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Unformatted text preview: 48 Chapter 5 CHEMICAL UNITS AND MASS
©2004, 2008, 2011 Mark E. Nob/e We now return to the matter of mass. Or, actually, the mass of matter. We last left mass in the minuscule world of the atom in Chapter 2 when we first introduced atomic
mass. Now we are going to discuss masses of polyatomic things, including elemental forms and
compounds. We begin with the atomic mass unit, u, but later we will come into the practical world of the
gram. Mass is one of the most important measurements in chemistry and the mass of a sample allows
us to calculate the number of molecules in the sample. Why do we want to? When we do chemical
reactions, a specific number of molecules (or whatever chemical units) reacts with a specific number of
other molecules. We need to know how many things are in a sample, but we cannot count individual
chemical units because they are too small and too many. We also cannot directly measure u's since that
unit is too small by itself. So, somewhere along the road, we've got to come up to grams (or some other
unit) which we can measure easily. We've got to have a connection between grams of sample and how
many chemical units are in the sample. The overall result is an important one: we cannot directly count
how many molecules are in a sample, but we can measure how many. And, as we discussed in Chapter
1, we can only measure to some limit of certainty; that means we will be doing sigfigs for numbers of
molecules and such. 5.1 Still stuck on u. In Chapter 2, we discussed atomic mass units with respect to a single atom and also to an "average"
atom. The average atom took into account the different isotopes and this gave us the atomic mass in u
or the atomic weight without the u. Now we do polyatomics. For these, we use "molecular mass" or
"formula mass", for which the units are again u. The terms "molecular weight" and "formula weight" will
again be a relative value, the same number but without the unit u. The molecular mass is the average mass of a single molecule in u's. This is easy to do: it's just the
sum of the masses of the atoms in the molecule. Let's do an example: find the molecular mass of
selenium difluoride, Ser. All that you need to do is add up the atomic masses of one Se and two F's, as
shown below at left. Another example: find the molecular mass of sulfur trioxide, $03. This is obtained
from the sum of the atomic masses of one 5 and three O's, as shown at right. SeF2 Se 78.96 u so3 5 32.07 u
F 19.00 u 0 16.00 u F 19.00 u 0 16.00 u 116.96 u 0 1600 U 80.07 u This means that the average mass of one molecule of SeF2 is 116.96 u and the average mass of one
molecule of SO3 is 80.07 u. (I keep saying average because we are using atomic masses, most of which
are averages. This is a minor detail, and I won't continue saying average.) For network compounds, there are no simple molecules and so the term formula mass applies. A
formula mass is the sum of the atomic masses of the formula unit. For example, the formula mass of
potassium chloride, KCI, is derived from the mass of one K and one CI. The formula mass of silicon
dioxide, SiOZ, is derived from the mass of one Si and two O's. KCI K 39.10 u SiO2 Si 28.09 u
Cl 35.45 u 0 16.00 U 7455 u w 60.09 u Since KCI is ionic, some of you might think that we should be using the mass of one K+ ion and the mass
of one Cl' ion. That's a good point. The masses of these separate ions differ from their atomic masses
by the mass of one electron (which is not much anyway). It turns out that this difference drops out for
the ionic compound as a whole. KCI K+ 39.10 u minus one e' mass
Cl‘ 35.45 u plus one e‘ mass
74.55 u, same as above This will ALWAYS be true for formula masses of ionic compounds: you just take the atomic masses off the
Periodic Table and don't worry about who's got how many electrons. Chapter 5: Chemical Units and Mass 49 Now that I've just told you that a network compound goes by "formula mass", I will now tell you that
we don't normally make the distinction in routine use. We use the term "molecular mass" generically.
Thus, many chemists might also say the molecular mass of KCI is 74.55 u even though it's a network ionic
compound. If your instructor requires the distinction, then follow it. TECHNICALITY. There's a technical point I need to make about sigfigs. Different instructors handle
molecular masses a bit differently for sigfig purposes. Notice that I set up each of these examples as a
simple addition. I used the sigfig rule for addition/subtraction which goes by the decimal place but, in real
life, when you plug these into your calculator, you'll often be using the multiplication key. For example,
the molecular mass of SO3 could be entered on the calculator as 3 x 16.00 + 32.07. 50 is this really
addition or is this really multiplication for sigfig purposes? I'll be treating it as an addition problem, but
you need to know what your instructor calls it. Does it matter? It can: sometimes it affects the roundoff
decision. I can illustrate this with a very simple example, P4. This is an allotrope of phosphorus which
I mentioned in Chapter 2. What's the molecular mass for P4? Addition roundoff method Multiplication roundoff method
P4 P 30.97 u P4 4 x 30.97 u = 123.9 u
P 30.97 u (Rounded to four sigfigs
P 30.97 u according to x/+ rule.
P 30.97 u The 4 is exact, so you go by
12338 u 30.97, which has four sigfigs.) (Rounded to second decimal
according to +/— rule) The two values reflect a different limit of certainty. Since I'm doing addition method, I would give the
mass as 123.88 u, but another instructor could be doing multiplication and use the value 123.9 u. Which
is correct? Within the sigfig system, both can be considered correct but remember that the sigfig system
is itself a compromise system. It is not unusual for two ways to give the same answer but with different
numbers of sigfigs. Your job is to be clear on how your instructor wants you to handle these. Does it
even matter? Sometimes. Not always. But, yes, sometimes. OK, this ends the technicality. Let me make a point about formulas with parentheses. I'll illustrate with calcium dihydrogen
phosphate, Ca(H2PO4)2. If you're into gardening, you may have used this stuff. It's one ofthe compounds
in "superphosphate" fertilizer which is sold in stores. The formula unit consists of one calcium ion and tw_o
dihydrogen phosphate ions. The subscript two outside the parentheses means two times everything
inside. For the mass of the formula unit, we add the masses of one Ca (40.08 u), Egg H's (1.008 u each),
tw_o P's (30.97 u each) and eight O's (16.00 u each). The formula mass is the sum, 234.05 u. 5.2 We jump to the real world. In our real world we can see an object and we can measure its mass in grams or whatever. An atom
is incredibly tiny, however, and any real sample has a zillion atoms. Actually, several zillions. I described
this in Chapter 2 but this is a really big point so I'm repeating it. Atoms are incomprehensiny small. A
real sample contains an incomprehensibly huge number of them. This is what Nature has given us:
immensely huge numbers of infinitesimally tiny objects. In science many years ago, they established a numerical unit for working with these massively huge
numbers. A "numerical unit" is just a unit which means a number. The most common example of a
numerical unit in common usage is a dozen. A dozen is 12. That's 12 of anything. Eggs, roses, donuts,
etc. (Years ago, a dozen donuts was 13.) Another numerical unit is gross. A gross is 144. Of anything.
Dozen and gross are numerical units. Chemists also have a numerical unit for dealing with atoms and molecules, but the number itself is
incredibly huge. Big. REALLY BIG. The unit is called a "mole". I admit that I've often thought this word
sounds totally unimpressive for a number of humongous proportion, but that's the name they gave it and
we're stuck with it. The mole is abbreviated mol. (The abbreviation knocks off one whole letter. Some
abbreviation.) The numerical value of the mole is 6.02214179 x 1023. Why such a strange number?
Well, it's not straightforward. They set the system up based on a measured number of atoms in exactly
12 grams of 12C. There are reasons for that and we don't need to go there. The net result is that it's not
a simple, whole number. It's also not an exact number because it's a measured value. So it's not like
a dozen, which is defined to be exactly twelve. For a mole, there are sigfigs to consider. In typical work,
it is rounded to 6.022 x 1023, four sigfigs. That's the way we'll use it most of the time. 50 Chapter 5: Chemical Units and Mass mol = 6.022 x 1023
As we saw at the end of Chapter 1, such a relationship can be written as two conversion factors.
23
mol OR 6.022 x 10
6.022 x 1023 mol By the way, the numerical value 6.022 x 1023 is called "Avogadro's number" after Avogadro who had
nothing to do with the number itself. Actually, Avogadro was a lawyer before going into sciences. Big
number. Lawyer. Go figure. 6.022 x 1023 is huge. It's immense. It's very difficult to comprehend the size of that number.
6.022 x 1023 of anything you can see is unthinkable. Consider one mole of the smallest thing you can
see. How about a human hair, only a fraction of an inch long? That's not a lot, but a mole of those would
weigh about a trillion tons, even without the dandruff. Now, apply this immensely huge number to an
infinitesimally tiny molecule and you get something reasonable: one mole of water molecules occupies
1.2 tablespoons. I said it upstairs and I'll say it again: THIS IS WHAT NATURE HAS GIVEN US: IMMENSELY HUGE NUMBERS OF INFINITESIMALLY TINY
OBJECTS. You must catch the significance here! The mole is the numerical unit which connects a huge number of
infinitesimally tiny molecules to our real world and realistic sample sizes. The mole is our basic, routine
unit for "how many". This is what we commonly use day to day. There is another consequence of this mole unit which allows us to connect atomic mass units to
grams. A mole of u's is a gram. 9 = 6.022 x 1023 u
This likewise can be presented in two ways. 23
9 OR 6.022 X 10 u 6.022 x 1023 u g
The mole is our important, numerical link to how many atoms and to their mass in grams. This brings us to "molar mass". Molar mass is the mass in grams of one mole of anything: it could
be a mole of atoms or a mole of molecules or a mole of formula units. This will change the way we deal
with mass. Until now, everything we did with mass was in u for one unit: the atomic mass of one atom
or the molecular mass of one molecule or the formula mass of one formula unit. All of those were u's for
one individual unit. Now, molar mass is the SAME NUMBER but in grams for one mole of those units.
Compare this to our previous examples. SeF2 116.96 u for one molecule 116.96 9 for one mole of molecules
SO3 80.07 u for one molecule 80.07 g for one mole of molecules KCI 74.55 u for one formula unit 74.55 g for one mole of formula units
SiO2 60.09 u for one formula unit 60.09 g for one mole of formula units
P4 123.88 u for one molecule 123.88 9 for one mole of molecules
Ca(H2PO4)2 234.05 u for one formula unit 234.05 9 for one mole of formula units A key point from all of the above examples is that the number of u for one molecule (or one formula
unit) is the SAME NUMBER as the number of grams for one mole of those molecules (or one mole of those
formula units). Why does the SAME NUMBER apply to both values? Because everything else drops out.
I'll show you using selenium difluoride. Here's the full dimensional analysis string for calculating the molar
mass of SeF2 from its molecular mass. molecular mass molar mass r—————\ r—*—ﬁ
116.96 u X g X 6.022 x 1023 molecules of 5er = 116.96 9 molecule of Ser 6.022 x 1023 u mol of 5er mol of SeF2 Notice that "molecule(s) of Ser", "u" and even "6.022 X 1023" all cancel and drop out. (Go ahead and
cross them out.) The net result is that for ﬂy formula unit, the number of u for one unit is the SAME
NUMBER as the number of grams for one mole of those units. Chapter 5: Chemical Units and Mass 51 Although the numbers are the same, you need to keep in mind the distinction between the terms.
Atomic mass, molecular mass and formula mass are for one unit. Molar mass is for one mole of those
units. Remember this. A few more examples for emphasis, please. Ne 20.18 u for one atom 20.18 g for one mole of atoms
CO2 44.01 u for one molecule 44.01 g for one mole of molecules
(NH4)ZSO4 132.15 u for one formula unit 132.15 g for one mole of formula units In all cases, these mass relationships can be represented by either of two fractional forms for use
as conversion factors. I won't show this for all examples above, but I'll just do the selenium difluoride
example. One molecule of SeFZ: 116.96 u Conversion factors: OR M molecule 116.96 u One mole of SeFZ: 116.96 9
Conversion factors: M OR m—OI—
mol 116.96 9 The g/mol and mol/g conversions are used extensively, as you will see. Let's move on to sample sizes other than one mole. Consider this problem: how many moles of tin
atoms are in a sample of 17.49 9 Sn? From the Periodic Table, the molar mass for tin is 118.7 9, so our sample of tin has some fraction
of a mole of atoms. We need to find out just how much. We can execute this problem by dimensional
analysis (although you can also use some other way if you want). The molar mass for tin means that mol Sn = 118.7 9 Sn
which can be taken as ﬂ OR —m—°———
mol 118.7 9 Our calculation is a one—step problem starting from g and going to mol. We need the conversion factor
on the right with mol upstairs and g downstairs in order to cancel 9 Sn. path: 9 Sn ' mol Sn
17.49 9 Sn x m = 0.1473 mol Sn
118.7 9 Sn From this we see that our sample of 17.49 g Sn corresponds to 0.1473 mol Sn. Remember that a mol is just a number. It's always a mol of something. Just like a dozen is always
a dozen of something. In general, for calculations such as this, we deal with a mole of formula units,
whatever the formula unit may be. This may not always be clear, so let me give some pointers on how
to interpret a "mole of formula units". It depends on whether you have a molecular compound or a
network compound. I'll illustrate these with some examples that we've worked with. > IfI write "mol Sn", it means a mole of Sn formula units. Tin is a metal and it exists as a metallic
network; the formula unit is one atom of tin in the network. Thus, "mol Sn" means a mole of tin
atoms in the network. > IfI write "mol Ser", it means a mole of Ser formula units. This is a molecular compound, so the
formula unit is one molecule. Thus, "mol SeF2" means a mole of those molecules. > IfI write "mol KCI", it means a mole of KCI formula units. KCI is an ionic compound, so this is an
ionic network. The formula unit is one K+ cation and one Cl‘ anion. Thus, "mol KCI" means a
mole of {K' + Cl‘} formula units. > IfI write "mol SiOZ", it means a mole of SiO2 formula units. SiO2 is a covalent network compound
and the chemical unit is the network. Thus, "mol SiOz" means a mole of {Si + 2 O} formula units. Keep these pointers in mind. Although mol is our common unit for "how many", you can still do individual units by bringing in
Avogadro. As an illustration, let's keep the tin example from above but we'll change the question: how
many Sn atoms are in 17.49 9 Sn? 52 Chapter 5: Chemical Units and Mass To do this, we must execute one more step beyond mole. We must take mole to individual atoms
by way of Avogadro. path: 9 Sn ’ m0 Sn ’ atoms Sn
23
17.49 9 Sn x M x W = 8373 x 1022 atoms Sn
118.7 9 Sn mol Sn Notice that the answer is a huge number: there's an extremely large number of atoms in just a modest
sample size. The grams/moles conversion is one of the most common conversions you'll do. You'll need to be able
to do it both ways, 9 ' mol and mol ' g. You need to use molar mass as the conversion factor. Here
are two Examples, one in each direction, using carbon dioxide, C02. The molar mass means mol CO2 = 44.01 9 CO2
for which you can write the following.
44.01 9 OR mol
mol 44.01 9
Set this up as for the Sn example earlier.
path: 9 CO2 ’ mol C02
3.08 9 CO2 m°' C02 = 0.0700 mol co2 44.01 9 CO2 The same conversion factors apply but now you need the one with mol in the denominator. path: mol CO2 ' 9 CO2
1.621 mol c02 x w = 71.34 9 CO2
mol CO2 I cannot tell you enough: the gramstomoles and molestograms conversions are immensely
important. These are so important that it must become automatic for you to know how to do both ways.
Get used to this. 5.3 Percent composition There's another aspect of masses which I need to bring up. This aspect goes back quite a few years
in chemistry but it also ties to some of the methods which chemists today use in order to identify compounds. We start with the notion of "percent composition". Percent composition is the percent by mass of
an element in a compound. This gives a relative distribution of the mass of a compound among its
different elements. I'll illustrate using sulfur trioxide, $03; from earlier in this Chapter, we know that this
compound has a molecular mass of 80.07 u and a molar mass of 80.07 g. For purposes of percent
composition, we can base the calculation on one molecule and work with u or we can base the calculation
on one mole and work with grams. It doesn't matter: the numbers will come out the same. Let's do one
molecule for now and work with u. Of the 80.07 u in one molecule, 32.07 u were from the sulfur atom. What percent is that? Percent composition of sulfur: mass of S: 32.07 u — 100% = 40.05%
total mass: 80.07 u Chapter 5: Chemical Units and Mass 53 This tells us that 40.05% of the mass of one SO3 molecule is derived from the sulfur atom in the molecule. You can do this for each element in a compound. For the oxygens in S03, you count all three O's.
Of the 80.07 u in one 503 molecule, 3 x 16.00 u were from the three oxygen atoms. What percent is
that? Percent composition of oxygen: mass of 0's: 3 x 16.00 u
total mass: 80.07 u This tells us that 59.96% of the mass of the molecule is derived from the oxygen atoms. 100% = 59.95% Notice that the u drops out of the calculation. If you work with grams, then 9 will also drop out. In
the end, everything comes out the same, so it doesn't matter which way you go. You can make all of this into an equation. For any one element Q in some formula, the percent
composition of Q is determined by the following. percent composition = x 1000/0 mass of formula unit That's the u version. You can use 9 instead. Notice how this equation relates exactly to what we did for
the sulfur trioxide example. Here, you can do one. Example 3. Find the percent composition of phosphorus in calcium dihydrogen phosphate,
Ca(H2PO4)2. We did its formula mass earlier: 234.05 u. Plug in.
number of atoms of P x atomic mass of P l l X 100% = 1 mass of formula unit The answer is 26.46%. If you didn't get that answer, then figure out what happened. Did you remember
the two outside the parentheses? The percent compositions for the other elements in calcium dihydrogen
phosphate are 17.12% Ca, 1.723% H and 54.69% 0. Go ahead and check those numbers, too. Ideally, the percent compositions for all elements in a compound should add to exactly 100%. We
deal in the real world of measurements and uncertainty, however, and things aren't always ideal. Go back
to the SO3 example. Add the percent compositions together for the S and the 0's. You get 100.00%.
OK, fine. Now add the percent compositions together for the elements in calcium dihydrogen phosphate.
What did you get? You don't get exactly 100.00%. These things can happen. They're not wrong.
They're just not perfect. Percent compositions are usually measured using certain kinds of instrumentation. These processes
are called elemental analysis. There are several important uses of elemental analysis. For example, when
chemists prepare a new compound, they are required to provide experimental proofto support their claim.
There are different ways of doing this for different kinds of compounds, but elemental analysis has been
one of the most common methods. Let's say you claim to make a new compound and you claim to know
what its formula is. You conduct an elemental analysis and measure the percent composition of one or
more elements. You compare these experimental results to the calculated percent compositions using
the same approach we did above. If you have a good match, this helps to support your claim. If not, you
goofed. I'll give you a real example. In my research group we prepared a new compound which we
proposed to have the lengthy formula MoZC4oH73N206P3SS. (Don't ask me to name it.) We believed this
to be the correct formula based on different instrumental evidence. Elemental analysis was done for
carbon, hydrogen and nitrogen. (You don't have to do all the elements in the compound.) From the 54 Chapter 5: Chemical Units and Mass formula, you can calculate what the percent compositions should be. The experimental results were
41.0% C, 6.6% H and 2.3% N. They're close, and that's considered a satisfactory match for such a
complicated case. 5.4 An empirical approach Another use of elemental analysis is to work backwards: start with percent composition and work
towards the formula. This is useful when dealing with an unknown compound. Unfortunately, this has
limitations since it cannot give you the true formula by itself, but it does provide valuable clues for
ultimately identifying many compounds. Let's do an example. We'll look at a cold case: refrigerants. These are the fluids which make refrigerators and air
conditioners operate. There are many different types and sizes of refrigerators and air conditioners, and
there are many different refrigerants used depending upon the application and design. Many of the
refrigerants which are used in applications for the general public are compounds of carbon and fluorine
with hydrogen and/or chlorine atoms also in the molecule. Many of these compounds had been used for
refrigeration (and for various sorts of other things, too) for years until they started getting into the upper
atmosphere and messing with the ozone layer. Well, not all of them. In the past, many were called
Freons, but Freon is a trademark name; nowadays they're generically called refrigerant orjust R. There's
a bunch of these compounds and the industry uses a number to identify them. (These are not their real
chemistry names.) Previously, R—12, whose chemical formula is CFZCIZ, was one of the most widely used
in automobile air conditioners; that one was fairly hazardous to the ozone layer and it was banned years
ago although it may yet be in some older cars. Nowadays, automobile air conditioners use R—134a which
is C2H2F4. Many home air conditioners use R—22 which is CHFZCI; it's being phased out also and newer
refrigerants based on mixtures are coming into use. It turns out that the chlorine atoms in the molecules
render these compounds especially damaging to the ozone layer, so the compounds with chlorine are the
ones being phased out fastest. Some of these compounds are also greenhouse gases which adds to the
environmental problems. By the way, if you want to know how air conditioners work, you can find this in Section 35.4. OK, let's get back to elemental analysis. Let's say you have an unknown refrigerant compound and
you need to find out which one it is. You conduct an elemental analysis and find that the compound is
composed of the elements carbon, fluorine and chlorine. The percent compositions are 14.1% C, 44.6%
F and 41.5% CI. What is the formula of the compound? Notice that the numbers don't add to exactly
100.0%. Again, measurements aren't always perfect, but they should be close. Here's the general gist. Percent composition reflects the distribution of the masses of the atoms of
the different elements in the compound. From these mass distributions, we can derive a relative ratio of
the numbers of atoms of the different elements. This ratio is our target. We start with the relative mass distributions as given by the percent compositions.
14.1% C : 44.6% F : 41.5% CI Now, pick a sample size. Any size would work, but most people would choose 1 g or 100 g when dealing
with percents. I pick 100. g. 50, in 100. g of this compound, there are zillions of molecules, each
containing one or more atoms of C, F and Cl. By the measured percent compositions, my 100. 9 sample
is composed of 14.1 g total C mass, 44.6 9 total F mass and 41.5 9 total Cl mass scattered among all the
molecules. This gives a mass ratio for the elements. 14.1gC : 44.6gF : 41.5gCl We want to know how many moles each of these corresponds to, so we bring in molar mass for each
element. 14.1gC I 44.6g F I 41.59Cl
12.01 g C/mol C ' 19.00 g F/mol F I 35.45 g Cl/mol Cl Now divide each to get the mole ratio.
1.17 mol C : 2.35 mol F : 1.20 mol Cl Remember that these are total mole amounts for the various atoms present in the sample of 100. g. For
a formula, we need a whole number ratio of atoms. Frequently by this stage you can see such a whole
number ratio; you may already see in this case that we have a 1:2: 1 ratio, but the numbers don't always
work out this well. When that happens, choose the one element with the fewest moles and find the ratio Chapter 5: Chemical Units and Mass 55 of each other element to that one. I'll illustrate this procedure. Here, C has the fewest moles, so we'll
find the numbers of F and of Cl relative to C. F's per C
F 2.35 C 1.17 Since measurements aren't always exact, we take 2.01 as meaning the whole number two. So there are
two F atoms per C atom in the molecule. = 2.01 Cl's per C 9 = ———1'20 = 1.03 C 1.17 We take 1.03 as meaning the whole number one. So there is one Cl atom per C atom in the molecule.
We now have our target ratio of atoms in the molecule. 1C:2F:1C Unfortunately, this is as far as we can go with percent composition. No, this is not yet the formula and
we still don't know what the compound is. Why? Because many formulas could have this ratio: CFZCI,
C2F4CIZ, C3F6CI3, C4F8CI4, etc. They all have this ratio. This is the limitation which I mentioned above:
percent composition will get you a ratio but it won't guarantee you a true formula. You need more
information. Fortunately, we can usually obtain more information by other methods. I'll show you one
such method, butI need to pause and make two points first. Our 1:2:1 example here was fairly easy to see, but not all are that easy. Some compounds have
screwy ratios. Let's say you calculate a ratio of 1.00: 1.50 for two elements. Remember the ratio still has
to end up as whole numbers, which means 2:3 in this case. Another example could be 1.00:1.67, which
corresponds to 3:5. Of course, there are many other ratios. Some can be tricky. Be careful. The second point to note involves a new term. Once you derive the smallest, wholenumber ratio
of the elements in the compound, it can be written in formula format. Our 1:2:1 ratio above could be
written as CFZCI. This formula format is called an "empirical formula". Unfortunately it looks like a regular
chemical formula and this leads to confusion. An empirical formula is the smallest, whole—number ratio
of elements in a compound. An empirical formula may or may not be the true chemical formula. The
possibilities I mentioned upstairs, CFZCI, C2F4C2, C3F6C3 and C4F8CI4, all have the same empirical formula,
CFZCI. This is quite general, although the possibilities may not always be real compounds. For another
example, C2H4, C3H6, C4H8, CsHm, etc., all have the same empirical formula (CH2), and all of these all real
but different compounds. A key point to introduce here is that a compound's true chemical formula is some whole—number
multiplier of its empirical formula. I can write this as follows. true chemical formula = multiplier x empirical formula By the way, that multiplier can be one; this means that the empirical formula is indeed the same as the
true chemical formula. For example, the chemical formula for water and the empirical formula for water
are the same, HZO. OK, back to where we were. I said there are experimental ways of connecting the element ratio
(empirical formula) to the true chemical formula. One of the easiest is molar mass, since molar masses
for many things can be measured. The molar mass from the real chemical formula is related to the mass
of the empirical formula by the same multiplier as above. molar mass = multiplier >< mass of empirical formula
I'll illustrate with the CH2 examples which I just mentioned. empirical formula [——————— possible molecular formulas ——
CH2 C2H4 C3Hs C4Hs CSHIO
multiplier: 2 3 4 5
molar masses: 14.03 g 28.05 g 42.08 g 56.10 g 70.13 g The bottom line is this: the formula multiplier is the same as the molar mass multiplier. If you can
measure the molar mass, then you can determine the multiplier. 56 Chapter 5: Chemical Units and Mass Returning to our CFZCI example, let's say we measured the molar mass to be 173 g. What is the true
chemical formula? The mass of the empirical formula, CFZCI, is 85.46 g. The molar mass, 173 g, is then
a multiple of this. 173g = multiplier x 85.469 The multiplier calculates to be 2.02, but it must be a whole number; we conclude this to be two. This
means that the true chemical formula is two times the empirical formula. The final answer is C2F4CIZ. Remember: percent composition can get you an empirical formula, but that may or may not be the
true chemical formula. Other information, such as molar mass, is required to connect the empirical
formula to the true chemical formula. Let's bring in another problem and work it out from scratch. Let's do a sugar. We talked about
sugars in Chapter 3, and I mentioned that they can have different isomers which have the same chemical
formula. We can't do isomers here, but we can solve for a chemical formula. Example 4. Elemental analysis of an unknown sugar gave percent compositions of 40.6% C and
6.56% H; the rest is O. The molar mass was measured to be 150. g. What is the true formula of this
sugar? What to do? We first need to determine the empirical formula from the percent compositions. Then,
we connect this empirical formula to the true formula using the given molar mass. For the empirical formula part, we have 40.6% C and 6.56% H as given directly. The percent
composition for O is not given directly, but we know that percent compositions add to 100.0%. 100.0% = 40.6% + 6.56% + ?°/o forO From this we find 52.8% for 0. Now, proceed just like the refrigerant problem earlier. It goes pretty
much the same. I'll leave blanks for you to fill in. The percent compositions are relative mass distributions. Based on a sample size of 100. 9, this
gives a mass ratio for the elements. 40.6gC : 6.56gH : 52.890
We convert the mass ratio to mole ratio, bringing in the molar masses of the elements. 40.6 g c _ 6.56 g H _ 52.8 g 0
12.01 g C/mol c ' 1.008 g H/mol H ' 16.00 g O/mol o Solve for each: 3.38 mol C : mol H : mol 0
By now, you may or may not see how these mole numbers relate to a whole number ratio. We'll work through the full process regardless. Ofthe three, which mole value is the minimum? (Clue: C is not
the minimum.) Set up the atom ratios, putting the minimum in the denominator. C's per 5
E = 3'38 = If you're confused by
what to fill in, look
back at the refrigerant
I I problem.
5 per 5
Now, what whole number ratios are represented by this outcome? C : H : 0 That provides your empirical formula. Write the empirical formula here. That was the hard part; now you're almost done. Relate your empirical formula to the true formula
by a multiplier. The multiplier comes from the formula mass relationships. Chapter 5: Chemical Units and Mass 57 molar mass = multiplier >< mass of empirical formula 1509 = multiplier x g Fill in the blank for the mass of the empirical formula. Then re—arrange the
equation and solve for the multiplier. Enter it at right. Now, multiply your multiplier times your empirical formula, and you
get the true formula. Write your final answer here for the true formula. You're done. Would you like to check your answer? Go look up ribose or xylose, which are two
isomers with this formula. You can find these in a reliable source online or you can even try a dictionary. Problems 1. True or false.
a. Avogadro's number is an exact, numerical value.
b. One gram equals one mol of atomic mass units.
c. The empirical formula of butane is CZHS.
cl. The empirical formula of disulfur dichloride is SCI. 2. What is the molecular weight or formula weight for each of the following?
a. Bi(OH)3 b. (NH4)2CO3 c. Asz(CH3)4 3. What is the molecular mass or formula mass for each of the following?
a. copper(II) permanganate b. dichlorine hexaoxide 4. What is the molar mass for each of the following?
a. CH4S b. Pt(NO3)2 c. HTeOF5 5. What is the molar mass for each of the following?
a. selenium tetrafluoride b. cobalt(II) perchlorate 6. Citric acid has the formula C6H807. What is the mass of 0.05043 mol of citric acid?
7. Fruit sugar (fructose) has the formula C6H1206. How many moles are present in 8.006 g of fructose? 8. What is the percent composition of sulfur in each of the following?
a. CH4SO3 b. tetraphosphorus trisulfide 9. A binary, covalent compound has the following percent compositions: 23.4% B and 76.6 0/o Cl.
a. What is the empirical formula?
b. If the molar mass is 185 g/mol, what is the molecular formula? 10. A covalent compound of H, Si and F is 4.105% H and 57.20% Si.
a. What is the empirical formula?
b. If the molar mass is 98.21 g/mol, what is the molecular formula? ...
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This note was uploaded on 01/12/2012 for the course CHEM 201 taught by Professor Hoyt during the Fall '07 term at University of Louisville.
 Fall '07
 Hoyt
 Chemistry

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