Download Chemistry Basics - Mr. Grays Physical Science Class

The Scientific Method
The SCIENTIFIC METHOD is the logical way in which a scientist goes about trying to solve a
problem. A student needs to understand the scientific method, and the words that are used to
describe the process. The actual descriptions of the steps of the scientific method will vary from
text to text, but the underlying process remains the same. The steps of the scientific method are
shown below:
1.
2.
3.
4.
5.
6.
State the problem.
Collect observations.
Form a hypothesis.
Test the hypothesis.
Form a theory.
Modify a theory.
Now, let us review a real life example of how you might use the scientific method. Suppose you
notice an area in your front lawn where the grass is not growing correctly. The rest of your lawn
has thick, green grass, but this one area has very sparse grass. This, then, is your problem.
1. State the problem. "Grass won't grow in that area of my lawn!"
You would then go outside and look at that area. What makes that area different from the areas
where the lawn is growing nicely? Does one area get more or less sun? What is the soil like?
Compare as many likely factors that you can think of.
2. Collect observations. "The sparse area is surrounded by several evergreen trees,
which drop needles and block much of the sunlight. The soil appears just as rich as the
soil in other areas, but the pH is lower. All areas seem to be getting similar amounts of
water. The temperature in the shaded area is lower than the areas that are not shaded."
Based on the information that you gathered, and your knowledge of Biology, you are ready to
form a hypothesis. Remember, a hypothesis is an educated guess. It is only your background
knowledge in this subject that separates a true hypothesis from what would merely be a guess.
Now, considering the observations you made, you might decide that pH of the soil in the sparse
area is the problem. You form a hypothesis and put it in what is called "if . . . then" format.
3. Form a hypothesis. "If the pH of the soil was higher, then my grass would grow
properly."
Now you want to design an experiment that can be used to test your hypothesis. It is important
that your experiment be controlled, that you keep all conditions between groups the same, except
for that condition which you are testing. It is also important that you conduct your experiment on
several different samples, so that your results may prove conclusive.
4. Test the hypothesis. "I took 200 small pots and used them to grow 200 samples of
grass. I split the 200 pots into 5 groups of 40, and I adjusted the pH of the soil with
calcium oxide (lime) until the five groups had pH readings of; 3,5,7,9, and 11
respectively. In all of the samples I used the same amount and type of soil and the same
type and number of grass seeds. Each sample was kept in the same room with identical
conditions as far as light, temperature and water."
If you conduct your experiment carefully, you will probably find differences between the groups
of grass that you grew. If don't see anything that leads you to believe that the higher pH would
cause growth problems in your lawn, then you may reject your original hypothesis and form a new
one, maybe one that is based on the difference in sunlight. If your experiment supports your
hypothesis, then you may be on to something, but more testing would be required before you could
say for sure.
In real life, by the time you were done with the above experiment, it may be winter and you
would no longer be worried about your lawn. It is not a realistic way of solving this problem, when
it would be much easier to ask a gardener about the problem, or read more about lawn care.
Although the experiment may not carry over realistically, the scientific method does. You would
still want to change only one thing at a time, when trying to improve the grass in that area. The
lesson is that all problems should be approached in a logical manner.
Measuring Matter
Chemistry, the branch of Science involving the study of matter, is a quantitative subject.
This means that measurements and calculations involving matter are central to the subject at
hand. Matter is sometimes defined as "anything that has mass and volume." Another
definition of matter is "anything that has the property of inertia." This lesson deals with some
of the most basic and important concepts of Chemistry, in that the concepts involved are
necessary to understand the definitions of matter.
Each time that you do a laboratory activity this year you will be collecting data. Mass and
volume are probably the most common types of quantitative measurements that you will take,
so it is important for you to have a clear understanding of what you are measuring. When you
know the mass and volume of a substance, you can find its density. Weight is not something
that you will often have to worry about in Chemistry, but since many students confuse the
concept with mass, I will also explain weight here. The unit called The Mole will also be
introduced here.
Inertia - The inertia of an object represents its ability to resist changes in its motion. This
change in motion could be in terms of speed, or in terms of direction. If you are in a car that
stops suddenly, you continue to move forward, because of your inertia. When a car makes a
sharp turn, you might feel yourself moving to one side, due to your inertia.
Mass - The balances that you use in laboratory measure mass, not weight. Mass is
sometimes defined as the amount of matter in an object. 10.0 grams of gold would contain
twice as many gold atoms as 5.0 grams of gold. Students sometimes confuse mass and
volume because the term "massive" can mean "large" in English. In Chemistry, mass has
nothing to do with size. The SI unit for mass is the kilogram (kg).
Conservation of Mass - One of the basic scientific laws of Chemistry is called the Law of
Conservation of Mass. The law states that matter is neither created nor destroyed in an
ordinary chemical reaction. Now, it turns out that mass and energy can be converted into one
another, in a nuclear reaction, but (unless something goes horribly wrong) we will not be
carrying out any nuclear reactions in the Chemistry lab. This means that no matter what we
do in the Chemistry lab, the mass that we put into the reaction will be the same as the mass
that we get out of the reaction. In other words, in each chemical reaction that we carry out in
lab, the mass of the products will be equal to the mass of the reactants.
Volume - Volume is the amount of space that an object takes up. When you buy a 2-liter
bottle of soda, the soda takes up 2 liters of space. A 200 cm3 sample of gas is twice as large
as a 100 cm3 sample of gas. The volume of liquids can be measured using graduated
cylinders, beakers and flasks. You can determine the volume of regularly shaped objects with
a meter stick. Objects that have an irregular shape are often measured through what is called
"the water displacement method." This means that you determine the volume of the object by
finding out how much water it displaces. In Chemistry we often measure volume in milliliters
(ml) or cubic centimeters (cm3). It is important for you to know that milliliters and cubic
centimeters are equivalent units, so 1 ml = 1 cm3. There are 1000 milliliters in one liter.
Density - Density is the amount of matter in a given unit of volume. It can be measured in
grams per cubic centimeter (g/cm3). It is a measure of how tightly packed the atoms of a
substance are. When we say that ice is less dense than water, we mean that the water
molecules are more tightly packed when they are in the liquid state. The formula for
determining density is:
Submarines, scuba divers and many types of fish can alter their depth in the water by
adjusting their density.
Weight - Weight is a measure of the force of gravitational attraction between two objects, one
of which is usually the earth. The weight of a certain object can change as it moves closer or
further away from the earth. On the moon, objects weigh about 1/6th of what they weigh on
earth. Mass, on the other hand, does not change with location. To gain or lose mass an object
must gain or lose atoms!
The Mole - As you might imagine, we will not be able to measure the mass or volume of
individual atoms or molecules in the Chemistry lab. Molecules are so small that a single
drop of water contains billions and billions of them. Just like eggs are grouped in dozens, and
other items are grouped in grosses, atoms and molecules are grouped in moles. We may not
be able to measure the mass of one water molecule in lab, but we can measure the mass of one
mole of water molecules. In fact, one mole of water has a mass of 18.0 grams. How many
items make up a mole of items? 6.02 x 1023! A mole of helium atoms would be 6.02 x 1023
atoms. A mole of carbon dioxide molecules would be 6.02 x 1023 molecules. Don't let the
big number scare you. You will get use to it!
Classification of Matter
Anything that has mass and volume is matter. Matter is also defined as anything with the
property of inertia. All of the solids, liquids and gases that you may encounter in your daily
life would be classified as some type of matter. You are familiar with the taxonomy of living
things from Biology. Now you will learn how scientists classify matter that makes up
everything.
Table 1-4a Classification of Matter
Matter
Anything with mass and volume.
Substance
Matter with constant composition
Element
Substance
made up of
only one type
of atom
Examples gold, silver,
carbon,
oxygen and
hydrogen
Compound
Two or more
elements that are
chemically
combined
Mixture
Matter with variable composition
Heterogeneous Mixture
Mixtures that are made
up of more than one
phase
Examples - water, Examples - sand, soil,
carbon dioxide,
chicken soup, pizza,
sodium
chocolate chip cookies.
bicarbonate,
carbon monoxide
Homogeneous Mixtures
Also called solutions.
Mixtures that are made
up of only one phase
Examples - salt water,
pure air, metal alloys,
seltzer water.
Definitions:
Substance - A material with a constant composition. This means that the substance is the
same no matter where it is found. NaCl, H2O, Ne, CO2, and O2 are all substances, because
their composition will be the same no matter where you find them. All elements and all
compounds are defined as substances.
Elements - Elements are substances that are made up of only one type of atom. At this time,
there are 113 known elements, most of which are metals. The symbols shown on the periodic
table represent the known elements. Even atoms are made up of smaller particles, but they
are not broken down by ordinary chemical means.
Compounds - Compounds are substances that are made up of more than one type of atom.
Water, for example, is made up of hydrogen and oxygen atoms. Carbon dioxide is made up
of carbon and oxygen atoms. Table salt is made up of sodium and chlorine. Compounds
differ from mixtures in that they are chemically combined. Unlike elements, compounds can
be decomposed, or broken down by simple chemical reactions.
Phase - A phase is any region of a material that has its own set of properties. In a chocolate
chip cookie the dough and the chips have different properties. Therefore they represent
separate phases. Pure gold, which is an element, would only contain one phase. Italian
dressing would clearly represent several phases, while a solution of salt water may only
contain one phase.
Homogeneous Materials - Any material that contains only one phase would be considered
homogeneous. Elements like hydrogen, compounds like sugar, and solutions like salt water,
are all considered homogeneous because they are uniform. Each region of a sample is
identical to all other regions of the same sample.
Mixtures - Mixtures are made up of two or more substances that are physically combined.
The specific composition will vary from sample to sample. Some mixtures are so well
blended that they are considered homogeneous, being made up of only one phase. Other
mixtures, containing more than one phase, are called heterogeneous.
Solutions - Solutions are a special type of homogeneous material, because unlike compounds,
the parts of a solution are physically, not chemically, combined. When you mix a glass of salt
water, the salt does not chemically react with the water. The two parts just mix so well that
the resultant solution is said to be uniform. Ice tea, coffee, metal alloys, and the air we
breathe are some examples of solutions.
Solutions are made up of two parts: The solute, which gets dissolved, and the solvent,
which does the dissolving. In the case of salt water, salt is the solute and water is the
solvent.
Heterogeneous mixtures - Heterogeneous mixtures are made up of more than one phase and
they can be separated physically. The aforementioned chocolate chip cookie, a tossed salad,
sand, and a bowl of raisin bran cereal are all examples of obvious heterogeneous mixtures.
Physical and Chemical Properties and
Changes
The properties of a substance are those characteristics that are used to identify or describe
it. When we say that water is "wet", or that silver is "shiny", we are describing materials in
terms of their properties. Properties can be divided into the categories of physical properties
and chemical properties. Physical properties are readily observable, like; color, size, luster,
or smell. Chemical properties are only observable during a chemical reaction. For
example, you might not know if sulfur is combustible unless you tried to burn it.
Another way of separating kinds of properties is to think about whether or not the size of
a sample would affect a particular property. No matter how much pure copper you have, it
always has the same distinctive color. No matter how much water you have, it always freezes
at zero degrees Celsius under standard atmospheric conditions. Methane gas is combustible,
no matter the size of the sample. Properties, which do not depend on the size of the sample
involved, like those described above, are called intensive properties. Some of the most
common intensive properties are; density, freezing point, color, melting point, reactivity,
luster, malleability, and conductivity.
Extensive properties are those that do depend on the size of the sample involved. A
large sample of carbon would take up a bigger area than a small sample of carbon, so volume
is an extensive property. Some of the most common types of extensive properties are; length,
volume, mass and weight.
Pieces of matter undergo various changes all of the time. Some changes, like an increase
in temperature, are relatively minor. Other changes, like the combustion of a piece of wood,
are fairly drastic. These changes are divided into the categories of Physical and Chemical
change. The main factor that distinguishes one category form the other is whether or not a
particular change results in the production of a new substance.
Physical changes are those changes that do not result in the production of a new
substance. If you melt a block of ice, you still have H2O at the end of the change. If you
break a bottle, you still have glass. Painting a piece of wood will not make it stop being
wood. Some common examples of physical changes are; melting, freezing, condensing,
breaking, crushing, cutting, and bending. Special types of physical changes where any object
changes state, such as when water freezes or evaporates, are sometimes called change of state
operations.
Chemical changes, or chemical reactions, are changes that result in the production of
another substance. When you burn a log in a fireplace, you are carrying out a chemical
reaction that releases carbon. When you light your Bunsen burner in lab, you are carrying out
a chemical reaction that produces water and carbon dioxide. Common examples of chemical
changes that you may be somewhat familiar with are; digestion, respiration, photosynthesis,
burning, and decomposition.
Energy and Chemical Reactions
Energy is usually defined as "the ability to do work." In this definition, the term work is
not being used like we use it in English. So, in order to understand energy, you must
understand work. Work is defined as "the result of a force acting on a body and producing
motion." If you push a desk across the floor you are doing work on the desk because you are
exerting a force that causes motion. You use energy to do the work. The amount of work
done is equal to the force used multiplied by the displacement of the object (W=Fxd). Greater
amounts of work require greater amounts of energy. The SI unit of work or energy is the
joule (J). You will learn much more about work in your study of Physics.
The two basic categories of energy are potential energy and kinetic energy.
Potential Energy
It is probably easiest to think of potential energy as "stored" energy. It is also defined as
"energy of position."
Gravitational potential energy is "stored" energy that an object has due to its weight and
its position with reference to some other point. A bowling ball has more gravitational
potential energy sitting on a shelf, than does a ping-pong ball sitting on the same shelf. The
same bowling ball would have even more gravitational potential energy, with reference to the
floor, if it were on a higher shelf.
Chemical energy is stored in foods and fuels, and can be released when these compounds
undergo chemical reactions. You probably remember, from Biology, how energy is released
from glucose during the process of respiration, as shown below:
C6H12O6 + 6O2 ---> 6H20 + 6CO2 + ENERGY
Kinetic Energy
Kinetic Energy is defined as "the energy of motion." A fast moving car has a great deal of
kinetic energy, based on both its mass and velocity (speed). When a car crashes into the back
of another car, it transfers some of its kinetic energy into the car in front of it. When a car
going 60 mph hits a parked car, the parked car does not move away at 60 mph. In a future
lesson, we will discuss where the rest of the kinetic energy goes.
Billiard balls are good models for kinetic energy as well. Before the break shot, the balls
have no kinetic energy with reference to the table. Energy comes from the moving cue stick
to set the balls in motion.
Thermal Energy is defined as the energy that a substance has due to the chaotic motion of its
molecules. Molecules are in constant motion, and always possess some amount of kinetic
energy. In fact, when you measure the temperature of an object, you are measuring the
average kinetic energy of the molecules of that object. Does that mean when something has a
temperature below 0oC it has negative kinetic energy? Look for the answer to that question
in lesson 2-9.
Conservation of Energy
Similar to the law of conservation of mass, the law of conservation of energy states that
"energy is conserved", or "energy can neither be created nor destroyed." Like the
aforementioned law, this law does not hold true in the case of nuclear reactions, but it does
hold true for the reactions that we encounter in our everyday life.
You know of course that the engine does not create the energy that powers your car. The
engine is a machine that allows the stored chemical energy of gasoline to be transformed into
mechanical energy that drives the wheels of the car. Energy is not always found in a
convenient form, so many of man's inventions are designed to transform one type of energy to
another. Below are a few examples of what I mean.
1. Different types of stoves are used to transform the chemical energy of the fuel (gas, coal,
wood, etc.) into heat energy.
2. Solar collectors can be used to transform solar energy into electrical energy.
3. Wind mills make use of the kinetic energy of the air molecules, transforming it to
mechanical or electrical energy.
4. Hydroelectric plants transform the kinetic energy of falling water into electrical energy.
Now, if energy is always conserved, why does it seem that energy is sometimes lost? For
example, a "break shot" in a game of billiards causes the balls to bounce around on the table
for a period of time. We transfer kinetic energy from the cue stick - to the cue ball - to the
other balls. Eventually the balls on the table stop moving. If energy were conserved,
wouldn't the balls continue to move? Well, energy is lost, but it is not destroyed. Some of the
kinetic energy is transformed into various types of energy which the billiard balls can't make
use of. A large amount of the kinetic energy is turned into heat energy because of the friction
between each ball and the surface of the table. That is why more expensive pool tables are
made with certain materials, which will cause less friction
Chemical Reactions
A chemical reaction is a series of changes that results in the production of one or more new
substances. These chemical changes, which were introduced to you in lesson 1-5, are always
accompanied by a change in energy. That means that either energy is given off during the
reaction, or energy is taken in.
Reactions that release energy are called exothermic. In this type of reaction, the products
have less potential chemical energy than the reactants, because energy was given off in the
form of heat. When you stand next to a barbecue grill, you feel the heat being released by the
combustion reaction that is taking place around the burners. The reaction of the propane gas
found in grills is shown below:
C3H8 + 5O2 ---> 4H2O + 3CO2 + energy
propane + oxygen yields water + carbon dioxide + energy
Reactions, which take in energy, are called endothermic. In this type of reaction, the
products have more potential chemical energy than the reactants. Think of the chemical
reaction that takes place in "cold-packs." A seal is broken that separates two containers with
the plastic bag. As the contents from the separate containers begin to react, energy is
absorbed from the surroundings. If you place the cold-pack on your body, your body begins
to supply some of the energy that is required to get the reaction going. What you experience
as "cold" has to do with the temperature of that area of your body changing as heat flows to
the cold-pack.
Some exothermic reactions require some energy to get them started, but then they release
more energy than they originally took in. Think of the fact that a match requires initial
energy, provided by the friction between it and the sandpaper on the matchbook, to start
burning. Once the match starts burning, it releases more energy than it took in, so the reaction
is still exothermic. The products still have less potential chemical energy than the reactants.
The initial energy that is required to get the reaction to begin is called activation energy.
Heat vs. Temperature
Heat and temperature are two concepts that are often confused. They are related to each other
because they are both related to the concept of thermal energy, as discussed in lesson 1-6. As you
may recall, thermal energy is the energy that a substance possesses due to the, kinetic energy, or
motion of its molecules. Three factors affect the amount of thermal energy that a substance has:
1) Temperature - Temperature is a measure of the average kinetic energy of the molecules of a
substance. An increase in temperature results in an increase in the kinetic energy of the molecules
and an increase in thermal energy. It is fair to say that temperature and thermal energy vary
directly, but they are not the same thing.
2) Mass - Mass is a measure of the amount of matter in a substance, as you recall from lesson 1-3.
It makes sense that a more massive sample will have more thermal energy than a smaller sample, if
all other factors are equal. Imagine the difference in total energy between a spoonful of boiling
water and a vat of boiling water.
3) Specific Heat Capacity - Each material is able to "hold" a certain amount of thermal energy at a
given temperature, due to what we call its specific heat. Think of the wide range of temperatures
that your feet encounter during a day at the beach. The water may seem cold while the sand feels
quite warm. The wood on the boardwalk may feel comfortable, but the blacktop in the parking lot
is burning hot. Things will heat up at different rates, due, in part, to their different specific heat
values.
So, as you see, temperature is one of the factors that affects the thermal energy of a substance.
What is heat? Heat is the transfer of thermal energy from a hotter to a colder object. What we
think of as "cold" is really the absence of heat. An object with at a higher temperature can release
more heat than the same object at a lower temperature, but temperature is only one of the factors
that affect the amount of heat an object can transfer.
The factors that affect the amount of heat are the same as the factors that affect thermal energy,
for reasons that should now be clear to you. Thermal energy is only measurable as heat, during
heat transfer. The amount of heat transferred can be found according to the following formula:
amount of heat transferred = mass x change in temperature x specific heat
It is important to note that the symbols that are used for the formula will vary from textbook to
textbook, but the values that they represent never change. One way to write the heat transfer
formula is shown below:
q = m(T)Cp
Where q = heat transferred, T = the change in temperature and Cp = the specific heat.
This formula will be used in a later lesson, but you should begin to understand the concept now.
Going back to our earlier example, picture a spoonful of 100.0 oC water and a vat filled with water,
also at 100.0 oC. Which would you rather have spilled on you? The water in the vat can transfer
much more heat, despite the fact that its temperature is no higher that the water in the spoon. This
should help you understand that heat and temperature are not the same thing at all.
Elemental Names and Symbols
As of this date, there are about 112 known elements. Each element is made up of its own type
of atom. Not all atoms of an element are identical, but each atom of an element has the same
number of protons. Protons are subatomic particles found in the nucleus of an atom, and the
number of protons in an atom is called the atomic number. Each element has its own unique atomic
number, as well as its own name and symbol.
The names of the elements have various origins. Some are named, in Greek or Latin, for
properties that they possess. Some are named in honor of scientists who made important
discoveries in the field of Chemistry. Others are named for the place where they were first
discovered. Some are even named after gods of mythology.
Since there are only 26 letters in our alphabet, and over 100 elements, combinations of letters had
to be used in order to give each element its own unique symbol. Each symbol is made up of one
capital letter, which may be followed by 0-2 lower case letters. Some elemental symbols, like
oxygen with the symbol "O", will be easy for you to remember. Other elemental symbols, like
sodium with the symbol "Na", will be a bit harder to commit to memory. Unfortunately for us,
some symbols (like Na) are derived from the Latin name for the element.
Chemistry students are often required to memorize some, or all, of the elemental symbols. There
is an important reason for this, and it will save you and your teacher a great deal of class time in the
future. Check with your teacher to see which elemental symbols you will need to memorize for an
upcoming quiz. You can print out and use the chart below, but remember to check with your
teacher to see if it is complete, or if there are other symbols that you need to know. Also, as you go
about studying the symbols, keep in mind that the case of the letter is very important.
Now, be sure to check out the worksheets and the online quizzes!
Table 1-8a
Select Elements and Their Symbols
Aluminum Al
Francium
Argon
Ar
Germanium Ge
Plutonium Pu
Barium
Ba
Gold
Au
Potassium K
Beryllium Be
Helium
He
Radium
Ra
Boron
B
Hydrogen
H
Radon
Rn
Bromine
Br
Iodine
I
Rubidium
Rb
Cadmium Cd
Iron
Fe
Selenium
Se
Calcium
Ca
Krypton
Kr
Silicon
Si
Carbon
C
Lead
Pb
Silver
Ag
Cerium
Ce
Lithium
Li
Sodium
Na
Cesium
Cs
Magnesium Mg
Strontium
Sr
Chlorine
Cl
Manganese Mn
Sulfur
S
Chromium Cr
Mercury
Hg
Tin
Sn
Cobalt
Neon
Ne
Titanium
Ti
Co
Fr
Phosphorus P
Copper
Cu
Nickel
Ni
Curium
Cm
Nitrogen
N
Fluorine
F
Oxygen
O
Zinc
Zn
The International System of Measurements
(SI)
In 1960, the Eleventh General Conference on Weights and Measures was held in Paris.
They adopted a universal system of measurement units called Le Systeme International
d'Unites (French), which is a revised version of the metric system. This International System,
or SI, as it is commonly referred to, is used for commerce and Science around the world.
There are seven SI base units. Everything that is measurable, can be measured by these
base units, or by units derived from these bases. The table below shows the bases, their
international symbols, and what they are used to measure.
Table 2.1a SI Base Units
Base Quantity
Name of unit
Symbol
Length
Meter
m
Mass
Kilogram
kg
Time
Second
s
Electrical Current
Ampere
A
Temperature
Kelvin
K
Amount of Substance
Mole
mol
Luminous Intensity
Candela
cd
Units that are made up of some combination of SI base units are called Derived Units.
Table 2.1b shows some of the derived units that are common in Science.
Table 2.1b Derived Units
Base Quantity
Common Units
Volume
dm3
Density
kg/dm3
Acceleration
m/s2
Force
kg x m/s2
Prefixes are used with the base units in order to increase or decrease the value that they
represent. All of the prefixes represent some factor of 10, and they can be used with any of
the SI base units. Table 1.3 represents some of the most common prefixes, their symbols,
and the number that is used to multiply the base factor by.
Table 2.1c SI Prefixes
Prefix
Symbol
exa-
E
Multiply the base by
1 000 000 000 000 000 000
peta-
P
1 000 000 000 000 000
tera-
T
1 000 000 000 000
giga
G
1 000 000 000
mega
M
1 000 000
kilo
k
1000
hecto-
h
100
deca-
da
10
deci-
d
0.1
centi
c
0.01
milli-
m
0.001
micro-
u
0.000 001
nano-
n
0.000 000 001
pico-
p
0.000 000 000 001
femto-
f
0.000 000 000 000 001
atto-
a
0.000 000 000 000 000 001
Your teacher can tell you which units and prefixes are most important for the course you are
studying. You must become comfortable with converting units and applying prefixes.
Last Modified
Accuracy, Precision and Uncertainty in
Measurement
There is no such thing as a perfect measurement. Each measurement contains a degree of
uncertainty due to the limits of instruments and the people using them. In laboratory
exercises, students are expected to follow the same procedure that scientists follow when they
make measurements. Each measurement should be reported with some digits that are certain
plus one digit with a value that has been estimated.
For example, if a student is reading the level of water in a graduated cylinder that has lines
to mark each milliliter of water, then he or she should report the volume of the water to the
tenth place (i.e. 18.5 ml.) This would show that the 18 mls are certain and the student
estimated the final digit because the water level was about half way between the 18 and 19
mark.
Two concepts that have to do with measurements are accuracy and precision.
The accuracy of the measurement refers to how close the measured value is to the true or
accepted value. For example, if you used a balance to find the mass of a known standard
100.00 g mass, and you got a reading of 78.55 g, your measurement would not be very
accurate. One important distinction between accuracy and precision is that accuracy can be
determined by only one measurement, while precision can only be determined with multiple
measurements.
Precision refers to how close together a group of measurements actually are to each other.
Precision has nothing to do with the true or accepted value of a measurement, so it is quite
possible to be very precise and totally inaccurate. In many cases, when precision is high and
accuracy is low, the fault can lie with the instrument. If a balance or a thermometer is not
working correctly, they might consistently give inaccurate answers, resulting in high precision
and low accuracy.
A dartboard analogy is often used to help students understand the difference between
accuracy and precision. Imagine a person throwing darts, trying to hit the bull's-eye. The
closer the dart hits to the bull's-eye, the more accurate his or her tosses are. If the person
misses the dartboard with every throw, but all of their shots land close together, they can still
be very precise.
You must strive for both accuracy and precision in all of your laboratory activities this
year. Make sure that you understand the workings of each instrument, take each measurement
carefully, and recheck to make sure that you have precision. Without accurate and precise
measurement your calculations, even if done correctly, are quite useless.
Significant Digits or Figures
Significant digits, which are also called significant figures, are very important in Chemistry.
Each recorded measurement has a certain number of significant digits. Calculations done on these
measurements must follow the rules for significant digits. The significance of a digit has to do
with whether it represents a true measurement or not. Any digit that is actually measured or
estimated will be considered significant. Placeholders, or digits that have not been measured or
estimated, are not considered significant. The rules for determining the significance of a digit will
follow.
Rules For Significant Digits
1. Digits from 1-9 are always significant.
2. Zeros between two other significant digits are always significant
3. One or more additional zeros to the right of both the decimal place
and another significant digit are significant.
4. Zeros used solely for spacing the decimal point (placeholders) are
not significant.
Recognizing significant digits will become much easier over time, as you continue to practice
the rules. Below are some examples, which show the number of significant digits in a group of
numbers, and an explanation why the digits are significant.
Table 1.1 Examples of Significant Digits
EXAMPLES
# OF SIG. DIG.
453 kg
3
5057 L
4
5.00
3
0.007
1
COMMENT
All non-zero digits are
always significant.
Zeros between 2 sig. dig.
are significant.
Additional zeros to the
right of decimal and a sig.
dig. are significant.
Placeholders are not sig.
Alternate Rule for Significant Digits
Here is an alternate rule for determining significant digits that Mr. McNamara taught me last
year. He, in turn, learned it from a show on television. If I could credit the person who made it
up, I would. The rule is really a "trick", which might allow students to get the correct answers
without really understanding the concepts. I would recommend that students only use this as a
secondary method, for the purpose of checking their answers.
When you look at the number in question, you must determine if it has a decimal point or not.
If it has a decimal, you should think of "P" for "Present". If the number does not have a decimal
place, you should think of "A" for "Absent".
Example, for the number 35.700, think "P", because the decimal is present.
For the number 6500, you would think "A", because the decimal is absent.
Now, the letters "A" and "P" also correspond to the "Atlantic" and "Pacific" Oceans,
repectively. Now, assume the top of the page is North, and imagine an arrow being drawn
toward the number from the appropriate coast. Once the arrow hits a nonzero digit, it and all of
the digits after it are significant.
Example 1. How many significant digits are shown in the number 20 400 ? (remember that we
use spaces, rather than commas, when writing numbers in Science.
Well, there is no decimal, so we think of "A" for "Absent". This means that we imagine an
arrow coming in from the Atlantic ocean, as shown below;
20 400 


 







 
Let's look at one more example.
Example 2. How many significant digits are shown in the number 0.090 ?
Well, there is a decimal, so we think of "P" for "Present". This means that we imagine an
arrow coming in from the Pacific ocean, as shown below;
 0.090
The first nonzero digit that the arrow will pass in the 9, making it, and any digit to the right
of it significant.
Answer - There are 2 significant digits in the number 0.090
Here are the significant digits, shown in boldface. 0.090
Each number that we record as a measurement contains a certain number of significant digits,
which show accurate or estimated digits. When we do calculations our answers cannot be more
accurate than the measurements that they are based on. We must be careful to follow the
following rules whenever we perform calculations in Chemistry class.
Multiplying and Dividing
RULE: When multiplying or dividing, your answer may only show as many significant
digits as the multiplied or divided measurement showing the least number of significant
digits.
Example: When multiplying 22.37 cm x 3.10 cm x 85.75 cm = 5946.50525 cm3
We look to the original problem and check the number of significant digits in each of the original
measurements:
22.37 shows 4 significant
digits.
3.10 shows 3 significant
digits.
85.75 shows 4 significant
digits.
Our answer can only show 3 significant digits because that is the least number of significant
digits in the original problem.
5946.50525 shows 9 significant digits, we must round to the tens place in order to show only 3
significant digits. Our final answer becomes 5950 cm3.
Adding and Subtracting
RULE: When adding or subtracting your answer can only show as many decimal places as
the measurement having the fewest number of decimal places.
Example: When we add 3.76 g + 14.83 g + 2.1 g = 20.69 g
We look to the original problem to see the number of decimal places shown in each of the
original measurements. 2.1 shows the least number of decimal places. We must round our
answer, 20.69, to one decimal place (the tenth place). Our final answer is 20.7 g
The Factor-Label Method or Dimensional
Analysis
During the course of this year you will be required to solve many types of problems that
involve units. Often, you will be required to change from one unit to another. This is not
actually very hard, because it is something that you do in your day-to-day life. For example,
the running time of a movie may be listed as 90 minutes. You could convert that to hours
using a conversion factor. A conversion factor is an expression for the relationship between
units. In this case you would be using the relationship 1 hour = 60 minutes. You can set up
a factor-label problem as shown below:
Step 1. Show what you are given on the left, and what units you want on
the right.
Step 2. Insert the required conversion factors to change between units.
In this case we need only one conversion factor, and we show it as the fraction,
1hr/60min. We put the units of minutes on the bottom so that they will cancel
out with the minutes on the top of the given.
Step 3. Cancel units where you can, and solve the math.
Of course, most of us can do the above calculation in our heads. This is because we are
very familiar with the units and the conversion factors involved. Not all conversions will be
that easy, but if you follow the steps correctly, there should be little chance for mistake.
Follow the example below.
Example 1. A student determines that the density of a certain material is 4.46 g/cm3.
What would be the density of this material in g/L?
Well, in order to solve this problem you must remember that 1000 cm3 = 1L. Then follow
the same steps as the previous problem.
Step 1. Show what you are given on the left, and what units you want on
the right.
Step 2. Insert the required conversion factors to change between units.
Note that I have changed the "look" of the fractions to show the
cancellation of units more clearly.
Step 3. Cancel units where you can, and solve the math.
Answer - 4460 g/L (note that we are showing the correct number of significant digits.)
Example 2. Imagine that water is leaking from a container, at a rate of 1.2 ml/hour. If this
rate does not change, how many liters of water will be lost in a week?
We can make a list of the conversion factors that we will need.
1 L = 1000 ml
24 h = 1 day
7 day = 1 week
Step 1. Show what you are given on the left, and what units you want on
the right.
Step 2. Insert the required conversion factors to change between units.
Step 3. Cancel units where you can, and solve the math.
We must round to two significant digits, as shown in the original problem.
Answer - 0.20 L/week
Calculating with Scientific Notation
Scientific notation is simply a method for expressing, and working with, very large or
very small numbers. It is a short hand method for writing numbers, and an easy method for
calculations. Numbers in scientific notation are made up of three parts: the coefficient, the
base and the exponent. Observe the example below:
5.67 x 105
This is the scientific notation for the standard number, 567 000. Now look at the number
again, with the three parts labeled.
5.67 x 105
coefficient
base
exponent
In order for a number to be in correct scientific notation, the following conditions must be
true:
1. The coefficient must be greater than or equal to 1 and less than 10.
2. The base must be 10.
3. The exponent must show the number of decimal places that the decimal needs to be
moved to change the number to standard notation. A negative exponent means that the
decimal is moved to the left when changing to standard notation.
Changing numbers from scientific notation to standard notation.
Ex.1 Change 6.03 x 107 to standard notation.
remember, 107 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10 000 000
so,
6.03 x 107 = 6.03 x 10 000 000 = 60 300 000
answer = 60 300 000
Instead of finding the value of the base, we can simply move the decimal seven places to
the right because the exponent is 7.
So, 6.03 x 107 = 60 300 000
Now let us try one with a negative exponent.
Ex.2 Change 5.3 x 10-4 to standard notation.
The exponent tells us to move the decimal four places to the left.
so, 5.3 x 10-4 = 0.00053
Changing numbers from standard notation to scientific notation
Ex.1 Change 56 760 000 000 to scientific notation
Remember, the decimal is at the end of the final zero.
The decimal must be moved behind the five to ensure that the coefficient is less than
10, but greater than or equal to one.
The coefficient will then read 5.676
The decimal will move 10 places to the left, making the exponent equal to 10.
Answer equals 5.676 x 1010
Now we try a number that is very small.
Ex.2 Change 0.000000902 to scientific notation
The decimal must be moved behind the 9 to ensure a proper coefficient.
The coefficient will be 9.02
The decimal moves seven spaces to the right, making the exponent -7
Answer equals 9.02 x 10-7
Calculating with Scientific Notation
Not only does scientific notation give us a way of writing very large and very small
numbers, it allows us to easily do calculations as well. Calculators are very helpful tools,
but unless you can do these calculations without them, you can never check to see if your
answers make sense. Any calculation should be checked using your logic, so don't just
assume an answer is correct. This page will explain the rules for calculating with scientific
notation.
Rule for Multiplication - When you multiply numbers with scientific notation, multiply
the coefficients together and add the exponents. The base will remain 10.
Ex 1. Multiply (3.45 x 107) x (6.25 x 105)
first rewrite the problem as:
(3.45 x 6.25) x (107 x 105)
Then multiply the coefficients and add the exponents:
21.5625 x 1012
Then change to correct scientific notation and round to correct significant digits: 2.16
x 1013
NOTE - we add one to the exponent because we moved the decimal one place to the
left.
Remember that correct scientific notation has a coefficient that is less than 10, but greater
than or equal to one.
Ex. 2. Multiply (2.33 x 10-6) x (8.19 x 103)
rewrite the problem as: (2.33 x 8.19) x (10-6 x 103)
Then multiply the coefficients and add the exponents: 19.0827 x 10-3
Then change to correct scientific notation and round to correct significant digits 1.91
x 10-2
Remember that -3 + 1 = -2
Rule for Division - When dividing with scientific notation, divide the coefficients and
subtract the exponents. The base will remain 10.
Ex. 1 Divide 3.5 x 108 by 6.6 x 104
rewrite the problem as:
3.5 x 108
--------6.6 x 104
Divide the coefficients and subtract the exponents to get:
0.530303 x 104
Change to correct scientific notation and round to correct significant digits to get: 5.3
x 103
Note - We subtract one from the exponent because we moved the decimal one place to
the right.
Rule for Addition and Subtraction - when adding or subtracting in scientific notation,
you must express the numbers as the same power of 10. This will often involve changing
the decimal place of the coefficient.
Ex. 1 Add 3.76 x 104 and 5.5 x 102
move the decimal to change 5.5 x 102 to 0.055 x 104
add the coefficients and leave the base and exponent the same: 3.76 + 0.055 = 3.815 x
104
following the rules for rounding, our final answer is 3.815 x 104
Rounding is a little bit different because each digit shown in the original problem must be
considered significant, regardless of where it ends up in the answer.
Ex. 2 Subtract (4.8 x 105) - (9.7 x 104)
move the decimal to change 9.7 x 104 to 0.97 x 105
subtract the coefficients and leave the base and exponent the same: 4.8 - 0.97 = 3.83 x
105
round to correct number of significant digits: 3.83 x 105
Problem Solving with the 5-Step Method
A Chemistry student is required to solve many different types of problems. Despite
the variety of problems, some general practices will help you when solving any type of
problem. Good problem solving strategies will allow you to tackle many types of
problems, and to develop the confidence that you will need to work at a faster pace. We
will be covering what we call the "5-step method" of problem solving. There are other
methods, but this works as well as any other.
The steps for the 5-step method are as follows;
1. Write down the "given" or the known information. For this step, look over the
question and take out the information that has been provided. This includes any
"constants" or information that the problem assumes that you know, or at least know to
look up. For example, you may be asked to solve a problem which involves knowing the
density of copper. The problem may not actually give you the density of copper, but you
may have that information on a reference table. You might think, "how am I supposed
to know to look up information that is not mentioned in the problem!?!" The truth is, it
is not as bad as it seems. When you use the 5-step method, you will realize when you
don't have enough information to solve a problem. That will be your key that you are
missing a constant.
2. Determine and write down the unknown variable. This is one of the easier steps.
Most people can read a question and determine what the unknown is, or what the
question is asking for.
3. Choose an appropriate equation. You may or may not have a reference sheet with
equations on it when you need to solve a problem. In certain testing situations, you may
have to come up with the equations from memory. In any case, the process of selecting
the appropriate equation involves selecting one that includes some or all of the variables
that you have been given, and only contains one unknown. The unknown is not always
the one that you are looking for in your final answer, if the particular problem involves
more than one equation. If you can come up with an equation that contains the variable
that your question is asking for, and it is the only unknown in the equation, then the
problem can be solved with the one equation.
4. Isolate the unknown in the equation. This involves manipulating the equation
algebraically, so that the only thing on one side of the equal sign represents the physical
quantity that you are solving for. Do this before substituting values for any of the
variables. If you notice more than one unknown in your equation, go back and look at
your reference tables for constants.
5. Plug the known values into the equation, solve for the unknown, round and add
units. Remember to round your final answer according to the rules of significant digits,
and include units.
Now let us see an example using the 5-step method to solve problems.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Step 1. Write down the "given" or the known information.
Ah, I see that this is a problem involving the volume, or amount of space occupied by a
wooden block. I will start by writing the word "given" in my word space. Below this, I
will list what I know, assigning appropriate variables to what I have been given.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Given
V = 258 cm3
W = 21.0cm
H = 13.8 cm
2. Determine and write down the unknown variable.
It is easy to determine the unknown variable in this example. The question clearly
states, "What is the length of a wood block?" To the right of where I wrote the "givens"
in my work space, I will write the word "find" and list the appropriate variable for my
unknown.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Given
V = 258
cm3
W=
21.0cm
H = 13.8
Find
L=?
cm
3. Choose an appropriate equation.
The appropriate equation comes easily to mind. To find the volume of a rectangle, I
need to use:
Volume = Length x Width x Height
or
V=LxWxH
You might notice that the units that come with the values you have been given are often
helpful in determining your equation. The fact that we have cm3 for one value and just
cm for two other values suggests that multiplication has occured.
Now I will write the word "formula" to the right of my other work. Below that, I will
write the formula in its standard form.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Formula
Given
Find
V = 258 cm3
L=?
V=LxWxH
W = 21.0cm
H = 13.8 cm
4. Isolate the unknown in the equation.
Avoid the temptation to plug numbers into the equation now, as most Science
teachers will probably want you to isolate the unknown first. Rewrite the equation with
the unknown on one side of the equal sign.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Given
V = 258 cm3
W = 21.0cm
H = 13.8 cm
Find
L=?
Formula
V=LxWxH
V
L = ---------WxH
5. Plug the known values into the equation, solve for the unknown, round and add units.
Let's rewrite the working equation at the top of our workspace, and show all of our
work below it. Remember to work with units. Once you solve the problem, you must
round according to the rules for significant digits.
Example 1. What is the length of a wood block with a volume of 258 cm3, if the
width of the block is 21.0cm and the height is 13.8 cm?
Given
Find
V = 258 cm3
L=?
W = 21.0cm
H = 13.8 cm
V
258 cm3
L = ---------- = ------------------------- =
WxH
21.0cm x 13.8 cm
V=LxWxH
Formula
V
L = ---------WxH
L = 0.890269151 cm
Answer. Length = 0.890 cm
Note - We rounded our final answer to 3 significant digits because the lowest number of
significant digits in the problems was 3.
Last Modified
Density Calculations
Density is an important intensive property, which can be used to help determine the
identity of an unknown substance. While the mass or the volume of a substance will vary
from sample to sample, the density will remain the same at a given temperature. As you
know, the density of a substance is a measure of how much mass is present in a given unit of
volume. The formula is shown below:
In laboratory exercises, it is easy to find the mass and volume of most solids, so it is
common to solve for density. When dealing with gases, however, it is often easy to find the
volume, but very hard to find the mass. By looking up the density of a known gas in a
reference table, and using the experimental volume, you can calculate the mass using the
equation above. As in any algebraic expression, we can solve for any of the three variables as
long as the other two variables are given or known.
Solving For Density
When solving for density, you would use the formula exactly as it appeared above. Here is
an example where density is the unknown, and the steps for solving the problem:
1. A student determines that a piece of an unknown material has a mass of 5.854 g and a
volume of 7.57 cm3. What is the density of the material, rounded to the correct number of
significant digits?
First: Write the correct formula at the top of your page, and list the knowns and the
unknowns.
D=?
M= 5.854 g
V = 7.57 cm3
Second: Substitute the known values in the problem
Third: Calculate your answer, including units
D = 0.77331571994 g/cm3
Fourth: Round to the correct number of significant figures
D = 0.773 g/cm3
Solving For Mass
When solving for mass, we must take the original formula, and isolate the unknown like so:
Multiply both sides by v
The "v's" cancel out
Isolate for mass
m=vxD
Here is an example where we must solve for mass being the unknown
2. Iron has a known density of 7.87 g/cm3. What would be the mass of a 2.5 dm3 piece of
iron?
Notice that the density is given in the units g/cm3, but the volume is given in the units dm3.
Therefore, this problem requires an additional step.
First: Change the question so that the volume is given in the same units as the density.
Use the factor label method:
So: Iron has a known density of 7.87 g/cm3. What would be the the mass of a 2.5 dm3 piece
of iron?
2500 cm3
Second: Write the original formula for density, and then isolate the unknown (mass). List the
"knowns" and the "unknown"
Original Formula
Adjusted Formula
m=vxD
D = 7.87 g/cm3
m=?
v = 2500 cm3
Third: Substitute the known values in the problem
m = 2500 cm3 x 7.87 g/cm3
Fourth: Calculate the answer including units
m = 2500 cm3 x 7.87 g/cm3
m = 19675 g
Fifth: Round to the correct number of significant figures
m = 2.0 x 104 g
Note - we use scientific notation to express the correct number of significant digits.
Solving For Volume
When solving for volume, we must take the original formula, and isolate the unknown like so:
Original Formula
Multiply both sides by volume
m=vxD
Divide both sides by density
Adjusted formula
Now, here is an example of a density problem where volume is the unknown
3. Mercury has a density of 13.5 g/cm3. How much space would 50.0 g of mercury occupy?
First: Write the original formula for density, and then isolate the unknown (Volume). List
the "knowns" and the "unknown".
Original Formula
Multiply both sides by volume
m=vxD
Divide both sides by density
Adjusted formula
D = 13.5 g/cm3
M = 50.0 g
V=?
Second: Substitute the known values in the problem
Third: Calculate your answer, including units
v = 3.70370370. . . cm3
Fourth: Round to the correct number of significant figures
v = 3.70 cm3
Percent Error
Students often assume that each measurement that they make in the laboratory is true and
accurate. Likewise, they often assume that the values that they derive through
experimentation are very accurate. However, sources of error often prevent students from
being as accurate as they would like. Percent error calculations are used to determine how
close to the true values, or how accurate, their experimental values really are.
The value that the student comes up with is usually called the observed value, or the
experimental value. A value that can be found in reference tables is usually called the true
value, or the accepted value. The percent error can be determined when the true value is
compared to the observed value according to the equation below:
Let's look at an example of how the formula would be used in a real-life situation.
Ex. 1 A student measures the mass and volume of a piece of copper in the laboratory and
uses his data to calculate the density o the metal. According to his results, the copper has a
density of 8.37 g/cm3. Curious about the accuracy of his results, the student consults a
reference table and finds that the accepted value for the density of copper is 8.92 g/cm3.
What would be the student's percent error?
Solution - Step 1. Determine which values are known.
The students result, or the observed value = 8.37 g/cm3.
The accepted, or true value = 8.92 g/cm3.
Step 2. Substitute these values in the percent error calculation, as shown below:
Step 3. Solve for the unknown, and round to correct significant digits.
Percent Error = -6.17%
Please note that the negative sign does not mean that the error was less than zero, which
would be impossible. It shows that the student's calculated value was actually too low.
Many teachers may ask you to report the absolute value of your answer.
Temperature Conversions
In your everyday life and in your study of Chemistry, you are likely to encounter three different
temperature scales. When you watch the weather report on the news, they will report the
temperature on one scale, yet you measure temperature in the laboratory on a different scale. Many
Chemistry equations must be done using yet another temperature scale. Clearly, you can see the
importance of the use of units when reporting temperature. You can also see the need, for a student
of Science, to be able to convert temperatures from one scale to another. This page is designed to
help you do just that.
The Fahrenheit Scale - The Fahrenheit scale is the scale that is used when they report the weather
on the news each night. It is probably the temperature scale that you are most familiar with, if you
live in the United States. The thermometers that you have in your house, for uses such as;
swimming pools, cooking, bath tubs, or reading body temperature, are all likely to be in Fahrenheit.
In Canada and most other countries, the news will report the temperature on the Celsius scale.
The Celsius Scale - The Celsius scale, is commonly used for scientific work. The thermometers
that we use in our laboratory are marked with the Celsius scale. The Celsius scale is also called the
Centigrade scale because it was designed in such a way that there are 100 units or degrees between
the freezing point and boiling point of water. One of the limitations of the Celsius scale is that
negative temperatures are very common. Since we know that temperature is a measure of the
kinetic energy of molecules, this would almost suggest that it is possible to have less than zero
energy. This is why the Kelvin scale was necessary.
The Kelvin Scale - The International System of Measurements (SI) uses the Kelvin scale for
measuring temperature. This scale makes more sense in light of the way that temperature is
defined. The Kelvin scale is based on the concept of absolute zero, the theoretical temperature at
which molecules would have zero kinetic energy. Absolute zero, which is about -273.15 oC, is set
at zero on the Kelvin scale. This means that there is no temperature lower than zero Kelvin, so
there are no negative numbers on the Kelvin scale. For certain calculations, like the gas laws,
which you will be learning soon, the Kelvin scale must be used.
Figure 2-9a Comparison of Temperature Scales
Set Points
water boils
Fahrenheit
212
Celsius
100
Kelvin
373
body temperature
98.6
37
310
water freezes
32
0
273
absolute zero
-460
-273
0
There will be times when you need to be able to convert a temperature from one scale to
another. In real life, you might be in a country where temperature is reported in Celsius and you
will want to convert that into Fahrenheit, in order to figure out if you need to wear a sweater. In
your laboratory work, you may need to change from Celsius to Kelvin in order to calculate the
volume that a gas might occupy at standard temperature and pressure. The table below will show
you the formulas that you can use to change from one scale to another.
Figure 2-9b Temperature Conversion Formulas
Conversion
Formula
Example
Celsius to Kelvin
K = C + 273
21oC = 294 K
Kelvin to Celsius
C = K - 273
313 K = 40 oC
Fahrenheit to Celsius
C = (F - 32) x 5/9
89 oF = 31.7 oC
Celsius to Fahrenheit
F = (C x 9/5) + 32
50 oC = 122 oF
* Note, to change back and forth between Fahrenheit and Kelvin is a two step process.
Heat Transfer Calculations
In lesson 1-6, you were introduced to the concept of heat. That lesson described the energy of
chemical reactions, and explained the difference between endothermic and exothermic reactions.
Chemists measure the heat given off or taken in during a chemical reaction to determine the energy
of a specific chemical or physical reaction. In this lesson, you will learn to calculate the amount of
heat transferred during a physical or chemical change.
Scientists use a device called a calorimeter to measure the transfer of heat during a physical or
chemical change. You will be using a device like this when you conduct the Calorimetry
Laboratory. It is essential that you understand the calculations required in that activity, so that you
can benefit from the learning opportunity that the laboratory makes available to you.
As you should know by now, the formula for heat transfer calculations is:
amount of heat transferred = mass x change in temperature x specific heat
I will remind you that the symbols that are used for the formula will vary from textbook to
textbook, but the values that they represent never change. One way to write the heat transfer
formula is shown below:
q = m(T)Cp
Where q = heat transferred, DT = the change in temperature and Cp = the specific heat.
The SI units for heat transferred are joules, however calories are still often used for problems
involving water. You should memorize the conversion factor; 4.18 J = 1 cal. The units for
specific heat are joules/grams x degrees Celsius (J/g x oC) or Calories/grams x degrees Celsius
(cal/g x oC). Temperature is usually given in degrees Celsius.
You will solve these problems logically and algebraically. Logically, meaning you will strive to
understand the logic of performing each step, and that you will check to make sure that your answer
makes sense. As in any algebra problem, you will only have one unknown. The rest of the
information will be provided for you. We will work through one example of each of the possible
types of heat transfer problems that you will be responsible for. Then you can move on to try the
worksheets and the online quiz programs.
Type 1. Heat Transferred (q) is the unknown:
Ex. Aluminum has a specific heat of 0.902 J/g x oC. How much heat is lost when a piece of
aluminum with a mass of 23.984 g cools from a temperature of 415.0 oC to a temperature of 22.0
o
C?
Step 1: First read the question and try to understand what they are asking you. Can you picture a
piece of aluminum foil that is taken out of an oven. Imagine the aluminum losing heat to its
surroundings until the temperature goes from 415.0 oC to 22.0 oC.
Step 2: Write the original formula.
q = m(T)Cp
Step 3: List the known and unknown factors. Looking at the units in the word problem will help
you determine which is which.
q=?
m = 23.984 g
T = (415.0 oC - 22.0 oC) = 393.0 oC
Cp = 0.902 J/g x oC
(remember, they asked for the change in temperature)
Step 4. Substitute your values into the formula
q=?
m = 23.984 g
T = (415.0 oC - 22.0 oC) = 393.0 oC
Cp = 0.902 J/g x oC
q = m(T)Cp
q = 23.984 g x 393.0 oC
x 0.902 J/g x oC
Step 5. Cross out units where possible, and solve for unknown.
q = 23.984 g x 393.0 oC
x 0.902 J/g x oC
q = 8501.992224 J
Step 6. Round to the correct number of significant digits and check to see that you answer makes
sense.
q = 8.50 x 103 J
Our answer makes sense because joules (J) are acceptable units for q, and the value should be
positive based on the wording of the question.
Type 2. mass (m) is the unknown:
Ex. The temperature of a sample of water increases by 69.5 oC when 24 500 J are applied. The
specific heat of liquid water is 4.18 J/g x oC. What is the mass of the sample of water?
Step 1: First read the question and try to understand what they are asking you. Energy is being
used to change the temperature of a sample of water by 69.5 oC. What size sample of water would
require 24 500 J to make that change?
Step 2: Write the original formula, and then modify it isolate the unknown.
q = m(T)Cp
q=
m(T)Cp
--------------(T)Cp (T)Cp
m = q/(T)Cp
Step 3: List the known and unknown factors. Looking at the units in the word problem will help
you determine which is which.
q = 24 500 J
m=?
T = 69.5 oC
Cp = 4.18 J/g x oC
Step 4. Substitute your values into the formula.
q = 24 500 J
m=?
T = 69.5 oC
Cp = 4.18 J/g x oC
m = q/(T)Cp
m = 24 500 J/69.5 oC x 4.18 J/g x oC
Step 5. Cross out units where possible, and solve for unknown.
m = 24 500 J/69.5 oC x 4.18 J/g x oC
m = 84.3344463184 g
Step 6. Round to the correct number of significant digits and check to see that you answer makes
sense.
m = 84.3 g
Our answer makes sense because grams are the correct units for mass, and the value should be
positive.
Type 3. change in temperature (T) is the unknown:
Ex. 850 calories of heat are applied to a 250 g sample of liquid water with an initial temperature of
13.0 oC. Find a) the change in temperature and b) the final temperature. (remember, the specific
heat of liquid water, in calories, is 1.00 cal/g x oC.)
Step 1: First read the question and try to understand what they are asking you. Here they are
heating up a sample of water. They want to know how many degrees increase will result from 850
calories of heat. Further, they want to know the final temperature of the water, which will simply
be equal to the initial temperature + the change in temperature.
Step 2: Write the original formula, and then modify it isolate the unknown.
q = m(T)Cp
q = m(T)Cp
---- --------------m Cp m Cp
T = q/m x Cp
Step 3: List the known and unknown factors. Looking at the units in the word problem will help
you determine which is which.
q = 850 cal
m = 250 g
T = ?
Cp = 1.00 cal/g x oC
Step 4. Substitute your values into the formula
q = 850 cal
m = 250 g
T = ?
Cp = 1.00 cal/g x oC
T = q/m x Cp
T = 850 cal/250 g x 1.00 cal/g x oC
Step 5. Cross out units where possible, and solve for unknown.
T = 850 cal/250 g x 1.00 cal/g x oC
Answer to step a) T = 3.4 oC
Answer to step b) final temperature = 13.0 oC + 3.4 oC = 16.4 oC
Step 6. Round to the correct number of significant digits and check to see that you answer makes
sense.
Answers are already rounded correctly. They make sense because they show the correct units for
temperature and because the final temperature is higher than the initial temperature, as it should be.
Type 4. Specific Heat (Cp) is the unknown:
Ex. When 34 700 J of heat are applied to a 350 g sample of an unknown material the temperature
rises from 22.0 oC to 173.0 oC. What must be the specific heat of this material?
Step 1: First read the question and try to understand what they are asking you. Specific heat is a
concept that some students struggle with. The question is about finding the specific heat by seeing
how much the temperature changes when a certain amount of heat is applied. Metal heats up faster
than water because it has a low specific heat. If a material has a low specific heat, the temperature
change will be greater for a given amount of heat, when all other things are equal.
Step 2: Write the original formula, and then modify it to isolate the unknown.
q = m(T)Cp
q = m(T)Cp
--- ------------m(T) m(T)
Cp = q/m(T)
Step 3: List the known and unknown factors. Looking at the units in the word problem will help
you determine which is which.
q = 34 700 J
m = 350 g
T = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?
Step 4. Substitute your values into the formula
q = 34 700 J
m = 350 g
T = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?
Cp = q/m(T)
Cp = 34 700 J/350 g x 151.0 oC
Step 5. Cross out units where possible, and solve for unknown.
Cp = 34 700 J/350 g x 151.0 oC
Cp = 0.65657521286 J/g x oC
Step 6. Round to the correct number of significant digits and check to see that you answer makes
sense.
Cp = 0.66 J/g x oC
Our answer is logical, and the units are correct.
Development of The Atomic Theory
Development of the Atomic Theory
The concept of atoms, as you will learn, is quite old. The ancient Greeks had an atomic theory
more than 2000 years ago. It is interesting to note that the idea originated from philosophy, and
was based on reason rather than data. It is also interesting to note that, for many scientists, the final
"proof" of the existence of atoms was provided by a, then-unknown 26 year old man named Albert
Einstein. The story of the atom makes for interesting reading, and it involves a huge cast of
characters whose lives spanned thousands of years. I have found some excellent essays on the
internet. Follow the links below and enjoy.
I. The Greek Concept of Atomos - by John L. Park
II. Greek Theory and Roman Practice - by James A. Plambeck
III. Middle Ages Through Alchemy - by James A. Plambeck
IV. Two Centuries of Transition - by James A. Plambeck
The Modern Atomic Theory
John Dalton, an English chemist, might be called "the father of the modern atomic theory."
During the early 1800's, Dalton's interests in Meteorology and gases lead him to read the works of
Antoine Lavoisier and Joseph Proust. Lavoisier had stated the law of conservation of mass, that
the mass of materials before a chemical reaction takes place is exactly equal to the mass of the
materials after the reaction is completed. Proust had observed the law of definite proportions,
stating that the proportion by mass of the elements in a given compound is always the same.
Dalton felt that the findings of these men gave strong support to the idea of atoms. He formulated
an atomic theory that would include the observations found by Lavoisier and Proust.
Dalton's Atomic Theory
1) All elements are composed of atoms, which are indivisible and
indestructible particles.
2) All atoms of the same element are exactly alike; in particular, they all
have the same mass.
3) All atoms of different elements are different; in particular, they have
different masses.
4) Compounds are formed by the joining of atoms of two or more
elements. In any compound , the atoms of the different elements in the
compound are joined in a definite whole-number ratio, such as 1 to 1, 2 to
1, 3 to 2, etc.
Much has happened since the time of Dalton, which has made it necessary to update his atomic
theory. We currently believe that all elements are composed of atoms, but we know that those
atoms are not indestructible. Atoms are split in nuclear reactions, and they are made up of even
smaller particles. We also know that atoms of the same element can have different masses, when
they represent different isotopes. Despite these differences, much of Dalton's atomic theory
remains useful to this day.
Development of the Atomic Model
To borrow an example from Albert Einstein, imagine if you had never seen a clock or a watch
before, and someone gave you an intricate Swiss timepiece. Imagine studying the motion of the
hands, but never being allowed to remove the watch face and see the mechanisms which produced
the sychronized movements. If you thought about it long enough, you might be able to come up
with a model to explain the motion of the hands, but you could never be sure that your model was
an accurate depiction of what was going behind the face of the watch. In fact, if someone was to
come along with a better explanation for the motion of the hands, you would be forced to update
your model.
Our atomic model has much in common with the imaginary watch from the above example. We
can't base our model on actual observations of atoms, because they are too small to be seen with our
most sensitive instruments. Instead, we must come up with a model of an atom that can account for
and explain observations that we can actually see. As new observations are made, we are forced to
update our model to accommodate them. As a result, our model of the atom has evolved over time,
and we must accept the fact that it is likely to change again in the future.
The story so far . . .
Democritus may not have been the first of the
ancient Greeks to suggest an atomic theory, this
distinction goes to his teacher Leucippus, but his
name is often associated with the first atomic theory,
because of his support of it. To Democritus, atoms
were completely solid, homogeneous, indestructible
objects.
Democritus
c460-371 BC
Joseph John Thomson subjected cathode rays to
magnetic and electric fields and showed that the
beam was deflected as would be expected for
negatively charged particles. He calculated the ratio
of the electron's charge to its mass. On April 30,
1897, Thomson announced that the cathode rays
consisted of negatively charged particles, which
represented fundamental particles of matter. He
was not the first person to suggest that these
particles existed, nor did he coin the term
"electron", yet he is generally credited with the
discovery of the electron. He was awarded with the
Nobel Prize in Physics in 1906.
J.J. Thomson
1856-1940
J.J. Thomson is also remembered for his "plumpudding" model of the atom, which suggested a solid
atom with positively and negatively charged particles
evenly distributed throughout the mass of the atom.
Link One - The original paper in which J.J. Thomson announces his discovery of
the electron to the world.
Link Two - Thomson on the number of corpuscles (electrons) in an atom.
Link Three - Thomson on the structure of the atom.
Link Four excerpts from Thomson's Nobel prize address
Lord Ernest Rutherford
1871-1937
Ernest Rutherford, who was once a student of
Thomson's, is credited with discovering that most of
the atom is made up of "empty space." In 1909 he
and his assistants conducted the "gold foil"
experiment, from which he concluded that "the
greater part of the mass of the atom was
concentrated in a minute nucleus." In this model,
the positively charged nucleus was surrounded by a
great deal of "empty space" through which the
electrons moved.
Link One - Geiger's paper on the gold foil
Link Two - Rutherford describing the gold foil experiment
Link Three - Rutherford's paper on the structure of the atom
In 1909, Robert Millikan conducted his "oil-drop"
experiment which allowed him to measure the
charge on an electron. Combining his results with
those of Thomson, Millikan found the mass of the
electron to be 9.11x10-28 g. He was awarded with the
Nobel Prize in physics in 1923.
Robert Millikan
1868-1953
In 1913, Niels Bohr proposed improvement to
Rutherford atomic model. For this reason, the
planetary model of the atom is sometimes called the
Rutherford-Bohr model. Bohr added the idea of
fixed orbits, or energy levels for the electron
traveling around the nucleus. This model allowed
for the idea that electrons can become "excited" and
move to higher energy levels for brief periods of
time.
Niels Bohr
1913-1963
Link One - Bohr's address on the spectrum of hydrogen
Link Two - An article on atomic structure written by Niels Bohr
Lord Rutherford predicted the existence of the
neutron is 1920. Walter Bothe obtained evidence of
the neutron in 1930. However it was James
Chadwick, who repeated Bothe's work, who is
known as the discoverer of the neutron. He found
these uncharged particles with essentially the same
mass as the proton. He was awarded the Nobel Prize
in physics in 1935.
James Chadwick
1891-1974
Link One - A letter on the possible existence of the neutron
Link Two - Chadwick's paper on the discovery of the neutron
Although there is something attractive about the
idea of an atom being very much like a tiny solar
system, the planetary model of the atom was found
to be inadequate. Planck's quantum theory had
illustrated the "particle-like" properties of waves.
Louis de Broglie suggested that particles might have
properties of waves. The result of this investigation
is sometimes called the wave-particle duality of
nature. This duality, which states that particles act
like waves and waves like particles, applies to all
waves and all particles. However, the more massive
the particles, the less obvious the wave properties.
Electrons, having very little mass, exhibit significant
wave-like properties.
Heisenberg pointed out that it is impossible to
know both the exact position and the exact
momentum of an object at the same time. Applying
this concept to the electron we realize that in order
to get a fix on an electron's position at any time, we
would alter its momentum. Any attempt to study the
velocity of an electron will alter its position. This
concept, called the Heisenberg Uncertainty principle,
effectively destroys the idea of electrons traveling
around in neat orbits. Any electron that is subjected
to photons will have its momentum and position
affected.
Werner Heisenberg
The Charge-Cloud Model
Experiments conducted in the 1920's, 1930's and
1940's continued to point out problems with the
planetary model of the atom. These experiments,
which will be discussed in next chapter, lead to the
development of the charge-cloud model. The chargecloud model, which is also called the quantummechanical model, does not attempt to describe the
path of each electron in a fixed orbit. Scientists now
describe the possible positions of electrons in terms
of probability. Computers can calculate the points
in space that an electron has the highest probability
of occupying. These points can be connected to form
a three-dimensional shape. Electrons are
characterized in terms of the three-dimensional
shapes that their probability fields define. The sum
total of the various paths of electrons, traveling at
very high speeds, is described as the electron cloud.
Atomic Structure
Our current model of the atom is called the charge-cloud model, the orbital model, the wavemechanical model, or the quantum-mechanical model. According to this model of the atom, the
positively charged protons and the neutral neutrons are still located in the nucleus of the atom. The
electrons, no longer thought of as locked into "fixed" orbits, are collectively located in an area
called the electron cloud. The boundaries of this cloud are set by the probability of finding each
electron in given areas. These electrons, moving at extremely high speeds, effectively occupy the
entire area of the cloud, in the same way that moving fan blades effectively occupy the entire area
through which they pass.
I. Protons
Protons are positively charged subatomic particles that are found, along with neutrons, in the
nucleus of the atom. Protons, along with neutrons, make up most of the mass of the atom. The
mass of a single proton is about 1.67265 x 10-24 grams, or 1.0073 u (atomic mass units). Although
the positive charge of the proton is equal to the negative charge on the electron, one proton has as
much mass as around 1840 electrons. The elements on the periodic table are arranged in order of
increasing number of protons (see atomic number below.) A hydrogen atom has one proton and a
helium atom has two.
II. Atomic Number
The number of protons in the nucleus of an atom is called its atomic number. The atomic
number, which is given the symbol Z, is what determines the identity of an element. All atoms of
the same element have the same number of protons and the same atomic number. Atoms of
different elements, by definition, will have a different number of protons and therefore, different
atomic numbers. Elements with the atomic numbers from 1 to 112 have been identified so far.
One of the numbers found in each elemental box on the periodic table will be the atomic number.
Unlike the mass number, the atomic number is always a whole number.
III. Neutrons
The neutron is a neutral particle that is found in the nucleus of most atoms. Although the
neutron has no charge, it does contribute to the mass of the atom. Each neutron has a mass of about
1.67495 x 10-24 grams, or 1.0087 u. The most common type of hydrogen, called protium, has no
neutrons. Deuterium, another form of hydrogen, has one proton and one neutron in the nucleus of
each atom. Tritium, the third form of hydrogen, has two neutrons and one proton in each nucleus.
IV. Mass Number
The vast majority of the mass of an atom is found in the nucleus. The mass of a proton or a
neutron is approximately 1 u (atomic mass unit). It would take around 1840 electrons to equal the
mass of one proton. For this reason, the masses of the electrons are not considered when
calculating the mass number of an atom. The mass number, which is given the symbol A in
elemental notation, consists of the total number of protons and neutrons in the nucleus of the atom.
V. Isotopes
Although all atoms of the same element have the same number of protons, they can have a
different number of neutrons. Atoms of the same element with different numbers of neutrons are
called isotopes. The three forms of hydrogen discussed in the above section on neutrons represent
different isotopes of hydrogen. Isotopes are often identified by mass number. For example,
carbon-12 would be carbon with a mass number of 12, while carbon-14 has a mass number of 14.
As in the case of carbon-14, some isotopes of certain elements are unstable, which means that they
undergo radioactive decay.
VI. Atomic Mass
The atomic masses shown on the periodic table represent a weighted average based on the
relative abundance of each isotope of a particular atom. Although some books and some teachers
still refer to atomic mass as "atomic weight", this is not considered correct.
V. Quantum Numbers
As you now know, scientists no longer think of electrons following the fixed orbits described by
Bohr's planetary model of the atom. Rather, electrons are thought to effectively take up the entire
space around the nucleus, out to a certain distance. Quantum numbers are used to describe the
allowable values of certain physical quantities of an electron's behavior.
The first quantum number, also called the principle quantum number, describes the radius of the
electrons orbit. The principle quantum number is designated by the letter n, and its value
corresponds to the numbered energy levels of the Bohr atom. So, an electron with a n value of 4
will be found in the fourth energy level.
The second quantum number, called the angular momentum quantum number, is given the letter
l. This quantum number may have a value ranging from zero to n-1, and thus is limited by the
value of the principle quantum number. The second quantum number gives us the type of
sublevel. A sublevel with l=0 is an s sublevel. l=1 designates a p sublevel. d sublevels have an l
value of 2. f sublevels have an l value of 3.
The third quantum number, designated by the letter m, defines the spatial orientation of the
orbital. The value of m will range from +l to -l. thus, an s orbital can have only a value of 0 for m,
but a p sublevel can have a value of -1, 0 or +1.
The fourth quantum number, which is given the letter s, describes the spin on the electron as
either clockwise or counterclockwise. The Pauli exclusion principle states that no two electron in
an atom can have the same set of four quantum numbers. Therefore, if two electrons occupy the
same orbital, they must have opposite spins.
The Periodic Table
The Origin of the Periodic Table
When a person is confronted with a large number of items, it is only natural to look for
similarities that can be used to develop a classification scheme. A person who collects
baseball cards may group his cards according to team or position. Biologists classify all living
organisms in a five-kingdom classification system, based on similar characteristics. Early
Chemists studied the group of known elements, and tried to come up with logical
classification systems based on what they knew.
An early attempt was made by a German Chemist named Johann Dobereiner, in 1817.
Dobereiner noticed that there were several groups of three elements which shared certain
properties. For example; chlorine, bromine and iodine all share antiseptic properties.
Dobereiner called each group of three elements with similar properties a "triad". He also
noticed that when he arranged the elements of a triad by atomic mass, the middle element
had an atomic mass that fell close to the middle of the other two atomic masses.
Almost fifty years later, an English Chemist named John Newlands proposed an updated
classification system. Newlands had noticed that when the 49 known elements were arranged
in order of increasing atomic masses, certain properties would repeat every eighth element.
He arranged the elements into seven groups of seven elements and called his system the "law
of octaves."
In the 1860's a German Chemist named Lothar Meyer was developing a periodic table
based on the idea that when the elements were arranged by atomic mass, certain properties
were repeated periodically.
In 1869 Dimitri Mendeleev, a Russian Chemist, published the first periodic table. It had
eight columns, and it contained blank spaces for elements that Mendeleev predicted must
exist, although they had not yet been discovered. His predictions turned out to be correct.
New elements were discovered, and they fit into the spaces that he had left in his periodic
table.
It was not until 1914 that Henry Moseley corrected the periodic law, based on his
discovery of the atomic numbers of several elements. This modern periodic law states, "The
periodic properties exhibited by the elements are a function of the atomic numbers."
The Structure of the Periodic Table
The Periodic table holds so much more information than most people realize. The average
person can see that the table shows the atomic number and atomic mass of each element, but
a student of Chemistry learns that there is much more there, for someone who knows how to
read the table correctly. In this lesson, you will learn more about the structure of the periodic
table, so that you will be able to extract some more of this additional information.
Unlike Mendeleev's table, the modern Periodic Table is arranged according to atomic
number. Remember that it is the atomic number, or nuclear charge, that determines the
identity of the element. The horizontal rows of elements on the periodic table are called
periods. The vertical columns are called groups or families. By looking at the column that an
element is found in, you can predict the valence shell configuration with a good deal of
accuracy. By looking at the period that the element is found in, you can determine the energy
level which the valence shell is found. Look at the examples shown on the table below;
Table 3-4a
O
Na
Zr
KEY
s - section
d - section
p - section
f - section
Sodium (Na) is in the third period (row), so its valence shell is in the third energy level.
Furthermore, it is in the first column of the s section, so its valence configuration is 3s1.
Zirconium (Zr) is in the fifth period, so its valence shell is in the fifth energy level. It is in
the second column of the d section, but the d electrons are never in the valence shell.
Zirconium has a valence configuration of 5s2, because 5s2 is filled before its last electron (4d2)
is filled. There are a few exceptions, but most of the transition metals will have two valence
electrons.
Oxygen (O) is found in the second period, and in the fourth column of the p section.
Because the p is filled after the s electrons, oxygen's valence shell would be 2s2 2p4.
Metals, Nonmetals and Metalloids
The elements of the periodic table belong in three basic categories; metals, nonmetals and
semimetals (or metalloids). We will discuss the criteria that Chemists use to classify the
elements in the next lesson, but for now you should be able to locate the areas on the periodic
table below;
Table 3-4b
KEY
Metals
Semimetals
Nonmetals
As you can see, the vast majority of the elements are considered metals. Some of the
elements that are considered metals would not surprise you. For example, the average person
would think of copper, gold, and iron as metals, and indeed, chemists do also. However,
some of the elements that are considered metals by Chemists may surprise you. Did you
know that chemists consider calcium and potassium metals? Sodium does not look at all like
what most people think of as metallic, it is soft and white, yet a Chemist considers it a metal.
Below you will see a table that summarizes the distinguishing characteristics of metals,
nonmetals and metalloids.
Table 3-4c
*General characteristics of metals, nonmetals and metalloids
Metals
Nonmetals
Metalloids




Hard and Shiny
3 or less valence
electrons
Form + ions by losing
eGood conductors of
heat and electricity




Gases or dull, brittle
solids
5 or more valence
electrons
Form - ions by
gaining ePoor conductors of
heat and electricity




Appearence will vary
3 to 7 valence electrons
Form + and/or - ions
Conduct better than
nonmetals but not as
well as metals
* Remember that these are general characteristics, there are exceptions.
Elemental Families and Groups
Elements that are found in the same group (column) tend to have similar properties,
because they have similar valence shell configurations. This is not as true of the elements that
are found in columns that contain metals, nonmetals and metalloids, like columns 14 and 15.
Some of the groups contain elements that are similar enough to be given a family name. The
family names that you should know are shown on the table below;
Table 3-4d
Key
Alkali
Metals
Alkaline
Earth Metals
Transition
Metals
Halogens
Noble Gases
Elemental Notation and Isotopes
Elemental Notation
Each element has one or more atoms associated with it. In each case, every atom of a particular
element has the same number of protons. However, a particular element may have several different
types of atoms, with different number of neutrons. As a result, each atom of a particular element
has the same atomic number, but a group of these atoms may have several different mass
numbers. The atomic number represents the number of protons in the atom's nucleus, and it is
this number that determines the identity of the element. Because the protons are the charged
particles in the nucleus, the atomic number is also called the nuclear charge. The mass number
represents the total number of protons and neutrons in an atoms nucleus.
Elemental Notation is a shorthand way of writing information about a particular type of element,
isotope or atom. An example of the accepted form of elemental notation is shown below:
The large "X" represents where you will find the atom's elemental symbol. The mass
number, which is given the symbol "A", is located in the upper left-hand corner. The
atomic number (also called nuclear charge), which is given the symbol "Z", is found in the
lower left-hand corner.
Now, let's look at an example of how this notation might actually appear.
Here we see a carbon atom with an atomic number of 6 (like all carbon atoms) and a
mass number of 12. Not all carbon atoms have a mass number of 12. You may have
heard of carbon-14, which would have two more neutrons than this carbon atom,
accounting for the higher mass
Although less common, additional information may be located on the right side of the elemental
notation, as shown below:
This notation shows quite a bit of information. We know that it represents a sample of
calcium atoms, based on the elemental symbol "Ca". The mass number, found in the
upper left corner, is 40. The atomic number, found in the lower left corner, is 20. The
"+2", shown in the upper right corner, represents the charge on the atoms of the
sample. Here we see a group of atoms which have lost two electrons each. The number
"5", seen in the lower right corner, represents the number of atoms in the sample.
Isotopes are atoms with the same atomic number, but different mass numbers. In other words,
isotopes have the same number of protons, but a different number of neutrons. The three isotopes,
or forms of hydrogen are shown below:
Here we see three atoms of hydrogen. Each atom of hydrogen has one proton in its
nucleus, so the atomic number of each is "1". The mass number of each of these atoms
varies. The hydrogen on the top has 2 neutrons in addition to the 1 proton in the
nucleus, for a total mass number of 3. The second atom has 1 neutron in addition to the
1 proton in the nucleus, for a total mass number of 2. The bottom atom has no
neutrons, only 1 proton in its nucleus, for a mass number of one. A graphic
representing the three isotopes of hydrogen is shown below.
Now let us demonstrate how you can tell the number of protons, neutrons and electrons in a
particular atom by reading the elemental notation.
Elemental
Symbol
# of Protons
(equal to Z)
3
9
8
56
# of Neutrons
(equal to A-Z)
4
10
8
80
# of electrons
(equal to Z
modified by
charge
number)
3
9
10
54
Lewis Dot Diagrams
Electron dot diagrams, which are also called Lewis dot diagrams, are very useful tools in
Chemistry. They will give you the ability to determine the type(s) of covalent bonds that an
element may make in certain situations. They can also be used to predict the type of ion that an
atom might make when it forms an ion. Each dot diagram consists of; an elemental symbol, which
represents the kernel of the atom, and a group of 1-8 dots which shows the configuration of the
outer-most electron shell of the atom, also called the valence shell.
Below is an example of the proper Lewis dot diagram for the element oxygen.
The "O" in the example above represents the kernel of the atom, that is the nucleus and all of the
electrons, except those in the valance (outer) shell. Each of the four "sides" of the symbol
represents an orbital in the outermost energy level of the atom. Since each orbital can hold only
two electrons, the sides of the dot diagram can only hold up to two dots. The six dots show the
configuration of the of the valence electrons.
To make a Lewis dot diagram, you need to know how many electrons are in the valence shell. If
you don't know off hand, you would start by writing the electron configuration. You should
remember that the electron configuration of oxygen is 1s2 2s2 2p4. The six electrons shown on the
dot diagram are those in the second principle energy level. You fill in one valence electron on
each side of the elemental symbol, and then double up as many sides as you need to in order to
include each one. Remember that each side can only hold up to two dots!
Below you will see an example of the order of filling in the dots on a dot diagram for an element
with eight valence electrons. Please note that you can place the first two dots on any side, but the
rest of the dots should be placed in either a clockwise or counter clockwise manner, with no side
receiving two dots until each side gets one.
Step One
Step Two
Step Three
Step Four
Step Five
Step Six
Step Seven
Step Eight
By looking at the electron dot diagram for oxygen we can see that oxygen has two unpaired
electrons, so it has two electrons available for standard covalent bonds.
These unpaired electrons might make two single covalent bonds, as is the case in water
(H2O). Or, they might make one double covalent bond, as the case of magnesium oxide (MgO).
When Lewis dot diagrams are used for compounds, "x's" are often used to substitute for the dots of
one or more elements in order to show which electrons came from which element.
Let us look at the Lewis dot diagrams for both oxygen and hydrogen as free elements, and then
at water as a compound.
Now we will show electron configurations and the dot diagrams for a few more elements.
Remember that the dots represent the electrons that are found in the valence shell, or energy level
with the highest value for n (principle quantum number). For these examples, I will color the
valence shell of each element black.
Lithium, Z = 3
Electron Configuration
1s2 2s1
Lewis Dot Diagram
Sulfur, Z = 16
Electron Configuration
1s2 2s2 2p6 3s2 3p4
Lewis Dot Diagram
Krypton, Z = 36
Electron Configuration
Lewis Dot Diagram
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
Elemental Symbol Review
Earlier this year, you were introduced to elemental symbols, and you were encouraged to
memorize many of the most common ones. The reason that you will want to memorize these
symbols, is because it will help you to follow the class lessons in the weeks to come. If you need to
stop what you are doing, and figure out what a particular symbol is, you may not be able to keep up
with the class. You will be using these symbols in special combinations, called chemical formulas.
The better you know the elemental symbols, the easier it will be to learn the skills that follow.
As of this date, there are about 112 known elements. Each element is made up of its own type
of atom. Not all atoms of an element are identical, but each atom of an element has the same
number of protons. Protons are subatomic particles found in the nucleus of an atom, and the
number of protons in an atom is called the atomic number. Each element has its own unique atomic
number, as well as its own name and symbol.
The names of the elements have various origins. Some are named, in Greek or Latin, for
properties that they possess. Some are named in honor of scientists who made important
discoveries in the field of Chemistry. Others are named for the place they where first discovered.
Some are even named after gods of mythology.
Since there are only 26 letters in our alphabet, and over 100 elements, combinations of letters had
to be used in order to give each element its own unique symbol. Each symbol is made up of one
capital letter, which may be followed by 0-2 lower case letters. Some elemental symbols, like
oxygen with the symbol "O", will be easy for you to remember. Other elemental symbols, like
sodium with the symbol "Na", will be a bit harder to commit to memory. Unfortunately for us,
some symbols (like Na) are derived from the latin name for the element.
Chemistry students are often required to memorize some, or all, of the elemental symbols. There
is an important reason for this, and it will save you and your teacher a great deal of class time in the
future. Check with your teacher to see which elemental symbols you will need to memorize for an
upcoming quiz. You can print out a use the chart below, but remember to check with your teacher
to see if it is complete, or if there are other symbols that you need to know. Also, as you go about
studying the symbols, keep in mind that the case of the letter is very important.
Table 5-1a
Select Elements and Their Symbols
Aluminum Al
Francium
Fr
Phosphorus P
Argon
Ar
Germanium Ge
Plutonium Pu
Barium
Ba
Gold
Au
Potassium K
Beryllium Be
Helium
He
Radium
Ra
Boron
B
Hydrogen
H
Radon
Rn
Bromine
Br
Iodine
I
Rubidium
Rb
Cadmium Cd
Iron
Fe
Selenium
Se
Calcium
Ca
Krypton
Kr
Silicon
Si
Carbon
C
Lead
Pb
Silver
Ag
Cerium
Ce
Lithium
Li
Sodium
Na
Cesium
Cs
Magnesium Mg
Strontium
Sr
Chlorine
Cl
Manganese Mn
Sulfur
S
Chromium Cr
Mercury
Hg
Tin
Sn
Cobalt
Co
Neon
Ne
Titanium
Ti
Copper
Cu
Nickel
Ni
Zinc
Zn
Curium
Cm
Nitrogen
N
Fluorine
F
Oxygen
O
Oxidation Numbers
The chemical formula for water is H2O. Carbon Dioxide is CO2. Why does oxygen combine in
different ratios, in different compounds? Do Chemistry students need to memorize the chemical
formulas for each of the millions of known compounds? Is there a way to predict the ratio by
which elements will combine in a given situation? Fortunately, that is what oxidation numbers are
for.
You probably recall learning about ions in Biology. An ion is a charged particle formed when a
neutral atom or group of atoms gain or lose one or more electrons. When a single atom forms an
ion, as in the case of Al+3, it is called a monatomic ion. When of group of atoms that are
covalently bonded together form an ion, as in the case of NH4+, it is called a polyatomic ion.
Sometimes ions with opposite charges are attracted together and will form ionic compounds.
Table Salt, NaCl is such a compound formed from Na+ ions and Cl- ions. Neutral atoms can also
form compounds when they join together, as in the case of water (H2O). However, since these
compounds are not composed of ions, they are called molecular compounds. You will learn more
about these types of compounds in lesson 5-3.
Regardless as to whether a compound is made up of ions or not, each atom in the compound has
an apparent charge. This apparent charge, called the oxidation number, represents the charge that
an atom would have if electrons were transferred completely to the atom with the greater attraction
for them in a given situation. These oxidation numbers can be used to predict the ratio by which
atoms will combine when they form compounds.
The following rules help us assign the oxidation number of elements:
Table 5-2a - Predicting Oxidation Numbers
1. In free elements (that is, in uncombined state), each atom has an oxidation number
of zero. Ex. In O2, the oxidation number of each oxygen atom is zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on
the ion. Ex. The oxidation number of Ca2+ is +2.
3. All alkali metals (elements in column 1of the periodic table, with the exception of
hydrogen) have an oxidation number of +1. Ex. The oxidation numbers of Li, K, and
Na will always be +1.
4. All alkaline earth metals (elements in column 2 of the periodic table) have an
oxidation number of +2. Ex. The oxidation number of Ba is +2.
5. The oxidation number of Aluminum (Al) is always +3.
6. The oxidation number of oxygen in most compounds (such as H2O and CO2) is -2. In
hydrogen peroxide (H2O2) and peroxide (O22-) oxygen shows a -1 oxidation number.
7. The oxidation number of hydrogen is +1, except when in is bonded to a metal as a
negative ion, in which case it is -1. Ex. H2O shows hydrogen as +1. NaH shows
hydrogen as -1.
8. When halogens (elements in column 17 on the periodic table) form negative ions,
they will have an oxidation number of -1. Ex. NaCl and CaCl2 both show chlorine with
a -1 oxidation number.
9. In a neutral molecule, the sum of the oxidation numbers of all of the atoms must be
zero. Ex. In H2O, each hydrogen is +1 and the oxygen is -2. So, (2 x +1) + (-2) = 0.
10. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion
must be equal to the net charge of the ion. Ex. In the polyatomic ion known as
hydroxide (OH-), the oxygen is -2 and the hydrogen is +1. So, (-2) + (+1) = -1, the same
as the charge on the hydroxide ion (OH-)
Now, in time you will find it easy to predict many oxidation numbers, as you become more
familiar with the periodic table and the rules above. Until that time, you should make use of
reference tables that list the oxidation numbers of common ions. Depending on your teacher, he or
she may allow you to make use of such tables for quizzes and exams. For your convenience, I will
provide examples of these tables below. Keep in mind that the table that your teacher uses may
differ from the ones provided below.
Table 5-2b - Oxidation Numbers of Some Common Monatomic Ions
CHARGE
+1
+2
+3
+4
NONE
-1
-2
ION
Aluminum (Al)
X
Argon (Ar)
X
Barium (Ba)
X
Bromide (Br)
X
Cadmium (Cd)
X
Calcium (Ca)
X
Cesium (Cs)
X
Chloride (Cl)
X
Fluoride (F)
X
Hydride (H)
X
Hydrogen (H)
X
Iodide (I)
Lithium (Li)
X
X
Magnesium (Mg)
X
Neon (Ne)
X
Oxide (O)
X
Potassium (K)
X
Sodium (Na)
X
Silver (Ag)
X
Strontium (Sr)
Sulfide (S)
X
X
Zinc (Zn)
X
Now, some elements show different positive oxidation numbers, in different situations. The
stock system, which you will learn more about in lessons 4-3 and 4-4, uses Roman numerals to
show the oxidation number of the element. For example, Lead(II) is lead with an oxidation number
of +2. Chromium(III) is Chromium with an oxidation number of +3. The oxidation number for
these types of elements will always be positive. I provide a table below, but once you understand
the stock system you will not need the table any longer.
Table 5-2c - Stock System
Oxidation Numbers of Metals with Multiple Oxidation States
CHARGE
ION
+1
+2
+3
Chromium(III)
+4
X
Cobalt(II)
X
Copper(I)
X
Copper(II)
X
Iron(II)
X
Iron(III)
X
Lead(II)
X
Lead(IV)
X
Manganese(II)
X
Mercury(II)
X
Nickel(II)
X
Tin(II)
X
Table 5-2d - Oxidation Numbers of Some Common Polyatomic Ions
CHARGE
ION
+1
Acetate, (CH3COO-)
Ammonium (NH4+)
+2
-1
X
X
Carbonate (CO32-)
Chlorate (ClO3-)
-2
X
X
-3
Chromate (CrO42-)
Cyanide (CN-)
X
X
Dichromate (Cr2O72-)
X
Hydroxide (OH-)
X
Hypoclorite (ClO-)
X
Iodate (IO3-)
X
Nitrate (NO3-)
X
Nitrite (NO2-)
X
Oxalate (C2O42-)
X
Perchlorate (ClO4-)
X
Permanganate (MnO4-)
X
Peroxide (O22-)
X
Phosphate (PO43-)
X
Silicate (SiO32-)
X
Sulfate (SO42-)
X
Sulfite (SO32-)
X
Tartrate (C4H4O62-)
X
Tetraborate (B4O72-)
X
2-
Thiosulfate (S2O3 )
X
Writing Chemical Formulas
A chemical formula is a combination of elemental symbols and subscript numbers that is used
to show the composition of a compound. Depending of the type of compound that the formula
represents, the information that it provides will vary slightly. Before we go about learning how to
write chemical formulas, it is important that you clearly understand the difference between
molecular compounds and ionic compounds.
Ionic compounds are composed of charged ions that are held together by electrostatic forces. A
typical type of ionic compound, called a binary compound because it is made up of two elements,
will be composed of metallic positive ions (cations) and nonmetal negative ions (anions).
Another type of ionic compound, called a ternary compound as it contain three elements, is
composed of monatomic ions and polyatomic ions. When dealing with ionic formulas it is very
important to remember that the formula does not show how the compound actually exists in nature.
It only shows the ratio by which the individual ions combine. For example, the ionic formula for
calcium chloride is CaCl2. Since calcium chloride is an ionic compound, this formula does not
mean that there are actually two chlorine atoms floating around attached to one calcium atom.
Ionic compounds are actually continuous, lacking the discrete units that make up a sample of a
molecular substance. Rather, the formula shows that a sample of calcium chloride contains twice
as many chlorine atoms as calcium atoms. Remember that ionic compounds are not molecules, so
the formula CaCl2 is said to represent one formula unit of calcium chloride.
Molecular compounds are held together by covalent bonds, or shared pairs of electrons.
Molecular formulas do show these molecules as they actually exist as discrete units in nature.
When we say that the molecular formula of water is H2O, we can see that the molecules of water
are made up of three atoms, two hydrogen atoms are covalently bonded to each oxygen atom. A
special type of chemical formula, called an empirical formula, shows the composition of a
molecule not as it actually exists, but in a simple whole number ratio. The difference between
empirical and molecular formulas will be explained in lesson 5-5.
This lesson will concentrate on writing simple chemical formulas when given a formula name.
In learning how to write chemical formulas, you will make use of the oxidation numbers that you
learned about in lesson 5-2. For your convenience, print out the tables from lesson 5-2 before you
continue with this lesson, as they will be referred to from time to time.
Writing Ionic Formulas
I. Binary Compounds - Binary compounds are compounds that are composed of only two
elements. When you write the formulas for binary compounds, they will consist of two elemental
symbols, and they may also have one or two subscript numbers, if the elements don't combine in a
one to one ratio. You are probably familiar with the formula NaCl for table salt. This formula
shows no subscripts because one ion of Na will be present for each ion of Cl, in any sample of table
salt.
You will be given the name of a binary compound and you will be expected to be able to write the
proper formula for the compound. There will be two sources of information for writing the correct
formula. The compounds name will give you the elements that make up the compound. The
oxidation numbers of the ions involved will show you the ratio by which they combine. Let's go
through an example;
Example 1. Write the correct formula for Barium Fluoride.
Step one - Write the symbols for the elements in the compound. If you need to review the
elemental symbols, see lesson 5-1. Note that the ending "ide" is used for fluoride to show that it is
a negative ion of fluorine.
Barium = Ba
Fluoride = F
Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or some similar
table), and write them as superscripts to the right of the elemental symbols. Note that when no
number accompanies a charge symbol, as in the case of fluoride below, they charge value is
understood to be "1".
Barium = Ba2+
Fluoride = F-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
In this case, (2+) + 2(-1) = 0. So, two fluoride ions will cancel out one barium ion. Since it would
take two fluoride ions (each with a charge of negative one) to cancel out one barium ion (with a
charge of plus two) we use a subscript of two after the symbol for fluorine to show the ratio.
BaF2
If this seems confusing to you, it will get simpler over time.
Example 2. Write the proper formula for the ionic compound lithium bromide.
Step one - Write the symbols for the elements in the compound. Note that the ending "ide" is used
for bromide to show that it is a negative ion of bromine.
Lithium = Li
Bromide = Br
Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or some similar
table), and write them as superscripts to the right of the elemental symbols. Note that when no
number accompanies a charge symbol, as in the case of fluoride below, they charge value is
understood to be "1".
Lithium = Li+
Bromide = Br-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
In this case, (+1) + (-1) = 0. so, one lithium ion will cancel out the charge of one bromide ion. This
means that the two elements will combine in a one to one ratio, and know subscripts will be needed.
LiBr
II. Ternary Compounds - Ternary compounds are composed of three different elements. The
most common types of ternary compounds consist of a metallic cation (positive ion) and a
polyatomic anion (negative ion). The only common polyatomic ion with a positive charge is the
ammonium ion. At any rate, To write these formulas you will want to have reference tables with
the information provided on tables 5-2b and 5-2d.
Example 1. Write the proper chemical formula for potassium hydroxide.
Step one - Write the symbols for the monatomic and polyatomic ions in the compound. You will
find the symbol potassium on table 5-2b. Hydroxide is a polyatomic ion, which will be found on
table 5-2d. Eventually you will recognize the name of a polyatomic ion, but for now if you can't
find an ion on one table, look on the other.
Potassium = K
Hydroxide = OH
Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or some
similar tables), and write them as superscripts to the right of the elemental symbols.
Potassium = K+
Hydroxide = OH-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
Parenthesis must be used if you need more than one of a polyatomic ion. In this case, (+1) + (-1) =
0. So, only one of each ion is used. No subscripts are necessary. If you needed more than one
hydroxide ion, it would be put in parenthesis with the subscript on the outside.
KOH
Note the importance of upper and lower case
Example 2. Show the correct formula for Calcium Nitrate.
Step one - Write the symbols for the monatomic and polyatomic ions in the compound.
Calcium = Ca
Nitrate = NO3
Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or some
similar tables), and write them as superscripts to the right of the elemental symbols.
Calcium = Ca2+
Nitrate = NO3-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
Parenthesis must be used if you need more than one of a polyatomic ion. In this case (+2) + 2(1) = 0. We need to show two nitrate ions in our formula. The subscript is put on the outside of the
parenthesis to show that the entire polyatomic ion is doubled.
Ca(NO3)2
The correct use of parenthesis will seem hard at first, but you must master this skill with practice!
III. The Stock System - Some elements, like iron and lead, have more than one oxidation number.
If you were given a compound name like lead chloride, you would not know if you should used an
oxidation number of +2 or +4 for the lead. The stock system is used to specify which form of an
element, that shows multiple oxidation numbers, is used in a particular compound. A roman
numeral is shown after the name of the positive ion (cation) to indicate the oxidation number of the
positive ion.
Example 1. Show the correct formula for lead(IV) nitrate.
Step one - Write the symbols for the ions in the compound.
Lead = Pb
Nitrate = NO3
Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and 5-2d, or
some similar tables). The positive ion will have a positive oxidation number equal to the roman
numeral. Write the numbers as superscripts to the right of the elemental symbols.
Lead = Pb4+
Nitrate = NO3-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
Parenthesis must be used if you need more than one of a polyatomic ion.
Pb(NO3)4
Example 2. Show the correct formula for Copper(II) Fluoride
Step one - Write the symbols for the ions in the compound.
Copper = Cu
Fluoride = F
Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and 5-2d, or
some similar tables). The positive ion will have a positive oxidation number equal to the roman
numeral. Write the numbers as superscripts to the right of the elemental symbols.
Copper = Cu2+
Fluoride = F-
Step three - Use the correct combination of ions to produce a compound with a net charge of zero.
Parenthesis must be used if you need more than one of a polyatomic ion.
CuF2
Writing Molecular Formulas
I. Binary Molecular Compounds - The standard method for naming binary molecular compounds
has changed over the years. Currently, the stock system is commonly used for naming molecular
compounds. Names like "carbon dioxide", "carbon monoxide", and "dinitrogen pentoxide" are
really remnants of an older system that used prefixes to identify the number of elements involved.
When you are writing the formula for a molecular compound using the stock system, you will not
really notice any difference from the methods described above, until you study bonding. You
should be aware that you are not dealing with ions when you are working with molecular formulas,
rather you are looking up what might be called the apparent charge on each atom.
Example 1. Write the correct formula for nitrogen(IV) oxide.
Step one - Write the symbols for the elements involved.
Nitrogen = N
Oxide = O
Step two - Use the roman numeral as the apparent charge of the first element. Find the apparent
chart of the second element by looking on reference tables such as 5-2a.
Nitrogen = N4+
Oxide = O2-
Step three - Determine the ratio by which the elements will bond to show a net charge of zero. Use
subscripts to indicate the number of atoms of each element present. In this case, (+4) + 2(-2) = 0.
NO2
Example 2. Write the correct formula for nitrogen(III) oxide.
Step one - Write the symbols for the elements involved.
Nitrogen = N
Oxide = O
Step two - Use the roman numeral as the apparent charge of the first element. Find the apparent
charge of the second element by looking on reference tables such as 5-2a.
Nitrogen = N3+
Oxide = O2-
Step three - Determine the ratio by which the elements will bond to show a net charge of zero. Use
subscripts to indicate the number of atoms of each element present. In this case, 2(+3) + 3(-2) = 0.
N2O3
II. Other Molecular Formulas - There are other types of molecular formulas, besides binary,
which you will eventually be required to write. These lessons will be presented at other times.
Naming Compounds
As first mentioned in an early lesson, there are two main types of compounds, ionic and
molecular. Some of the compounds that you will learn about this year will require special systems
for naming, and we will learn about them at a later time. For example, we will learn how to
correctly name the various types of acids and bases when we study the chapters on acids and bases.
In this lesson you will learn enough to name most of the compounds that you will come in contact
with in your laboratory activities this year. References will be made to the tables from lesson 5-2,
so it would be wise to have them handy as you go over this material.
.
I. Binary Compounds. As you know, binary compounds consist of only two elements. The
formula for a binary compound may contain more than two letters, but it will contain only two
capital letters. When naming a binary compound, regardless of whether it is ionic or molecular,
follow the following steps:
1. Write the name of the element represented by the first symbol in the formula.
2. Write the name of the element represented by the second symbol in the formula, but change the
ending of the element's name to "ide".
3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done.
4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names.
Example 1. What is the correct name for the compound AlBr3 ?
Step 1. Write the name of the element represented by the first symbol in the formula.
aluminum
Step 2. Write the name of the element represented by the second symbol in the formula, but change
the ending of the element's name to "ide". In this case, bromine becomes bromide.
aluminum bromide
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done. Aluminum always has an oxidation number
of +3, therefore there is no need for a roman numeral. Our answer is;
aluminum bromide
Example 2. What is the correct name for the element NiS ?
Step 1. Write the name of the element represented by the first symbol in the formula.
nickel
Step 2. Write the name of the element represented by the second symbol in the formula, but change
the ending of the element's name to "ide". In this case, sulfur becomes sulfide.
nickel sulfide
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done. Nickel forms oxidation numbers of +2, +3
and +4, so we must go to the next step.
Step 4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names. We check the oxidation number of sulfide and find that it is -2.
If one nickel is canceling out one sulfur than the apparent charge on the nickel must be +2. (+2) +
(-2) = 0.
nickel(II) sulfide
Example 3. What is the correct name for the compound Fe2O3 ?
Step 1. Write the name of the element represented by the first symbol in the formula.
iron
Step 2. Write the name of the element represented by the second symbol in the formula, but change
the ending of the element's name to "ide". So, oxygen becomes oxide.
iron oxide
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done. Iron can be +2 or +3, so we must go on to
step 4.
Step 4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names. We know that oxygen is -2 in this case. Since we have 3 atoms
of oxygen, each with a charge of -2, then the total negative charge is -6. We must have +6 to
balance out the -6. Since there are two iron atoms to make up a total of +6, each must be +3.
2(+3) + 3(-2) = 0.
iron(III) oxide
II. Ternary Compounds - Ternary compounds contain three elements. The only type of ternary
compounds that we will learn how to name in this chapter are those that consist of one polyatomic
ion and one monatomic ion. The vast majority of these types of compounds consist of a positive
monatomic ion and a negative polyatomic ion. For this type of compound you follow the steps
below:
Step 1. Write the name of the element represented by the first symbol in the compound.
Step 2. Write the name of the polyatomic ion, without changing the ending.
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done.
Step 4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two names.
Example 1. Name the compound Ca(CN)2 ?
Step 1. Write the name of the element represented by the first symbol in the compound.
Calcium
Step 2. Write the name of the polyatomic ion, without changing the ending.
Calcium Cyanide
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done. Calcium is always +2, so the final answer is
as below:
Calcium Cyanide
Example 2. What is the name of the compound Fe(NO3)2 ?
Step 1. Write the name of the element represented by the first symbol in the compound.
iron
Step 2. Write the name of the polyatomic ion, without changing the ending.
iron nitrate
Step 3. Check a reference table to determine the number of positive oxidation numbers that the first
element forms. If it only forms one then you are done. Iron can be +2 or +3, so we must go on to
step 4.
Step 4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two names. Nitrate shows an oxidation number of -1. Since there are two nitrate ions
in the compound, the total negative charge is -2. Therefore, the iron must be +2. (+2) + 2(-1) = 0.
iron(II) nitrate
Special Exception: The Ammonium ion (NH4 +) is a positive polyatomic ion. When it combines
with a negative monatomic ion, you change the ending of the negative ion to "ide". When it
combines with a negative polyatomic ion, you just name both ions.
(NH4)2S is called ammonium sulfide
NH4OH is called ammonium hydroxide
Molecular and Empirical Formulas
By now you should know that a molecular formula is a group of elemental symbols, and
possibly subscript numbers, which represent the composition of a molecule. The molecular
formula shows you how many of each atom can be found in a certain molecule. Some examples of
molecular formulas, that you are most likely familiar with, are shown in table 5-5a.
Table 5-5a - Some Molecular Formulas and the Information they Convey
Formula
Composition
H2O
Total of 3 atoms in each molecule - 1 oxygen and 2 hydrogen
O2
Total of 2 atoms in each molecule - 2 oxygen atoms
CO2
Total of 3 atoms in each molecule - 1 carbon and 2 oxygen
Total of 24 atoms in each molecule - 6 carbon, 12 hydrogen, and 6
oxygen.
C6H12O6
While the molecular formula shows the molecule as it actually exists, the empirical formula for
a compound shows the simplest whole number ratio for the elements in the compounds. To find the
empirical formula for a compound, find the largest whole number that can be divided into each of
the subscripts evenly. If no number will go into any of the subscripts evenly, then the empirical
formula and the molecular formula for the compound are the same. Table 5-5b shows examples of
some empirical and molecular formulas.
Table 5-5b - Comparing Empirical and Molecular Formulas
Compound
Molecular Formula
Empirical Formula
Water
H2O
H2O
Hydrogen Peroxide
H2O2
HO
Glucose
C6H12O6
CH2O
Methane
CH4
CH4
Ethane
C2H6
CH3
Octane
C8H18
C4H9
Molecular and Formula Mass
As you might expect, the term molecular mass refers to the mass of a molecule. Fortunately,
calculating the molecular mass is no harder than understanding the concept. The masses shown on
the periodic table are the atomic masses. To determine the mass of a molecule, simply add up all
of the atomic masses for the atoms that make up the molecule. As a general rule, whenever you
take mass information of the periodic table, round to three significant digits. Here are two
examples of molecular mass calculations.
Table 5-6a. Determining the Molecular Mass
Molecular Mass of CO2
C = 12.0 u x 1 atom = 12.0 u
O = 16.0 u x 2 atoms = 32.0 u
-------Total
=
44.0 u
Molecular Mass of C6H12O6
C = 12.0 u x 6 atoms = 72.0 u
H = 1.01 u x 12 atoms = 12.1 u
O = 16.0 u x 6 atoms = 96.0 u
-------Total
=
180.1 u
As you can see, to find the molecular mass of carbon dioxide we simply look up the atomic
masses of carbon and oxygen. Multiply oxygen's atomic mass by two because there are two
oxygen atoms in each molecule of carbon dioxide. Then add the masses together. The units are
atomic mass units with the symbol (u).
Finding the formula mass of an ionic compound is just as easy, in fact, the calculation is exactly
the same. The only difference has to do with terminology. We don't call ionic compounds
molecules, so you can't find the molecular mass of an ionic compound. Instead, we do the exact
same calculation, but we call the results the formula mass of the compound. The table below
shows how to calculate the formula mass of two ionic compounds.
Table 5-6b, Determining the Formula Mass
Formula Mass of CuSO4
Formula Mass of Ca(OH)2
Cu = 63.5 u x 1 atom = 63.5 u
Ca = 40.1 u x 1 atom = 40.1 u
S = 32.1 u x 1 atom = 32.1 u
O = 16.0 u x 2 atoms = 32.0 u
O = 16.0 u x 4 atoms = 64.0 u
-------
H = 1.01 u x 2 atoms = 2.02 u
--------
Total
Total =
=
159.6 u
74.1 u
Using Coefficients with Formulas
You have learned that the subscript numbers in a chemical formula represent the number of
atoms in one molecule or in one formula unit of an ionic compound. Now you will learn about the
other numbers, called coefficients, that you often see to the left of a chemical formula. While a
subscript number acts as a multiplier for a single element (unless there are parenthesis), a
coefficient number acts as a multiplier for all of the atoms in the entire compound. As with
subscripts, when no number is present then "1" is understood. Look at the example below:
CO2
Here we have one molecule of carbon dioxide. The subscript 2 in the formula above only
pertains to the oxygen in the compound. The total number of atoms in the compound is 3.
Now let us put a coefficient in front of the molecule and see how that changes things.
5 CO2
The coefficient 5 refers to the entire molecule. It shows that there are 5 molecules of carbon
dioxide. Since each molecule is made up of 3 atoms, the total number of atoms is now 15. There
are 5 carbon atoms and 10 oxygen atoms.
Now, for an example with parenthesis;
Ba(NO3)2
Here we have one formula unit of the ionic compound, barium nitrate. We say "formula unit"
instead of "molecule" because ionic compounds don't form molecules. The subscript 3 pertains to
the oxygen, showing 3 oxygen atoms for each polyatomic ion of nitrate. The subscript 2 is a
multiplier for everything in the parenthesis, because it is showing that there are two nitrate ions for
every barium ion. The total number of atoms for each formula unit of barium nitrate is 9. There
are; 1 barium atom, 2 nitrogen atoms, and 6 oxygen atoms.
Now let's put a coefficient in front of the formula unit and see how it changes the tally:
3 Ba(NO3)2
Now we have 3 formula units of barium nitrate. The 3 coefficient acts as a multiplier for the
entire compound. If there are 9 atoms in one formula unit of barium nitrate, then there must be 27
atoms in three formula units. There are: 3 barium atoms, 6 nitrogen atoms and 18 oxygen atoms.
Table 5-7a will show you some more examples of coefficients and formulas. Study the table to
make sure that you understand each atomic tally.
Table 5-7a
Atomic Tallies for Specific Quantities of Molecules and Ionic Compounds
Given
2 NaNO3
6 C2H6
3 (NH4)2S
2 atoms Na
12 atoms C
6 atoms N
2 atoms N
36 atoms H
24 atoms H
Atomic Tally
6 atoms O
Total Atoms 10
3 atoms S
Total Atoms 48
Total atoms 33
Percentage Composition
When you are asked to determine the percentage composition of a compound, it should be
understood that this refers to the percentage by mass. In other words, the percentage composition
of water shows what percentage of the mass of a water molecule is made up of hydrogen and what
percentage of the mass is made up of oxygen. Percentage composition by mass does not tell us the
relative rate by which the atoms combine, or what portion of the volume is made up by an
individual element. Percentage composition, as shown here, refers to mass.
Finding the percentage composition of a compound should be very simple. You know how to
determine the percentage composition of other things by number, and the process used here is just
the same. Look at two examples of word problems.
Example 1. Homer has a box of assorted doughnuts. 3 of these doughnuts are chocolate. What
percentage of Homer's doughnuts are chocolate? (note, we are talking about total number, not mass,
in this case.)
number of chocolate doughnuts
% of doughnuts that are chocolate = --------------------------------- x 100
Total number of doughnuts
3
% of doughnuts that are chocolate = ----------- x 100
12
% of doughnuts that are chocolate = 25%
Example 2. Jim goes to a school that has 258 total students. If 94 of the students are girls, what
percentage of the student body is made up of girls?
number of girls
% of the student body that is made up by girls = -------------------------- x 100
Total number of students
94
% of the student body that is made up by girls = -------------------------- x 100
258
% of the student body that is made up by girls = 36.4%
In both cases, you found the percent composition by putting dividing one part by the total and
multiplying by 100.
part
percent composition = --------------------- x 100
whole
Finding the percentage composition by mass works the same way. You divide the mass that one
element contributes to the compound by the mass of the entire compound. Let's look at the formula
and two examples.
partial mass from element
% of the mass of the compound that is made up by an element = -------------------------- x 100
total mass of the compound
Example 1. What percentage of the mass of carbon dioxide (CO2) is made up by the carbon?
Solution:
first find the mass of the total compound.
C = 12.0 u x 1 atom = 12.0 u
O = 16.0 u x 2 atoms = 32.0 u
-------44.0 u
next use the formula:
partial mass from carbon
% of the mass of CO2 that is made up by carbon = -------------------------- x 100
total mass of the CO2
12.0 u
% of the mass of CO2 that is made up by carbon = ------------------------- x 100
44.0 u
% of the mass of CO2 that is made up by carbon = 27.3%
Answer = 27.3%
Now, it is more typical to be asked what the percentage composition of the entire compound is.
In the example above, you can assume that if carbon makes up approximately 27% of the mass of
carbon dioxide then oxygen makes up about 73%, for the total must be 100%. Example 2, below,
shows the typical wording for this type of problem. The solution will follow.
Example 2. What is the percentage composition of glucose (C6H12O6) ?
solution:
find the mass of the entire molecule,
C = 12.0 u x 6 atoms = 72.0 u
H = 1.01 u x 12 atoms = 12.1 u
O = 16.0 u x 6 atoms = 96.0 u
---------180 u
Then use the formula for each element in the compound:
partial mass from element
% of the mass of the compound that is made up by an element = -------------------------- x 100
total mass of the compound
72.0 u
% for Carbon = ---------------- x 100 = 40.0%
180 u
12.1 u
% for Hydrogen = ---------------- x 100 = 6.7 %
180 u
96.0 u
% for Oxygen = ---------------- x 100 = 53.3%
180 u
One way to check your answer is to make sure that all of the percentages add up to approximately
100%. (i.e. 40.0% + 6.7% + 53.3% = 100%) Your total may be off by a few tenths of a percent,
due to rounding.
Determining an Empirical Formula from
Number of Moles
One of the interesting aspects of our study of moles is that it will give you the ability to
determine the empirical formula of a compound formed in our laboratory, by calculating the molar
ratio by which the elements combine. It will also allow you to determine the empirical formula of a
compound after you determine a compounds percentage composition by mass. This lesson will go
over the math involved in these types of procedures.
I. Determining the empirical formula of a compound from mass.
In an upcoming laboratory activity, you will be asked to experimentally determine the empirical
formula of a compound that you produce from the process of burning magnesium. This synthesis
reaction will produce magnesium oxide. Your purpose will be to determine the empirical formula
of the magnesium oxide by analyzing the molar ratio by which the elements combine.
Ex. 1. What is the empirical formula for a compound if an 8.1 g sample contains 4.9 g of
magnesium and 3.2 g of oxygen?
Solution: Remember that a molar ratio is a ratio of the number of moles, not the mass. We are
not looking for a ratio of mass to mass, but of moles-moles. Our first step is to determine how
many moles of each element are found in this compound. We do that be dividing the mass of the
given element by its molar mass:
mass of that element in the sample
# of moles of an element = ------------------------------------molar mass of the element
Given: mass of magnesium = 4.9 g
molar mass of magnesium = 24.3 g
mass of oxygen = 3.2 g
molar mass of elemental oxygen = 16.0 g (note we use elemental, not diatomic oxygen)
4.9 g
# of moles of magnesium = -------------- = 0.20 moles
24.3 g/mole
3.2 g
# of moles of oxygen = --------------- = 0.20 moles
16.0 g/mole
This problem is easy because, as you can see, the elements are combining in a 0.20 to 0.20 ratio, or
in simpler terms, a 1:1 ratio. This means that for every one atom of magnesium, there is one atom
of oxygen in the compound. This gives us an empirical formula of MgO.
Answer: MgO
Now, let us try a problem that is a little bit harder.
Example 2. Calculate the empirical formula of a compound that is made from 1.67 g of cerium and
4.54 g of iodine.
Solution: We will solve the problem the same way as example 1, with one exception. After we
find out the number of moles of each element present, we will convert are results into a simple,
whole number ratio.
mass of that element in the sample
# of moles of an element = ------------------------------------molar mass of the element
Given: mass of cerium = 1.67 g
molar mass of cerium = 140 g
mass of iodine = 4.54 g
molar mass of iodine = 127 g
1.67 g
# of moles of cerium = ---------- = 0.0119 moles
140 g/mole
4.54 g
# of moles of iodine = ------------- = 0.0357 moles
127 g/mole
Now, to turn the number of moles into a simple whole number ratio, divide the smaller number of
moles (0.0119) into both values:
# of moles of Ce = 0.0119 moles
-------------- = 1
0.0119 moles
# of moles of I = 0.0357 moles
--------------- = 3
0.0119 moles
This means that for every one atom of Ce there are 3 atoms of I, so our empirical formula must be
CeI3.
Answer: CeI3
II. Determining empirical formula from percentage composition.
This type of problem is essentially the same as the problems described above, with one slight
difference. In these problems you start with a percentage composition instead of mass. However,
if you assume that you are studying a 100g sample, you can easily change percentages to grams.
Then solve the problems exactly as shown above.
Example 1. Determine the empirical formula of a compound that contains 36.5% sodium, 25.4%
sulfur, and 38.1% oxygen.
Solution: By assuming that we can study a 100g sample of the compound, we can change % to
grams. so:
Example 1. Determine the empirical formula of a compound that contains 36.5g sodium, 25.4g
sulfur, and 38.1g oxygen.
Now we can solve them by finding the molar ratio by which the elements combine.
mass of that element in the sample
# of moles of an element = ------------------------------------molar mass of the element
Given: mass of sodium = 36.5 g
molar mass of sodium = 23.0 g
mass of sulfur = 25.4 g
molar mass of sulfur = 32.1 g
mass of oxygen = 38.1 g
molar mass of oxygen = 16.0 g
36.5 g
# of moles of sodium = ------------- = 1.59 moles
23.0g/mole
25.4g
# of moles of sulfur = --------------- = 0.791 moles
32.1 g/mole
38.1 g
# of moles of oxygen = --------------- = 2.38 moles
16.0 g/mole
Now, find the simplest whole number ratio by dividing the smallest number of moles into all three
values.
# of moles of Na = 1.59 moles
-------------- = 2.01
0.791 moles
# of moles of S = 0.791 moles
-------------- = 1
0.791 moles
# of moles of O = 2.38 moles
-------------- = 3.01
0.791 moles
The ratio shows that 2 atoms of sodium combine with 1 atom of sulfur and 3 atoms of oxygen, so
our answer is Na2SO3.
Answer: Na2SO3
III. Determining the molecular formula from the empirical formula and the molecular mass.
In this variation you are given the empirical formula and the molecular mass of a compound and
asked to determine the molecular mass. This is accomplished by dividing the molecular mass by
the mass shown in the empirical formula. The result is a whole number that is used as a multiplier
for all of the subscripts in the empirical formula.
Example 1. What is the molecular formula of a compound with an empirical formula of CH2 and a
molecular formula of 56.0 u?
Solution: Find the mass of the empirical formula (CH2):
C = 12.0 x 1 atom = 12.0 u
H = 1.01 x 2 atoms = 2.02 u
---------14.0 u
next, divide that number into the molecular mass:
56.0 u
------- = 4
14.0 u
Now use that number, 4, as a multiplier for the subscripts in the empirical formula:
CH2 x 4 = C4H8
Answer = C4H8
Molarity
I am sure that you have some experience with the concept of solution concentrations.
If you ever made or drank a liquid made from a powdered mix, such as ice tea or hot cocoa,
you probably are familiar with the difference between what is called a "weak" solution or a
"strong" solution. Even if you don't drink coffee, you may have heard a relative complain
about his or her coffee being too strong or to weak. Molarity is simply a measure of the
"strength" of a solution. A solution that we would call "strong" would have a higher
molarity than one that we would call "weak".
Now, a solution is made up of two parts. The solute is what gets mixed into the
solution, like powdered drink mix. The solvent is that which does the dissolving, like water
in the case of ice tea. Let us suppose that we have a powdered ice tea mix that calls for 4
scoops of powder for every 2 quarts of water. We might say that the normal recipe for the
ice tea mix looks like:
4 scoops of powder
To make ice tea of "normal" strength = ----------------------2 quarts of water
If we wanted to make the ice tea twice as strong as normal, we could use 8 scoops of
powder with 2 quarts of water, or 4 scoops of powder with 1 quart of water:
8 scoops of powder
4 scoops of powder
To make ice tea twice the "normal" strength = -------------------- or ---------------------2 quarts of water
1 quart of water
Molarity is just like the strength of the solution above, except that it is more exact.
"Scoops" are not very accurate measuring devices, and they don't need to be when you are
making ice tea. However, when you are doing any type of quantitative analysis in the lab,
you want to be as accurate as possible. Therefore, the formula for Molarity will be more
exact than the formula for ice tea mix.
# of moles of solute
Molarity = ---------------------Liters of solution
The unit for molarity is M and is read as "molar". (i.e. 3 M = three molar)
Molarity problems vary quite a bit. Pay careful attention to the wording of the problem,
and focus on what you are given and what the problem is asking for. Start with the original
molarity formula shown above, but be prepared to modify it when the need arises. I show
you several examples of molarity problems in order to introduce you to the methods
involved in solving different types of problems.
I. Basic molarity problems where the molarity is the unknown.
Example 1. What is the molarity of a 5.00 liter solution that was made with 10.0 moles of
KBr ?
Solution: We can use the original formula. Note that in this particular example, where the
number of moles of solute is given, the identity of the solute (KBr) has nothing to do with
solving the problem.
# of moles of solute
Molarity = ---------------------Liters of solution
Given: # of moles of solute = 10.0 moles
Liters of solution = 5.00 liters
10.0 moles of KBr
Molarity = -------------------------- = 2.00 M
5.00 Liters of solution
Answer = 2.00 M
Example 2. A 250 ml solution is made with 0.50 moles of NaCl. What is the Molarity of
the solution?
Solution: In this case we are given ml, while the formula calls for L. We must change the
ml to Liters as shown below:
250 ml
1 liter
x -------- = 0.25 liters
1000 ml
To avoid confusion, I will usually make the unit change right in the original question:
Example 2. A 250 ml 0.25 L solution is made with 0.50 moles of NaCl. What is the
Molarity of the solution?
Now, solve the problem as you solved example 1.
# of moles of solute
Molarity = ---------------------Liters of solution
Given: Number of moles of solute = 0.50 moles of NaCl
Liters of solution = 0.25 L of solution
0.50 moles of NaCl
Molarity = --------------------- = 2.0 M solution
0.25 L
Answer = 2.0 M solution of NaCl
II. Basic molarity problems where volume is the unknown.
This is similar to when we studied density, we have a formula with three possible
unknowns. When the molarity of the solution and the number of moles of solute are given,
but the volume is unknown, we must adjust our original formula to isolate the unknown
variable. Observe:
# of moles of solute
Molarity = ---------------------Liters of solution
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
Molarity x Liters of solution = # of moles of solute
----------------------------Molarity
Molarity
# of moles of solute
Liters of solution = -------------------Molarity
Example 1. What would be the volume of a 2.00 M solution made with 6.00 moles of LiF?
Solution:
# of moles of solute
Liters of solution = -------------------Molarity
Given: # of moles of solute = 6.00 moles
Molarity = 2.00 M (moles/L)
Liters of solution = 6.00 moles
----------2.00 moles/L
Answer = 3.00 L of solution
Now, you must also be prepared for the fact that the number of moles is not always
given to you. Sometimes you will be given the mass of the solute and you will need to
determine the number of moles by dividing the mass given by the Molar mass of the solute.
In these cases, use the formula below.
mass given
# of moles = ----------------Molar mass
Example 2. What is the volume of 3.0 M solution of NaCl made with 526g of solute?
Solution:
First find the molar mass of NaCl.
Na = 23.0 g x 1 ion per formula unit = 23.0 g
Cl = 35.5 g x 1 ion per formula unit = 35.5 g
---------58.5 g
Now find out how many moles of NaCl you have:
mass of sample
# of moles = ----------------Molar mass
Given: mass of sample = 526 g
Molar mass = 58.5 g
526 g
# of moles of NaCl = -----------58.5 g
Answer: # of moles of NaCl = 8.99 moles
Finally, go back to your molarity formula to solve the problem:
# of moles of solute
Liters of solution = -------------------Molarity
Given: # of moles of solute = 8.99 moles
Molarity of the solution = 3.0 M (moles/L)
8.99 moles
# of Liters of solution = ------------3.0 moles/L
Final Answer = 3.0 L
III. Basic molarity problems where the number of moles is the unknown.
Of course, the total number of moles used in the creation of a solution might be unknown
to you. However, given the molarity and the volume of the solution, you can determine the
number of moles of solute. Observe:
# of moles of solute
Molarity = ---------------------Liters of solution
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
# of moles of solute = Molarity x Liters of solution.
Example 1. How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3 of a
5.0M solution?
Notice that the volume is given in cm3. Since there are 1000 cm3 in 1 liter, 500 cm3 must be
equal to 0.500 liters. Make that change right in the problem.
Example 1. How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3
0.500 L of a 5.0M solution?
Now you are ready to solve.
Solution:
# of moles of solute = Molarity x Liters of solution.
Given: Molarity = 5.0 M (moles/L)
Volume = 0.500 L
# of moles of CaCl2 = 5.0 moles/L x 0.500 moles
Answer = 2.5 moles of CaCl2
Notice that the identity of the solute does not work into the math of the problem. However,
if the wording was different, it would. Observe example # 2.
Example 2. How many grams of CaCl2 would be used in the making of 5.00 x 102 cm3 of a
5.0M solution?
In this case, what they are looking for is different. You could start to solve this problem
the same way you did example 1, but the end would require you to change the number of
moles of CaCl2 to the mass of CaCl2. You would use the formula below.
mass of sample
# of moles = ----------------Molar mass
mass of sample
# of moles x Molar mass = ----------------- x Molar mass
Molar mass
mass of sample = # moles of solute x Molar mass
Given: # of moles of solute = 2.5 moles (from our answer to example 1.)
Molar mass of solute (CaCl2) = 111 g/mole (from the periodic table)
Mass of CaCl2 = 2.5 moles x 111 g/mole
Answer: Mass of CaCl2 = 280 g (when rounded correctly)
Now that you have seen some of the types of problems that can be solved with these
formulas, go on and practice these calculations using the worksheets below. Make sure that
you can adjust each formula to fit a given situation. Avoid memorizing formulas and trying
to work the problems mechanically. Use the correct units as a way to check your answer.
When you feel ready, take the random online quiz to see how you are doing.
The Kinetic Theory of Gases
The Kinetic Theory of Gases is one of the most interesting topics in Chemistry. If you
come to truly understand the concepts in this chapter, it will change the way you look at the
world around you. One of the great things about this topic is that it explains some of the
phenomena that you encounter in your everyday life. For example, have you ever inflated a
pool float until it was firm, thrown it into a cold pool, and then wondered why the float then
seemed like it was not fully inflated? Do you know why a basketball seems flat after it has
spent the night in a cold garage? What determines the time it takes to smell the perfume of
a woman who walks past you? How do hot air balloons work? How can a small barbeque
tank hold enough propane to cook with all summer long? All of these questions can be
answered by someone who has studied this chapter.
Gases have special properties that liquids and solids don't have. The molecules that
make up the gas are free to move about, and a gas will take up the size and shape of its
container. Knowing the volume of a gas tells you very little about the quantity of matter,
because any sample of gas will fill its container. If you have a ten-gallon tank on your
barbeque, it is always technically full! In order to have an idea of the amount of matter that
a sample of gas represents, you need to know the temperature and the pressure of the gas.
Ideal Gases - Consider how different a gas is from a solid. In a gas, the size of the
sample has very little to do with the size of the actual atoms that make up the gas itself.
Even in relatively dense gas samples, the space in between the molecules will be much
larger than the molecules themselves. When we do math problems involving gases, we treat
the particles as point masses, or particle with mass but no volume. Ideal gases differ from
real gases in another important way. In real gases, there will be an attraction between the
particles involved. These attractions are often minor and we ignore them when we do math
problems involving gases. It is important to remember the differences between real gases
and ideal gases. It is also interesting to note that real gases will act most like ideal gases at
low pressure and high temperature, when the gas sample is less dense.
Pressure - You may recall that pressure is defined as a force over an area. In Chemistry,
pressure is often measured in kilopascals (kPa), millimeters of mercury (mm of Hg), or
atmospheres (atm). For convenience sake, a standard atmospheric pressure has been set at
101.3 kPa, which is also equal to 760 mm of Hg and 1.0 atm. As a student of Chemistry
you should be aware of the following constants and conversions:
Standard Atmospherice Pressure = 101.3 kPa = 760 mm of Hg = 1.0 atm
1 kPa = 7.50 mm of Hg
Temperature - Many of the thermometers that are used in Chemistry laboratories are
marked with the Celsius scale. However, when we do math problems involving the
temperature, volume and pressure of gases, we must use the Kelvin scale. The reason for
this is the fact that it is possible to have negative numbers on the Celsius scale, and that
would cause problems when measuring the volume of a gas at low temperatures. In order to
do any gas law calculations involving temperatures, you must first convert the temperature
to Kelvin. As a reminder, the conversions between Kelvin and Celsius are shown below.
Co + 273 = K
K - 273 = Co
For convenience, standard temperature has been set at 273 Kelvin, which is equal to
0oC. Standard temperature and pressure is abbreviated as STP. Conditions will vary from
laboratory to laboratory and from day to day. You will often be called upon to adjust the
volume of the gas that you collected in your own lab to STP, meaning standard conditions
for pressure and temperature. Remember the information below.
STP = 101.3 kPa and 273 K (or any equivalent values, i.e. 1 atm and 0oC)
Boyle's Law
You Probably started experimenting with Boyle's Law when you were a small child.
When you squeeze a balloon, you might notice that the harder you push, the harder it seems
to push back. When you lie back on an inflatable mattress, or pool float, it compresses up to
a point and then seems to stop. This is because as you decrease the volume of a confined gas,
the pressure that it exerts increases. This relationship, called Boyle's Law, is summarized by
the statement:The volume of a sample of gas is inversely proportional to its pressure, if
temperature remains constant.
When two variables are inversely proportional, like pressure and volume in the example
above, the product of the two variables will always remain constant. Because of the
relationship between the pressure and volume of a gas sample at constant temperature, if you
double the value of one, you divide the other by two. The chart below will demonstrate the
inverse relationship between the volume and pressure of a gas. Imagine a gas sample trapped
in a cylinder which allows you to adjust the pressure. Notice how the pressure changes cause
the volume to change, while the product of the two variables will remain a constant (K).
Table 7-3a Data for a Sample of Gas at Constant Temperature and Varying Pressure.
Trial
Pressure
Volume
Formula
Calculation
1
100 kPa
40 cm3
PV=K
100 kPa x 40 cm3 = 4000 kPa x
cm3
2
50 kPa
80 cm3
PV=K
50 kPa x 80 cm3 = 4000 kPa x
cm3
3
200 kPa
20 cm3
PV=K
200 kPa x 20 cm3 = 4000 kPa x
cm3
4
400 kPa
10 cm3
PV=K
400 kPa x 10 cm3 = 4000 kPa x
cm3
5
25 kPa
160 cm3
PV=K
25 kPa x 160 cm3 = 4000 kPa x
cm3
Boyle's law is sometimes used to determine the volume that a gas would have at another
pressure. If you were to collect a sample of gas under the atmospheric conditions in your lab
on a given day, you might want to mathematically determine what the volume of the gas
would be under different conditions. The formula that can be used to calculate the affects of
pressure changes on the volume of a gas at constant temperature is shown below:
P1V1 = P2V2
Were P = Pressure and V =Volume
Example 1 - A sample of gas collected in a 350 cm3 container exerts a pressure of 103 kPa.
What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature
remains constant.)
Solving:
Write the original formula:
P1V1 = P2V2
Then list what is given and what is unknown.
P1 = 103 kPa
V1= 350 cm3
P2 = 150 kPa
V2 = ?
Next, predict what should happen. The pressure is going up by nearly 1/3, so the volume
should go down by a bit less than 1/3.
Now, Adjust the original formula to isolate the unknown, solve and round to the correct
number of significant digits.
a) P1V1 = P2V2
b) P1V1 = P2V2
-------
--------
P2
P2
c) V2 = P1V1
------P2
V1= 350 cm3
V2 = 103 kPa x 350 cm3
------------------------150 kPa
P2 = 150 kPa
V2 = 240.333333 cm3
V2 = ?
V2 = 240 cm3
P1 = 103 kPa
Finally, check to see that the results match your prediction. The volume did go down by close
to 1/3.
Of course, the original formula can be solved for a different unknown. For example, you
can determine what the pressure would have to be in order to end up with a certain volume.
Example 2 - A sample of neon has a volume of 239 cm3 at 2.00 atm of pressure. What
would the pressure have to be in order for the gas to have a volume of 5.00 x 102 cm3?
Solving:
Write the original formula:
P1V1 = P2V2
Then list what is given and what is unknown.
P1 = 2.00 atm
V1= 239 cm3
P2 = ?
V2 = 5.00 x 102 cm3
Next, predict what should happen. You want the volume to more than double, so the
pressure would have to be less than half the original.
Now, Adjust the original formula to isolate the unknown, solve and round to the correct
number of significant digits.
a) P1V1 = P2V2
b) P1V1 = P2V2
-------- -------V2
V2
c) P2 = P1V1
----------
V2
V1= 239 cm3
P2 = 2.00 atm x 239 cm3
-----------------------5.00 x 102 cm3
P2 = ?
P2 = 0.956 atm
V2 = 5.00 x 102 cm3
P2 = 0.956 atm
P1 = 2.00 atm
Finally, check to see that the results match your prediction. The pressure would have to be
less than 1/2.
Dalton's Law
The total pressure in a container is the sum of the partial pressures of all the gases in the
container. That may sound like common sense to you, but that is Dalton's Law of partial
pressures. As simple as it may sound, it is one of the most useful of the gas laws in real life. A
common method of gas collection in the laboratory involves displacing water from a bottle, so
that you know when the bottle is full of an invisible gas. The gas that is left in the bottle will
not be pure, it will be a mixture that contains a certain amount of water vapor. To find the
pressure of the dry gas alone, we need to subtract out the pressure of the water vapor. This is
one way that Dalton's law of partial pressure can be used. In this form, the formula that we
use looks like this:
Pdry gas = Ptotal - Pwater vapor
Where P = Pressure
In order to solve the problem in a real-life situation, you need a reference table that shows
the pressure of water vapor at various temperatures. Your textbook probably has such a
table, but for convenience, I have provided one below. There will be a link at the bottom of
the page where you can print a table out to use on exams.
Table 7-4a Vapor Pressure of Water
Temperature Pressure
oC
kPa
0
0.6
Temperature Pressure
oC
kPa
20
2.3
Temperature Pressure
oC
kPa
30
4.2
3
0.8
21
2.5
32
4.8
5
0.9
22
2.6
35
5.6
8
1.1
23
2.8
40
7.4
10
1.2
24
3.0
50
12.3
12
1.4
25
3.2
60
19.9
14
1.6
26
3.4
70
31.2
16
1.8
27
3.6
80
47.3
18
2.1
28
3.8
90
70.1
19
2.2
29
4.0
100
101.3
Now we will demonstrate how one problem of this type would be solved.
Example 1. A sample of hydrogen gas is collected over water at 14.0 oC. The pressure of the
resultant mixture is 113.0 kPa. What is the pressure that is exerted by the dry hydrogen
alone?
Solving:
Write the formula you will need:
Pdry gas = Ptotal - Pwater vapor
Look up the vapor pressure of water at 14.0 oC on table 18-4a: 1.6 kPa
List what is known and unknown:
Pdry gas = ?
Ptotal = 113.0 kPa
Pwater vapor = 1.6 kPa
Substitute and solve:
Pdry gas = Ptotal - Pwater vapor
Pdry gas = 113.0 kPa - 1.6 kPa
Pdry gas = 111.4 kPa
Another way that you will be will use this gas law is to simply determine the partial
pressure of another gas, other than water vapor, in a mixture. These types of problems are
very easy, and it would be best to look at the formula below:
Ptotal = P1 + P2 + . . . Pn
Where P1, P2, and Pn are the partial pressures of the gases involved.
Below is an example of how this formula would be used:
Example 2. A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure of 278
kPa. If the partial pressures of the oxygen and the hydrogen are 112 kPa and 101 kPa
respectively, what would be the partial pressure exerted by the nitrogen.
Ptotal = P1 + P2 + . . . Pn
278 kPa = 112 kPa + 101 kPa + Pnitrogen
Pnitrogen = 278 kPa - (112 kPa + 101 kPa)
Pnitrogen = 65 kPa
Charles's Law
An inflatable pool float may seem quite firm as it sits on a deck in the hot sun. However,
minutes after you toss to float into the cold pool, the same float may seem under-inflated.
You may suspect that the float has developed a slow leak, but that may not be the most likely
explanation for the apparent loss of air pressure. It may be that Charles's law is responsible.
Charles's law, discovered by Jacques Charles, states that the volume of a quantity of gas, held
at constant pressure, varies directly with the Kelvin temperature.
Gases expand as they are heated and they contract when they are cooled. In other words,
as the temperature of a sample of gas at constant pressure increases, the volume increases. As
the temperature goes down, the volume decreases as well. The mathematical expression for
Charles's law is shown below:
V1/T1 = V2/T2
Remember that Charles's law calculations must be done in the Kelvin scale.
Example 1. A 250 cm3 sample of neon is collected at 44.0 oC. Assuming the pressure remains
constant, what would be the volume of the neon at standard temperature?
Solution:
First change the Celsius temperature to Kelvin.
K = oC + 273
K = 44.0 oC + 273
K = 317
Now list the given quantities and the unknown.
T1 = 317 K
V1 = 250 cm3
T2 = 273 K (standard temperature in Kelvin
V2 = ?
Now predict the results. The temperature is going down, so the volume must go down as well.
Write the original formula and then isolate the unknown: a)
b)
c)
V1 V2
---- = ---T1
T2
T2 x V1 V2 x T2
---- = ---T1
T2
V2 = T2 x V1
-----------
T1
Now, substitute, solve and round to correct significant digits.
T1 = 317 K
V1 = 250 cm3
T2 = 273 K (standard temperature in Kelvin
V2 = ?
V2 = T2 x V1
-----------
T1
V2 = 273 K x 250 cm3
----------------------317 K
V2 = 215.2996845 cm3
V2 220 cm3
Finally, we check that our prediction was correct. The volume did go down.
Example 2. A sample of oxygen gas has a volume of 2.73 dm3 at 21.0 oC. At what
temperature would the gas have a volume of 4.00 dm3?
Solution:
First change the Celsius temperature to Kelvin.
K = oC + 273
K = 21.0 oC + 273
K = 294 K
Now list the given quantities and the unknown.
T1 = 294 K
V1 = 2.73 dm3
T2 = ?
V2 = 4.00 dm3
Now predict the results. The temperature must go up in order for the volume to go up.
Write the original formula and then isolate the unknown: a)
b)
Now, substitute, solve and round to correct significant digits.
T1 = 294 K
V1 = 2.73 dm3
V1 V2
---- = ---T1
T2
T2 = T1 x V2
----------V1
T2 = ?
V2 = 4.00 dm3
T2 = T1 x V2
-----------
V1
T2 = 294 K x 4.00 dm3
----------------------2.73 dm3
T2 = 430.7692308 K
T2 431 K
Finally, we check that our prediction was correct. The temperature did go up.
Combined Gas Law Problems
There are many situations that would call for the use of more than one of the gas laws.
Perhaps the most common situation involves using both Charles's law and Boyle's law on the
same gas sample. Because pressure and temperature will change from day to day and from
location to location, it is common to mathematically adjust the volume of a gas to standard
temperature and pressure. In such a situation, the values for STP are used for T2 and P2.
The formula for the combined gas law is shown below:
V1P1T2
V2 = ----------P2T1
Example 1. A 350 cm3 sample of helium gas is collected at 22.0 oC and 99.3 kPa. What
volume would this gas occupy at STP?
Solving:
First, you must change the Celsius temperature to Kelvin.
A 350 cm3 sample of helium gas is collected at 22.0 oC 295 K and 99.3 kPa. What volume
would this gas occupy at STP?
Now, list the givens and the unknown
V1 = 350 cm3
P1 = 99.3 kPa
T1 = 295 K
V2 = ?
P2 = 101.3 kPa (standard pressure)
T2 = 273 K (standard temperature)
Now, substitute the values into the equation, which is already set up for volume 2 as the
unknown.
V2 = V1P1T2
----------P2T1
V2 = 350 cm3 x 99.3 kPa x 273 K
-------------------------------------101.3 kPa x 295 K
V2 = 317.5034718 cm3
V2 = 320 cm3
Density of Gases
As you now know, unlike solids and liquids, the volume and density of a gas will change in
a significant way as its temperature and or pressure changes. Sometimes, you will be asked to
calculate the density that a gas would have at a given temperature and pressure. The key to
this type of problem is to realize that the mass of the gas will not change, just because the
pressure and temperature do. You simply do a combined gas law problem and then divide
the given mass by the new volume.
Example 1. A sample of gas with a mass of 125 mg occupies a volume of 213 cm3 at 21.0 oC
and 103.5 kPa. What will the density of the gas be at STP?
First, let us change the temperature to Kelvin, as we do with all gas law problems.
Example 1. A sample of gas with a mass of 125 mg occupies a volume of 213 cm3 at 21.0 oC
294 K and 103.5 kPa. What will the density of the gas be at STP?
Next, we list all of the givens and the unknown.
m = 125 mg
V1 = 213 cm3
P1 = 103.5 kPa
T1 = 294 K
V2 = ?
P2 = 101.3 kPa (standard pressure)
T2 = 273 K (standard temperature)
Now, write the formula and solve for the unknown:
V2 = V1P1T2
----------P2T1
V2 = 213 cm3 x 103.5 kPa x 273 K
--------------------------------------101.3 kPa x 294 K
V2 = 202 cm3
Next, use the formula for density, where:
m = 125 mg
V = 202 cm3
D = m/V
D = 125 mg/202 cm3
Density of the gas at STP = 0.619 mg/cm3
Balancing Chemical Equations
According to the law of conservation of mass, a chemical equation must be balanced. This
means that the total number of atoms on the reactant side must be equal to the total number
of atoms on the product side. This really involves three skills; interpreting a chemical
formula, determining whether or not a chemical equation is balanced, and balancing the
equation.
I. Interpreting a Chemical Formula.
If you can read a chemical formula correctly, you can check the balance on a chemical
equation. One of the biggest problems for new Chemistry students is correctly reading the
number of atoms inside parenthesis. Let us practice this skill first, with a couple of examples.
Example 1. How many atoms of each element are there in one formula unit of ammonium
sulfide?
Ammonium Sulfide is (NH4)2S
Remember that a subscript pertains only to the element that precedes it, unless it precedes
parenthesis, in which case it is a multiplier for each element in the parenthesis. In the
example above, the subscript 4 only pertains to hydrogen, while the subscript 2 acts as a
multiplier for both nitrogen and hydrogen, giving us as a final tally;
(NH4)2S
2 atoms of nitrogen;
8 atoms of hydrogen;
and 1 atom of sulfur.
Example 2. How many atoms of each element are there in one formula unit of barium
nitrate?
Barium Nitrate = Ba(NO3)2
Now, the subscript 3 pertains only to the oxygen, but the subscript 2 becomes a multiplier for
each element in the parenthesis. Therefore;
Ba(NO3)2
1 atom of barium;
2 atoms of nitrogen;
and 6 atoms of oxygen.
II. Checking the Balance of a Chemical Equation.
When we write chemical equations for a chemical reaction, we use special numbers called
coefficients to represent multiple molecules or formula units. For example;
6H2O
As before, the subscript 2 pertains only to the hydrogen. However, the coefficient 6
pertains to every element in the compound, whether or not they are found in parenthesis.
The 6 tells us that there are six molecules of water, with a total of 12 atoms of hydrogen and 6
atoms of oxygen. Once again, a coefficient pertains to every element in the compound,
regardless of parenthesis. You will need to keep this in mind when you check the balance of
an equation.
Example 1. Determine if the following reaction is balanced or not.
Ca(OH)2(cr) ---> CaO(cr) + H2O(g)
Let us make an organized tally table and compare both sides of the equation;
Ca(OH)2(cr) ---> CaO(cr) + H2O(g)
Reactant Side
Product Side
Ca
O
H
1 atom
2 atoms
2 atoms
Elements
Ca
O
H
1 atom
2 atoms
2 atoms
Elements
As you can see, this reaction is balanced, so no coefficients are necessary.
Example 2. Check the balance on the following chemical reaction;
Ca(OH)2(aq) + HCl(aq) ---> CaCl2(aq) + H2O(l)
Ca(OH)2(aq) + HCl(aq) ---> CaCl2(aq) + H2O(l)
Reactant Side
Product Side
Ca
O
H
Cl
Ca
O
H
Cl
1 atom
2 atoms
3 atoms
1 atom
1 atom
1 atom
2 atoms
2 atoms
As you can see, this reaction is not balanced. You are not allowed to change any
subscripts, but coefficients may be added in order to obtain balance.
III. Balancing Chemical Equations
Balancing chemical equations is a skill that only develops with practice, but for starters, look
at the tally above. Notice that you need more Cl on the reactant side. What would the tally
look like if we add a coefficient of 2 to the HCl on the reactant side?
Ca(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + H2O(l)
Ca(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + H2O(l)
Reactant Side
Product Side
Ca
O
H
Cl
Ca
O
H
Cl
1 atom
2 atoms
4 atoms
2 atoms
1 atom
1 atom
2 atoms
2 atoms
Now we need more oxygen and more hydrogen on the product side. Let's add a coefficient of
2 to the H2O on the product side and check the balance again.
Ca(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
Ca(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
Reactant Side
Product Side
Ca
O
H
Cl
Ca
O
H
Cl
1 atom
2 atoms
4 atoms
2 atoms
1 atom
2 atom
4 atoms
2 atoms
Now the equation is balanced.
Example 2. Write a balanced chemical equation for the reaction below;
Propane reacts with oxygen gas to yield carbon dioxide and water.
First, you need to be able to turn a word equation into a chemical equation. The one above
would become;
C3H8(g) + O2(g) ----> CO2(g) + H2O(g)
Now, let us tally the information in a table:
C3H8(g) + O2(g) ----> CO2(g) + H2O(g)
Reactant Side
Product Side
C
H
O
C
H
O
3 atoms
8 atoms
2 atoms
1 atom
2 atoms
3 atoms
Well, a quick look shows us that we will need more hydrogen and more carbon on the right
hand side. Let us start by multiplying the number of hydrogen on the product side by four,
giving us a total of 8 atoms of hydrogen. Be aware that this will also change the number of
oxygen atoms on the product side. Let us look at how a coefficient of 4 in front of water
changes things.
C3H8(g) + O2(g) ----> CO2(g) + 4H2O(g)
C3H8(g) + O2(g) ----> CO2(g) + 4H2O(g)
Reactant Side
Product Side
C
H
O
C
H
O
3 atoms
8 atoms
2 atoms
1 atom
8 atoms
6 atoms
Now we have a match with the number of hydrogen atoms. Let us balance the carbon atoms
next, because in order to change the carbon atoms on the product side, it will also affect the
number of oxygen atoms. We need to multiply the number of carbon atoms on the product
side by three, so we will place a coefficient of three in front of the carbon dioxide and check
the tally again.
C3H8(g) + O2(g) ----> 3CO2(g) + 4H2O(g)
C3H8(g) + O2(g) ----> 3CO2(g) + 4H2O(g)
Reactant Side
Product Side
C
H
O
C
H
O
3 atoms
8 atoms
2 atoms
3 atom
8 atoms
10 atoms
Now, we have matched the number of atoms for two of the elements. A subscript of 5 in front
of the oxygen on the reactant side should finish the job.
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(g)
Reactant Side
Product Side
C
H
O
C
H
O
3 atoms
8 atoms
10 atoms
3 atom
8 atoms
10 atoms
We have achieved proper balance! In practice, the process is not nearly as long and tedious
as this may have appeared. Once you gain some experience, you will find that you can
balance these equations quickly and painlessly. Start practicing with the worksheets below,
and be sure to browse the links for more information.
Classification of Chemical Reactions
Chemists have identified millions of different compounds, so there must be millions of
different chemical reactions to form them. When scientists are confronted with an
overwhelming number of things, they tend to classify them into groups, in order to make
them easier to study and understand. One popular classification scheme for chemical
reactions breaks them up into five major categories or types. Some of these types have been
given more than one name, so you need to learn them all. Even if your teacher prefers one
name over another, you need to recognize each name, as you may encounter different names
in different places.
Types of Chemical Reactions:
1. Synthesis (also called Direct Combination) - A synthesis reaction involves two or more
substances combining to make a more complex substance. The reactants may be elements or
compounds, and the product will always be a compound. The general formula for this type of
reaction can be shown as;
A
+
B
---->
AB
or
element or compound + element or compound -----> compound
Some examples of synthesis reactions are shown below;
2H2(g) + O2(g) ----> 2H2O(g)
C(s) + O2(g) ----> CO2(g)
CaO(s) + H2O(l) ----> Ca(OH)2(s)
2. Decomposition (also called Analysis) - In a decomposition reaction, one substance is
broken down into two or more, simpler substances. This type of reaction is the opposite of a
synthesis reaction, as shown by the general formula below;
AB
---->
A
+
B
or
Compound ------> element or compound + element or compound
Some examples of decomposition reactions are shown below;
C12H22O11(s) ----> 12C(s) + 11H2O(g)
Pb(OH)2(cr) ----> PbO(cr) + H2O(g)
2Ag2O(cr) ----> 4Ag(cr) + O2(g)
3. Single Displacement (also called Single Replacement) - In this type of reaction, a neutral
element becomes an ion as it replaces another ion in a compound. The general form of this
equation can be written as;
In the case of a positive ion being replaced: A + BC ----> B + AC
or
In the case of a negative ion being replaced: A + BC ----> C + BA
in either case we have;
element + compound ----> element + compound
Some examples of single displacement reactions are shown below:
Zn(s) + H2SO4(aq) ----> ZnSO4(aq) + H2(g)
2Al(s) + 3CuCl2(aq) ---> 2AlCl3(aq) + 3Cu(s)
Cl2(g) + KBr(aq) ----> KCl(aq) + Br2(l)
4. Double Displacement (also called Double Replacement) - Like dancing couples, the
compounds in this type of reaction exchange partners. The basic form for this type of
reaction is shown below;
AB + CD ----> CB + AD
or
Compound + Compound ----> Compound + Compound
Some examples of double displacement reactions are shown below;
AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)
ZnBr2(aq) + 2AgNO3(aq) ----> Zn(NO3)2(aq) + 2AgBr(cr)
H2SO4(aq) + 2NaOH(aq) ----> Na2SO4(aq) + 2H2O(l)
5. Combustion - When organic compounds like propane are burned, they react with the
oxygen in the air to form carbon dioxide and water. The reason why these combustion
reactions will stop when all available oxygen is used up is because oxygen is one of the
reactants. The basic form of the combustion reaction is shown below;
hydrocarbon + oxygen ----> carbon dioxide and water
Some examples of combustion reactions are;
CH4(g) + 2O2(g) ----> 2H2O(g) + CO2(g)
2C2H6(g) + 7O2(g) ----> 6H20(g) + 4CO2(g)
C3H8(g) + 5O2(g) ----> 4H2O(g) + 3CO2(g)
The Mole
The mole concept is one of the most feared and misunderstood concepts in all of chemistry.
Many adults who cringe at recalling Chemistry will tell you that they never understood that "mole
stuff." The funny thing about it is that the mole is really a very simple concept. I think that many
people never really understand it because it is not always presented clearly. If explained correctly, I
feel that the mole can be an easy concept to master.
You know that there are 12 items in a dozen. No matter what the item, a dozen is equal to 12.
A gross is another unit of grouping. There are 144 items in a gross. A score, another set group, is
equal to 20 items. You can have a score of years or a score of rocks, but it will always be 20 items.
Now, a mole is a unit of grouping, just like these examples. The only difference is that a mole
represents a large number of items, 602 000 000 000 000 000 000 000 (or 6.02 x 1023) items to be
more specific. What the items are doesn't matter. You can have a mole of molecules, a mole of
ions, or a mole of stars. The number of items in a mole will always be 6.02 x 1023 . This number
is known as Avogadro's number.
Why such a large number for the mole? Well, why do they sell eggs in a dozen? Maybe
because no one wants to buy just one egg, and if you buy fifty, some will go bad before you eat
them. The reason we need so many items in a mole may be because we need to group molecules in
very large groups in order to be able to get a measurable reading on our balances. We can't find the
mass of one atom, or even one gross of atoms, on our laboratory balances, the instruments are not
sensitive enough. We can, however, find the mass of one mole of atoms on our balance.
The real reason for packing 6.02 x 1023 items into a mole is because there are 6.02 x 1023 u (or
atomic mass units) in one gram. This allows us to use the mass numbers on the periodic table for
both the mass of an atom (atomic mass) and the mass of a mole of atoms (molar mass), we only
need to change the units. Table 5-2a will demonstrate what this idea:
Table 9-1a Atomic and Molar Masses
Element and Symbol
Atomic Mass - Mass of 1
Atom
Molar Mass - Mass of 6.02 x
1023 Atoms
Carbon - C
12.0 u
12.0 g
Helium - He
4.00 u
4.00 g
Copper - Cu
63.5 u
63.5 g
Potassium - K
39.1 u
39.1 g
At this point, some students might say "Hey, I thought that a mole is always 6.02 x 1023? How can
the molar mass of carbon and helium be different?" That is like saying, "How can the weight of a
dozen elephants be different than the weight of a dozen ants? Shouldn't they both be 12?" It is the
number of items that is always the same, not the mass or weight or size of those items.
Of course the periodic table can be used to determine the molar mass of molecules and formula
units as well. If the molecular mass is found in atomic mass units (u), the molar mass of that
molecule will have the same value with the unit grams (g). Table 5-2b has some examples.
9-1b Molecular and Molar Masses
Compound Name and
Molecular Formula
Molecular Mass - Mass of
one Molecule
Molar Mass - Mass of 6.02 x
1023 Molecules
Water - H20
18.0 u
18.0 g
Carbon Dioxide - CO2
44.0 u
44.0 g
Glucose - C6H12O6
180 u
180 g
The mole allows us to do many of the important calculations that are required for quantitative
analysis of samples in the lab. Below we will go over examples of several types of these
calculations. When we use moles in calculations we will abbreviate the units to mol. When we
show the molar mass of a substance we will show the units in g/mole (read "grams per mole")
Molar Conversions
I. CHANGING MASS TO NUMBER OF MOLES
We know that we can find the molar mass of a substance from the periodic table. We also know
that we can find the mass of a sample by using a balance in the laboratory. By dividing the mass of
a sample by its molar mass, we get the number of moles of the substance.
mass of the sample
# of moles of the substance = --------------------------Molar mass of substance
Example 1. If you find the mass of a sample of glucose (C6H12O6) to be 90.0 g, how many moles of
glucose do you have?
Solution - We find the Molar mass of glucose the same way as we would find its molecular mass.
We look up the masses of each atom on the periodic table, multiply by the number of atoms present
and add the total.
Molar mass of C6H12O6 is:
Carbon = 12.0 g x 6 atoms
= 72.0 g
Hydrogen = 1.01 g x 12 atoms = 12.1 g
Oxygen = 16.0 x 6 atoms
= 96.0 g
--------180.1 g or 180 g (1.80 x 102 g)
Now use the formula:
mass of the sample
# of moles of the substance = --------------------------Molar mass of substance
Given: The mass of the sample is 90.0 g
The Molar mass of the substance is 180 g/mol
90.0 g
# of moles of glucose = ------------------- = 0.500 mol
180 g/mol
Answer - 0.500 moles of glucose.
II. CHANGING NUMBER OF MOLES TO MASS
If we are given the number of moles of a substance, we can convert that into a mass, based on
our knowledge of Molar mass. We can use the formula from the section above, and adjust it for the
new unknown, which will be mass of the sample:
mass of the sample
# of moles of the substance = --------------------------Molar mass of substance
To isolate the mass of the sample, we must multiply both sides by Molar mass of a substance:
mass of the sample x Molar mass of substance
# of moles of the substance x Molar mass of substance = --------------------------Molar mass of substance
So we are left with the formula:
mass of the sample = # of moles of the substance x Molar mass of the substance.
Example 1. A certain laboratory procedure requires the use of 0.100 moles of magnesium. How
many grams of magnesium would you mass out on the balance?
Solution, from the period table we get the Molar mass of magnesium as 24.3 g. We then place that
information in the formula below:
mass of the sample = # of moles of the substance x Molar mass of the substance.
Given: The # of moles of the substance = 0.100 mol
The Molar mass of the substance = 24.3 g/mol
mass of the sample = 0.100 mol of magnesium x 24.3 g/mol
Answer = 2.43 g of magnesium
III. CHANGING NUMBER OF MOLES TO NUMBER OF
PARTICLES
This is like asking you to change 3 dozen doughnuts into total number of doughnuts. A very
simple calculation that you can do in your head. We know that a dozen is always 12, so 3 x 12 =
36. Changing number of moles to number of particles should not be much harder. We know that
one mole is always Avogadro's number (6.02 x 1023). So, we just need to multiply the number of
moles by Avogadro's number.
Total number of particles = number of moles x 6.02 x 1023
Example 1. How many molecules of carbon dioxide are found in 2.50 moles of carbon dioxide?
Solution: We use the formula:
Total number of particles = number of moles x 6.02 x 1023
Given: Number of moles = 2.50 mol
Total number of molecules of CO2 = 2.50 mol x 6.02 x 1023 molecules/mol
Answer = 1.51 x 1024 molecules of CO2
IV. CHANGING NUMBER OF PARTICLES TO NUMBER OF MOLES
This calculation is obviously the opposite of the calculation shown in the section above. If I told
you I had 30 doughnuts and I asked you how many doughnuts there are, you could divide 30 by 12
and get 2.5 dozen. The logic is the same for determining number of moles.
# of moles = total number of particles
-----------------------6.02 x 1023
Example 1. How many moles of O2 are represented by 7.45 x 1024 molecules of O2?
Notice that it does not matter what compound the question asks for. Just as 30 of any item would
represent 2.5 dozen, 7.45 x 1024 of any molecule will represent a set number of moles. There is no
need to use the periodic table for this type of problem
Solution: Use the formula:
# of moles = total number of particles
-----------------------6.02 x 1023
Given: The total number of particles = 7.45 x 1024 molecules
# of moles of O2 = 7.45 x 1024 molecules
----------------------------6.02 x 1023 molecules/mol
Answer = 12.4 moles of O2
V. CHANGING NUMBER OF PARTICLES TO MASS
This is an example of what might be called a "two-step problem". The first step is to change
the number of particles to number of moles, as shown in section IV of this page. Then change the
number of moles into mass, as shown in section II of this page.
Example 1. What would be the mass of 3.75 x 1021 atoms of iron?
Solution: First change number of atoms to moles with the formula:
# of moles = total number of particles
-----------------------6.02 x 1023
Given: The total number of particles = 3.75 x 1021 atoms
# of moles of iron = 3.75 x 1021 atoms
--------------------------6.02 x 1023 atoms/mol
Answer to step one = 0.00623 moles of iron.
Now, change number of moles to mass using the equation:
mass of the sample = # of moles of the substance x Molar mass of the substance.
Given: The number of moles of the substance = 0.00623 mol
The Molar mass of iron (from periodic table) = 55.8 g/mol
mass of the Fe sample = 0.00623 mol x 55.8 g/mol
Final Answer: Mass of the iron = 0.348 g
VI. CHANGING MASS TO A NUMBER OF PARTICLES
We can, of course, do the opposite of the procedure described in step V, in order to change a
given mass into the number of particles. This would also be a two-step operation. First we would
change the mass to number of moles, as described in step I. Then we would change number of
moles to number of particles as shown in step III.
Example 1. How many water molecules would be found in a 54.0 gram sample of water?
Solution. First we would change the mass of the sample to number of moles. Find the Molar mass
of water:
Molar mass of H2O
H = 1.01g x 2 atoms = 2.02 g
O = 16.0 g x 1 atom = 16.0 g
-------Molar mass of H2O
= 18.0 g
Now we use the formula:
mass of the sample
# of moles of the substance = --------------------------Molar mass of substance
Given: The mass of the sample is 54.0 g
The Molar mass of the substance is 18.0 g/mol
54.0 g
# of moles of water = --------------18.0 g/mol
Answer to step one = 3.00 moles of water.
Now, change the number of moles of water to number of particles using the formula:
Total number of particles = number of moles x 6.02 x 1023
Given: The number of moles of water = 3.00 mol
The Total number of molecules of water = 3.00 mol x 6.02 x 1023 molecules/mol
Final Answer = 1.81 x 1024 molecules of water.
VII. VARIATIONS
Please be aware that the six types of problems described above are only some of the major types
of problems that you might be asked. You must read each question carefully and determine what
the question is asking for. Changing one or more words in the question can completely change the
answer, as shown in the two examples below:
Example 1. How many molecules would be found in a 36.0 gram sample of water?
Example 2. How many atoms would be found in a 36.0 gram sample of water?
Do you see the difference? Example 2 requires you to have your answer in atoms. Since there are
three atoms (two hydrogen and one oxygen) in one molecule of H2O, the answer would be three
times as great as the answer to example 1.
There are many other possible variations to these types of questions. This is why I encourage you
to solve each problem logically instead of mechanically. Memorizing formulas is not enough, you
need to be able to reason what the question is actually asking for.
Mass-Mass Problems
The coefficients in chemical equations provide us with some important information. They
show us the molar ratio relationships that exist between the reactants and products. It is very
important for you to remember that the coefficients show the ratio of particles-particles and
of moles-moles, not mass-mass. For example, look at the reaction shown below;
2H2(g) + O2(g) ----> 2H2O(g)
The coefficients tell us that 2 molecules of hydrogen react with 1 molecule of oxygen to
form 2 molecules of water. They also tell us that 2 moles of hydrogen will react with 1 mole
of oxygen to form 2 moles of water. These coefficients do not tell us the ratio of mass-mass,
meaning 2 grams of hydrogen do not react with one gram of oxygen to form 2 grams of
water! That would not make sense!
However, knowing the molar relationships between the substances involved allow us to
calculate the mass relationships. This quantitative study of chemical reactions is called
stoichiometry. Mass-Mass problems generally involve the following steps:
1. Changing the mass given to the number of moles with the formula;
mass given
# of moles = -----------------------molar mass
2. Determine the number of moles of the unknown by comparing the molar ratio;
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
3. Change the number of moles of the unknown to mass with the formula;
mass = # of moles x molar mass
Example 1. How many grams of water are produced when 7.00 grams of oxygen react with
an excess of hydrogen according to the reaction shown below?
2H2(g) + O2(g) ----> 2H2O(g)
First, realize that the "excess" reactant has nothing to do with the actual math. If you had to
start from an unbalanced equation, the "excess" reactant would be used in the balancing, but
once an equation is balanced, the "excess" is just extra information. You can cross it out at
this point;
Example 1. How many grams of water are produced when 7.00 grams of oxygen react with
an excess of hydrogen according to the reaction shown below?
2H2(g) + O2(g) ----> 2H2O(g)
Next, identify which is the "given" and which is the unknown. Remember, the substance that
they give you information about is called the "given." The substance they are asking you
about is the "unknown." So;
Given
Unknown
2H2(g) + O2(g) ----> 2H2O(g)
7.00g
xg
Now, solve the problem according to the steps that were described above.
1. Changing the mass given to the number of moles with the formula;
mass given = 7.00g
molar mass of oxygen = 32.0g/mole
mass given
# of moles = -----------------------molar mass
7.00g
# of moles = -----------------------32.0g/mole
# of moles of oxygen = 0.219 mole
2. Determine the number of moles of the unknown by comparing the molar ratio;
Number of moles oxygen = 0.219 moles
coefficient of given (oxygen) = 1
coefficient of unknown (water) = 2
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
0.219
X
------- = ------1
2
Number of moles of water (unknown) = 0.438 mole
3. Change the number of moles of the unknown to mass with the formula;
# of moles of water = 0.438 mole
molar mass of water = 18.0g/mole
mass = # of moles x molar mass
mass of water = 0.438 mole x 18.0 g/mole
mass of water = 7.89 g
Our final answer, 7.89 grams of water are produced when 7.00 grams of oxygen react with an
excess of hydrogen. Do you know where the additional 0.89 grams of mass come from?
Example 2. How many grams of sulfuric acid are required to react completely with 15.0
grams of zinc in a single displacement reaction?
Here you are not given a balanced reaction, so the first step is to call upon many of the
Chemistry skills which you should have mastered by now to derive the chemical equation;
Zn(s) + H2SO4(aq) ----> ZnSO4(aq) + H2(g)
Now that the equation is balanced, you can label the given and the unknown and cross out the
excess information;
given
unknown
Zn(s) + H2SO4(aq) ----> ZnSO4(aq) + H2(g)
15.0g
Xg
Next, go through the three steps for solving mass-mass problems;
1. Changing the mass given to the number of moles with the formula;
mass given = 15.00g
molar mass of zinc = 65.4 g/mole
mass given
# of moles = -----------------------molar mass
15.00g
# of moles = ------------------------
65.4g/mole
# of moles of zinc = 0.229 mole
2. Determine the number of moles of the unknown by comparing the molar ratio;
Number of moles zinc = 0.229 moles
coefficient of given (zinc) = 1
coefficient of unknown (sulfuric acid) = 1
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
0.229
X
------- = ------1
1
Number of moles of sulfuric acid (unknown) = 0.229 mole
3. Change the number of moles of the unknown to mass with the formula;
# of moles of sulfuric acid = 0.229 mole
molar mass of sulfuric acid = 98.1g/mole
mass = # of moles x molar mass
mass of sulfuric acid = 0.229 mole x 98.1 g/mole
mass of sulfuric acid = 22.5 g
Our final answer is that it would take 22.5 grams of sulfuric acid to completely react with 15.0
grams of zinc. Can you figure out how many dm3 of hydrogen that would produce at STP?
Mixed Mass-Volume Problems
Mixed mass-volume problems include mass-volume problems and volume-mass problems.
In order to become proficient at solving mixed mass-volume problems, it is imperative that
you solve problems logically as opposed to mechanically. Many students are able to do all of
the Math work involved in these problems, but they still get them wrong on exams, because
they fail to identify the exact type of problem that it is. If you go through the motions of the
calculations, and never check to see if they make sense, you are missing an important chance
to catch a mistake. Read each problem carefully, try to understand it, and think about what
you are doing with each calculation. Does it make sense to divide a mass by the molar volume
of a gas? Would you multiply the number of moles by the molar mass of helium to get its
volume? Avoid foolish mistakes and these problems will become quite easy, with practice.
For many types of mixed mass-volume problems, you will follow these same basic steps;
1. Start with a balanced chemical equation.
2. Identify and label the given and the unknown, cross out the rest.
3. Turn the given quantity into moles by;
A) dividing by molar mass in the event that a mass is given, or
B) dividing by molar volume in the event that a volume is given.
4. Determine the number of moles of the unknown that will by produced, by comparing the
molar
ratio shown in the balance chemical equation.
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
5. Change the number of moles of the unknown to the units that the question asked for by;
A) multiplying by the molar mass, if the question asks for grams, or
B) multiplying by the molar volume (22.4 dm3/mole) if the question asks for dm3.
It should be noted that not all questions are going to fit into this neat format. You need to
be able to make adjustments based on any special circumstances that arise in a problem.
What will you do if the question asks for cm3 instead of dm3? What will you do if the
question gives you number of particles, instead of mass or volume? These special
circumstances are exactly why you need to strive to understand these problems, not just solve
them mechanically.
Example 1. At STP, what would be the volume of the carbon dioxide produced when 90.0
grams of glucose react with an excess of oxygen?
1. Start with a balanced chemical equation.
C6H12O6(s) + 6O2(g) ----> 6H2O(g) + 6CO2(g)
2. Identify and label the given and the unknown, cross out the rest.
Given
Unknown
C6H12O6(s) + 6O2(g) ----> 6H2O(g) + 6CO2(g)
90.0 g
X dm3
3. Turn the given quantity into moles by;
A) dividing by molar mass in the event that a mass is given, or
B) dividing by molar volume in the event that a volume is given.
mass given
# of moles = ---------------------molar mass
90.0 g
# of moles glucose = ---------------------180.1 g/mole
# of moles of glucose = 0.500 moles
4. Determine the number of moles of the unknown that will by produced, by comparing the
molar ratio shown in the balance chemical equation.
moles of the given (glucose) = 0.500 moles
moles of the unknown (carbon dioxide) = X
coefficient of the given (glucose) = 1
coefficient of the unknown (carbon dioxide) = 6
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
0.500 moles X moles
------------ = ------------1
6
X = 3.00 moles
# of moles of carbon dioxide = 3.00 moles
5. Change the number of moles of the unknown to the units that the question asked for by;
A) multiplying by the molar mass, if the question asks for grams, or
B) multiplying by the molar volume (22.4 dm3/mole) if the question asks for dm3.
# of moles of carbon dioxide = 3.00 moles
molar volume of a gas at STP = 22.4 dm3/mole
volume of a gas at STP = # of moles of the gas x molar volume of gases at STP (22.4 dm3/mole)
volume of the carbon dioxide gas at STP = 3.00 moles x 22.4 dm3/mole
volume of carbon dioxide at STP = 67.2 dm3
Our final answer, 67.2 dm3 of carbon dioxide are produced when 90.0 grams of glucose react
with an excess of oxygen gas.
Volume-Volume Problems
As I stated in the lesson 9-3, the coefficients in chemical equations provide us with
some important information. They show us the molar ratio relationships that exist
between the reactants and products. It is very important for you to remember that the
coefficient show the ratio of particles-particles and of moles-moles, not mass-mass.
For example, look at the reaction shown below;
2H2(g) + O2(g) ----> 2H2O(g)
The coefficients tell us that 2 molecules of hydrogen react with 1 molecule of oxygen
to form 2 molecules of water. They also tell us that 2 moles of hydrogen will react
with 1 mole of oxygen to form 2 moles of water. These coefficients do not tell us the
ratio of mass-mass, meaning 2 grams of hydrogen do not react with one gram of
oxygen to form 2 grams of water! That would not make sense!
However, when dealing with both a gaseous given and a gaseous unknown, we can
take advantage of a special property of gases to make our calculations much simpler.
Avogadro's law tells us that "at equal temperatures and pressures, equal volumes of
different gases contain the same number of particles." This lead to the idea of a
standard for the volume of one mole of gas. By now, probably know that one mole of
any gas at standard temperature and pressure has a volume of 22.4 dm3. This
constant, 22.4 dm3, is called the molar volume of a gas. Each element or compound
has its own individual molar mass, but the molar volume of all gases is the same. This
makes stoichiometry problems involving only gases much easier to do. To
demonstrate, I will show you an example of solving a volume-volume problem the long
way, and then the same problem with the shortcut.
To solve Volume-Volume problems the long way, we would use a modified version
of the three steps from last lesson:
1. Change the volume of the given gas to number of moles by using the formula;
volume of gas at STP
# of moles = ----------------------------------------------molar volume of gases (22.4 dm3/mole)
2. Determine the number of moles of the unknown by comparing the molar ratio;
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
3. Change the number of moles of the unknown gas to volume at STP by using the
formula;
volume of a gas at STP = # of moles of gas x molar volume of gas (22.4 dm3/mole)
Now, if you are a good Math student you can probably see why we can take a
shortcut. Steps 1 and 3 call for opposite operations (x and / by 22.4) so they will cancel
each other out! For the sake of the students who don't see that right away, I will
demonstrate the Math that you would end up doing if you followed the three steps
above.
Example 1. What volume of oxygen gas would react with 35.0 dm3 of hydrogen gas at
STP, according to the equation below?
2H2(g) + O2(g) ----> 2H2O(g)
First, label the given and the unknown, and cross out the other information, which is
only useful when balancing the equation;
given
unknown
2H2(g) + O2(g) ----> 2H2O(g)
35.0 dm3
Xdm3
Now, we will follow the three steps shown above. Remember, this will be the long way
to do things. I am only doing this for demonstration purposes!
1. Change the volume of the given gas to number of moles by using the formula;
Volume of given gas (hydrogen) at STP = 35.0 dm3
Molar volume of gases = 22.4 dm3
volume of gas at STP
# of moles = ----------------------------------------------molar volume of gases (22.4 dm3/mole)
35.0 dm3
# of moles = -----------22.4 dm3/mole
# of moles of hydrogen gas = 1.56 moles
2. Determine the number of moles of the unknown by comparing the molar ratio
(from balanced equation);
Number of moles of given (hydrogen) = 1.56 moles
Coefficient of given (hydrogen) = 2
Coefficient of unknown (oxygen) = 1
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
1.56 moles
X moles
------------ = -----------2
1
2x = 1.56 moles
# of moles of the unknown (oxygen) = 0.780 moles
3. Change the number of moles of the unknown gas to volume at STP by using the
formula;
# of moles of oxygen (unknown) = 0.780 moles
Molar volume of a gas at STP = 22.4 dm3/mole
volume of a gas at STP = # of moles of gas x molar volume of gas (22.4 dm3/mole)
Volume of the oxygen gas at STP = 0.780 moles x 22.4 dm3/mole
Volume of the oxygen gas at STP = 17.5 dm3
So, our final answer is that 17.5 dm3 of oxygen gas would react with 35.0 dm3 of
hydrogen gas, at STP, according to the reaction;
2H2(g) + O2(g) ----> 2H2O(g)
Do you notice that the ratio of volume-volume is identical to the ratio of molesmoles! This is because the opposite operations that I mentioned earlier. In step one,
we divided by 22.4. In step three, we multiplied by 22.4. These steps cancel each
other out, so they do not need to be shown! When you are working with only gases, at
equal temperature and pressure, all you have to do is compare the ratios shown in a
balanced chemical equation. In other words, the coefficient in a chemical equation
show the ratio of particles-particles, moles-moles, and volume-volume (when dealing
with gases at the same temperature and pressure.) To solve volume-volume problems,
use the formula below;
volume of the given
volume of the unknown
---------------------------------- = ---------------------------------coefficient of the given
coefficient of the unknown
Let us go back and solve the original problem the easy way:
Example 2. What volume of oxygen gas would react with 35.0 dm3 of hydrogen gas at
STP, according to the equation below?
2H2(g) + O2(g) ----> 2H2O(g)
First, identify the given and the unknown, and cross out the extra information;
given
unknown
2H2(g) + O2(g) ----> 2H2O(g)
35.0 dm3
Xdm3
volume of the given (hydrogen) = 35.0 dm3
volume of the unknown (oxygen) = X dm3
coefficient of the given (hydrogen) = 2
coefficient of the unknown = 1
volume of the given
volume of the unknown
---------------------------------- = ---------------------------------coefficient of the given
coefficient of the unknown
35.0 dm3
Xdm3
---------- = --------2
1
2X = 35.0 dm3
(X) volume of the oxygen = 17.5 dm3
So, we get the same answer with much less work! Imagine how much time this will
save you on tests where other students are solving volume-volume problems the long
way! Please remember -
***The above shortcut only works with volume-volume
problems!***
LABORATORY SAFETY CHECKLIST
Read the following list and check off each item only when you thoroughly understand it. Ask your instructor to
clarify any points that you do not understand. It is your responsibility to know and to adhere to each of the
safety procedures on the list.
GENERAL
Be prepared to work when you arrive at the laboratory. Familiarize yourself with the procedures before
beginning the activity. Be ready to work in a mature and responsible manner.
Know the location of the emergency equipment and how to use them.
Perform only those lab activities assigned by your teacher. Never do anything in the laboratory that is not
called for in the laboratory procedure or by your teacher. Never work alone in the lab. Do not engage in
playful and\or potential hazardous activities.
Follow all instructions, both written and oral, carefully.
Work areas should be kept clean and tidy at all times. Only lab manuals and notebooks should be brought to
the work area. Other materials will be left on the back windowsill.
Clothing should be appropriate for working in the lab. Jackets, ties, or other loose garments should be
removed.
Eye protection must be worn at all times. Safety goggles are required for every laboratory exercise. Contact
lenses should not be worn in labs unless absolutely necessary.
Eating and drinking are never allowed in the lab room. Wash your hands frequently when working with
chemicals, and at the end of each lab activity.
Set up apparatus as described in your procedures or by your teacher. Always use the prescribed instruments
for handling equipment.
Report all accidents to the teacher immediately.
Return all materials to their proper locations after each lab activity. Nothing should be left on the counter
tops. Clean and wipe dry all work surfaces at the end of the class.
HANDLING CHEMICALS
Read and double check labels on reagent bottles before removing any reagent. Take only as much reagent as
you need.
Do not return unused reagent to stock bottles.
Avoid touching chemicals with your hands. If chemicals do come in contact with your hands, wash them
immediately.
Never smell any chemicals unless instructed to do so by your teacher. Never put your face near the mouth of
a container that is holding chemicals.
Notify your teacher if you have any medical problems that might relate to lab work, such as allergies or
asthma.
HEATING SUBSTANCES
Exercise extreme caution when using a gas burner. Keep your head, hands and clothing away from the flame.
Always turn the burner off when it is not in use.
Do not bring any substance in contact with a flame unless instructed to do so.
Never look into a container that is being heated.
When heating a substance in a test tube, make sure the the mouth of the tube is not pointed at yourself or
anyone else.
Never leave unattended anything that is being heated or visibly reacting.
Keep water away from hot plates.
LABORATORY SAFETY AGREEMENT
I have read and understand the information contained in the handout entitled "Laboratory Safety Checklist" and
I agree to abide by the rules and procedures described there, and explained by my teacher. I will also abide by
any other rules and regulations provided by my Chemistry teacher. I will do my part to maintain a safe working
environment in the Laboratory.
____________________________________
(Name - Please print clearly)
_____________________________________
(Signature)
______________
(Date)
_____________________________________
(Signature of parent or guardian)
Laboratory Equipment
Students are required to learn the names and functions of various types of equipment that they
will use in Chemistry laboratory activities. This knowledge will come in handy when designing
or conducting experiments during the course of the year. This page shows some of the most
common items, and describes what they are used for. Images from this page can also be used for
laboratory reports or web pages that you will make during the course of the year.
Each student is responsible for becoming familiar with the laboratory room. You need to know
where to find equipment when working on an activity. When you are done with an activity, clean
your materials and return them to the proper drawers.
Beam Balance
Used to find the mass of
various materials.
Electronic Balance
Used to find the mass of
materials. This particular
balance is not as precise as
the beam balance
Bunsen Burner
Shown with rubber hose
connected to gas jet. Used to
heat materials in lab.
Sparker
Used to ignite the Bunsen
burner.
Gas Jet and Outlet
The hose from the Bunsen
burner connects to the jet.
Hot Plate
Used for heating liquids.
Ring Stand
Iron Ring
Wire Gauze
Used in many lab activities Often attached to ring stand to Often placed over the iron
as the support for other
use as support for a beaker. ring, to provide a "stage" for
apparatus.
a beaker.
Typical Setup
Ring stand shown with iron
ring, wire gauze and a
beaker. The Bunsen burner
would go below the "stage".
Clamp
May be attached to a ring
stand and be made to hold a
test tube or thermometer.
Tongs
Used to handle hot beakers
and other glassware
Funnel
May be placed in an iron
ring. Used for filtration or
the delivery of liquids.
Massing Tray
Chemicals are not placed
directly on a balance. These
trays hold the reagents.
Scoop
Used to transfer chemicals
from one vessel to another.
Goggles
Must be worn for each lab
activity.
Trough
Used for lab activities that
require a basin for water.
Sink
Wash all glassware after each
activity.
Test Tube Rack
Holds many test tubes.
Test Tube
Used for many activities
which require multiple
reagents or solutions.
Pipestem Triangle
Can be placed on an iron ring
to provide a stage for a
crucible.
Beaker
Erlenmeyer Flask
Volumetric Flask
Probably the most common May be used to hold liquids, Often used when solutions of
vessel for holding liquids in instead of beakers, when a
specific concentration are
the laboratory.
smaller opening is preferred.
being made.
Graduated Cylinder
Used to measure the volume
of liquids.
Watch Glass
Used to evaporate off the
liquid part of a solution.
Dropper Pipet
Used to transfer small
amounts of liquid.
Glass Stirring Rod
Used to stir liquids.
U Tube
Titration Setup
Useful as a "salt bridge" when Typical setup for conducting
making batteries.
acid-base neutralization
reactions.
Experimental Design
You will be called upon to design several experiments during the course of this year.
These experiments will differ from your typical laboratory activities in several ways.
Instead of following pre-designed steps, you will need to write procedures for the
experiment. The success of the activity will depend upon the quality of the procedures that
you write.
Designing the experiment is the first, and most important step, because a poor set of
procedures will render the experiment invalid. Conducting an invalid lab activity is like
going fishing without catching any fish, it is not time well spent. If you don't catch any
fish, there is little difference between standing around doing nothing and standing around
with a fishing pole. Similarly, carrying out invalid procedures is a clear waste of time.
Another thing to keep in mind as you design your experiment is that it must be
replicable. This means that your design and your report must be so clear that anyone can
carry out the experiment exactly as you have. That is why a clear, concise laboratory report
is an essential follow up to each experiment. This page will offer some advice about
designing these experiments, in the format of your typical laboratory report.
Purpose: The purpose of the activity will be given to you by the instructor. Without a
clear idea of what the purpose of the experiment is, you would be unable to complete the
other steps of experimental design. Keep the purpose foremost in your mind as you design
your experiment. No matter what else occurs, the purpose of the experiment must be
achieved. If you don't understand the purpose as given by the instructor, ask questions until
you do. It is imperative that you understand the purpose before you proceed.
Sample Purpose Statements:
1. "To determine the specific heat of a sample of copper."
Materials: The materials section of your laboratory report will list all of the materials that
are needed to carry out your experimental design. Select the appropriate equipment to
satisfy your needs (see lesson 1-5). List everything that you intend to use, and be specific.
If your design will call for a beaker, state the size of the beaker that you recommend. If
your design calls for a reagent, state the mass, or size of the sample, required. The material
section serves the same purpose as a shopping list for someone who is going to try a new
recipe. It will allow a person to get everything in order before they begin the procedures.
Sample Material Section:
250 ml beaker, 100 ml graduated cylinder, Bunsen burner, sparker, ring stand, wire gauze,
iron ring, calorimeter, two thermometers, tongs, a precut sample of pure copper (5 - 10
grams).
Procedure: As stated earlier, the procedure section is the most important. Without proper
procedures, the rest of the activity is a waste of time. Be prepared for people to critique,
and look for flaws in, your procedure. Carrying out flawed procedures can hurt your
reputation in the scientific community, or at least in your Chemistry classroom. I will offer
some tips to keep in mind when writing your procedure sections.
1. Be Specific: The recipe analogy is a good one. A cookbook will not have recipes that
say things like "put some flour in a bowl and add some water." Recipes are very specific,
because they are designed for others to read and follow. Your procedure section should be
just as specific. Someone following your procedures to the letter should expect the same
results that you observed.
2. Be Concise: Don't elaborate on steps that are self-evident. There is no need to go over
the use of a balance in every procedure section. Massing an object is a procedure unto
itself, and you can assume that the reader is familiar with it. It is sufficient to instruct the
reader to "take 5.50 grams of . . ." How to mass the reagent is up to him or her.
3. Be Clear: Sometimes a procedure will call for the use of more than one of a certain
vessel or object. Use letters to indicate which one you are speaking of. If you need three
250ml beaker, instruct the reader to label them "A", "B", and "C" respectively.
Sample Procedure Section:
1. Mass the sample of copper and record your findings.
2. Assemble your ring stand, iron ring and wire gauze as shown in figure 1-1.
3. Add approximately 200 ml of water into your 250 ml beaker of water, and place it on
your wire gauze "stage".
4. Ignite your Bunsen burner with your sparker and adjust the flame to the proper height.
5. Place the Bunsen burner under the wire gauze and begin heating the water.
6. Drop your sample of copper into the beaker.
7. Check to make sure that both of your thermometers are reading the same temperature.
8. Use one thermometer to monitor the temperature of your hot water.
9. Use the graduated cylinder to measure out exactly 100.0 ml of water, then pour it into
your calorimeter.
10. Use your second thermometer to monitor the temperature of the water in the
calorimeter.
11. Let your hot water bath reach and maintain a temperature of approximately 100.0 oC for
about five minutes. Then record this temperature as Cu T1.
12. Record the temperature of the water in the calorimeter as H2O T1.
13. Use tongs to quickly and carefully transfer the copper from the hot water bath to the
calorimeter.
14. Monitor the temperature in the calorimeter until the temperature stops rising. record
this final high temperature as both Cu T2 and H2O T2.
15. Repeat the experiment as time allows.
Data Section: When you collect data, you must organize it carefully, or you might not be
able to make sense of it at a later date. Design a data table that will compliment your
procedure section. Your data collection table should have a spot for each piece of data that
your procedure calls for. Headers can be used to help the reader find where a particular
piece of data should be recorded. Your data table should be constructed prior to carrying
out the procedures.
Sample Data Section:
The Specific Heat of Copper.
Data Table
Trial 1
A) Mass of Copper Sample
Mass Cu
______g
Trial 2
Mass of Cu
______g
Trial 3
Mass of
Cu______g
B) Initial Temperature of Cu T1
Copper in Hot Water Bath. ______oC
Cu T1 ______oC Cu T1 ______oC
C) Initial Temperature of
Water in Calorimeter.
H2O T1
______oC
H2O T1 ______oC H2O T1 ______oC
D) Final Temperature of
Copper in Calorimeter.
Cu T2
______oC
Cu T2 ______oC Cu T2 ______oC
E) Final Temperature of
Water in Calorimeter.
(same as Cu).
H2O T2
______oC
H2O T2 ______oC H2O T2 ______oC
Analysis: The analysis of your data is extremely important. If you misinterpret your data,
you will come up with unfounded conclusion which other readers may catch. The analysis
of your data may include calculations and/or graphing. When you do calculations, you
must show each formula that you use, and each step in your calculation.
Sample Analysis Section: Only a portion is shown (with made up data)
Calculations
A) Change in Temperature of the Water in the Calorimeter
Formula:
Trial 1
Trial 2
T = H2O T2 - H2O T1
T = H2O T2 - H2O T1
Given:
H2O T2 = 43.5 C
H2O T1 = 23.5 oC
H2O T2 = 39.3 oC
H2O T1 = 23.5 oC
Solve:
T = 43.5 oCoC =
T = 39.3 oC - 23.5 oC =
T = 20.0 o
T = 15.8 oC
Answer:
o
B) Total Heat Gained by the Water in the C lorimeter
Formula:
Trial 1
Trial 2
q = m(T)Cp
q = m(T)Cp
Given:
m = 100.0 g
T = 20.0 oC
Cp = 4.18 J/g x oC
m = 100.0 g
T = 15.8 oC
Cp = 4.18 J/g x oC
Solve:
q = 100.0g x 20.0 oC x
4.18J
---------------------g x oC
q = 100.0g x 15.8 oC x
4.18J
---------------------g x oC
q = 8360 J
q = 6600 J
Answer:
Conclusion: Each instructor may require a different type of conclusion section. You may
have questions to ask and answer, or you may have an essay to write. When you design
your own experiment, your conclusion should show that you have met the original purpose
as described by the instructor. In labs that deal with any type of a quantitative result, you
may be asked to calculate your percent error (See lesson 2-7). You may also be asked to
explain possible sources of error. Often, your instructor can estimate how well you
performed in lab, simply by reading your conclusion essay.
Sample Conclusion: Only a Portion is Shown:
". . . As shown in my percent error calculation, my error was approximately 7 %. There
are many possible sources of error in this experiment, and I will discuss those that I feel are
the most likely:
1. Improper Insulation - The calorimeter that I used for this experiment was made of
Styrofoam, which is a good insulator, but, it is far from perfect. Some heat may have been
transferred to the air around the vessel, altering my experimental values for q.
2. Transferring Copper - When the copper was transferred from the hot water bath to the
calorimeter, heat was lost to the surrounding air, altering my values for q.
3. Transferring Water with Copper - When I transferred the copper from the hot water bath
to the calorimeter, a certain amount of water clung to the metal. This amount of water,
although very little, may have thrown off my actual values for the mass of the water.
etc.
That is all of the specific tips that I can think of at this time. Work with your instructor to
find out other ways of improving your experiments and the reports that are based on them.
Percentage Yield
You have learned from laboratory experience that there is sometimes a difference
between an experimental result and a mathematical result. Using stoichiometry, we can
mathematically determine the amount of a product that should be formed during an
experiment, yet we sometimes find that we don't end up with exactly the right amount of
product. We use sources of error to explain this difference, and we have even done percent
error calculations to calculate how far off we are from the expect results. Percentage yield
problems fall under this type of problem. They simply allow us to calculate what percent of
the expected product we are able to account for by the end of our experiment. The formula
that we use is;
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
Example 1. A student conducts a single displacement reaction that produces 2.755 grams of
copper. Mathematically he determines that 3.150 grams of copper should have been
produced. Calculate the student's percentage yield.
Solve:
actual amount of product: 2.755 g
expected amount of product: 3.150 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
2.755g
percentage yield = --------------- x 100
3.150g
percentage yield = 87.4603174 %
percentage yield = 87.46 %
Example 2. A student completely reacts 5.00g of magnesium with an excess of oxygen to
produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the
student's percentage yield?
Solve:
In this problem, you are given the actual amount of the product, but you are not given the
expected amount of the product. The second mass shown, 5.00g, is the mass of one of the
reactants. In order to determine the expected amount of the product in this problem, you
must begin with a mass-mass problem.
First, write a balanced chemical equation for the reaction:
2Mg(s) + O2(g) ----> 2MgO(s)
Now, label the given and the unknown and cross out the rest:
given
unknown
2Mg(s) + O2(g) ----> 2MgO(s)
5.00 g
Xg
Change the mass given to moles by dividing by the molar mass of Mg.
mass given: 5.00g
molar mass of Mg: 24.3 g
5.00g
Number of moles of Mg = ----------24.3 g/mole
Number of moles of Mg = 0.206 moles
Compare the molar ratio between the given and the unknown to determine the number of
moles produced.
Coefficient of Mg; 2
Coefficient of MgO: 2
Number of moles of Mg: = 0.206 moles
Number of moles of MgO: = ?
number of moles of given
number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
0.206 moles X moles
-------------- = ------------2
2
Number of moles of MgO produced = 0.206 moles
Now, change the number of moles of MgO produced to mass by multiplying by the molar
mass of MgO.
# of moles of MgO = 0.206 moles
Molar mass of MgO = 40.3g/mole
mass = # of moles x molar mass
mass of MgO = 0.206 moles x 40.3 g/mole
mass of MgO = 8.30 g
Now, you are ready to solve the percentage yield problem.
actual mass of MgO produced = 8.10 g
expected mass of MgO = 8.30 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
8.10 g
percentage yield = ------------ x 100
8.30 g
percentage yield = 97.6 %
Ratio and Proportion
Chemistry is a quantitative subject, meaning chemistry
deals with quantities: solids, liquids, and gases. The
science and study of chemistry can be broken down into
two groups, the information of Chemistry like chemical
reactions, moles, and the four states of matter, and
math problems such as balancing equations, mass-mass,
and molarity problems. A lot of these problems found in
chemistry contain ratios and proportions.
The word ratio is derived from the Latin word "ratio"
which means "computation." A ratio can be defined as
the relative size of two quantities expressed as the
quotient of one divided by the other.
Ex. The ratio of 7 to 4 is written: 7:4 or 7/4. To
solve this problem you would divide 4 from 7
To simplify a compound or an equation you take the
coefficients of each element or compound and reduce
them.
Ex. C6H12O6 --> 6:12:6 --> 1:2:1
Here are some examples on how to determine the ratio
from a chemical formula:
Ex. 2H2 + O2 --> 2H2O
From this example you can see that for every two
molecules of hydrogen gas there is one molecule of
oxygen gas. The ratio is 2:1. If you have fifty
molecules of oxygen gas then from this ratio I would
have one hundred molecules of hydrogen gas.
Ex. 2Al(NO3)3 + 3H2SO4 --> Al2(SO4)3 + 6HNO3
In this chemical formula you can see that aluminum
nitrate combined with hydrogen sulfate(sulfuric acid)
yields aluminum sulfate and hydrogen nitrate. From this
example you can see that for every two molecules of
aluminum nitrate there are three molecules of hydrogen
sulfate. The ratio is 2:3. If you have twenty molecules
of aluminum nitrate then you would have thirty
molecules of hydrogen sulfate.
The word proportion is derived from the Latin word
"proportio" which means "portion." A proportion can be
described as an equation which states that two ratios
are equal.
Ex. 4 = 5
8
10
To solve a proportion you cross-multiply. In this
example you multiply the 4 and the 10 together and the
5 and the 8 together.
Ex. 6 = x
5
10
To solve for x in this proportion you cross multiply the 6 and 10 together, and the 5 and x
together: 5x = 60. Then divide the 60 by 5 to get your answer: 60/5 = 12.
Ex. 4 = 8
x 16
To solve for x in this proportion you cross multiply the 4 and 16 together, and the 8 and x
together: 8x = 64. Then divide the 64 by 8 to get the value of x: 64/8 = 8.
Ex. x+1 = 6
x+2 x+4
To solve for x in this proportion you must crossmultiply. In this example you multiply the x+1 with the
x+4. When you multiply the two binomials together you
will get one trinomial. (x+1)(x+4) = x2+5x+4. You can
obtain this trinomial by using the F.O.I.L. method.
First multiply the x's from each binomial together.
Then multiply the outer numbers and variables
together(X x 4), and the inner numbers and variables
together(X x 1), and add the results: 4x and 1x
together. Finally you multiply the last numbers and
variables in the two binomials together (4 x 1). Your
final answer is x2+5x+4. Next you cross-multiply the 6
with the (x+2). Distribute the 6 through the binomial
(x+2).
You multiply the 6 with the X and then the 6 with the
2. You final answer is 6x+12. Set the two equations
equal to each other: x2+5x+4=6x+12. Move all the numbers
and variables together on one side,--> x2-x-8=0. Break
this equation into two binomials and you will have your
answer. This is what you should get:(x+4) and (x-4). X
is equal to 4.
Prefix System for Naming
Compounds
Hydrocarbon Prefixes
Number of Carbon Atoms
Prefix
1
meth-
2
eth-
3
prop-
4
but-
5
pent-
6
hex-
7
hept-
8
oct-
9
non-
10
dec-
To use the above table, the compound must be a hydrocarbon, that is, made up of carbon and
hydrogen atoms. First, you must count the number of carbon atoms. Next, you must match
the prefix with the number of carbon atoms. Then, add the suffix –ane onto the prefix. If the
atoms are linked in a ring, then you add another prefix, cyclo-.
Examples:
1. C6H12 = ____________
2. C2H6 = ___________
Give the number of carbon atoms
4. decane = ______________
5. octane = _______________
6. propane = ______________
Solutions: 1.) hexane, 2.) ethane, 3.) ten carbon 4.) eight carbon 5.) three carbon
For compounds which are not hydrocarbons, an outdated naming system used prefixes.
Although this system is outdated, you will still encounter it from time-to-time.
Prefix System for Atoms
Number of Atoms
Prefix
1
mono-
2
di-
3
tri-
4
tetra-
5
penta-
6
hexa-
7
hepta-
8
octa-
Again, you must first take the chemical equation and see if it is binary or not. By adding the
appropriate prefix above, you create the correct compound name. This will work in all
instances. For naming compounds with roman numerals, check out the link which is below on
help for naming compounds.
Examples:
1. CO2 = ______________
2. CO = _______________
3. K2S = _______________
4. SiS2= _______________
solutions: 1.) carbon dioxide 2.) carbon monoxide 3.) dipotassium sulfide 4.) silicon disulfide
Balancing Equations
All chemical formulas should be represented in a balanced form, meaning that the
number of atoms of each element are the same on both sides of the equation. In
order to do this, the coefficients must be so, so that the number of atoms of each
element is the same on both sides of the equation. The number of atoms of each
element can be determined by multiplying the coefficient of the element by the
subscript next to it. For example: 2Pb3O2 in this equation, there are 6 atoms of Pb
(lead) and 4 atoms of O (oxygen).
Balancing equations: Br3 + O2  BrO
In this equation we have on the left side (aka the reactant side), 3Br, 2O
On the right side (a.k.a the product side) we have
1Br, 1O
In order to balance this equation, we will adjust the coefficients (using trial and
error) to achieve a balanced form which is:
2Br3 + 3O2  6BrO
Here, the same number of atoms of Bromine are on the left side as they are on the
right side, and the same number of atoms of Oxygen are on the left side as they are
on the right side.
Also, sometimes parenthesis are used: (SO4)3
In this example, the subscript 3 applies to both of the elements inside the
parenthesis, so there are 3 atoms of Sulfur (S) and 12 atoms of Oxygen (O). The
number outside the parenthesis is multiplied by the number of atoms that the
element contains inside the parenthesis.
Other Examples
(see link at bottom of page for answers)
1. Cu +H2O  CuO + H2
2. Al(NO3)3 + NaOH  Al(OH)3 + NaNO3
3. KNO3  KNO2 +O2
4. Fe + H2SO4  Fe2(SO4)3 +H2
5. O2 + CS2  CO2 + SO2
6. Cu + Cl2  CuCl2
7. Mg + N2  Mg3N2
8. C + O2 CO2
9. CO + O2 CO2
10. K + H2O KOH + H2
11. NaOH + H2SO4 NaSO4 + H2O
12. Al + H2SO4 Al2(SO4)3 + H2
13. NH4NO2 N2 + H2O
14. NH3 + CuO Cu + N2 + H2O
15. C2H6 + O2 H2O + CO2
16. P4O10 + H2O H3PO4
Writing Ionic Formulas
When learning about Ionic Formulas and how one goes about constructing them,
one must remember that ions are are atoms or molecules that has gained or lost one
or more electron (e-). This number is key to understanding the concept of writing
ionic formulas.
There are many uses if uses of ions in chemistry there are ionic bonds, ionic
compounds, ionic radius, ionization constant, ionization energy, all are useful terms
and can be referred to by the definitions column on the left hand column,
When one wants to learn how to write an Ionic formulas one must carefully identify
the elements that are being done within the reaction, then we go about identifying
what the oxidation number is and if it is polyatomic.
Example:
Common table salt, NaCl, is made from sodium, Na and Cl. They both have
charges and are written as follows:
Na+1 Cl -1
To confirm that this formula is correct we observe its oxidation numbers (one
positive and the other negative) in this situation both oxidation numbers are equal,
thus there is no additional work to be done since they form to become a stable
compound (neutral) and the formula for table salt is agreed upon NaCl.
However when we get into harder degrees of writing these ionic reactions we have
to put together elements that have different charges and in order to make the
compound neutral we have to balance the charges and change the subscripts.
When writing the formula for a compound made up of aluminum ion and the sulfate
ion we write;
Al +3 and SO4 -2
To make the sum of the charges equal to zero, we must detect the least common
multiple of 3 and 2. The least common multiple is 6. It is necessary to have two Al +3
and three in the compound to maintain neutrality. Writing two aluminum ions in the
formula is simple.
(Al)2
For the sulfate , the entire polyatomic ion must be put in parenthesis and on the
outside include the three sulfate ions on the outside.
(SO4) 3
Thus, the aluminum sulfate has the formula Al2(SO4)3 (Parenthesis are used in a
formula only when you are expressing multiples of a polyatomic ion. If you only one
sulfate ion were needed in writing formula, parentheses would not be used.
Hints and Tips:
Note that the numbers that are in red are the ions' charges.
2. When forming a compound the positive must ALWAYS be placed before the negative NO
EXCEPTIONS!!
3. When a compound is unbalanced it becomes unbalanced and unstable.
4. the entire polyatomic ion must be placed in parenthesis to indicate that there are already a number
of ions required
Use of Oxidation Numbers
An ion is a charged atom or group of atoms. A monatomic ion consists of one atoms, while a
polyatomic ion consists of two or more atoms. The charge of a ion is called its oxidation
number. Oxidation numbers are very important in much of chemistry because many times
atoms do lose or gain electrons. When this happens, they become ions. However, oxidation
numbers are especially important when writing chemical formulas for ionic compounds. The
following is a step-by-step process of how one would write a chemical formula.
Example: Write a chemical formula for the compound made from sodium and chlorine
("sodium chloride").
Step 1: Look up the oxidation numbers for both sodium (Na) and chlorine (Cl) on a table.
You will see that the charge for Na is 1+, while the charge for Cl is 1-.
Step 2: Combine the ions so that the sum of the charges equals zero. In this case the charges
of the ions, when added together, naturally equal zero. Therefore, no subscripts need be
added.
1 + (1-) = 0.
Step 3: Having checked the oxidation numbers and made sure they add up to zero, you can
now just put the Na and the Cl together to form NaCl.
Writing chemical formulas is actually not a very complicated process. It is pretty
straightforward. However, the following example is a bit trickier because the use of a
subscript comes into play.
Example: Write a chemical formula for the compound made of the calcium ion and the
chloride ion.
Step 1: Look up the oxidation numbers for both the calcium (Ca) ion and the chloride (Cl)
ion. Note that chloride is just a name for the monatomic ion formed from chlorine. You will
see that the charge of Ca is 2+, while the charge of Cl is 1-.
Step 2: Combine the ions so that the sum of the charges equals zero. Here, however, a
subscript must be used. 2 + (1-) does not equal 0. Therefore, the subscript 2 must be added to
the Cl ion, because 2 + 2(1-) = 0.
Step 3: Having checked the oxidation numbers and made sure they add up to zero, you can
now just combine the Ca and the Cl2 to form CaCl2.
Example: Write a formula for the compound ammonium dichromate.
Step 1: Look up the oxidation numbers for both of these polyatomic ions. Ammonium, you
will see, has a +1 charge (NH4+). Dichromate, you will see, has a 2- charge (Cr2O7 2-).
Step 2: Combine the ions so that the sum of the charges equals zero. Here, however, a
subscript must be used. Also, parentheses must be used because you are using polyatomic
ions. You will need to add a subscript 2 to NH4 so that the sum of the charges equals zero.
Step 3: Place parentheses around the NH4 so that the whole NH4 has the subscript 2. Add on
the Cr2O7. Your final result will be (NH4)2Cr2O7.
The Concept of Moles
At first glance, moles sound like a complicated chemical concept. A quick read
through the textbook gives you a bunch of numbers, exponents, and chemical
formulas. In reality, though, the concept of moles is very easily understood when
taken step-by-step.
First off, a mole is simply a grouping unit and a way to make counting and
measuring small objects like atoms and ions A LOT easier. We already use this
system in our daily lives. For example…
dozen eggs=12 eggs
score people=20 people
A mole is the same as a dozen or a score and represents 6.02 x 1023 objects (which
can be atoms, ions, molecules, or formula units). So, a mole of atoms contains
6.02x1023 atoms just like a score of people means 20 people. This number is called
the Avogadro constant and is considered a SI standard just like the kilogram, the
meter, the second, etc. It was a convenient constant developed by Amadeo
Avogadro so that a number of atoms (or other particles) would have a mass in
grams equal to the mass of one atom (or another particle) in atomic mass units.
This constant applies to all elements as equal numbers of different atoms always
have the same mass ratio.
Now, one mole of particles has a mass in grams (g) equal to one particle in
atomic mass units (u). So if one mole of a particle has a mass of 4.02 grams, then a
single particle has a mass of 4.02 u. The mass of one mole of an object is called the
molar mass of that object.
E.g. One mole of iron=55.847 g
One atom of iron=55.847 u
The last step in understanding moles is converting them. Keep in mind during
the following examples that atoms/molecules/ions/formula units are
interchangeable. So in the first example, atom can be replaced with
molecule/ion/formula unit depending on the question or situation.
Atoms-Moles Conversions
If you had 48 eggs, how would you find how many dozens of eggs there were?
1.
1.
2.
3.
4.
A dozen=12 eggs
There are 48 eggs
48 12=4
48 eggs=4 dozens
The same concept can be applied to moles. If one had 6.02x1024 atoms, how many
moles of that atom are there?
1.
1.
2.
3.
4.
A mole=6.02x1023 atoms
There are 6.02x1024 atoms
6.02x1024 6.02x1023=10 moles
6.02x1024 atoms=10 moles
On the other hand, if you wanted to convert the number of particles from mole,
then do the opposite. Multiply 6.02x1023 to the number of moles to get the number
of particles.
Grams-Moles Conversions
If a mass of one egg was one gram, how many grams would a dozen eggs mass?
1.
1. An egg=1 gram
2.
3.
4.
5.
You must find the mass of a dozen eggs
A dozen=12 eggs
1 gram x 12 eggs=12 grams
A dozen=12 grams
Now if the mass of one molecule was 10 grams, how many grams would a mole of
molecules mass?
1.
1.
2.
3.
4.
5.
A molecule=10 grams
You must find the mass of one mole of molecules
A mole=6.02x1023 molecules
10 grams x 6.02x1023=6.02x1024
A mole=6.02x1024 grams
Volume of Gas-Moles Conversions
Plain and simple. Divide the volume of the gas at STP by 22.4dm3, the volume of
one mole any gas at STP. When converting moles to volume of the gas at STP,
simply multiply the number of moles by 22.4dm3.
Eg. What is the volume of 0.5 moles of carbon dioxide at STP?
0.5 x 22.4=11.2
The volume of carbon dioxide at STP is 11.2dm3 or 10dm3 with appropriate usage
of significant digits.
In actual mole conversion problems, you must find out the mass of a particle
yourself. To do so, find the atomic mass of the element in question. For example, if
you needed to convert Carbon, you would look for the atomic mass of C or 12.011.
For ease of calculation, limit the mass number to 3 significant digits. So for all
intents and purposes an atom of Carbon equals 12.0 grams. For the mass of
compounds, simply add all the atomic masses of the atoms involved. So, water
(H2O) will have the following mass:
H=1.01 grams x2 (since there are 2 H atoms)=2.02 grams
O=16.0 grams
2.02+16.0=18.02 grams=18.0 grams
H2O=18.0 grams
It might be helpful to copy down the table below if your chemistry instructor hasn’t
given it to you already.