Transcript INTRODUCTION and CHAPTER ONE

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INTRODUCTION
and
CHAPTER ONE
Read the Introduction and Chapter 1.
Chemistry is NOT a spectator sport.
Work out complete solutions for all
the bold numbered problems.
Welcome to the
World of
Chemistry
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Beginning Chemistry 200
• Read the book before class!
• Study the examples and do the
practice exercises.
• Be prepared for each class
session, lecture and laboratory.
• Don’t get behind.
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Outline for Chapter 1
• Definitions
–Homogenous and Heterogeneous
–Matter ?
–3 states of matter
–Chemical –vs.- Physical Change
• Calculations
–Density
–Temperature
»Kelvin  Celsius  Fahrenheit
»Fahrenheit  Celsius  Kelvin
–Significant Figures (Sig Figs)
Matter
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Mixtures
(a)Heterogeneous
(b)Suspension
(c)Homogenous
Cookie
Blood
Salt water
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The Language of Chemistry
• CHEMICAL ELEMENTS
- pure
substances that cannot be decomposed by
ordinary means to other substances.
Aluminum
Bromine
Sodium
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The Language of Chemistry
• The elements, their
names, and their
symbols are given
on the
PERIODIC
TABLE
• How many
elements are
there?
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The Periodic Table
Dmitri Mendeleev (1834 - 1907)
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Sodium
Find sodium, Na,
on the chart.
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• An atom is the smallest particle of
an element that has the chemical
properties of the element.
Copper
atoms on
silica
surface.
Find copper on the chart.
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The Atom
An atom consists of a nucleus (of protons
and neutrons) and electrons in space
about the nucleus.
Electron cloud
Nucleus
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CHEMICAL COMPOUNDS
are composed of atoms and so can
be decomposed to those atoms.
The red compound is
composed of
• Ni- Nickel
• C- Carbon
• O- Oxygen
• N- Nitrogen
Fixed composition
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MOLECULE
The smallest unit of a compound that
retains the chemical characteristics of
the compound.
MOLECULAR FORMULA
Composition of molecules
H2O
C8H10N4O2 - Caffeine
The Nature of Matter
Gold
Au
Mercury
Hg
Chemists are interested in the nature of
matter and how this is related to its atoms
and molecules.
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Graphite
layer
structure of
carbon
atoms
reflects
physical
properties.
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Chemistry & Matter
• We can explore the MACROSCOPIC world
— what we can see —
• Understand the PARTICULATE world
— we cannot see —
We can write SYMBOLS to describe these worlds.
A Chemist’s View
Macroscopic
Particulate
Symbolic- 2 H2(g) + O2 (g) --> 2 H2O(g)
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STATES OF MATTER
• SOLIDS — have rigid shape, fixed
volume. External shape can reflect the
atomic and molecular arrangement.
–Reasonably well understood.
• LIQUIDS — have no fixed shape and
may not fill a container completely.
–Not well understood.
• GASES — expand to fill their
container.
–Good theoretical understanding.
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THE THREE STATES OF
MATTER
Bromine (gas)
Aluminum (solid)
Water or H2O (liquid)
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KINETIC NATURE OF MATTER
Matter consists of atoms and molecules
in motion.
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Physical
Properties
What are some physical
properties?
–Color
–Melting and boiling
point
–Odor
–Conductivity
–Density
Physical Changes
Some physical changes
would be
– boiling of a liquid
– melting of a solid
– dissolving a solid in a
liquid to give a
homogeneous mixture
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DENSITY
- an important and useful
physical property
mass
(g)
Density =
volume (cm3)
Mercury
13.6 g/cm3
Gold
19.3 g/cm3
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Which is more dense?
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Relative Densities of the Elements
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Sig Figs
Sig Fig PPT
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mass
(g)
Density =
volume (cm3)
Problem: A piece of
copper has a mass of
57.54 g. It is 9.36 cm
long, 7.23 cm wide,
and 0.95 mm thick.
Calculate density
(g/cm3).
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SOLUTION
1. Get dimensions in common units.
1cm
. mm •
= 0.095 cm3
0 95
10 mm
2. Calculate volume in cubic centimeters.
(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm3
3.
Calculate the density.
6.4 cm3
= 9.0g/cm3
57.54g
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PROBLEM: Mercury (Hg) has a density
of 13.6 g/cm3. What is the mass of 95 mL
of Hg? In grams? In pounds?
Solve the problem using
DIMENSIONAL
ANALYSIS.
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PROBLEM: Mercury (Hg) has a density
of 13.6 g/cm3. What is the mass of 95 mL
of Hg?
First, note that 1
cm3 = 1 mL
Then, use dimensional analysis to
calculate mass.
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95 cm •
13.6 g
cm3
3
= 1.3 x 10 g
What is the mass in pounds?
See next slide
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The milliliter
and the cubic
centimeter are
equivalent.
Notice the units
of 10’s.
back
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PROBLEM: Mercury (Hg) has a density
of 13.6 g/cm3. What is the mass of 95 mL
of Hg?
What is the mass in pounds? (1 lb = 454 g)
1.3 x 103 g •
1 lb
= 2.8 lb
454 g
Density
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PROBLEM: An object weighing 15.67 g is
placed in water starting at 6.8 mL, then
displaces water to 20.2 mL. What is the
density of the object?
20.2
6.8
20.2mL - 6.8mL = 13.4mL
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PROBLEM: An object weighing 15.67 g is
placed in water starting at 6.8 mL, then
displaces water to 20.2 mL. What is the
density of the object?
1
3
15.67g
=
1.17g/cm
(20.2 - 6.8)ml
Chemical Properties and
Chemical Change
•Burning hydrogen (H2) in
oxygen (O2) gives H2O.
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Chemical Properties
• Similar to Physical Properties only
with reference to a Chemical
reactions
–Heat and or light produced
–Color
–Oder
Chemical Properties and
Chemical Change
• Burning hydrogen (H2) in
oxygen (O2) gives H2O.
• Chemical change or
chemical reaction
involves the
transformation of one or
more atoms or molecules
into one or more different
molecules.
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Chemical Change
2 Al + 3 Br2
Al2Br6
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Electrolyzing water
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UNITS OF MEASUREMENT
• We make QUALITATIVE
observations of reactions —
changes in color and physical
state.
• We also make QUANTITATIVE
MEASUREMENTS, which involve
numbers.
• Use SI units — based on the
metric system
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UNITS OF MEASUREMENT
Use SI units — based on the
metric system
length
(meter, m)
mass
(kilogram, kg,
and gram, g)
time
(second)
Units of Length
• 1 kilometer (km) = ? meters (m)
• 1 meter (m) = ? centimeters (cm)
• 1 centimeter (cm) = ? millimeter (mm)
• 1 nanometer (nm) = 1.0 x 10-9 meter
O—H distance =
9.4 x 10-11 m
9.4 x 10-9 cm
0.094 nm
94 pm
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Measurement
• Learn the prefixes in Table 1.4
• Other Relationships
1 cm3 = 1 mL = 0.001 L
1.00 lb = 454 g
1.00 in = 2.54 cm
1.06 qt = 1.00 L
Significant figures Page 47
Precision and accuracy Page 43
Examples
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Temperature Scales
• Fahrenheit
• Celsius
• Kelvin
Anders Celsius
1701-1744
Lord Kelvin
(William Thomson)
1824-1907
Temperature Scales
Notice that 1 Kelvin degree = 1 degree Celsius
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Calculations Using
Temperature
• Generally require temp’s in Kelvin
• T (K) = t (°C) + 273
• Body temp = 37 oC + 273 = 310. K
• Liquid nitrogen = -196 oC + 273 = 77 K
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In Class Problems
• A rectangular box has dimensions of
20.0 cm  15.0 cm  8.00 mm. Calculate
the volume of the box in liters.
• A standard sheet of paper has dimensions
of 8.5 inch by 11 inch. A sheet of paper
weighs on the average 0.150 g and has a
density of 0.710 g/cm3. Calculate the
thickness of the paper in cm.
• A gallon (3.78 L) of latex paint can cover
385 ft2 of the surface of a wall. What is
the average thickness of one coat of paint
(in micrometers)?
Sample problems
• Calculate the volume of 525 g of
mercury, d=13.534g/cm3.
• The melting point of tin is 505.5 K.
Calculate the Celsius temperature.
• Find the symbol for gold and the
element name for K.
• An iron sheet is 3.50 cm square and
has a mass of 15.396 g. The density
of iron is 7.87 g/cm3. Calculate the
thickness of the iron sheet in mm.
The End!
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Dimensional Analysis
“English-English”
(one conversion)
1) 6 in = ? ft
6 in
1 ft
= 0.5 ft
12 in
2) 3.5 gal = ? qt
3.5 gal 4 qt
= 14 qt
1 gal
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Dimensional Analysis
“Metric-Metric”
(one conversion)
1) 5.0 cm = ? mm
5.0 cm 10 mm
= 50. mm
1 cm
2) 4.0 dg = ? kg
4.0 dg 1 kg
-4 kg
=
4.0
x
10
4
10 dg
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Dimensional Analysis
“Metric-English”
(one conversion)
1) 200.0 cm = ? in
200.0 cm 1.00 in
=
2.54 cm
78.74 in
2) 34 qt = ? L
34 qt 1.00 L =
1.06 qt
32 L
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Dimensional Analysis
“English-English” (two or more conversions)
1) 6 in = ? mile
6 in
1 mile
1 ft
12 in 5280 ft
=
9 x 10-5 mile
=
450 oz
2) 3.5 gal = ? oz
3.5 gal
4 qt 32 oz
1 gal 1 qt
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Dimensional Analysis
“Metric-Metric” (two or more conversions)
1) 5.0 cm = ? km
5.0 cm 1 m
1 km
100cm 1000 m
= 5.0 x 10-5 km
2) 4 kg = ? pg
4 kg
103 g
1 kg
1012 pg
1g
=
4 x 1015 pg
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Dimensional Analysis
“Metric-English” (two or more conversions)
1) 200 m = ? in
200 m 100 cm 1.00 in
1 m
2.54 cm
=
8000 in
2) 34 qt = ? mL
34 qt
3 mL
10
1.00 L
1.06 qt 1 L
=
3.2 x 104 mL
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Dimensional Analysis
“Derived Unit Conversions”
Area
1) 8.0 ft2 = ? cm2
8.0 ft2
2
2
144 in2 (2.54) cm
=
2
2
1.00
in
1 ft
7400 cm2
2) 2.3 cm2 = ? nm2
2.3 cm2 1014 nm2
1 cm2
= 2.3 x 1014 nm2
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Dimensional Analysis
“Derived Unit Conversions”
Volume
1) 445 dm3 = ? mL
445 dm3 103 cm3 1 mL
1 dm3 1 cm3
= 4.45 x 105mL
2) 5 cm3 = ? mm3
5 cm3
103 mm3
1 cm3
=
5 x 103 mm3
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Dimensional Analysis
“Other Conversions Problems”
1) 100. km/hr = ? mile/hr
1 mile
100. km 105 cm 1.00 in 1 ft
1 km
5280 ft
hr
2.54 cm 12 in
= 62.1 mile/hr
2) 25 m/gal = ? nm/qt
25 m
gal
109 nm
1m
1 gal
4 qt
=
6.2 x 109 nm/qt
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Dimensional Analysis
“Other Conversions Problems” (continue)
3) Calculate the density of a material if 45
mL of it has a mass of 128 g.
128 g
= 2.8 g/mL
45 mL
4) Calculate the volume in mL of 2.5 g of a
material that has a density of 3.65 g/mL.
2.5 g
mL
3.65 g
=
0.68 mL
Dimensional Analysis
“Other Conversions Problems”
(continue)
5) Calculate the mass of 20 L of a material
that has a density of 8.54 g/mL.
20 L
103 ml
1L
8.54 g
ml
=
2 x 105 g
6) How many grams of gold are in 48 g of
an alloy that is 22.1% gold?
48 g alloy 22.1 g gold
=
100.0 g alloy
11 g gold
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Practice Problems
1) 6.45 m = ? cm
2) 12.4 kg = ? mg
3) 184 oz = ? g
4) 24 oz/hr = ? L/day
5) Determine the volume (in L) of 2 kg of sodium
chloride. (density = 2.17 g/mL)
6) How many grams of brass contain 50.0 g of
zinc? (This brass contains 15% Zn)
1) 645 cm 2) 1.24 x 107 mg 3) 5220 g 4) 17 L/day
5) 0.9 L 6) 330 g
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