CPA Chemistry: Summer Assignment
This year in chemistry students will be using a method for calculating known as dimensional
analysis. We use this in stoichiometry in chemistry as well as in metric conversions. In chemistry
we use the metric system in any measurements we need to make. This year you are going to
learn to make proper measurements and to convert in the metric system using dimensional
analysis. Please read the notes below and then perform the following calculations.
Metric Units and Conversions

Metric Units
a. Metric measurement is based upon a few fundamental, standard units of measure known as “base
units”.
i. Base units are units of measure that represent different types of measurement. The common
base units used in the metric system are as follows:
1. Meter (distance)
2. Liter (volume)
3. Gram (mass)
b. Each of the base units either represents a standardized measurement of the unit itself or is
accompanied by a prefix to show the scale of the measurement related to the standard of measure used
by the metric system.
i. The prefixes used to scale the measurements can be seen in the table below:
Metric Prefixes
Abbreviation
T
G
M
k
h
da
BASE
d
c
m
µ
n
p
Prefix Name
tera
giga
mega
kilo
hecto
deca
deci
centi
milli
micro
nano
pico
Value
1012
109
106
103
102
101
101
102
103
106
109
1012
Meaning
trillion
billion
million
thousand
hundred
ten
tenth
hundredth
thousandth
millionth
billionth
trillionth
*Please note that the values for all of the prefixes are shown as positive values due to the comparison of the base unit being smaller/larger than the prefix.*
Larger units

smaller units
Equalities
a. When comparing/converting between units within the same base unit, equalities can be used to relate
the units. The equalities will be built from the metric prefix table above.
b. Building equalities:
i. Building equalities is done by relating the value of a prefix to the base unit of measurement.
ii. The larger unit (prefix or base) will receive the number ‘1’ and the other unit of comparison will
receive the value of the prefix.
1. EX: 1 Tm = 1012 m
a. Explanation- Since ‘tera’ is located to the left of the base unit ‘meter’ it is larger
and will receive the number ‘1’. The base unit will then receive the value for
‘tera’.
i. This allows us to be able to reason that, “in 1 terameter, there are 1012
meters”.
2. EX: 1 g = 103 mg
a. Explanation- Since ‘gram’ is located to the left of the prefix ‘milli’ it is larger and
will receive the number ‘1’. The prefix will then receive the value for ‘milli’.
i. This allows us to be able to reason that, “in 1 gram, there are 103
milligrams”.
TEACHERS NOTE: ** Placing the number ‘1’ with the larger unit allows us to more easily recognize how large a value is
because we are familiar with meters, liters, and grams. We may never experience a unit on the scale of teraliters or
picoliters. So, this method allows us to gain an understanding for exactly how large or small a quantity may be. It also
eliminates the necessity for negative exponents in the table listed above.**
c. Metric Equalities – you will be required to have all of these memorized. Also, meters can be substituted
with any of the other base units mentioned on page 1 of these notes.
1 Tm = 1012 m
1 Gm = 109 m
1 dam = 101 m
1 Mm = 106 m
1 m = 101 dm
1 m =102 cm
1 m = 106 µm

1 m = 109 nm
1 km = 103 m
1 hm = 102 m
1 m = 103 mm
1 m = 1012 pm
Converting Units
a. We are going to use a process known as dimensional analysis to covert between different units given a
certain measurement.
i. The major focus of this process is cancelling units. This is also known as “factor-label method”
and sometimes learned as “unit multipliers”.
b. Examples:
1. How many grams are present in 61 dg?
a. Given: 61 decigrams
Unknown: # grams
61 𝑑𝑔 ×
1𝑔
101 𝑑𝑔
=
6.1 𝑔
2. How many nanoliters are present in 5 L?
a. Given: 5 liters
Unknown: # nanoliters
5𝐿 ×
109 𝑛𝐿
1𝐿
=
5 × 109 𝑛𝐿 𝑜𝑟 5,000,000,000 𝑛𝐿
**Prefix  base calculations are going to be one-step problems.**
** Prefix  prefix calculations should be done in two steps – convert to base unit first**
3. How many megameters are present in 9x108mm?
a. Given: 9x108mm
Unknown: # Mm
1𝑚
9 × 108 𝑚𝑚 × 103 𝑚𝑚 ×
1 𝑀𝑚
106 𝑚
= 9.0 × 10−1 𝑀𝑚 𝑜𝑟 0.90 𝑀𝑚
4. A vehicle was measured to have a mass of 1,129 kg. How many micrograms is the mass
of this vehicle?
a. Given: 1,129 kg
Unknown: # µg
1,129 𝑘𝑔 ×
103 𝑔
1 𝑘𝑔
×
106 µ𝑔
1𝑔
=
1.129 × 1012 µ𝑔
CPA Summer Practice Problems:
Convert the following showing ALL work. This will be handed in the first day of class. If you have
questions throughout the summer, please e-mail your teacher.
1) How many deciliters are present in 45 mL?
2) How many kilograms are present in 850 g?
3) How many centimeters are present in 175 mm?
4) How many kiloliters are present in 1.2 TL?
5) How many picometers are present in 2.4 cm?
6) How many grams are present in 4,500 mg?
7) How many micrometers are present in 25 cm?
8) How many decagrams are present in 0.005 kg?
9) How many centiliters are present in 0.075 L?
10) How many milligrams are present in 15 g?
11) How many nanometers are present in 0.0078 dm?
12) How many Megagrams are present in 25,600 g?