Copyright (c) 1996 by James H. Burness
Since a molecule is made up of two or more atoms, chemists use
the term molecular mass to refer to the mass of a molecule.
The molecular mass of Cl2 is 2(35.457 u) = 70.91 u.
Just as a given pair of Teddy Bear eyes never have a mass of 71g
(the "average" mass), a single chlorine molecule never
has a mass of 70.91 u. Nevertheless, using average molecular masses
will always give accurate results for any macroscopic-sized collection
of molecules.
Determine the molecular mass, to two decimal places, for each of the following (refer to a periodic table for the atomic masses):
Answers:
1. 2(1.008 u) + 16.00 u = 18.02 u
2. 12(12.011 u) + 22(1.008 u) + 11(16.00 u) = 342.31 u
3. 22.99 u + 35.45 u = 58.44 u
4. 39.10 u + 26.98 u + 2(32.07 u) + 8(16.00 u) + 12(18.02u) =
474.46 u
Note: Strictly speaking, since the compounds in 3 and 4 are ionic,
they don't exist as molecules (see p. 38.) It is better to use
the term formula mass, rather than molecular mass, for
ionic substances.
When molecules, rather than atoms, are placed in a mass spectrometer,
it usually happens that some of the molecular ions formed by electron
bombardment are unstable and break apart into fragment ions.
Some of the molecular ions, however, make it to the detector without
fragmenting. These ions have the same mass as the molecule (the
mass of the electron which is removed to form the ion is negligible,
as shown in the first table
on page 13), and they are therefore
called parent molecular ions. For example, the parent molecular
ion in the mass spectrum of chlorine gas is either (35Cl-35Cl)+,
(35Cl-37Cl)+, or (37Cl-37Cl)+.
The cluster of peaks for the parent molecular ion in the mass
spectrum of Cl2 is shown in Fig. 3
on page 17. Compare
this with Fig. 1 on page 11. The similarity of the two figures
shouldn't be surprising at all, now that we have discussed the
concept of atomic and molecular masses.
Figure 3. Mass Spectrum of Chlorine in the Region of the Parent Molecular
Ion
Since a mass spectrometer detects the fragments formed when a
molecule breaks apart, and since these fragments are determined
by the molecule's structure, the mass spectrum of a compound is
a fingerprint. In other words, no two substances have the same
mass spectrum. This is true even if they have the same formula.
Consider the mass spectrum shown below
for acetone, which has
a molecular formula C3H6O (and thus a molecular
mass of 58 u):
Figure 4. Mass Spectrum of Acetone
The peak at m/e=58 represents the parent molecular ion,
(CH3COCH3)+.
It is due to molecules of acetone, minus an electron, which reached
the mass spectrometer detector. But not all molecules strike the
detector intact; some of the molecules hit by the electrons are
fragmented, resulting in smaller mass pieces which reach the detector.
For example, the base peak, at m/e=43, has a mass which is 15
units less than that of the parent molecular ion.
Which species is giving rise to the peak at m/e=43?
Answer: There is a mass difference of 15 between 58 and
43. This corresponds to the mass of a methyl group, CH3
. Thus, the fragment at 43u corresponds to loss of a methyl group
from the parent molecule. The ion which causes the peak at
m/e=43, then, is (CH3CO)+. Since the
parent molecule can lose CH3 from either side of the
molecule, it isn't surprising that the peak at 43 is the base
peak.
The peak at m/e=15 is due, of course, to a methyl fragment ion.
The fragmentation of a molecule is analogous to hitting the Teddy
Bear heads with a baseball bat and looking at the pieces which
are left over. The "parent" species would be a complete
head which managed to stay together after the impact and would
have a mass equal to the sum of the masses of the components.
On the other hand, if an ear is knocked out of its slot, the resulting
"fragment" would have a mass 19 grams less than that
of a complete head, since each ear weighs 19g.
Consider now the molecule propanal, which has the same overall
formula (and thus the same molecular mass) as acetone. The arrangement
of atoms in the two molecules is, however, different (compare
the structures shown in the two figures of the mass spectra).
Compounds with the same formula but a different arrangement of
atoms are called isomers (don't confuse this term with
isotopes, which are atoms of the same type with different
numbers of neutrons).
The mass spectrum of propanal is shown in Fig. 5:
Figure 5. Mass Spectrum of Propanal
Which species are giving rise to the peaks at m/e=58, 57, and
29?
Answer: The peak at m/e=58 is due to the parent molecular
ion, (CH3CH2COH)+. The peak at
m/e=57 is due to the (CH3CH2CO)+
ion, and the base peak at m/e=29 arises from two ions: (CH3CH2)+
and COH+, both of which happen to have the same mass.
As you can see, the mass spectra of the two isomers are indeed
quite different. Chemists can examine the mass spectrum of a substance
and identify the fragment ions. Then they can see how the fragments
must have originally fit together, like the pieces of a jigsaw
puzzle. This means that the structure of an unknown molecule can
be deduced. Since mass spectra are like molecular fingerprints,
they can also be used to unambiguously identify countless known
substances, such as illegal drugs and gasoline additives.
Molecules, as well as the atoms which make them up, the mechanistic steps
by which they undergo reactions with other molecules, and the
fragments which are left over when we blast them apart, are real.
Unlike our hypothetical Teddy Bear head factory, they exist.
Now that we have discussed the topics of atomic and molecular
mass, we can move on the the central concept of the chapter --
the mole. The mole is, actually, central to most of the concepts
and calculations in chemistry, including stoichiometry, gases,
equilibrium and electrochemistry, just to name a few. Yet although
the mole is so critically important, it is extremely simple to
understand. If you understand the concept of a dozen, you will
grasp what a mole represents.
In order to appreciate why chemists use this concept, it is necessary
to remember that the atom is so tiny. In fact, the atom's minuscule
size was also the reason scientists invented the atomic mass unit.
Let's ask the following question: "How much does a carbon-12
atom weigh in grams?"
Recall that the mass of 12C is defined as exactly
12 u. It is logical that if one atomic mass unit is equivalent
to 1.6605 x 10-24 g, then 12 u has a mass twelve times
as much:
Now let's ask the question, "How many carbon-12 atoms would
we need to have for a total mass of exactly 12 grams?" Before
we determine this number, think about it for a moment. Atoms have
an extremely small mass. Thus, we would expect an incredibly huge
number of atoms to be present in a 12-gram sample, which would
fit in the palm of your hand. Algebraically, we can rephrase our
question as
Solving for x, we get x = 12g1.9925 x 10-23 g/atom
= 6.022 x 1023 atoms. Since the 12 is an exact
[defined] value, it doesn't limit the significant figures allowed
in the answer. The number we just calculated is, indeed, a very
large number. We'll discuss just how large it is shortly, but
first let's find the answer to another question:
"How many 'average' chlorine atoms are present in a sample
of chlorine atoms with a mass of 35.45 grams?"
First, we obtain the mass of an "average" chlorine atom
using chlorine's atomic mass:
Notice that this is the same answer as the number of atoms in
12 grams of 12C.
We are now in a position to define the mole:
The abbreviation for mole isn't much of one; we can drop the 'e'
and abbreviate it as "mol". Never use "m"
as the abbreviation, since this represents molality, a unit for
measuring the concentration of a solution.
Note that the definition of a mole refers to particles,
not to specific species like atoms or molecules. Thus, it is proper
to talk about a mole of atoms, a mole of molecules, a mole of
ions, a mole of electrons, etc. It is not proper, however,
to use the term mole for a heterogeneous or homogeneous mixture.
For example, it is incorrect to refer to a mole of air, since
air consists of a mixture of gases (one can, however, refer to
a mole of nitrogen or a mole of oxygen, since these are pure substances.)
A chemist will never say "one mole of a sodium chloride solution",
and the term "mole of concrete" doesn't make any sense
because concrete is a heterogeneous mixture.
The mole, then can be interpreted in two ways:
If you examine the calculation which we carried out above for
the carbon-12 and the chlorine atoms, you'll see that it is no
coincidence that the number of atoms worked out to be the same.
For example, if we substitute the left-hand side of Eq. 6
for
the value 5.886 x 10-23 in Eq. 7
, we get:
(x)(35.45)( 1.6605 x 10-24) = 35.45 The atomic weight
cancels, and this is why we got the same answer for carbon as
for chlorine. If we solve for x, we see that x is equal to the
reciprocal of 1.6605 x 10-24, which is 6.022 x 1023.
The latter number is so important that it is given a special name,
in honor of the Italian scientist Amadeo Avogadro.
It is interesting that the monumental contributions made by Avogadro
were made as a result of his interest in chemistry as a hobby.
He was a lawyer by profession.
The mole, then, simply represents a counting unit. There are 6.022
x 1023 of something in a mole, just as there are 12
of something in a dozen and 144 of something in a gross. The mole
is often referred to as the chemist's dozen. The only reason
that Avogadro's number is so big is that atoms are so small.
Here are some analogies to help you visualize just how tremendously
large this number really is:
In terms of doing stoichiometry calculations, the most useful
result of the foregoing discussion is that
One mole of "average" atoms (or molecules) has a
mass in grams equal to the atomic (or molecular) mass.
This mass is known as the substance's Molar Mass.
In other textbooks or chemical literature sources, you may see
molar mass referred to as atomic weight, molecular weight, or
formula weight, depending on whether the substance exists as single
atoms, molecules, or formula units of an ionic compound, respectively.
The term molar mass is preferable because it can be used for any
one of these three types of substances. Chemists use the adjective
"molar" to specify that the value of a particular property
is for one mole of a substance. Thus, molar volume is the volume
of a mole; molar conductivity is the conductivity of a mole of
a metal; and molar mass is simply the mass of a mole. The abbreviation
for molar mass is MM. The units of molar mass are grams/mole if
the value is shown alone, e.g. Sucrose, MM = 342.31 g/mole (see
the Concept Check on page 15.) If molar mass is stated verbally,
however, the units are grams, e.g. "The molar mass of sucrose
is 342.31 g", since the adjective "molar" already
specifies "per mole."
Refer to a periodic table as you examine each of the following
examples. Make sure that you understand each and every sentence.
Notice that the word "average" didn't need to be used for the second example above. This is because fluorine exists monoisotopically in nature as the 19F isotope.
How many peaks would you expect to find for the parent molecular
ion in the mass spectrum of F2? At which m/e value
would it (they) be found?
Answer: One peak would be expected, at m/e = 38 u.
In two moles of ethane, C2H6,
Answers:
As it turns out, our Teddy Bear analogy has another connection
to real life. The entire Teddy Bear scenario can actually be used
as a model for the formation of a chemical compound:
FA + EYE2 + EAR2 -------> FA(EYE)2(EAR)2
100 g + 71g (average) + 38 g
-------> 209 g (average)
C + Cl2 + F2 ----------> CCl2F2
12.011 g (average) + 70.9 g (average) + 38 g --------->
120.9 g (average)
Remember that the face had four points of attachment; likewise,
carbon always forms four bonds in its compounds. In the same way
that the eyes and ears were always supplied as pairs, the elements
chlorine and fluorine always exist in nature as diatomic molecules.
There are two differences between the analogy and the real chemical
equation: 1) the mass of carbon is not the same as the mass of
the face (a 12-g face wouldn't be very heavy!) and 2) we assumed
that all faces were identical, whereas carbon actually exists
as a mixture of two isotopes (C-12 and C-13, with abundances of
98.89% and 1.11%, respectively).
Notice that the equation above for the formation of CCl2F2
is already balanced. What is the best way to interpret the stoichiometric
coefficients in the balanced equation? We already learned that
a generic interpretation of a coefficient is a "unit",
but which unit should we use? A dozen? A gross? A billion molecules?
The best "unit" to use for the stoichiometric coefficient
is the chemist's dozen, the mole:
Stoichiometric coefficients in balanced chemical
equations represent ratios between the number of moles of reactants
and products in the equation.
The best interpretation of the equation above is "One mole
of carbon will react completely with one mole of chlorine and
one mole of fluorine to yield one mole of dichlorodifluoromethane."
If we had chosen to balance the equation with fractional coefficients,
e.g.
then we would read the equation as "One-half mole of carbon
reacts completely with one-half mole of chlorine and one-half
mole of fluorine to produce one-half mole of dichlorodifluoromethane.
Remember that the coefficients themselves say nothing about how
much of the reactants or products is actually present; we could
have 100 grams each of carbon, chlorine, and fluorine, but the
two interpretations above would not be any different. The ratios
of moles which react are always one-to-one whether we use integral
or fractional coefficients.
It is important to remember the distinction between a subscript
in a formula and a stoichiometric coefficient. For example, the
notations Cl2 and 2Cl each represents a total of 70.9
grams of chlorine and a total of two moles of chlorine atoms,
but the meanings are entirely different. The former means one
mole of diatomic chlorine molecules, for a total of two
moles of chlorine atoms, whereas the latter means two moles of
monatomic chlorine atoms. The important difference is that
the two moles of chlorine atoms in the former example are bonded
to each other (like a pair of Teddy Bear eyes) while the two moles
of chlorine atoms in the latter example are separate atoms. Refer
again to scenario 1
on page 5. The Cl2 would be analogous
to the reactant in step 1, while 2Cl would be analogous to the
product in the same step.
Now that you realize that stoichiometric coefficients represent
moles, you will be able to perform stoichiometric calculations.
Common examples of the sorts of stoichiometric questions which
chemists might ask are
The two most important things to remember about stoichiometric
calculations are that
We'll look at step 1 first, then use balanced equations to perform
a variety of stoichiometric calculations. Refer back to the concept
check on the very first page of this chapter. Remember that if
you were able to get the answer to this, then you can easily do
stoichiometric calculations.
Except for reactions known as redox (an abbreviation for oxidation-reduction) reactions, which can sometimes be complex, chemical equations are balanced by a trial-and-error procedure. Redox reactions are balanced by a special set of rules which will be considered in chapter 4. Some reactions are easy to balance by trial-and-error, while others are a little more time-consuming. The following guidelines may help you to balance equations more easily:
We'll look at two examples of common types of reactions and will
refer to each of the rules above, when necessary, as we balance
them.
Combustion is the reaction of a substance with oxygen. When metals react with oxygen, the corresponding metal oxides are usually formed (common exceptions are sodium, which reacts with oxygen to form sodium peroxide, Na2O2, instead of Na2O, and potassium, which reacts with oxygen to form potassium superoxide, KO2, instead of K2O.)
When hydrocarbons (compounds containing only carbon and
hydrogen) or compounds containing carbon, hydrogen, and oxygen
react with oxygen, the products are always carbon dioxide and
water.
Complete and balance the equation for the combustion of aluminum.
Answer:
The unbalanced equation is
Following rules 1 and 2, we begin with the product and save oxygen
until last. Thus, we balance aluminum, obtaining
We don't want to change the coefficient in front of the product,
since that will force us to change the "2" we just placed
in front of Al. Thus, we want to change the coefficient in front
of O2 so that we have three oxygen atoms on the left.
A coefficient of 3/2 will do the trick:
If we wanted to use simplest whole-number coefficients, we would have:
Complete and balance the equation for the reaction between lithium
and oxygen.
Complete and balance the equation for the combustion of ethyl
alcohol, C2H5OH.
Answer:
The unbalanced equation is
Notice that it is easier to collect the hydrogens together before
beginning. Following rules 1 and 2, we begin with the ethyl alcohol
and balance carbon:
Balancing hydrogen next gives:
The total number of oxygen atoms on the right side is seven, so
the final balanced equation is
Complete and balance the equation for the combustion of methane,
CH4. This reaction occurs when a gas furnace or a gas
stove is used for heating.
Answer:
Methane is a hydrocarbon. The unbalanced equation is
Following rules 1 and 2, we begin with the methane and notice
that carbon atoms are already balanced. Since hydrogen shows up
in only one product and oxygen shows up in two, it might be easier
to balance hydrogens next:
Oxygen is easily balanced:
When an acid and base react, the products of the reaction are
a salt and water. The water is formed as a result of the H+
from the acid reacting with the OH- from the base.
The salt, then, is composed of the cation from the base and the
anion from the acid.
Complete and balance the equation for the reaction between hypochlorous
acid and magnesium hydroxide.
Answer:
From the nomenclature rules you learned in chapter 2, you first
translate the names into chemical formulas and complete the reaction:
acid + base ------> salt
+ water
According to rule 1, we should begin with either the base or the
salt. Let's begin with the salt. Magnesium is already balanced.
We note that the hypochlorite ion stays intact from reactants
to products, so rule 3 allows us to balance the ion as a unit.
Two moles of hypochlorite ions on the right side require two moles
of hypochlorous acid on the left side:
Since the stoichiometric coefficients for the reactants are now
fixed, we can balance H and O. There are 4 hydrogens on the left,
which requires a 2 in front of water:
Since oxygens balance, we are finished. Note that we could have
balanced oxygen before hydrogen. But we shouldn't count the O
atom in the hypochlorite ion, since it was included when we balanced
the ion as a unit. Thus, going back to the equation before the
previous one, we would count two oxygens on the left (from the
magnesium hydroxide), requiring a 2 in front of water. This balances
hydrogen, giving us the same final equation as before.
It is easier to balance acid/base reactions if we realize that
the formula for water can be written HOH, showing that one mole
of hydroxide ion from the base is needed for each mole of hydrogen
ion from the acid. Refer again to the example above. Since one
mole of magnesium hydroxide contains two moles of hydroxide ion
and one mole of the hypochlorous acid contains only one mole of
hydrogen ion, it will require two moles of acid for every mole
of base, and two moles of water will form.
Complete and balance the equation for the reaction between phosphoric
acid and potassium hydroxide. The product of this reaction is
a common ingredient in commercial fertilizers.
Answer:
The formulas for the reactants are H3PO4
and KOH. The unbalanced reaction is
Since the acid has three H+ ions and the base has only
one OH- ion, three moles of base will be required for
each mole of acid, producing three moles of water:
Now that you know how to balance a reaction, let's discuss the
procedure for using a balanced reaction to perform calculations.
Review the steps for solving stoichiometry problems, on page 25.
We'll use a visual algorithm to implement each step and to help
you think about the chemical concepts as you solve the problem.
We'll first solve the mole-mole problem on page 24:
"How many moles of water are formed if five moles of hydrogen
gas react with excess oxygen gas?"
Step 1: Write the balanced equation for the reaction.
Step 2: Write known quantities above or below the formula
for the substance; moles always go below the formula, and
quantities which will be used to determine moles go above the
formula. Use a question mark to represent the unknown (desired)
quantity.
Step 3: Draw an arrow from the quantity above the formula
for the known to the number of moles shown below the formula.
Show the factor you'll use to convert from the former to the latter.
[Note: this step does not apply here, since we were given moles
of H2.]
Step 4: Draw an arrow from the number of moles of the
known substance to moles of the unknown, and write the mole ratio
above the arrow. Use this ratio to calculate moles of unknown
substance from moles of the known substance.
Since the number of moles of water formed is equal to the moles
of hydrogen which react, five moles of hydrogen will produce five
moles of water.
Step 5: Draw an arrow from moles of unknown to desired
quantity of unknown (above the formula). Show the factor to be
used to convert from moles to the desired quantity. [Note: this
step does not apply here, since we were asked to determine moles
of H2O.]
The answer to the problem, then, is that 5 moles of water
will be formed.
Let's use these same steps to perform the mass-mass calculation
on page 24:
"How many grams of oxygen are needed to completely convert
50 g of mercury to mercuric oxide (HgO)?"
Step 1: The balanced equation is
Steps 2,3: We list grams of mercury above the atomic symbol,
a question mark above the O2, then draw our conversion
arrow:
If there are 200 grams in a mole of mercury, then 50 grams is
obviously one-quarter of a mole. Recall the concept check at the
beginning of this document:
Your favorite candy costs $2.00 per pound. You have 50 cents.
How much candy can you buy?
Do you see a connection between the candy example and the mercury?
Chemistry really is common sense.
Step 4: Draw the conversion arrow between moles of mercury and
moles of oxygen, showing the mole ratio:
The stoichiometric coefficients show that only one mole of oxygen is needed for every two moles of mercury. Thus, to find moles of oxygen we simply divide moles of mercury by 2. Since the mole ratio is 2:1, you know that you'll be either multiplying or dividing moles of mercury by the number 2. How do you know which to do? The answer is common sense. If you had multiplied by two, you would have obtained a larger number of moles of oxygen than the number of moles of mercury you had, clearly contradicting the meaning of the stoichiometric coefficients. Our stoichiometry map now looks like this:
Step 5: Finally, we draw the conversion arrow for changing moles
of oxygen to grams, and complete the calculation:
Thus, it will require 4.0 grams of oxygen to completely convert
50 g of mercury to HgO. It might seem surprising that 50 grams
of mercury only requires 4 grams of oxygen, but remember that
chemical equations represent mole ratios. Since the molar
mass of Hg is so large (200g/mol), we still only have a fraction
of a mole even with 50 grams. In other words, most of the mass
of any sample of HgO is due to mercury. We'll explore the idea
of percentage composition in a section later in the chapter.
CONTINUED IN PART 3
RETURN TO PART 1
Last modified June 16, 1997