Chapter 2 – Analyzing Data
2.1 Units and Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.1 Units and Measurements
Chemists use an internationally
recognized system of units to
communicate their findings.
Objectives
• Know the first five of the seven base units in the SI
system (time, length, mass, temperature and amount of
substance) including their units, symbols, and the basic
physical objects or phenomena on which they are
defined.
• Convert between the Kelvin and Celsius temperature
scales.
• Distinguish between and give examples of base units
and derived units.
Section 2.1 Units and Measurements
Objectives (cont)
• Use the equation for density to solve for an unknown
quantity using the problem solving process described
on page 38 in example problem 2.1.
• Understand both the concepts and experimental
procedure involved in the Mini Lab on page 39
(Determine Density) and be able to answer the
analysis questions listed at the end of the lab.
• Know all the prefixes from giga to nano in table 2.2
including their symbols and associated powers of ten
and be able to use them in a numerical problem.
Section 2.1 Units and Measurements
Key Concepts
• SI measurement units allow scientists to report
data to other scientists.
• Adding prefixes to SI units extends the range of
possible measurements.
• To convert to Kelvin temperature, add 273.15 to the
Celsius temperature. K = °C + 273.15
• Volume and density have derived units. Density,
which is a ratio of mass to volume, can be used to
identify an unknown sample of matter.
Units of Measurement
SI - Systemè Internationale d’Unités
7 base units
Base unit is based on an object or an
event in the physical world
Base unit is independent of other units
The 7 SI Base Units (see Table 2.1)
Quantity
Unit
Time
second
Only base unit with prefix
Length
meter
Abbrev.
s
m
Mass
kilogram
kg
Temperature
kelvin
K
Amount of substance
mole
mol
Current
ampere
A
Luminous Intensity
candela
cd
Prefixes
To better describe range of possible
measurements, base units (and other
units) are modified by using prefixes
Correspond to a particular power of ten
See table 2.2 (following)
Memorize value and the symbol
(including upper /lower case) of
prefixes in 10-9 to 106 range
SI Prefixes – Page 33
Capital letters
Greek letter
SI Prefixes – Expanded
Base Unit - Time
The Second
In 1967, the 13th General Conference
on Weights and Measures first defined
the SI unit of time as the duration of
9,192,631,770 cycles of microwave
light absorbed or emitted by the
hyperfine transition of cesium-133
atoms in their ground state undisturbed
by external fields
Length
The Meter
Distance light travels through a vacuum
in 1/(299 792 458) of a second
Mass
kilogram (kg)
Only base unit whose
standard is a physical
object
Defined by platinumiridium metal cylinder
kept in Sèvres, France
Temperature
kelvin (K)
The kelvin is the fraction 1/273.16 of
the thermodynamic temperature of the
triple point of water
Temperature
Kelvin scale
• Unit kelvin (small k), abbreviation K (but
no degree sign)
• Absolute zero = zero K
Celsius scale
• Unit °C (with degree sign)
• T(C) = T(K) - 273.15
Kelvin scale important for certain
formulas we will use in later chapters
Temperatures – K vs C
Boiling point water
373.15 K 100.00 C
Freezing point water
273.15 K 0.00 C
Absolute zero
0.00 K -273.15 C
The Mole and Avogadro’s Number
Abbreviation is mol
SI base unit for amount of substance
Defined as number of representative
particles (carbon atoms) in exactly 12 g
of pure carbon-12
Mole of anything contains 6.022 X 1023
representative particles
6.022 X 1023 = Avogadro’s number
Picky Details - Units
Abbreviations are avoided
Proper
• s or second
• cm3 or cubic centimeter
• m/s or meter per second
Improper
• sec
• cc
• mps
Picky Details - Units
Unit symbols are unaltered in the plural
Proper
• l = 75 cm
Improper
• l = 75 cms
Derived Units
Defined by a combination of base units
Velocity
v = l/t
m/s
Volume
V=lll
m3
liter (L)
cubic decimeter dm3
Density
d = mass / V
kg/m3
Derived Unit - Density
Density = mass
volume
Most common unit is g/cm3
Will return to using density in a word
problem later in presentation
Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.2 Scientific Notation and Dimensional
Analysis
Scientists often express numbers in
scientific notation.
• Express numbers correctly in standard scientific
notation, convert them to and from numbers not
expressed in scientific notation, and perform standard
arithmetic operations using them.
Section 2.2 Scientific Notation and
Dimensional Analysis
Key Concepts
• A number expressed in scientific notation is written as
a coefficient between 1 and 10 multiplied by 10 raised
to a power. To add or subtract numbers in scientific
notation, the numbers must have the same exponent.
• To multiply or divide numbers in scientific notation,
multiply or divide the coefficients and then add or
subtract the exponents, respectively.
Scientific Notation
Expresses a number as a multiple of
two factors:
• 1  number  10
3.1 -7.9
• 10 raised to a power
103 10-7
0
 10 = 1
n
 10 > 1
n positive integer
-n < 1
 0 < 10
n positive integer
Scientific Notation
3.1  104 = 31000
3.1  10-4 = 0.00031
• Have moved decimal point 4 places in
both cases
Some exponents match one of the
standard SI prefixes
• 4.27  10-6 s = 4.27 s
(microseconds)
Scientific Notation
Addition/Subtraction
• Must be same power of ten
Multiplication/Division
• Multiply first factors
• Add exponents if multiplying
• Subtract exponents if dividing
Put result back into standard form
• z.yyy  10n z = any non-zero digit
• One digit in front of decimal point
Practice – Scientific Notation
General
Problems 11(a-h), 12(a-d) page 41
Problems 76(a-d), 77(a-d) page 62
Addition & Subtraction
Problems 13(a-d), 14(a-d) page 42
Problems 78(a-j) page 62
Multiplication & Division
Problems 15(a-d), 16(a-d) page 43
Problems 79(a-f) page 62
Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.3 Uncertainty in Data
Measurements contain uncertainties that
affect how a result is presented.
• Define and compare accuracy and precision and
correctly identify which term or terms apply to a given
value based on a description of how the value was
determined.
• Calculate the percent error associated with a given
measurement.
• Determine the number of significant figures
associated with a given number.
Section 2.3 Uncertainty in Data
Objectives (cont)
• Determine the appropriate number of significant
figures to record when using an analog measuring
device such as a ruler.
• Determine the number of significant figures
associated with a result obtained from simple
arithmetic operations (addition, subtraction,
multiplication, division) on numbers.
• Round a number to a specified number of significant
digits.
Section 2.3 Uncertainty in Data
Objectives (cont)
• Show all work for a problem following the ProblemSolving Process described on page 38 in example
problem 2.1 while utilizing dimensional analysis,
significant figures, rounding and standard algebra
skills.
Section 2.3 Uncertainty in Data
Key Concepts
• An accurate measurement is close to the accepted
value. A set of precise measurements shows little
variation.
• The measurement device determines the degree of
precision possible.
• Error is the difference between the measured value
and the accepted value. Percent error gives the
percent deviation from the accepted value.
error = experimental value – accepted value
Section 2.3 Uncertainty in Data (cont.)
Key Concepts
• The number of significant figures reflects the precision
of reported data.
• Calculations should be rounded to the correct number
of significant figures.
• The rules for determining the number of significant
figures in a number produced in a mathematical
operation are different for multiplication/division and
addition/subtraction.
Accuracy and Precision
Accuracy refers to agreement of
particular value with true value
(sometimes true value is difficult to
determine; may require appropriate
calibration)
Precision refers to degree of
agreement among several
measurements of same quantity
Calibration
[Formal definition – don’t need to know]
The set of operations which establish, under
specified conditions, the relationship
between values indicated by a measuring
instrument or measuring system, and the
corresponding standard or known values
derived from the standard.
Purpose is to determine and/or improve the
accuracy of the device being calibrated
Calibration - Example
To calibrate an electronic balance, might use
following approach:
a) Assign a value of zero to the reading of the
output of the balance when nothing is on the
balance pan
b) Assign a value equal to the value of a
calibration mass (e.g., 1000.00 g) to the
reading of the output of the balance when the
calibration mass is placed on the balance pan
c) Assume linear behavior of the reading of the
balance between these two calibration points
Calibration - Example
For calibration to be possible, need a calibration
mass (e.g., a 100.00 g mass)
The actual (true) value of the calibration mass
has to be determined by the
organization/company supplying the mass – this
ultimately requires tracing back to the primary
standard of mass (making comparisons that are
linked to the kilogram standard mass)
It is only through calibration that the accuracy of a
measurement device (i.e., a balance) can be
determined and improved
Accuracy and Precision
Refer to figure 2.10
Accuracy and Precision
Refer to figure 2.10
Precision and Accuracy
precise and accurate
precise but not accurate
Precision and Accuracy
random error
Not precise, individual
measurement not generally
accurate but average is
systematic error
May be precise (only
marginally for this example);
average not accurate
Percent Error
% Error = 100  (actual – measured)
actual
(actual – measured) = error
Can ignore plus or minus signs for now
– need only absolute value of error
Actual = “true value”
In some cases, true value unknown
(Often compare a predicted value from
some model with measured value)
Practice
Accuracy and Precision
Problems 46, 48, page 54
Problem 87 page 63
Percent Error
Problems 32-34 page 49
Problems 49, 51 page 54
Problems 93(a-d), 94(a-d) page 63
Significant Figures
2 different types of numbers:
Exact
Measured
Exact numbers have infinite precision
Measured - obtained from measuring
device, have error and limited precision
Measured number written to reflect both its
numerical value and the precision to which
it was measured
Significant Figures and
Measurement Precision
For lab data, sig figs determined by
precision of measurement device
For digital device, last number to right
in display is limit to precision
For analog device, precision limited by
the estimated digit obtained from
“eyeballing” reading between markings
Examples to follow
Balances and Precision
Student
 0.1 g
(digital)
Triple beam
 0.01 g
(analog)
Electronic
analytical
 0.0001 g
(digital)
Significant Figures
Mass of object measured on student
balance (precision ± 0.1g) is 23.6 g
This quantity contains 3 significant
figures, i.e., three experimentally
meaningful digits
If same measurement made with
analytical balance (precision ± 0.0001g)
, mass might be 23.5820 g (6 sig. fig.)
Significant Figures and Precision
2 measurements of mass of same
object
Same quantity described at two
different levels of precision or certainty
Significant Figures Determined by
Measuring Device
32.33 C
32.3 C
Significant Figures Determined by
Measuring Device
0.1 mL graduations
Can estimate to
0.01 mL
Reading 1.70 mL
3 significant figures
Significant Figures Determined by
Measuring Device
0.1 cm
graduations
Can estimate to
0.01 cm
Reading 5.22 cm
3 significant
figures
What Is Nail Length?
~6.33 cm
Learning Check
Length of wooden stick?
1) 4.5 cm
2) 4.54 cm
3) 4.547 cm
Measurement of Volume
Most accurate but only
useable for one volume
Graduated
Cylinder
Volumetric
Flask
Buret Volumetric Syringe
Pipet
Measurement
of volume using
buret
Read at bottom
of liquid curve
(called the
meniscus)
20.16 mL
±0.01mL
Significant Figures
Rules for recognizing which digits are
significant – see the “Problem Solving
Strategy” on top of page 51 and the
following slide
Recognizing Significant (Sig) Figures
1. Non-zero numbers always significant
•
72.3 (3 sf)
4.737x10-8 (4 sf)
2. Zeros between non-zero numbers are
significant
•
60.5 (3 sf)
7.3002x10-4 (5 sf)
3. All final zeros to right of decimal point
are significant
•
6.20 ( 3 sf)
5.47000x109 (6 sf)
4. Zeros acting as placeholders not
significant
•
0.00253 (3 sf) 43200 (3 sf)
Recognizing Significant (Sig) Figures
5. Counting numbers (integers) and
defined constants or relationships have
an infinite number of significant figures
(all are exact numbers or relations)
• 60 s = 1 minute
• 1 foot = 12 inches
• 6 molecules
• 
Significant Digits Practice
Determine number of significant digits in
each of following:
45.8736
6
All digits count
.000239
3
Leading 0’s don’t
.00023900
5
Trailing 0’s do
4.8000 104
5
Trailing 0’s count
48000
2
0’s don’t count w/o decimal
3.982106
4
All digits count
1.00040
6
0’s between digits count as
well as trailing in decimal form
How many significant figures in each of
following measurements?
5.13
3
100.01
5
0.0401
3
0.0050
2
220,000
2
1.90 x 103
3
153.000
6
1.0050
5
?
Learning Check
Classify each of following as an exact
or measured number or relationship
1 yard = 3 feet
Diameter of red blood cell = 6 x 10-4 cm
There are 6 hats on shelf
Gold melts at 1064°C
?
Ans: exact, measured, exact, measured
Practice
Significant Figures
Problems 35(a-d), 36(a-d), 37 p. 51
Problems 47, 50 page 54
Problems 85, 88 page 63
Rounding Rules
See “Problem Solving Strategy” on bottom
of page 52
d = last significant digit r = digit to right of d
1. r < 5 then d  d
2. r > 5 then d  d + 1
3. r = 5 and digit after r ≠ 0 then d  d + 1
4. r = 5 and digit after r = 0 or nothing then
d  d + 1 if d odd; d  d if d even
(d always ends up even using this rule)
(You may see variations on rule 4 in other places)
Rounding
Round following to 4 significant figures:
4965.03
4965
780,582
780,600
1999.5
2.000x103 5 dropped, = 5
0 dropped, <5
8 dropped, >5
Note: you must include 0’s
If wrote as 2000 would have 1 SF
Rounding
Round following to 3 significant figures:
1.5587
.0037421
1367
128,522
1.6683x106
1.56
.00374
1370
129,000
1.67x106
Practice
Rounding
Problems 38(a-d), 39(a-d) page 53
Problems 91(a-f) page 63
Sig Figures and Multiplication/Division
Answer must have same number of
significant figures as number with
fewest number of significant figures
24 x 3.26 x 5.774 = 451.75776  450
= 4.5 x 102
2 sig figs allowed
6.38 × 2.0 = 12.76  13 (2 sig figs)
Operations with Significant Figures –
Adding or Subtracting
When adding or subtracting, number of
decimal places in result should equal
smallest number of decimal places in any
term in sum
135 cm + 3.25 cm = 138 cm
0 digits
after dp
2 digits
after dp
0 digits
after dp
135 cm term limits answer to units decimal
value
Sig Figures and Addition/Subtraction
Answer must have same number of
digits to right of decimal point as value
with fewest number of digits to right of
decimal point (value with lowest
precision)
• Note: not concerned with # of sig
figures in numbers
11.0 + 5.7732 + 2.01 = 18.7832  18.8
Addition/Subtraction
Focus on least significant digit (digit
with least precision)
25.5
+34.270
59.770
59.8
32.72
- 0.0049
32.7151
32.72
320
+12.5
332.5
330
Last example – special case
Least significant digit prior to decimal
point (can use sci. notation to justify)
Addition/Subtraction – Special Case
Use scientific notation to handle case
where number has zeros in front of
decimal and no digits of any kind after
Zeros in front of decimal are
placeholders to indicate power of ten
320 + 12.5 = 3.2x102 + 0.125x102
Now following ordinary rule, only
allowed one digit after decimal point
3.325x102  3.3x102 = 330
Addition/Subtraction: No Decimal Point
135000 m + 3250 m = ????
Convert to scientific notation
1.35x105 m + 3.25x103 m
Convert to common exponent (the largest)
1.35x105 m + 0.0325x105 m
Apply standard rule regarding digits after dp
1.35x105 m + 0.0325x105 m = 1.38x105 m
13500 m + 3250 m = 138000 m
Addition/Subtraction – Special Case
82000 + 5.32 = [82005.32] = 82000
Special case (focus on last precise digit):
82000 = 8.2x104 5.32 = 0.000532x104
8.2x104 + 0.000532x104 = 8.2x104
Addition/Subtraction
6.8 + 11.934 = 18.734  18.7 (1 place
after decimal)
Result has 3 SF even though 6.8 has
only 2 SF
0.56 + 0.153 = [0.713] = 0.71
10.0 - 9.8742 = [0.1258] = 0.1
10 – 9.8742 = [0.1258] = 0
[1x101 – 0.98742x101 = 0.01258x101= 0]
Addition/Subtraction
Limiting term: one having largest value of
the least significant digit
Least
significant Answer
Problem
digit
234 + 34.65 = 268.65
1.642x106 + 23x106 =
24.642x106
100 + 34.56 = 134.56
150 + 28.57 = 178.57
1’s place
269
1’s place
25x106
100’s place
100
10’s place
180
Rounding Rules
If doing a multistep calculation, round off
after last step provided all steps are either
multiplication/division or addition/subtraction
(can’t be mixed)
Round after series of additions or
subtractions before doing a multiplication or
division
Sig Figures and Mixed Operations
Some calculations involve both
multiplication/division and addition/subtraction
Must round intermediate result prior to
switching to new category of operation
8.52 + 4.1586  18.73 + 153.2 =
= 8.52 + 77.89 + 153.2 = [239.61] = 239.6
(8.52 + 4.1586)  (18.73 + 153.2) =
= 12.68  171.9 = [2179.692] = 2.180x103
(2180 has 3 SF, not 4 as required)
Sig Figures and Mixed Operations
Calculate 5.000Tc – 25C, where Tc =
Celsius temperature, for Tk = 298.1 K
?
Not 100! (need 3 SF)
Tc = Tk – 273.15 = 298.1- 273.15 = [24.95]
Tc = 25.0 C 5.000Tc = 125C
125C – 25C = 1.00x102 C
Mixed Operations - Percent Error
% Error = 100  (actual – measured)
actual
Calculating % error always involves a mixed
operation – subtraction followed by division
Must round after subtraction prior to dividing
Actual =16.24 g Meas. = 15.8 g % error ?
Error = 16.24 g – 15.8 g = [0.44 g] = 0.4 g
% Error = 100  0.4 / 16.24 = [2.463] = 2%
Not: 100  0.44 / 16.24 = [2.709] = 2.7%
Mixed Operations - Percent Error
Note: Results shown in example
problem 2.5 on page 49 are incorrect
All answers in that example should
have one significant figure (i.e., 3%, not
3.14%)
Sig Figures and Mixed Operations
Accepted value for density of copper =
8.92 g/cm3. 3 experiments to measure
its density resulted in values of 8.74,
9.01, and 8.83 g/cm3. Calculate % error
of average value of experiments.
Avg = (8.74+9.01+8.83)/3 = 8.86 g/cm3
% Error=100(actual–measured)/actual
% Error = 100  0.06 g/cm3/8.92 g/cm3
% Error = 0.7% (1 SF)
Sig Figures and Mixed Operations
Result of performing the following?
5.00  (22  1.85)
?
22  1.85 = 20.15 rounds to 2.0x101
Units place is significant
2.0x101  5.00 = 1.0x102
Answer has 2 significant digits because
subtraction operation generated a 2
significant digit intermediate result
Practice
Rounding – Addition/Subtraction
Problems 40(a-b), 41(a-b) page 53
Problems 92(a,b,d) page 63
Rounding – Multiplication/Division
Problems 42(a-d), 43(a-d), 44 p. 54
Problems 92(c,e) page 63
Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis (Conversions)
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.2 Scientific Notation and Dimensional
Analysis
Scientists often solve problems using
dimensional analysis.
• Use dimensional analysis (aka “the factor-label
method” or conversion factor) in a numerical
problem to convert a given quantity to different
units.
Section 2.2 Scientific Notation and
Dimensional Analysis
Key Concepts
• Dimensional analysis uses conversion factors to solve
problems.
Dimensional Analysis
(Conversions / Factor-Label)
In dimensional analysis always ask 3
questions:
1. What data are we given?
2. What quantity do we need?
3. What conversion factors are
available to take us from what we
are given to what we need?
Dimensional Analysis
Also referred to as factor-label method
1. Start with given quantity with its units
2. Multiply by conversion factors until
desired final units are obtained
3. Make sure units cancel in converting
Example: convert 48 km to meters
48 km  1000 m
1 km
= 48000 m = 4.8  104 m
Dimensional Analysis
Convert quantity 2.3 x 10-8 cm to
nanometers (nm)
Determine conversion factors
Centimeter (cm)  Meter (m)
1 cm = 0.01 m = 1 x 10-2 m
Meter (m)  Nanometer (nm)
1 x 10-9 m = 1 nm
Dimensional Analysis
Setup equation so cm and m units cancel
out leaving only nm
 m   nm 
  
 
2.3  10 cm  
 cm   m 
8
Fill-in values for conversion factors and
solve equation
 0.01m  
1 nm 
  
  0.23nm
2.3  10 cm  
9
 1 cm   1  10 m 
8
Dimensional Analysis
Convert quantity 14 m/s to miles per
hour (mi/hr)
Determine conversion factors
Meter (m)  Kilometer (km)
Kilometer(km)  Mile(mi)
1 mile = 1.6093 km 1000m = 1 km
Seconds (s)  Minutes (min)
Minutes (min)  Hours (hr)
60 sec = 1 min
60 min = 1 hr
Dimensional Analysis
Setup equation so m, km, s, and min cancel
out leaving only miles and hours

14m / s  

km  
  
m 
mi  
  
km  
s  
  
min  
min 
 
hr 
Fill-in values for conversion factors and
solve equation
 1 km  
1 mi   60 s   60 min 
  
  
  
 
14 m / s  
 1000 m   1.6093km   1 min   1 hr 
 31m i / hr
Multiple Conversion Factors
Convert 100 km/h to m/s
100 km  1000 m  1 h  1 min
h
km 60 min 60 s
= 27.8 m/s
= 30 m/s (to have 1 SF)
Multiple Conversion Factors
Baseball thrown at 89.6 miles per hour
Speed in meters per second?
mile/h
m/h
m/s
1 mile = 1.609 km; 60 s = 1 min; 60 min = 1 h
mile
km
103 m
speed = 89.6 h  1.609 mile  km
= 1.44x105
m
m
1h
1 min
h  60 min  60 s = 40.0 s
3 SF
Units Raised to a Power
Conversion factor must also be raised to
that power
Area of circle = 28 in2
in2
1 in = 2.54 cm
Area in cm2?
cm2
(1 in)2 = (2.54 cm)2
2
2
(2.54
cm)
6.45
cm
2
Area= 28 in2 
=
28
in
(1 in)2
1 in2
Area = 1.8 x 102 cm2
2 SF
Unit Conversion w Powers: mL to cm3
liter (L) defined = cubic decimeter (dm3)
milliliter (mL) = 10-3 L
= 10-3 L  1 dm3/L
= 10-3 dm3
= 10-3 dm3  (10-1 m/dm)3
= 10-3 dm3  10-3 m3/dm3
= 10-6 m3
= 10-6 m3  (1 cm/10-2 m)3
= 10-6 m3  1 cm3/10-6 m3
= 1 cm3
Dimensional Analysis
Convert 31,820 mi2 to square meters
(m2)
Determine conversion factors
Mile (mi)  kilometer (km)
1 mile = 1.6093 km
kilometer (km)  meter (m)
1000 m = 1 km
Dimensional Analysis
Setup equation so mi2 and km2 cancel out
leaving only m2 (must have squares)

31,820m i  

2
2
2
km  
  
mi 
m 
 
km 
Fill-in values for conversion factors and
solve equation
2
2
 1.6093km   1000m 
  
 
31,820m i  
1 m i   1 km 

2
2
6
2




2
.
5898
km
1

10
m
2
10 22
10



31,820m i  


8
.
2407

10
8.24110
m
2
2



1m i
1 km

 

Unit Conversion with Powers
Convert 531 lb/ft3 to units of g/cm3
1 in = 2.54 cm, 12 in = 1 ft, 1 kg = 2.2046 lb
531 lb/ft3 x (1 kg/2.2046 lb) x (103 g/kg)
= 2.409x105 g/ft3
2.409x105 g/ft3 x (ft/12 in)3 x (in/2.54 cm)3
2.409x105 g/ft3 x (ft3/1728 in3) x (in3/16.39
cm3)
= 8.507 g/cm3
= 8.51 g/cm3
3 SF
Practice
Dimensional Analysis (Conversions)
Practice Problems 17(a-c), 18(a-b),
19(a-h), 20 page 45
Practice Problems 21-23 page 46
Problems 27, 28, 30 page 46
Problems 80(a-f), 81- 84 page 63
Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.1 Units of Measurement
Objectives
• Show all work for a problem following the ProblemSolving Process described on page 38 in example
problem 2.1 while utilizing dimensional analysis,
significant figures, rounding and standard algebra
skills.
Word Problems in Chemistry
Will need to solve variety of such
problems
Exists standard approach, which will
also be used in physics
Illustrated in text using problem
involving density
Word Problem Involving Density
Problem 2.1, page 38
Sample of aluminum placed
in 25 mL graduated cylinder
containing 10.5 mL water.
Water level rises to 13.5 mL.
Mass of aluminum sample?
density of Al = 2.7 g/cm3
(also see Table R-7, page 971 for data)
Required Problem-Solving Steps
Read problem; make sure you understand
what is being asked
Analyze problem to find unknown (mass)
State defining equation (d = m/V)
Re-write equation to solve for unknown
• mass = volume  density
• m = V  d (use of symbols preferred)
Show all intermediate work with units
Evaluate final answer – check that units and
significant figures are correct
Word Problem 2.1
Solution shown on following slide differs
from that shown on page 38
Symbols (m) are used in place of words
(mass)
Subscripts used to convey additional
information about the symbol (VAl instead of
“volume of sample”; VAl vs VW to distinguish
volume of aluminum from measured volume
of water)
Prefer use of symbols in your work
Subscripts optional
Review of Problem 2.1
Al=Aluminum W=Water f=final i=initial
dAl = 2.7 g/cm3
VAl = DVw = Vwf – Vwi (water displacement)
VAl = 13.5 mL – 10.5 mL = 3.0 mL
dAl = mAl / VAl (main formula; must include)
mAl = dAl  VAl (formula solved for unknown)
mAl = 2.7 g  3.0 mL = 8.1 g of Al
mL
You must show symbols, values, and units
in all your work using the above method
Practice
Working word problems
Practice Problems 1 – 3 page 38
Problem 9, page 39
Problems 66 – 69 page 62
Problems 104, 105, 107 page 64
Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard
problem solving technique illustrated
using density]
2.2 Scientific Notation and Dimensional
Analysis
2.3 Uncertainty in Data
2.4 Representing Data
Section 2.4 Representing Data
Graphs visually depict data, making it
easier to see patterns and trends.
Objectives
• Create graphs to reveal patterns in data.
• Interpret data presented in graphs.
• Identify dependent and independent variables
• Create a properly labeled line graph from supplied
data (one with a reasonable number of labeled tic
marks, axes with labels, a graph title, and data points
and trend lines that are easily viewed and interpreted).
Section 2.4 Representing Data
Objectives (cont)
• Determine the numerical value and units of the slope
of a straight line that is presented in a line graph.
• Define and distinguish between the processes of
interpolation and extrapolation and use them to obtain
predicted values from a trend line.
Section 2.4 Representing Data
Key Concepts
• Circle graphs show parts of a whole. Bar graphs show
how a factor varies with time, location, or
temperature.
• Independent (x-axis) variables and dependent (y-axis)
variables can be related in a linear or a nonlinear
manner. The slope of a straight line is defined as
rise/run, or ∆y/∆x.
• Because line graph data are considered continuous,
you can interpolate between data points or
extrapolate beyond them.
Pie Chart
Good for showing breakdown of
quantities that add to 100%
Bar Graph
Trends of quantities versus a discrete
variable (month, year, sample number,
etc.)
Line Graph
Trends of quantities versus a continuous
variable (mass, time, volume,
temperature)
Line Graph
Independent variable (variable
deliberately changed by experimenter)
on x axis
Dependent variable on y axis
Term “plot of A versus B” means that
B = independent variable (x)
A = dependent (y)
Line Graphs
Graph should have title, axes labels
with units, tick marks, data markers,
and legend (if more than one set of
data is plotted on same graph)
If least squares/best-fit line used, get
slope from
Slope = y2-y1 = Dy
x2-x1 Dx
or get directly from software
Pay attention to the units of the slope
Line Graphs
Graph
displaying both
data points and
best fit line
Negative slope
(Dy < 0)
Line Graph - Excel
In MS Excel, making a “line” chart (graph)
will create a graph with a discrete x axis –
not appropriate for most scientific uses
In Excel, make a scatter chart – it will have
the necessary continuous x axis
When plotting data, use point markers
(circles, etc.) only – do not connect points
When plotting lines use lines (solid, dashed,
etc.) but no point markers
If variable have units, make sure they
appear in axis labels – “Length (m)”
Interpolation / Extrapolation
If have trendline (does not have to be a
straight line), points on trendline curve
considered to be continuous
Interpolation – reading value from a
point on curve that falls between
recorded data points
Extrapolation - reading value from a
point on curve that extends beyond
recorded data points (potentially
dangerous, especially if overdone)
Interpolation / Extrapolation
Extrapolated
point at
elevation =
700 m
Interpolated
point at
elevation =
350 m
Practice
Line graphs
Problems 53, 54, 58 page 58
Problem 111 page 64
End of Chapter