68, 80
OWL:


Chapter 1 Exercises and Tutors – Optional
Introductory math problems and Chapter 1
Homework problems – Required (homework
#1; due by 9/13)
46