Chapter 1 Notes for Chemistry Exam

Chapter 1: Matter, Measurements, and Calculations
1.1 What is Matter?:
• Matter – anything that has mass and occupies space
• Mass – a measurement of the amount of matter in an object
• Weight – a measurement of the gravitational force acting on an
object
Mass is constant, weight varies depending on gravity

1.2 Properties of Matter and Changes in Matter:
The properties of matter are classified into two types:
• Physical – can be observed or measured without changing or
trying to change the composition of the matter in question
• Chemical – Properties matter demonstrates when attempts are
made to change it into new substances
Likewise, the changes matter undergoes is also classified into two
types:
• Physical Changes – Don’t change the composition of the
substance (reshaping, resizing)
• Chemical Changes – Convert matter into a new substance
(burning, digesting)
Dividing a substance into smaller and smaller amounts doesn’t change
its’ properties, so how small can you go?

1.3 A Model of Matter:
A molecule is the smallest part of a pure substance that has the
properties of that substance and is capable of a stable existence. It is the
limit of physical subdivision for a pure substance.
What if I do want to change the properties of the substance?
Can molecules be chopped into smaller pieces? - Yes, atoms
Atoms – The limit of chemical subdivision of matter, the basic building
block for all matter, the smallest particles of matter that can result from
chemical changes
John Dalton was the first to develop the atomic theory of matter. It
states:
— All matter is made up of tiny particles called atoms
— All atoms of a specific element are identical to each other and
different from atoms of a different element
— All compounds are combinations of atoms of two or more
elements
— Every molecule of a specific compound always contains the
same number of atoms of each kind of element found in the
compound
— In chemical reactions, atoms are rearranged, separated, or
combined, but are never created or destroyed

1.4 Classifying Matter:
Is it a pure substance or a mixture of substances?
Pure substances have constant compositions throughout, fixed
properties and can’t be separated into simpler substances
Mixtures are not constant throughout and can be separated into two or
more substances by physical means. Their properties vary with their
composition.
Pure substances and some mixtures (like salt water) can be homogenous
throughout.
Homogenous matter has a uniform appearance and the same properties
throughout.
A solution is a homogenous mixture of two or more substances.
Mixtures in which properties and appearance are not the same
throughout are called heterogeneous matter.
Element – A pure substance consisting of only one kind of atom.
Ex: gold (Au), oxygen (O2), graphite (C)
Compound – Pure substance consisting of two or more kinds of atoms.
Ex: water (H2O), carbon dioxide (CO2)
A compound can be chemically split into elements or other compounds.

1.6 The Metric System:
The official system of measurement used in science.
The metric system is a decimal system in which larger and smaller units
of quantity are all related to each other by factors of 10.
basic units of measurement – A specific unit from which other units
for the same quantity are obtained by multiplication or division.
As the amount increases or decreases, a prefix is added to the basic unit
to indicate a factor of 10 change from the basic unit.
Example: a 5K race is a distance of 5 kilometers, or 5 thousand meters.
The meter is the basic unit of distance in the metric system and the
prefix kilo means 1000, so 5 kilometers is 5,000 meters.
Common prefixes of the metric system (Table 1.2, pg 13):
mega (M) - 1 million times basic unit, 106 x basic unit
kilo (k) – 1 thousand times basic unit, 103 x basic unit
deci (d) – 1/10th the size of the basic unit, 10-1 x basic unit
centi (c) – 1/100th the size of the basic unit 10-2 x basic unit
milli (m) – 1/1000th the size of the basic unit 10-3 x basic unit
micro (μ) – 1/1,000,000 the size of the basic unit 10-6 x basic unit
nano (n) – 1/1,000,000,000 the size of the basic unit 10-9 x basic unit
pico (p) – 1/1,000,000,000,000 the size of the b-unit 10-12 x basic unit

Basic units for commonly used measurements: See table 1.3, pg 17
— Length – meter (m)
— Volume – cubic decimeter (dm3)
— Mass – kilogram (kg)
— Temperature – Kelvin (K)
— Energy – joule (J)
— Time – second (s)
All temperatures used in equations for this and all other science courses
must be converted into degrees Kelvin.
Depending on the scale, water boils at 212 oF, 100 oC, and 373 K; and
water freezes at 32 oF, 0 oC, and 273 K.
To convert between the different temperature scales:
oC = 5/9(oF-32) oF = 9/5(oC) + 32
oC = K – 273 K = oC + 273
Chem 120/121 Lecture Notes, Chapter 1. Friday, Jan. 16th
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1.7 Large and small numbers:
When using really big or small numbers, scientific notation is used.
Scientific notation shows numbers as a product of a nonexponential term
and an exponential term, M x 10n.
The nonexponential term, M, is a whole number between 1 and <10,
written with the decimal to the right of the first non-zero digit. This is
the standard position.
The exponential term is a 10 raised to a whole number exponent that
may be either positive or negative. The value of n is the number of
places the decimal must be moved from the standard position in M to be
at the original position in the number when written normally.
If n is positive, the original position is to the right of the standard
position (the original number is bigger) and if n is negative, the original
position is to the left of the standard position (the original number is
smaller)
To multiply numbers in scientific notation, multiply the nonexponentials
and add the exponents.
Example: (2.2 x 103)(4 x 105) = (2.2 x 4)3+5 = 8.8 x 108
To divide numbers in scientific notation, divide the nonexponentials and
subtract the exponents.
Example: (9 x 107)/(2 x 105) = (9/2)7-5 = 4.52

1.8 Significant Digits:
All measurements contain a certain amount of uncertainty. The uncertainties are
determined by the limits of the measuring device.
Significant figures are the numbers in a measurement that represent the certainty of
the measurement, plus one more number representing an estimate.
Rules for determining the significance of zeros:
1. Zeros not preceded by nonzero numbers are not significant. 0.0037, 2 SDs
2. Zeros located between nonzero numbers are significant. 2.003, 4 SDs
3. Zeros located at the end of a number are significant. 4.0, 2 SDs
When using a measured number in a calculation, the answer can’t have more
certainty than the least certain measured value and should be written to reflect an
uncertainty equal to the most uncertain measurement.
Rules for multiplying and dividing:
The answer must contain the same # of SDs as the quantity with the fewest SDs.
Example: 10.456647447 x 2.0 = 21; Ex. 2) 150.101 / 10.05 = 15.03
Rules for Adding and subtracting: The answer obtained must have the same # of
digits to the right of the zero as the quantity with the fewest number of places to
the right of the zero.
Example: 8.73 + 4 = 13; Ex. 2) 55.120 – 10.11005 = 45.010
Calculators don’t use SDs, you must know how many SDs to use!
Not all numbers used in calculations have uncertainty, such as exact numbers,
these numbers don’t determine the number of SDs to be used in calculations.
For example, 1000 meters = 1 kilometer. This is by definition, there is no
uncertainty in this statement, therefore these numbers are not considered when
deciding how many SDs to use in an answer.
Another example of exact numbers are counting numbers, such as a dozen eggs =
12 eggs. No uncertainty, 1 dozen = 12 by definition, not 11.9 or 12.23. Therefore
1 or 12 in this example would not be used to determine the number of SDs to use.

1.9 How to Use Units in Chemistry Calculations:
Most calculations in chemistry will involve converting one numerical
value of some unit into another numerical value of another unit.
Example: how many grams are in 16.8 pounds? Or how many yards are
in 7 kilometers?
The key to solving these problems is the simple 4 step method shown
below:
Step 1) Write down the known or given quantity, including the
numerical value and the units.
Step 2) Leave some space and set the known quantity equal to the units
of the unknown quantity.
Step 3) Multiply the known quantity by one or more factors, such that
the units of the factors cancel each other out and generate the units of the
unknown quantity.
Step 4) AFTER the units are correct, do the math to calculate the
numerical value that goes with the units.
Example: how many yards are in 7 kilometers?
The known quantity is 7 km.
7 km 1000 m 1.094 yd = 7658 yds
1 km 1 m
What about SDs? We can only use 1 SD based on the 7 km, so the
answer is 8,000 yds, or 8 x 103 yds.

1.11 Density of Matter:
Even though two objects may be the same size, they may have
very different weights. Likewise, a pillowcase stuffed with
socks would weigh a lot more than the same pillowcase stuffed
with pennies. These differences in mass and volume result from
the unique densities of the objects.
Density = mass of an object divided by its volume, D = m/V
Densities of solids are often written as g/cm3, Densities of
liquids are often written as g/mL. Either way is correct because
1 mL = 1 cm3
Two objects of roughly the same size, such as a racquetball and
a billiards ball, have drastically different masses. The billiards
ball feels much heavier than the racquetball. Since both
volumes are ~ the same and the mass of the BB is > mass of the
RB, the density of the BB must be > the density of the RB.
Given the formula D = m/V and the values for any two of the
three properties, you should be able to solve for the unknown
quantity by rearranging the equation to solve for the unknown.
Example: The density of iron metal is 7.2 g/cm3, if you have a
900 mg sample of iron, what is the volume of that sample?
Step one, rearrange the equation D = m/V, to solve for V.
Multiplying both sides by V gives VD = m. Now divide both
sides by D to give: V = m/D
Using the equation V = m/D, we know m = 900 and D = 7.2,
so: V = 900/7.2 = 125 cm3
WRONG!!!! Use must write down the units!!!!
900 mg cm3 = 125 mg cm3
7.2 g g
mg cm3/g are NOT the units of Volume! You must FIRST get
your units right and the math will do itself!
We are solving for V, or Volume. The units of Volume are cm3.
We must get rid of the mg and g in the answer, but how?
What is the relationship between mg and g? 1 gram = 1000
milligrams
Can we use this relationship to eliminate the mg and gram units
in our calculation?
900 mg cm3 1 g = 0.125 cm3
7.2 g 1000 mg
Is this the correct number of significant figures?
No, the answer should have 2 SDs since the density is given as
7.2 g/cm3
Therefore, the correct answer for the volume of 900 mg of iron
powder is 0.13 cm3