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Matter and Measurement
Chapter 1
Chapter 1
1
The Study of Chemistry
What is Chemistry?
Chemistry is the study of the properties and behavior of
matter.
Matter – anything that occupies space and has mass.
Chapter 1
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Classification of Matter
The basic difference between these states is the
distance between the “bodies.”
• Gas – bodies are far apart and in rapid motion.
• Liquid – bodies closer together, but still able to
move past each other.
• Solid – bodies are closer still and are now held
in place in a definite arrangement.
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3
Classification of Matter
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Classification of Matter
Chapter 1
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Classification of Matter
Pure Substances and Mixtures
Mixture – combination of two or more substances
in which each substance retains its own
chemical identity.
– Homogeneous mixture – composition of this mixture
is consistent throughout.
• Solution (Air, gasoline)
– Heterogeneous mixture – composition of this mixture
varies throughout the mixture.
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Classification of Matter
Separation of Mixtures
Mixtures can be separated by physical means.
– Filtration.
– Chromatography.
– Distillation.
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Classification of Matter
Separation of Mixtures
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Classification of Matter
Pure Substances and Mixtures
It is also possible for a homogeneous substance to be
composed of a single substance – pure substance.
• Element – A substance that can not be separated into
simpler substances by chemical means.
• Atom – the smallest unit of an element that retains a
substances chemical activity.
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Classification of Matter
Elements
• There are 114 elements known.
• Each element is given a unique chemical symbol
(one or two letters).
– Carbon C, Nitrogen N, Titanium Ti
– Notice that the two letter symbols are always capital letter
then lower case letter because:
• CO – carbon and oxygen
• Co – element cobalt
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Classification of Matter
Pure Substances and Mixtures
It is also possible for a homogeneous substance to
be composed of a single substance – pure substance.
• Compound – A substance composed of two or more
elements united chemically in definite proportions.
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Classification of Matter
Compounds
• Formed by combining elements.
• The proportions of elements in compounds are the
same irrespective of how the compound was formed.
Law of Constant Composition (or Law of Definite
Proportions):
– The composition of a pure compound is always the same,
regardless of its source.
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Properties of Matter
Physical and Chemical Changes
Physical Property – A property that can be measured
without changing the identity of the substance.
Example: color, odor, density
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Properties of Matter
Physical and Chemical Changes
Intensive properties – independent of sample size.
Extensive properties - depends on the quantity of the
sample (sample size).
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Units of Measurement
Density
Density – mass per unit volume of an object.
mass
Density 
volume
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Properties of Matter
Physical and Chemical Changes
Physical change – the change in the physical
properties of a substance.
– Physical appearance changes, but the substances
identity does not.
Water (ice)  Water (liquid)
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Properties of Matter
Physical and Chemical Changes
Chemical change (chemical reaction) – the
transformation of a substance into a chemically
different substance.
– When pure hydrogen and pure oxygen react
completely, they form pure water.
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Properties of Matter
Physical and Chemical Changes
Chapter 1
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Units of Measurement
SI Units
• There are two types of units:
– fundamental (or base) units;
– derived units.
• There are 7 base units in the SI system.
• Derived units are obtained from the 7 base SI units.
• Example:
units of distance
Units of velocity 
units of time
meters

seconds
 m/s
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Units of Measurement
SI Units
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Units of Measurement
SI Units
Chapter 1
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Units of Measurement
Temperature
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22
Units of Measurement
Temperature
Kelvin Scale
Used in science.
Same temperature increment as Celsius scale.
Lowest temperature possible (absolute zero) is zero Kelvin.
Absolute zero: 0 K = -273.15oC.
Celsius Scale
Also used in science.
Water freezes at 0oC and boils at 100oC.
To convert: K = oC + 273.15.
Fahrenheit Scale
Not generally used in science.
Water freezes at 32oF and boils at 212oF.
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Units of Measurement
Temperature
Converting between Celsius and Fahrenheit
C 
5
F - 32
9
F 
9
C  32
5
Chapter 1
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Units of Measurement
Volume
• The units for volume are given by (units of length)3.
– i.e., SI unit for volume is 1 m3.
• A more common volume unit is the liter (L)
– 1 L = 1 dm3 = 1000 cm3 = 1000 mL.
• We usually use 1 mL = 1 cm3.
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Units of Measurement
Mass
Mass is the measure of the amount of material in an
object.
– This is not the same as weight which is dependant on gravity.
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Uncertainty in Measurement
• All scientific measures are subject to error.
• These errors are reflected in the number of figures
reported for the measurement.
• These errors are also reflected in the observation
that two successive measures of the same quantity
are different.
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Uncertainty in Measurement
Precision and Accuracy
• Measurements that are close to the “correct” value are
accurate.
• Measurements which are close to each other are
precise.
• Measurements can be
– accurate and precise;
– precise but inaccurate;
– neither accurate nor precise.
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Uncertainty in Measurement
Precision and Accuracy
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29
Uncertainty in Measurement
Significant Figures
• The number of digits reported in a measurement
reflect the accuracy of the measurement and the
precision of the measuring device.
• The last digit to the right in a number is taken to be
inexact.
• In any calculation, the results are reported to the
fewest significant figures (for multiplication and
division) or fewest decimal places (addition and
subtraction).
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Uncertainty in Measurement
Significant Figures
• Non-zero numbers are always significant.
• Zeros between non-zero numbers are always significant.
• Zeros before the first non-zero digit are not significant. Zeros at
the end of the number after a decimal place are significant.
• Zeros at the end of a number before a decimal place are
ambiguous. For this book, it a decimal point is used the zeros
are significant.
– 10,300 has 3 significant figures.
– 10,300. has 5 significant figures.
• Physical constants are “infinitely” significant.
Chapter 1
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Uncertainty in Measurement
Significant Figures
• Multiplication / Division
– The result must have the same number of significant figures as
the least accurately determined data
Example:
12.512 (5 sig. fig.)
5.1 (2 sig. fig.)
12.512 x 5.1 = 64
Answer has only 2 significant figures
Chapter 1
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Uncertainty in Measurement
Significant Figures
• Addition / Subtraction.
– The result must have the same number of digits to the right of
the decimal point as the least accurately determined data.
Example:
15.152 (5 sig. fig., 3 digits to the right),
1.76 (3 sig. fig., 2 digits to the right),
7.1 (2 sig. fig., 1 digit to the right).
15.152 + 1.76 + 7.1 = 24.0.
24.0 (3 sig. fig., but only 1 digit to the right of the decimal point).
Chapter 1
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Uncertainty in Measurement
Rounding rules
• If the leftmost digit to be removed is less than 5,
the preceding number is left unchanged.
“Round down.”
• If the leftmost digit to be removed is 5 or greater,
the preceding number is increased by 1.
“Round up.”
Chapter 1
34
Dimensional Analysis
•
•
•
•
In dimensional analysis always ask three questions:
What data are we given?
What quantity do we need?
What conversion factors are available to take us from
what we are given to what we need?
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Dimensional Analysis
• Method of calculation using a conversion factor.
12 inches  1 foot
12 inches
1 foot
1
or
1
1 foot
12 inches
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Dimensional Analysis
Example: We want to convert the distance 8 in. to feet.
(12in = 1 ft)
 1 ft 
 
8 in  
 12 in 
Chapter 1
37
Dimensional Analysis
Example: We want to convert the distance 8 in. to feet.
(12in = 1 ft)
 1 ft 
  0.67 ft
8 in  
 12 in 
Chapter 1
38
Dimensional Analysis
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
First we will need to determine the conversion factors
Centimeter (cm)  Meter (m)
Meter (m)  Nanometer (nm)
Or
1 cm = 0.01 m
1 x 10-9 m = 1 nm
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Dimensional Analysis
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
1 cm = 0.01 m
1 x 10-9 m = 1 nm
Now, we need to setup the equation where the cm cancels and nm is
left.
 m   nm 
  
 
2.3  10 cm  
 cm   m 
8
Chapter 1
40
Dimensional Analysis
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
1 cm = 0.01 m
1 x 10-9 m = 1 nm
Now, fill-in the value that corresponds with the unit and solve the
equation.
 0.01 m  
1 nm 
  
 
2.3  10 cm  
9
 1 cm   1  10 m 
8
Chapter 1
41
Dimensional Analysis
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
 0.01 m  
1 nm 
  
  0.23 nm
2.3  10 cm  
9
 1 cm   1  10 m 
8
Chapter 1
42
Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
First we will need to determine the conversion factors
Mile (mi)  Meter (m)
Meter (m)  kilometer (km)
Or
1 mile = 1.6093km
1000m = 1 km
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Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
Now, we need to setup the equation where the cm cancels and nm is
left.
1 mile = 1.6093km 1000m = 1 km

31,820 mi  

2
km  
  
mi  
Chapter 1
m
 
km 
44
Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
Now, we need to setup the equation where the cm cancels and nm is
left.
1 mile = 1.6093km 1000m = 1 km
2
2
 km   m 
  
 
31,820 mi  
 mi   km 
Notice, that the units do not cancel, each conversion factor must be
“squared”.
2
Chapter 1
45
Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
2
2
 1.6093 km   1000 m 
  
 
31,820 mi  
1 mi   1 km 

2
Chapter 1
46
Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
2
6
2




2
.
5898
km
1

10
m
2
  
 
31,820 mi  
2
2
1 mi
1 km

 

Chapter 1
47
Dimensional Analysis
Convert the quantity from 31,820 mi2 cm to square meters (m2)
2
6
2




2
.
5898
km
1

10
m
2
10
2



31,820 mi  


8
.
2407

10
m
2
2
 

1
mi
1
km

 

Chapter 1
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Dimensional Analysis
Convert the quantity from 14 m/s cm to miles per hour (mi/hr).
Determine the conversion factors
Meter (m)  Kilometer (km)
Kilometer(km)  Mile(mi)
Seconds (s)  Minutes (min)
Minutes(min)  Hours (hr)
Or
1 mile = 1.6093 km 1000m = 1 km
60 sec = 1 min
60 min = 1 hr
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Dimensional Analysis
Convert the quantity from 14 m/s cm to miles per hour (mi/hr).
1 mile = 1.6093 km
60 sec = 1 min

14m / s  

km  
  
m 
1000m = 1 km
60 min = 1 hr
mi  
  
km  
Chapter 1
s
min
 
  
 
min
hr

 

50
Dimensional Analysis
Convert the quantity from 14 m/s cm to miles per hour (mi/hr).
1 mile = 1.6093 km
60 sec = 1 min
1000m = 1 km
60 min = 1 hr
 1 km  
1 mi   60 s
  
  
14m / s  
 1000 m   1.6093 km   1 min
Chapter 1
  60 min 
  
 
  1 hr 
51
Dimensional Analysis
Convert the quantity from 14 m/s cm to miles per hour (mi/hr).
1 mile = 1.6093 km
60 sec = 1 min
1000m = 1 km
60 min = 1 hr
 1 km  
1 mi   60 s
  
  
14 m / s  
 1000 m   1.6093 km   1 min
 31 mi / hr
Chapter 1
  60 min 
  
 
  1 hr 
52
End of Chapter Problems
4, 10, 14, 20, 26, 34, 42, 60
Chapter 1
53