Chapter 2: Scientific Measurement
Download
Report
Transcript Chapter 2: Scientific Measurement
Chapter 2:
Scientific Measurements
Chemistry: The Molecular Nature
of Matter, 6E
Jespersen/Brady/Hyslop
Properties
Characteristics used to classify matter
Physical properties
Can be observed without changing chemical
makeup of substance
Ex. Gold metal is yellow in color
Sometimes observing physical property causes
physical change in substance
Ex. Melting point of water is 0 °C
Measuring melting temperature at which
solid turns to liquid
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
2
Solids:
States of Matter
Fixed shape & volume
Particles are close together
Have restricted motion
Liquids:
Fixed volume, but take container shape
Particles are close together
Are able to flow
Gases:
Expand to fill entire container
Particles separated by lots of space
Ex. Ice, water, steam
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
3
States of Matter
Physical Change
Change from 1 state to another
Physical States
Important in chemical equations
Ex. 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)
Indicate after each substance with abbreviation
in parentheses
Solids = (s)
Liquids = (ℓ)
Gases = (g)
Aqueous solutions = (aq)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
4
Chemical Properties
Chemical change or reaction that substance
undergoes
Chemicals interact to form entirely different
substances with different chemical & physical
properties
Describes behavior of matter that leads to
formation of new substance
“Reactivity" of substance
Ex. Iron rusting
Iron interacts with oxygen to form
new substance
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
5
Learning Check: Chemical or
Physical Property?
Chemical Physical
X
Magnesium metal is grey
Magnesium metal tarnishes in air
X
X
Magnesium metal melts at 922 K
Magnesium reacts violently with
hydrochloric acid
Jespersen/Brady/Hyslop
X
Chemistry: The Molecular Nature of Matter, 6E
6
Your Turn!
Which one of the following represents a physical
change?
A. when treated with bleach, some dyed fabrics
change color
B. grape juice left in an open unrefrigerated
container turns sour
C. when heated strongly, sugar turns dark brown
D. in cold weather, water condenses on the inside
surface of single pane windows
E. when ignited with a match in open air, paper
burns
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
7
Intensive vs. Extensive Properties
Intensive properties
Independent of sample size
Used to identify substances
Ex. Color
Density
Boiling point
Melting point
Chemical reactivity
Extensive properties
Depend on sample size
Ex. volume & mass
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
8
Identification of Substances by
Their Properties
Ex. Flask of clear liquid in lab. Do you drink it?
What could it be?
What can we measure to determine
if it is safe to drink?
Density
Melting Point
Boiling Point
Electrical conductivity
Jespersen/Brady/Hyslop
1.00 g/mL
0.0 °C
100 °C
None
Chemistry: The Molecular Nature of Matter, 6E
9
Gold or “Fool’s Gold?”
Can test by heating in flame
A. Real gold
Nothing happens
B. Pyrite (Fool’s Gold)
Sputters
Smokes
Releases foul-smelling fumes
Due to chemical ability to react chemically with
oxygen when heated
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
10
Your Turn!
Which of the following is an extensive property?
A. Density
B. Melting point
C. Color
D. Temperature
E. Mass
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
11
Observations
Fall into 2 categories
1. Quantitative observations
Numeric data
Measure with instrument
Ex. Melting point, boiling point, volume, mass
2. Qualitative observations
Do not involve numerical information
Ex. Color, rapid boiling, white solid forms
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
12
Measurements Include Units!!
1. Measurements involve comparison
Always measure relative to reference
Ex. Foot, meter, kilogram
Measurement = number + unit
Ex. Distance between 2 points = 25
What unit? inches, feet, yards, miles
Meaningless without units!!!
2. Measurements are inexact
Measuring involves estimation
Always have uncertainty
The observer & instrument have inherent physical
limitations
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
13
International System of Units (SI)
Standard system of units used in scientific &
engineering measurements
Metric
7 Base Units
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
14
SI Units
Focus on 1st six in this book
All physical quantities will have units derived from
these 7 SI base units
Ex. Area
Derived from SI units based on definition of area
length × width = area
meter × meter = area
m × m = m2
SI unit for area = square meters = m2
Note: Units undergo same kinds of mathematical
operations that numbers do!
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
15
Learning Check
What is the SI unit for velocity?
Velocity (v)
distance
Velocity units
time
meters
seconds
m
s
What is the SI unit for volume of a cube?
Volume (V) = length × width × height
V = meter × meter × meter
V = m3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
16
Your Turn!
The SI unit of length is the
A. millimeter
B. meter
C. yard
D. centimeter
E. foot
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
17
Table 2.2 Some Non-SI Metric Units Commonly
Used in Chemistry
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
18
Table 2.3 Some Useful Conversions
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
19
Decimal Multipliers
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
20
Using Decimal Multipliers
Use prefixes on SI base units when number is too
large or too small for convenient usage
Only commonly used are listed here
For more complete list see Table 2.4 in textbook
Numerical values of multipliers can be
interchanged with prefixes
Ex. 1 mL = 10–3 L
1 km = 1000 m
1 ng = 10–9 g
1,130,000,000 s = 1.13 × 109 s = 1.13 Gs
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
21
Laboratory Measurements
4 common
1.
2.
3.
4.
Length
Volume
Mass
Temperature
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
22
Laboratory Measurements
1. Length
SI Unit is meter (m)
Meter too large for most
laboratory measurements
Commonly use
Centimeter (cm)
1 cm = 10–2 m = 0.01 m
Millimeter (mm)
1 mm = 10–3 m = 0.001 m
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
23
2. Volume (V)
Dimensions of (length)3
SI unit for Volume = m3
Most laboratory
measurements use V in
liters (L)
1 L = 1 dm3 (exactly)
Chemistry glassware
marked in L or mL
1 L = 1000 mL
What is a mL?
1 mL = 1 cm3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
24
3. Mass
SI unit is kilogram (kg)
Frequently use grams (g) in laboratory as more
realistic size
1
1 kg = 1000 g
1 g = 0.1000 kg =
1000
g
Mass is measured by comparing weight of sample
with weights of known standard masses
Instrument used = balance
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
25
4. Temperature
Measured with thermometer
3 common scales
A.Fahrenheit scale
Common in US
Water freezes at 32 °F and boils at 212 °F
180 degree units between melting & boiling
points of water
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
26
4. Temperature
B. Celsius scale
Rest of world (aside from U.S.) uses
Most common for use in science
Water freezes at 0 °C
Water boils at 100 °C
100 degree units between melting & boiling
points of water
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
27
4. Temperature
C. Kelvin scale
SI unit of temperature is kelvin (K)
Note: No degree symbol in front of K
Water freezes at 273.15 K & boils at 373.15 K
100 degree units between melting & boiling points
Only difference between Kelvin & Celsius scale is
zero point
Absolute Zero
Zero point on Kelvin scale
Corresponds to nature’s lowest possible
temperature
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
28
Temperature Conversions
How to convert
between °F and °C?
tF
9 F
t 32 F
5 C C
Ex. 100 °C = ? °F
tF
9 F
100
5 C
C 32 F
tF = 212 °F
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
29
Temperature Conversions
Common laboratory thermometers are marked in
Celsius scale
Must convert to Kelvin scale
T K (t C 273.15
C)
1K
1 C
Amounts to adding 273.15 to Celsius
temperature
Ex. What is the Kelvin temperature of a solution at
25 °C?
T K ( 2 5 C 273.15
Jespersen/Brady/Hyslop
C)
1K
1 C
= 298 K
Chemistry: The Molecular Nature of Matter, 6E
30
Learning Check: T Conversions
1. Convert 100. °F to the Celsius scale.
tF
tC
9 F
t 32 F
5 C C
tC tF
5
C
100 F 32 F
9 F
5
C
32 F
9 F
= 38 °C
2. Convert 100. °F to the Kelvin scale.
We already have in °C so…
T K (t C 273.15
C)
1K
1 C
( 38 273.15
C)
1K
1 C
TK = 311 K
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
31
Learning Check: T Conversions
3. Convert 77 K to the Celsius scale.
T K (t C 273.15
C)
1K
t C (T K 273.15 K)
1 C
1 C
1K
t C ( 7 7 K 273.15 K)
1 C
1K
= –196 °C
4. Convert 77 K to the Fahrenheit scale.
We already have in °C so
tF
9 F
( 196
5 C
C) 32 F
Jespersen/Brady/Hyslop
= –321 °F
Chemistry: The Molecular Nature of Matter, 6E
32
Your Turn!
In a recent accident some drums of uranium
hexafluoride were lost in the English Channel. The
melting point of uranium hexafluouride is 64.53 °C.
What is the melting point of uranium hexafluoride
on the Fahrenheit scale?
A. 67.85 °F
B. 96.53 °F
tF
C. 116.2 °F
D. 337.5 °F
E. 148.2 °F
tF
9 F
t 32 F
5 C C
9 F
64 . 53
5 C
Jespersen/Brady/Hyslop
C 32 F
Chemistry: The Molecular Nature of Matter, 6E
33
Uncertainties in Measurements
Measurements all inexact
Contain uncertainties or errors
Sources of errors
Limitations of reading instrument
Ways to minimize errors
Take series of measurements
Data clusters around central value
Calculate average or mean values
Report average value
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
34
Limits in Reading Instruments
Consider 2 Celsius thermometers
Left thermometer has markings
every 1 °C
T between 24 °C & 25 °C
About 3/10 of way between marks
Can estimate to 0.1 °C = uncertainty
T = 24.3 0.1 °C
Right thermometer has markings
every 0.1 °C
T reading between 24.3 °C & 24.4 °C
Can estimate 0.01 °C
T = 24.32 0.01 °C
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
35
Limits in Reading Instruments
Finer graduations in markings
Means smaller uncertainties in measurements
Reliability of data
Indicated by number of digits used to represent it
What about digital displays?
Mass of beaker = 65.23 g on digital balance
Still has uncertainty
Assume ½ in last readable digit
Record as 65.230 0.005 g
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
36
Significant Figures
Scientific convention:
All digits in measurement up to & including
1st estimated digit are significant.
Number of certain digits plus 1st uncertain digit
Digits in measurement from 1st non-zero
number on left to 1st estimated digit on right
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
37
Rules for Significant Figures
1. All non-zero numbers are significant.
Ex. 3.456
has 4 sig. figs.
2. Zeros between non-zero numbers
are significant.
Ex. 20,089
or
2.0089 × 104
has 5 sig. figs
3. Trailing zeros always count as significant if
number has decimal point
Ex. 500.
or
5.00 × 102
has 3 sig. figs
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
38
Rules for Significant Figures
4. Final zeros on number without decimal point
are NOT significant
Ex. 104,956,000
or
1.04956 × 108
has 6 sig. figs.
5. Final zeros to right of decimal point are
significant
Ex. 3.00
has 3 sig. figs.
6. Leading zeros, to left of 1st nonzero digit, are
never counted as significant
Ex. 0.00012
or
1.2 × 10–4
has 2 sig. figs.
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
39
Learning Check
How many significant figures does each of the
following numbers have?
scientific notation # of Sig. Figs.
1. 413.97
2. 0.0006
4.1397 × 102
5
6 × 10–4
1
3. 5.120063
4. 161,000
5. 3600.
5.120063
7
1.61 × 105
3
3.6 × 103
2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
40
Your Turn!
How many significant figures are in 19.0000?
A. 2
B. 3
C. 4
D. 5
E. 6
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
41
Rounding to Correct Digit
1. If digit to be dropped is greater than 5, last
remaining digit is rounded up.
Ex. 3.677 is rounded up to 3.68
2. If number to be dropped is less than 5, last
remaining digit stays the same.
Ex. 6.632 is rounded to 6.63
3. If number to be dropped is 5, then if digit to left of
5 is
a.Even, it remains the same.
Ex. 6.65 is rounded to 6.6
b.Odd, it rounds up
Ex. 6.35 is rounded to 6.4
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
42
Scientific Notation
Clearest way to present number of significant
figures unambiguously
Report number between 1 & 10 followed by correct
power of 10
Indicates only significant digits
Ex. 75,000 people attend a concert
If rough estimate?
Uncertainty 1000 people
7.5 × 104
Number estimated from aerial photograph
Uncertainty 100 people
7.50 × 104
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
43
Learning Check
Round each of the following to 3 significant
figures. Use scientific notation where needed.
1.37.459
37.5 or 3.75 × 101
2.5431978
5.43 × 106
3.132.7789003
133 or 1.33 × 102
4.0.00087564
8.77 × 10–4
5.7.665
7.66
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
44
Accuracy & Precision
Accuracy
How close measurement is to true or
accepted true value
Measuring device must be calibrated
with standard reference to give
correct value
Precision
How well set of repeated
measurements of same quantity
agree with each other
More significant figures = more
precise measurement
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
45
Significant Figures in Calculations
Multiplication and Division
Number of significant figures in answer =
number of significant figures in least precise
measurement
Ex. 10.54 × 31.4 × 16.987 = 5620 = 5.62×103
4 sig. figs. × 3 sig. figs. × 5 sig. figs = 3 sig. figs.
Ex. 5.896 ÷ 0.008 = 700 = 7×102
4 sig. figs. ÷ 1 sig. fig. = 1 sig. fig.
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
46
Your Turn!
Give the value of the following calculation to the
correct number of significant figures.
635 . 4 0 . 0045
2 . 3589
A. 1.21213
B. 1.212
C. 1.212132774
D. 1.2
E. 1
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
47
Significant Figures in Calculations
Addition and Subtraction
Answer has same number of decimal places as
quantity with fewest number of decimal
places.
Ex.
4 decimal places
12.9753
Ex.
319.5
+ 4.398
336.9
1 decimal place
3 decimal places
1 decimal place
397
– 273.15
124
0 decimal places
2 decimal places
0 decimal place
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
48
Your Turn!
When the expression,
412.272 + 0.00031 – 1.00797 + 0.000024 + 12.8
is evaluated, the result should be expressed as:
A. 424.06
B. 424.064364
C. 424.1
D. 424.064
E. 424
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
49
Exact Numbers
Number that come from definitions
12 in. = 1 ft
60 s = 1 min
Numbers that come from direct count
Number of people in small room
Have no uncertainty
Assume they have infinite number of significant
figures.
Do not affect number of significant figures in
multiplication or division
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
50
Learning Check
For each calculation, give the answer to the correct
number of significant figures.
1.10.0 g + 1.03 g + 0.243 g = 11.3 g or
1.13 × 101 g
2.19.556 °C – 19.552 °C =
3.327.5 m × 4.52 m =
1.48 × 103 m
4.15.985 g ÷ 24.12 mL =
Jespersen/Brady/Hyslop
0.004 °C or
4 × 10–3 °C
0.6627 g/mL
or 6.627 g/mL
Chemistry: The Molecular Nature of Matter, 6E
51
Learning Check
For the following calculation, give the answer to the
correct number of significant figures.
(71.359 m 71.357 m )
1.
(3.2 s × 3.67 s)
(0 .0 0 2 m )
2
(11.7 4 4 s )
= 2 × 10–4 m/s2
(13.674 cm × 4.35 cm × 0.35 cm )
2.
(856 s + 1531.1 s)
(20.8 1 8 6 6 5 cm
3
)
(2 3 87.1 s)
Jespersen/Brady/Hyslop
= 0.87 cm3/s
Chemistry: The Molecular Nature of Matter, 6E
52
Your Turn!
For the following calculation, give the answer to
the correct number of significant figures.
(1 4 .5 cm 1 2 .3 3 4 cm )
(2 .2 2 3 cm 1 .0 4 cm )
A. 179 cm2
(1 7 8.8 4 3 c m )
B. 1.18 cm
(1 .183 c m )
2
C. 151.2 cm
D. 151 cm
E. 178.843 cm2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
53
Dimensional Analysis
Factor-Label Method
Not all calculations use specific equation
Use units (dimensions) to analyze problem
Conversion Factor
Fraction formed from valid equality or equivalence
between units
Used to switch from one system of measurement
& units to another
Given
× Conversion
Quantity
Factor
Jespersen/Brady/Hyslop
= Desired
Quantity
Chemistry: The Molecular Nature of Matter, 6E
54
Conversion Factors
Ex. How to convert a person’s height from 68.0 in
to cm?
Start with fact
2.54 cm = 1 in.
Dividing both sides by 1 in. or 2.54 cm gives 1
2.54 cm
1 in.
1 in.
1 in.
=1
2.54 cm
2.54 cm
1 in.
2.54 cm
=1
Cancel units
Leave ratio that equals 1
Use fact that units behave as numbers do in
mathematical operations
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
55
Dimensional Analysis
Now multiply original number by conversion
factor that cancels old units & leaves new
Given
× Conversion = Desired
Quantity
Factor
Quantity
68.0 in.
2.54 cm
1 in.
= 173 cm
Dimensional analysis can tell us when we have
done wrong arithmetic
68.0 in.
1 in.
2.54 cm
= 26.8 in2/cm
Units not correct
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
56
Using Dimensional Analysis
Ex. Convert 0.097 m to mm.
Relationship is
1 mm = 1 × 10–3 m
Can make 2 conversion factors
1 mm
1 10
3
1 10
m
3
m
1 mm
Since going from m to mm use one on left.
0 . 097 m
1 mm
1 10
3
m
Jespersen/Brady/Hyslop
= 173 cm
Chemistry: The Molecular Nature of Matter, 6E
57
Learning Check
Ex. Convert 3.5 m3 to cm3.
Start with basic equality
1 cm = 0.01 m
Now cube both sides
Units & numbers
(1 cm)3 = (0.01 m)3
1 cm3 = 1 × 10–6 m3
Can make 2 conversion factors
1 cm
1 10
3 .5 m
3
6
3
m
3
1 10
or
1 cm
1 10
6
1 cm
3
6
m
Jespersen/Brady/Hyslop
3
m
3
3
3.5 × 106 Cm3
Chemistry: The Molecular Nature of Matter, 6E
58
Non-metric to Metric Units
Convert speed of light from 3.00×108 m/s to mi/hr
Use dimensional analysis
1 min = 60 s
1 km = 1000 m
3 .00 10
8
m
s
1 . 08 10
hr
12
m
60 min = 1 hr
1 mi = 1.609 km
60 s
1 min
1 km
1000 m
60 min
1 hr
1.08 × 1012 m/hr
1 mi
1.609 km
6.71 × 108 mi/hr
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
59
Your Turn!
The Honda Insight hybrid electric car has a gas
mileage rating of 56 miles to the gallon. What is
this rating expressed in units of kilometers per liter?
1 gal = 3.784 L
A. 1.3 × 102 km L–1
B. 24 km L–1
C. 15 km L–1
D. 3.4 × 102 km L–1
1 mile = 1.609 km
56
mi
gal
1 gal
3.784 L
1.609 km
1 mi
E. 9.2 km L–1
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
60
Law of Multiple Proportions
If 2 elements form more than 1 compound they
combine in different ratios by mass
Same mass of 1 element combines with different
masses of 2nd element in different compounds
Experimentally hard to get exactly same mass of
1 element in 2 or more experiments
Can use dimensional analysis to calculate
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
61
Applying Law of Multiple Proportions
Titanium forms 2 different compounds with
bromine. In compound A we find that 4.787 g of
Ti are combined with 15.98 g of bromine. In
compound B we find that 6.000 g of Ti are
combined with 40.06 g of bromine. Determine
whether these data support the law of multiple
proportions.
Analysis
Need same mass of 1 element & compare
masses of 2nd element
6.000 g Ti for each
How much Br?
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
62
Applying Law of Multiple Proportions
Know:
In compound A:
In compound B:
4.787 g Ti ⇔ 15.98 g Br
6.000 g Ti ⇔ 40.06 g Br
Must find:
6.000 g Ti ⇔ ? g Br (compound A)
Solution:
6.000 g Ti
Compare
1 5.98 g Br
= 20.03 g Br
4 . 787 g Ti
4 0.06 g Br
20 . 03 g Br
2
1
Ratio of small whole numbers
Supports law of multiple proportions
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
63
Density
Ratio of object’s mass to its volume
density
mass
volume
d
m
V
Intensive property (size independent)
Determined by taking ratio of 2 extensive
properties (size dependent)
Frequently ratio of 2 size dependent properties
leads to size independent property
Sample size cancels
Units
g/mL or g/cm3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
64
Learning Check
A student weighs a piece of gold that has a
volume of 11.02 cm3 of gold. She finds the mass
to be 212 g. What is the density of gold?
d
d
m
V
212 g
11 . 02 cm
3
19.3 g/cm3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
65
Density
Most substances expand slightly when heated
Same mass
Larger volume
Less dense
Density slightly as T
Liquids & Solids
Change is very small
Can ignore except in very precise calculations
Density useful to transfer between mass &
volume of substance
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
66
Learning Check
1. Glass has a density of 2.2 g/cm3. What is the
volume occupied by 22 g of glass?
V
m
d
22 g
2 . 2 g/ cm
3
10. g/cm3
2. What is the mass of 400 cm3 of glass?
m d V 2 . 2 g/ cm
Jespersen/Brady/Hyslop
3
400 . cm
3
880 g
Chemistry: The Molecular Nature of Matter, 6E
67
Your Turn!
Titanium is a metal used to make artificial joints. It
has a density of 4.54 g/cm3. What volume will a
titanium hip joint occupy if its mass is 205 g?
A. 9.31 × 102 cm3
B. 4.51 × 101 cm3
C. 2.21 × 10–2 cm3
V
205 g
4 .54 g cm
3
D. 1.07 × 10–3 cm3
E. 2.20 × 10–1 cm3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
68
Your Turn!
A sample of zinc metal (density = 7.14 g cm-3) was
submerged in a graduated cylinder containing water.
The water level rose from 162.5 cm3 to 186.0 cm3
when the sample was submerged. How many grams
did the sample weigh?
mass density volume
A. 1.16 × 103 g
B. 1.33 × 103 g
volume ( 186.0 cm
23.5 cm
C. 23.5 g
D. 1.68 × 102 g
mass 7.14 g cm
3
3
162 . 5 cm )
3
3
23.5 cm
3
E. 3.29 g
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
69
Specific Gravity
Ratio of density of substance to density of water
specific gravity
density of substance
density of water
Unitless
Way to avoid having to tabulate densities in all
sorts of different units
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
70
Learning Check
Concentrated sulfuric acid is sold in bottles with a label
that states that the specific gravity at 25 °C is 1.84.
The density of water at 25 °C is 0.995 g cm–3. How
many cubic centimeters of sulfuric acid will weigh 5.55
kilograms?
Analysis:
5.55 kg sulfuric acid = ? cm3 sulfuric acid
Solution:
density sulfuric acid = specific gravity × density water
dsulfuric acid = 1.84 × 0.995 g/cm3 = 1.83
V sulfuric
acid
m
d
5 . 5 5 kg
0 .995 g/cm
Jespersen/Brady/Hyslop
3
5.58 cm3
Chemistry: The Molecular Nature of Matter, 6E
71
Your Turn!
Liquid hydrogen has a specific gravity of
7.08 × 10–2. If the density of water is 1.05 g/cm3 at
the same temperature, what is the mass of hydrogen in
a tank having a volume of 36.9 m3?
d sulfuric
acid
A. 7.43 × 10–2 g
specific
d sulfuric
B. 2.74 g
gravity
7.08 10
sulfuric
2
acid
dH
2O
1.05 g/cm
3
acid
= 7.43 × 10–2 g/cm3
C. 274 g
D. 2.74 × 106 g
m sulfuric
E. 2.61 × 106 g
Jespersen/Brady/Hyslop
acid
7.43 10
36 . 9 m
2
3
g/cm
3
100 cm
1m
Chemistry: The Molecular Nature of Matter, 6E
3
72
Importance of Reliable Measurements
To trust conclusions drawn from measurements
Must know they are reliable
Must be sure they are accurate
Measured values must be close to true values
Otherwise can’t trust results
Can’t make conclusions based on those results
Must have sufficient precision to be meaningful
So confident that 2 measurements are same
for 2 samples
Difference in values must be close to uncertainty in
measurement
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
73
Learning Check
You have a ring? Is it made of 24K gold?
Calculate density & compare to known
Density of 24 K gold = 19.3 g/mL
Use inaccurate glassware
Volume of ring = 1.0 mL
Use kitchen balance
Mass of ring = 18 1 g
Anywhere between 17 & 19 g
Density range is 17 – 19 g/mL
Could be 24 k gold or could be as low as 18K gold
(density = 16.9 g/mL)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
74
Learning Check (cont)
Use more precise laboratory balance
Mass of ring = 18.153 0.001 g
Use more precise glassware
Volume of ring = 1.03 mL
Density of ring = 18.153 g/1.03mL = 17.6 g/mL
Calculate difference between d24K gold & dring
19.3 g/mL – 17.6 g/mL = 1.7 g/mL
Larger than experimental error in density
~ 0.1 g/mL
Conclude: ring NOT 24 K gold!
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
75