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
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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
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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)
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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
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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
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X
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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
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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
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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
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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
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Your Turn!
Which of the following is an extensive property?
A. Density
B. Melting point
C. Color
D. Temperature
E. Mass
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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
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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
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International System of Units (SI)
 Standard system of units used in scientific &
engineering measurements
 Metric
 7 Base Units
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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!
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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
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Your Turn!
The SI unit of length is the
A. millimeter
B. meter
C. yard
D. centimeter
E. foot
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Table 2.2 Some Non-SI Metric Units Commonly
Used in Chemistry
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Table 2.3 Some Useful Conversions
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Decimal Multipliers
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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
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Laboratory Measurements
 4 common
1.
2.
3.
4.
Length
Volume
Mass
Temperature
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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Your Turn!
How many significant figures are in 19.0000?
A. 2
B. 3
C. 4
D. 5
E. 6
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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