Transcript GHW#3-Chapter-1

Chemistry 100(02) Fall 2014
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311
Phone 257-4941
Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m
Tu,Th,F 8:00 - 10:00 a.m. Or by appointment
Test Dates:
September 29, 2014 (Test 1): Chapter 1 & 2
October 20,
2014 (Test 2): Chapter 3 & 4
November 12, 2014 (Test 3) Chapter 5 & 6
November 13, 2014 (Make-up test) comprehensive:
Chapters 1-6 9:30-10:45:15 AM, CTH 328
CHEM 100 Fall 2014 LA Tech
Text Book & Resources
REQUIRED :
Textbook: Principles of Chemistry: A Molecular Approach, 2nd
Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase
the Mastering Chemistry
Group Homework, Slides and Exam review guides and sample
exam questions are available online: http://moodle.latech.edu/
and follow the course information links.
OPTIONAL :
Study Guide: Chemistry: A Molecular Approach, 2nd EditionNivaldo J. Tro 2nd Edition
Student Solutions Manual: Chemistry: A Molecular Approach, 2nd
Edition-Nivaldo J. Tro 2nd
Chapter 1. Matter, Measurement, and Problem
Solving
1. 1
1 .2
1 .3
1 .4
1 .5
1 .6
1 .7
1 .8
Atoms and Molecules…………………………………..
1
The Scientific Approach to Knowledge……………..
3
The Classification of Matter……………………………
5
Physical and Chemical Changes and Physical and Chemical
Properties……………………………………..
9
Energy: A Fundamental Part of Physical and Chemical
Change…………………………………………………….. 12
The Units of Measurement……………………………...
13
The Reliability of a Measurement………………………
20
Solving Chemical Problems…………………………….
27
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Chapter 1. KEY CONCEPTS
• What is chemistry?
• Scientific Method.
• Properties of the three states of
matter
• Physical changes and
properties.
• Chemical change and
properties.
• Categories of matter.
• Elements and Compounds
• Atomic symbols
• Chemical Elements and
properties
• Chemical Symbolism
• Separating Mixtures.
• Scientific Measurement
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• Prefixes of SI units
• Macro, micro and nano-scales
• Conversion factors.
• Factor label method.
• Uncertainty and significant
figures
• Temperature Conversions.
• Density Calculations.
• Three chemical Laws
• Dalton's atomic theory
• Interpreting chemical formulas
and chemical reaction.
• Concept of mole
• Gram to mole conversion
Units
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SI UNITS OF MEASUREMENT
Five Basic Units
Length
meter (m)
Mass
kilogram (kg)
Time
second (s)
Temperature
kelvin (K)
Amount
mole (mol)
6.02 x 1023 units
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The Units of Measurement
1) Give the name and abbreviation of the SI
Unit for:
a) Length
b) Mass
c) Time
d) Amount of substance
e) Temperature
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Metric System
How to change measurements to reasonable numbers
Prefix
Decimal
Equivalent
Symbol
Power of 10
mega-
M
1,000,000
Base x 106
kilo-
k
1,000
Base x 103
deci-
d
0.1
Base x 10−1
centi-
c
0.01
Base x 10−2
milli-
m
0.001
Base x 10−3
micro-
m or mc
0.000 001
Base x 10−6
nano-
n
0.000 000 001
Base x 10−9
pico
p
0.000 000 000 001
Base x 10−12
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Macroscale, Microscale, and Nanoscale
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2) Give the abbreviation for the following units
and describe what they are used to measure:
a) cubic centimeter
c) nanoseconds
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b) micrometer
d) millimole
3) Give the name and the abbreviation
(without looking in the book) of the SI or
metric prefix for:
a) 10-12
b) 106
c) 10-9
d) 10-2
e) 10-3
f) 109
g) 103
i) 10-6
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Uncertainty in a Measurement
types of errors, random and systematic.
Careless measurements
Low resolution instruments
Calibration errors
• The last digit is an
estimate.
• .
Measurement ≈ 26.13 cm
12
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Uncertainty and Significant Figures
All measurements involve some uncertainty.
Scientists write down all the digits that have no
uncertainty plus one additional uncertain digit.
If an object is reported to have a mass = 6.3492
g, the last digit (“2”) is uncertain ( it is probably
close to 2, but may be 4, 1 … etc).
There are five significant figures in this number.
All the digits are meaningful.
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Uncertainty and Significant Figures
To find the number of significant figures:
Read a number from left to right and count all digits,
starting with the first non-zero digit.
All digits are significant except those zeros that are
used to position a decimal point (“placeholders”).
significant
0.00034050
placeholders
significant
5 sig. figs.
Scientific Notation (3.4050 x 10-4)
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Significant Figure Primer
• All non-zero digits are significant.
– Example: 536 has three significant figures.
• Zeros between non-zero digits are also significant.
– Example: 6703 has four significant figures.
• Place holder zeros:
– Zeros to the left of a non-zero digit are NOT significant.
• Example: 0.0043 has two significant figures.
• Example: 0.0600 has three significant figures.
– Zeros to the right or after a non-zero digit to the decimal
point are NOT significant.
• Example: 7000 has only one significant figure.
• Example: 32040 has four significant figures.
• Example: 50.0 has three significant figures; all the
zeros are significant.
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Uncertainty and Significant Figures
Examples
Number
2.12
4.500
Sig. figs.
3
4
Comment
The zeros are not placeholders. They are
significant.
0.002541
4
0.00100
3
500
1, 2, 3 ?
The zeros are placeholders (not significant).
500.
3
Adding a decimal point is one way to show that the
zeros are significant.
5.0 x 102
2
No ambiguity.
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Only the last two zeros are significant.
Ambiguous. If a number lacks a decimal point the
zeros may be placeholders or may be significant.
The Reliability of a Measurement
4) Express the following numbers in scientific
notation with the appropriate number of
significant figures:
a) 10980000000
b) 414100
c) 0.000095162
d) 746.5 x 107
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Rounding
Look at the 1st non-significant digit (the digit
after the last one retained). If it:
is > 5, round the last retained digit up by 1.
is < 5, make no change.
equals 5, and the 2nd non-significant digit is:
absent, round the last retained digit up by 1.
odd, round the last retained digit up by 1.
even, make no change.
Consider rounding 37.663147 to 3 significant figures.
last retained
digit
2nd nonsignificant digit
1st non-significant digit
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It rounds up to 37.7
Rounding
Examples
Round the following numbers to 3 significant figures:
1st non-sig. 2nd non-sig.
Rounded
Number
digit
digit
Number
2.123
2.123
2.12
51.372
51.372
51.372
51.4
131.5
131.5
132.
24.752
24.752
24.752
24.7
24.751
24.751
24.751
24.8
0.06744
0.06744
0.0674
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Can you define accuracy vs. precision?
Precision = the degree of exactness of a
measurement that is repeatedly recorded.
Accuracy = the extent to which a measured
value agrees with a standard value
• Which set is more precise?
18.2 , 18.4 , 18.35
17.9 , 18.3 , 18.85
16.8 , 17.2 , 19.44
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Can you hit the bull's-eye?
Three
targets with
three arrows
each to
shoot.
How do
they
compare?
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Both
accurate
and
precise
Precise
but not
accurate
Neither
accurate
nor
precise
Unit Conversion Calculation
Speed of light is 3.00 x 108 m s-1 .
Convert the speed of light to miles per
year (1 mile = 1.61 km).
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Using Conversion Factors
5) Convert 78.01 inches into:
a) feet
b) meters
6) Convert 15.42 meters into:
a) kilometers
b) micrometers
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Significant Figures and Scientific
Notation
Example: 301 in scientific notation is 3.01 x 102.
NOTE: The decimal point was moved two
places
to the left.
Example: 0.0301 in scientific notation is 3.01 x 10-2.
NOTE: The decimal point was moved two
places to the right.
Both of these value indicate THREE significant
figures.
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Significant Figures in Add/Sub
The answer you report in a problem should only
include significant digits.
Addition and subtraction
Find the number of digits after the decimal point (adp) in
each number.
answer adp = smallest input adp.
Example
Add:
adp = 3
17.245
adp = 4
+ 0.1001
Rounds to: 17.345 (adp = 3)
17.3451
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Addition and Subtractions Examples:
Examples:
2 3 .4 6 7 in
+ 3 1 3 .2 1 in
3 3 6 .6 7 7 in
457
cm
0 . 6 8 cm
4 5 6 . 3 2 cm
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but you would report
336.68 in
but you would report
456 cm
Significant Figures Add/Sub
Example
Subtract 6.72 x 10-1 from 5.00 x 101
Write the numbers down with the same power of 10:
5.00 x 101
– 0.0672 x 101
adp = 2
adp = 4
4.9328 x 101
Rounds to: 4.93 x 101
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adp = 2
Significant Figures Mul/Div
Multiplication and Division
Find the number of significant figures (sig. fig.) in each number.
Answer has sig. fig = smallest input sig. fig.
Example
Multiply 17.425 and 0.1001
17.245
sig. fig. = 5
x 0.1001
sig. fig. = 4
1.7262245
Rounds to: 1.726
sig. fig. = 4
Example
= 14,221.402734
Multiply 2.346 x 12.1 x 500.99
Rounds to:
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1.42 x 104 (3 sig. fig.)
7) Perform the following calculations and give
the answer in the correct number of
significant figures.
a) 30.84 + 9.74
Scientific notation:
Answer:
b) 30.84 + 9.74486
Scientific notation:
Answer:
c) 145 + 1.54 x 10-6
Scientific notation:
Answer:
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7) Perform the following calculations and give
the answer in the correct number of
significant figures.
2 adp 2 adp
a) 30.84 + 9.74
Scientific notation:
2 adp 5 adp
b) 30.84 + 9.74486
Scientific notation:
0 adp 8 adp
c) 145 + 1.54 x 10-6
Scientific notation:
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2 adp
Answer: 40.58
4.058 x 101
2 adp
Answer: 40.58486
4.058 x 101
0 adp
Answer: 145.00000154
1.45 x 102
7) Perform the following calculations and give
the answer in the correct number of
significant figures.
d) 40.79 - 1.18432
Scientific notation:
Answer:
e) 1.43 x 0.848
Scientific notation:
Answer:
f) (7.601x107) x (8.09x10-4) Answer
Scientific notation:
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7) Perform the following calculations and give
the answer in the correct number of
significant figures.
2 adp 5 adp
2 adp
d) 40.79 - 1.18432
Answer: 41.97432
Scientific notation:
4.197 x 101
Multiplication
3 sfg 3 sfg
3 sfg
e) 1.43 x 0.848
Answer: 1.21264
Scientific notation: :
1.21 x 100 = 1.21
4 sfg
3 sfg
3 sfg
f) (7.601x107) x (8.09x10-4) Answer 61492.09 =615
Scientific notation:
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6.15 x 102
Temperature Scales: Comparison
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Temperature Conversions
Human body temperature is 98.6 oF.
Convert this temperature to oC and
K scale
oC
= 5/9 (98.6 - 32) = 5/9 (66.6) = 37.0
Convert the
scale 100/180
oC-->
Shift the scale
to zero
K = 37.0 oC +273.15 = 310.2 K
Shift the scale
to zero K
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Temperature Convesions
oF
-- > oC ; C = 5/9 (F - 32)
oC -- > oF ; F =9/5 C + 32
oC -- > K ; K = C + 273.15
8) Convert 98.6 °F into:
a) °C
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b) K.
Problem Solving by
Factor Label Method
– State question in mathematical form
– Set equal to piece of data specific to the problem
– Use conversion factors to convert units of data
specific to problem to units sought in answer
– Other names used Unit Conversion Method
or dimensional (Unit) Analysis
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Common Conversion Factors
Length
1 kilometer
= 1000 m = 0.62137 mile
1 inch
= 2.54 cm (exactly)
1 angstrom (Å) = 1 x 10-10 m
1 gallon
= 1 x 10-3 m3
= 1000 cm3 = 1000 mL
= 1.056710 quarts
= 4 quarts = 8 pints
1 amu
1 pound
1 ton (metric)
1 ton (US)
= 1.6606 x 10-24 g
= 453.59237 g = 16 ounces
= 1000 kg
= 2000 pounds
Volume 1 liter (L)
Mass
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Solving Chemical Problems
9) Convert 75 miles per hour into: m s-1.
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Solving Chemical Problems
10) Convert 100 m2 into cm2
11) Convert 1 m3 into cm3
.
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Significant Figures and
Mathematical Operations
Mathematical operations dictate the reporting of
significant figures in an answer.
1. Multiplication and Division
The least precise measured value determines the
number of significant figures in the reported answers.
2. Addition and Subtraction
The value with the smallest decimal measurement
determines the answer’s significant figure.
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Solving Chemical Problems
13) Perform the following mathematical operations
and give the answer with the correct number of
significant figures
a) (2.481 x 12.74) + 0.27
:
2.69
b)
(4.73 x 10-4) - (72.85)
(872.3) - (0.305)
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=
=
Density
Density = mass (g)
volume (cm3)
An INTENSIVE physical property
The physical property does not depend on
amount of substance.
The physical properties of mass and volume that
determine a substance’s density are EXTENSIVE.
Extensive physical properties are dependent on
amount.
Densities of liquid and gases are affected by
temperature.
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Density Calculations
PROBLEM:
Mercury (Hg) has a density of 13.6 g/cm3. What is the
mass of 95 mL of Hg in grams? In pounds?
Strategy:
1. Convert mL to cm3.
2. Solve for mass (in grams) using density
relationship.
3. Convert grams to pounds.
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A Density Calculation
PROBLEM:
Mercury (Hg) has a density of 13.6 g/cm3. What is the mass of 95 mL
of Hg in grams? In pounds?
Density
=
mass (g)
volume (cm3)
Step 1:
95 mL x (1cm3/1mL) = 95 cm3
Step 2:
13.6 g/cm3 = x / 95 cm3
x = 1.29 x 103 g, but report 1.3 x 103 g
Step 3:
1.3 x 103 g x (1 lb/454 g) = 2.9 lb
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Solving Chemical Problems
12) Aluminum block weighs 14.2 g and has a density of
2.70 g cm-3. Calculate the volume of the block.
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Problem:
The density of octane (C8H18) is 7.00 lb/gal.
a) What is density in mg/cm3?
b) What is the mass in grams of 1.25 liters of
octane?
Strategy:
1. Convert 7.00 lbs to mg.
2. Change gallons to cm3.
3. Determine the density of octane in mg/cm3.
4. Convert 1.25 L to mL.
5. Determine the mass of octane in 1.25 L using
density.
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