Lecture Notes

The mole is used so commonly in chemistry to measure the amount of an element or compound that it is sometimes elevated to the level of a concept. In fact, it is just a definition, in the same way that the term ream is defined as 500 sheets of paper or a dozen is defined as 12. Because the mole is part of the everyday language of chemistry we must become entirely familiar with its use. We begin with its origin in atomic weights.

Atomic Weights

In the 18th and 19th centuries, chemists began to explore weight relationships between the elements. During that period, combining weights were determined. (The term mass is preferred, but we will frequently use weight and mass interchangeably.) Let's pretend that we are working in a 19th-centruy lab and have just carried out a reaction between zinc and sulfur. Specifically, we mixed 65 grams of elemental zinc with 40 grams of elemental sulfur and heated them in a crucible for several hours. We then dissolved the excess sulfur in methylene chloride. The material that remained after the excess sulfur has been removed showed no sign of elemental zinc and weighed 97 grams.

Problem One
What is the weight of sulfur combined with zinc?

a) 40 g            b) 8 g            c) 32 g

Incorrect
Since sulfur remained after the reaction was finished, all of it could not have reacted with the zinc.

Incorrect
Go back and make sure that you are thinking about the final weight of the compound, and the fact that all of the zinc was consumed.

Correct
Because there was no zinc left at the end of the reaction, all 65 g of it must have been chemically consumed. The final material, which we know contained only zinc and sulfur weighed 97 g. Therefore, 97 - 65 = 32 g of the sulfur was consumed.

Thus, in this reaction the combining weights of zinc and sulfur are 65 g and 32 g. The significance of these combining weights became clear in the early 1800's when Dalton proposed that all matter consists of atoms and that reactions involve the combinations of these atoms in whole number ratios. Dalton also assumed that all atoms of the same element have the same mass (an assumption we now know is incorrect).

If we know that one atom of sulfur combines with one atom of zinc in forming the compound zinc sulfide, we can assume that if 32 g of sulfur react with 65 g of zinc, then an atom of zinc is 65/32 times heavier than an atom of sulfur. The following chemical equation represents this reaction:

Zn + S → ZnS

In this equation the symbols for the elements are used to represent the elemental forms, and the compound is written as ZnS to indicate that there is a one to one ratio of zinc and sulfur atoms. We can read the equation as "One atom of elemental zinc reacts or combines with one atom of elemental sulfur to form the compound zinc sulfide."

Problem Two
Suppose that it were true that one atom of zinc combines with two atoms of sulfur. Which of the following represents this reaction?

a) Zn + S → ZnS            b) Zn + S → ZnS2            c) Zn + 2 S → ZnS2

Incorrect
This equation represents the formation of a compound that contains one atom of sulfur for every atom of zinc.

Incorrect
The compound on the right side of the equation correctly expresses the one to two ratio of zinc and sulfur atoms. However, the equation is not balanced. There are two sulfur atoms on the right side of the equation, but only one sulfur on the left side. Chemical equations are just like mathematical equations; there must be an equality of atoms.

Correct
The compound on the right side of the equation correctly expresses the one to two ratio of zinc and sulfur atoms. It also tells us that the presence of two atoms of sulfur on the right side of the equation require two atoms of sulfur on the left.

Problem Three
Assuming that the same weights of each element--65 g zinc and 32 g sulfur--were consumed in this reaction of one atom of zinc with two atoms of sulfur, what would be the combining weight of sulfur?

a) 16 g            b) 64 g            c) 32 g

Correct
The 32 g of sulfur now represents two atoms of sulfur relative to one of zinc. Consequently, each sulfur atom would have to weigh half as much - 16 g.

Incorrect
If two atoms of sulfur combine with one of zinc, this answer would require that 128 g of sulfur would combine with 65 g of zinc.

Incorrect
Because the ratio of zinc to sulfur atoms is 1 to 2, the 32 g that have reacted now represent two sulfur atoms.

Atoms of different elements combine with sulfur (and other elements) in different ratios. For example, silicon combines with two sulfur atoms (the formula of silicon sulfide is SiS2); two atoms of sodium combine with one atom of sulfur (the formula of sodium sulfide is Na2S). Thus, it is important to determine the ratio of atoms in each compound. We will not pursue this here, but methods such as the use of the combining volumes of gases can provide this kind of information.

So, let us assume that eventually we put together a list of relative weights of atoms, such as the following:

Hydrogen 5
Carbon 60
Boron 50
Silver 535

Because hydrogen is the lightest element, let us assign it a weight of 1. Calculate the combining weights of the other elements relative to hydrogen. Your table should now look like this:

Hydrogen 1
Carbon 12
Boron 10
Silver 107

Problem Four
According to your new table, how much heavier is an atom of carbon than an atom of boron?

a) 12 times heavier            b) 12/10 times heavier            c) 10/12 times heavier

Incorrect
The table above tells us that carbon and hydrogen have relative atomic weights of 1 and 12. Thus, the average atom of carbon is 12 times heavier than the average atom of hydrogen.

Correct
The table above tells us that carbon and boron have relative atomic weights of 12 and 10. Thus, the average atom of carbon is heavier than the average atom of boron. The average carbon atom is 12/10 times heavier than the average atom of boron.

Incorrect
This answer would mean that an atom of carbon is lighter than an atom of boron. As you can see from the table that you constructed, carbon has the higher weight (12).

Problem Five

What mass of silver contains the same number of atoms as 12 g of carbon?

press for answer

107 g
The masses in your table represent the relative masses of the atoms of these elements. According to the table an average atom of silver is 107/12 = 8.9 times as massive as an average atom of carbon. Thus, in order to have the same number of atoms of silver we need a mass of carbon that is 107/12 times its mass. Since we have 12 g of carbon, we have 107/12 x 12 = 107 g silver.

Problem Six

What mass of silver contains the same number of atoms as 6 g of carbon?

press for answer

107/2 = 53.5 g silver
If you had trouble with this go back and redo the previous question. Then, try this one by proportion. That is, 6/12 = x/107.

Near the beginning of the 20th century, mass spectrometers, which allowed chemists to determine the mass of individual atoms, were produced. Several important observations followed the use of this instrument. First, it is was found that not all atoms of the same element have the same mass. (This is why we have used the phrase "average atom" above.) Atoms of the same element that have different masses are called isotopes. For example, there are two important isotopes of boron; both have 5 protons in the nucleusRemember that the nucleus of the atom contains protons, which are positively charged, and neutrons, which have no charge. These two nuclear particles are sometimes called nucleons and they have almost identical masses. The number of protons in the nucleus is called the atomic number of the element. This number distinquishes one element from another. For example, two elements, such as boron and nitrogen, could have the same number of neutrons or they could have the same number of electrons, but boron will always have 5 protons in the nucleus and nitrogen will always have 7 protons in the nucleus., but one has 5 neutrons in the nucleus while the other has 6 neutrons. The sum of the number of protons and neutrons in the nucleus is called the mass number of the isotope. This number is designated as a superscript before the symbolChemists use a type of shorthand to write and talk about the different elements. The element is represented by its chemical symbol, usually the first or first two letters in the element's name. For example, the symbol for fluorine is F, that for carbon is C. Sometimes the first two letters of the element's Latin name are used; such as Au for gold (from aurus) and Pb for lead (from plumbus). of the element. For example, the isotope of boron with 6 protons is designated as 11B.

The electron is the other particle in the atom that is of great interest to the chemist. The electrons are outside of the nucleus, they are negatively charged, and they have a much smaller mass than either the proton or the neutron. The charge on an electron is equal to, but opposite in sign, to the charge on a proton. Consequently, in a neutral atom the number of electrons is equal to the number of protons.

Periodic Chart

1   18
1
H
1.0079
2   13 14 15 16 17 2
He
4.0026
3
Li
6.941
4
Be
9.0122
  5
B
10.811
6
C
12.011
7
N
14.007
8
O
15.999
9
F
18.998
10
Ne
20.180
11
Na
22.990
12
Mg
24.305
3 4 5 6 7 8 9 10 11 12 13
Al
26.982
14
Si
28.086
15
P
30.974
16
S
32.065
17
Cl
35.453
18
Ar
39.948
19
K
39.098
20
Ca
40.078
21
Sc
44.956
22
Ti
47.867
23
V
50.942
24
Cr
51.996
25
Mn
54.938
26
Fe
55.845
27
Co
58.933
28
Ni
58.693
29
Cu
63.546
30
Zn
65.409
31
Ga
69.723
32
Ge
72.64
33
As
74.922
34
Se
78.96
35
Br
79.904
36
Kr
83.798
37
Rb
85.468
38
Sr
87.62
39
Y
88.906
40
Zr
91.224
41
Nb
92.906
42
Mo
95.94
43
Tc
(98)
44
Ru
101.07
45
Rh
102.91
46
Pd
106.42
47
Ag
107.87
48
Cd
112.41
49
In
114.82
50
Sn
118.71
51
Sb
121.76
52
Te
127.60
53
I
126.90
54
Xe
131.29
55
Cs
132.91
56
Ba
137.33
57 - 71 72
Hf
178.49
73
Ta
180.95
74
W
183.84
75
Re
186.21
76
Os
190.23
77
Ir
192.22
78
Pt
195.08
79
Au
196.97
80
Hg
200.59
81
Tl
204.38
82
Pb
207.2
83
Bi
208.98
84
Po
(209)
85
At
(210)
86
Rn
(222)
87
Fr
(223)
88
Ra
(226)
89 - 103 104
Rf
(261)
105
Db
(262)
106
Sg
(266)
107
Bh
(264)
108
Hs
(277)
109
Mt
(268)
110
Ds
(271)
111
Rg
(272)
 
 
  57
La
138.91
58
Ce
140.12
59
Pr
140.91
60
Nd
144.24
61
Pm
(145)
62
Sm
150.36
63
Eu
151.96
64
Gd
157.25
65
Tb
158.93
66
Dy
162.50
67
Ho
164.93
68
Er
167.26
69
Tm
168.93
70
Yb
173.04
71
Lu
174.97
  89
Ac
(227)
90
Th
232.04
91
Pa
231.04
92
U
238.03
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(262)

Problem Seven
Which of the following isotopes has 6 protons and 7 neutrons in the nucleus?

a) 7B            b) 13B            c) 12C            d) 13C

Incorrect
The superscript is the mass number, which is the total number of protons and neutrons in the nucleus, not just the number of neutrons.

Incorrect
The mass number is correct, but you have chosen the wrong element. Boron has only 5 protons in its nucleus.

Incorrect
You've got the right element, but the wrong mass number.

Correct

Problem Eight

Naturally occurring chlorine is made up of two isotopes, one with two more neutrons that the other. If the lighter isotopes has a mass number of 35, what is the correct symbol of the other isotope?

press for answer

37Cl
The symbol for chlorine is Cl and the correct isotope must have a number higher than 35 by 2.

Problem Nine

An isotope of some element has 70 neutrons in its nuclues and a mass number of 120. What is the element?

After the discovery of isotopes and a means for accurate measurement of the relative mass of these isotopes, it was decided that the mass of the 12C isotope should be defined as exactly 12.0000 atomic mass units (amu).

Problem Ten
How would you define one amu?

a) 12 g            b) one-twelfth of the mass of one 12C atom            c) 1 x 10-12 g

Incorrect
An amu must be much smaller than a gram, and 12 just happens to be the atomic mass of carbon as shown on the periodic chart.

Correct

Incorrect
Check out the question again. The definition doesn't even mention grams.

Protons and neutrons have very nearly the same mass. The mass of the electron is about 2000 times less than that of a proton or neutron.

Problem Eleven
What is the approximate mass of a proton (and a neutron) in amu units?

a) 12 amu            b) 6 amu            c) 1 amu

Incorrect
The mass of an entire 12C atom is 12 amu.

Incorrect
There are six protons, six neutrons, and six electrons in a 12C atom. The mass of a proton is almost the same as the mass of the neutron and therefore these 12 nucleons have a combined mass of 12 amu. Six amu is the approximate mass of 6 protons (or 6 neutrons).

Correct
There are six protons, six neutrons, and six electrons in a 12C atom. The weight of the electrons is so small relative to the protons and neutrons that it can be ignored. The mass of a proton is almost the same as the mass of the neutron and therefore these 12 nucleons have a combined mass of 12.00 amu. One proton (and one neutron) therefore has a mass of 1.00 amu.

Problem Twelve

What is the approximate mass of one 13C atom?

press for answer

13.00 amu

Problem Thirteen

What is the approximate mass of two atoms of 29Si?

press for answer

58.0 amu
Because each nucleon has a mass of 1.00 amu (and we can ignore the mass of the electrons if we confine our answer to 2 or 3 significant figures), one 29Si atom has a mass of 29.0 amu. Two atoms will be 2 x 29.0 = 58.0 amu.

Problem Fourteen

Suppose that it was decided to design a new system of atomic weights in which a 12C atom no longer weighed 12 amu, but rather the mass of a 12C atom would be defined as 1.00 amu. In this system, what would be the mass of a 16O atom?

press for answer

16/12 = 1.33 amu

Problem Fifteen
Naturally-occuring carbon is 99% 12C and 1% 13C. If you had 99 12C atoms and one 13C atom, what would be the mass of the 99 12C atoms and the one 13C atom?

a) 1188 amu and 13 amu            b) 99 amu and 1 amu            c) 1000 amu and 10 amu

Correct
The 99 12C atoms would have a mass of 99 x 12.00 = 1188 amu. The mass of the one 13C atom would be 1 x 13 = 13 amu.

Incorrect
There are 99 12C atoms and 1 13C atom, but each of these weighs more than one amu.

Incorrect
Who made up this answer?

Problem Sixteen

Now, determine the averageAn average, also called a mean, is calculated by dividing the sum of all of the data points in a set of values by the number of values. Suppose you had the following set of numbers that represent the length of time, in seconds, that you can hold your breath under water: 24, 29, 26, 28, and 28. The sum of these 5 numbers is 135, which we divide by 5 to get 27 s. This number is a better representation of the amount of time you are able to hold your breath than any of the individual numbers. mass of these one hundred atoms.

press for answer

12.01 amu
The total mass of the one hundred atoms is 1188 + 13 = 1201 amu. The average mass is 1201 amu/100 atoms = 12.01 amu per atom.

Sometimes, we may want to take a weighted average. Actually, all averages are weighted, but for an ordinary average every value in the set has the same weighting. For example, in the set above, each value counts 20% (one-fifth) toward the average. Indeed, we could determine the average by doing the calculation as follows:

0.20 x 24 + 0.20 x 29 + 0.20 x 26 + 0.20 x 28 + 0.20 x 28 =
4.8 + 5.8 + 5.2 + 5.6 + 5.6 = 27 s

Suppose that while you were measuring the times, you felt that the 24 second value was not a particularly good value. Perhaps this value was obtained while your brother did a cannonball in front of you. Likewise, one of the 28 second values was obtained when your chemistry teacher was walking by and probably was not a really representative value (you were trying to stay under water as long as possible). Because you don't want to exclude these values entirely, you decide to weight them differently. Perhaps you decide to weight the 24 second value only 5% (0.05) and one of the 28 second values you weight only 10% (0.10). These two points therefore contribute 15% (0.15) to the average. You want the weighting of the other values to be equal to one another so each of the other 3 must have a weighting of (100-15)/3 = 28.3%. Now your calculation will look like this:

0.05 x 24 + 0.283 x 29 + 0.283 x 26 + 0.15 x 28 + 0.283 x 28 =
1.2 + 8.2 + 7.4 + 4.2 + 7.9 = 28.9 s

Certainly, the different weightings have affected the mean.

A glance at a periodic chart will show you that the atomic mass of carbon is given as 12.01. Notice also, that there are no units on the atomic masses given on the chart. You now realize that every atomic mass shown on the periodic chart is an average for a large number of isotopes.

Problem Seventeen
Let's try the calculation of the atomic mass of argon for which there are three naturally occuring isotopes: 36Ar, 38Ar, and 40Ar. The percentages of these isotopes are: 0.34%, 0.07%, and 99.58%, respectively. Calculate the atomic mass of argon. Hint: a) You can choose the correct answer without having to use your calculator, and b) This is where you use the weighted average.

a) 33.33            b) 38.00            c) 39.98

Incorrect
This is the average of the percentages: (0.34 + 0.07 + 99.58)/3

Incorrect
This is the average of the masses of the three isotopes if they were weighted equally; that is, if there were the same number of atoms of each in naturally-occurring argon.

Correct
You can see from the low percentages of the isotopes with mass numbers of 36 and 38, that the mass must be very close to 40. In the calculation, the isotopes must be weighted according to their percentages (in decimal equivalents):
0.0034 x 36.00 + 0.0007 x 38.00 + 0.9958 x 40.00 = 39.98

Problem Eighteen

Chlorine exists as two isotopes: 35Cl with a mass of 34.97 and 37Cl with a mass of 36.95. The atomic weight of chlorine is 35.453. What is the percentage of each of the isotopes?

press for answer

75.6% 35Cl and 24.4% 37Cl.
In this problem you know the atomic weight, which is a weighted average of the masses of the two isotopes, but you do not know the percentages of each isotope. Because there are two isotopes, it might seem that there are two unknowns. In fact, the percentages of the two must add up to 100% and, therefore, if we let the percentage of 35Cl be x, the percentage of 37Cl will be 100 - x. After we convert the percentages to their decimal equivalents we have the equation:
34.97x + 36.95(1 - x) = 35.453
x = 75.6%
Notice that this problem illustrates the fact that we can not quite assume that the mass of an isotope can be found by multiplying the number of nucleons by 1.00. This slight discrepancy is due to several factors, including the fact that the masses of the proton and neutron are not exactly equal to 1.00000.

We now have two different interpretations of the atomic masses: 1) they represent the mass of the average atom of an element in amu's, given the convention that the mass of 12C is exactly 12 amu, and 2) they represent the relative combining masses of the elements.

Perhaps we should expand on the second interpretation. If you wanted to make the compound carbon dioxide from carbon and oxygen you first need to know that the formula of this compound is CO2. The subscripts after the symbol for each element give the ratio of atoms present in the compound. In the case of CO2, the subscript 2 for oxygen tells us that two oxygen atoms are present for every one atom of carbon (the absence of a subscript for carbon is equivalent to a subscript of 1). Next, you must find the atomic masses of carbon and oxygen--12 and 16, respectively. If the formula of the compound were CO, then you would merely need to combine 12 g of carbon with 16 g of oxygen in order to get the same number of atoms of each element. Or, you could combine 1.2 g of carbon with 1.6 g of oxygen, or you could combine 1.0 pound of carbon with 16/12 = 1.33 pound of oxygen, or 3.5 tons of carbon with 3.5 x 16/12 = 4.7 tons of oxygen. But, the formula of carbon dioxide is CO2, which tells us that we need twice as many oxygen atoms as carbon atoms. Therefore, if we want to use 12 g of carbon we must use 2 x 16 = 32 g of oxygen. If we want to start with 4 g of carbon then we must use 4 x (16/12) x 2 = 10.7 g of oxygen. Notice that the ratio of the atomic masses and the number of atoms of each element always play a role in the calculation.

Problem Nineteen
If you happen to have one pound of carbon, how many pounds of oxygen will be required to make CO2?

a) 2 lbs            b) 2.7 lbs            c) 1 lb

Incorrect
Carbon and oxygen react in a 1 to 2 atom ratio, not weight ratio.

Correct
1 lb x (2 x 16.00)/12 = 2.7

Incorrect
Reread the paragraph above.

Problem Twenty
If the compound SO2 contains 32 g of sulfur, how many grams of oxygen does it contain?

a) 16 g            b) 32 g            c) 64 g

Incorrect
The compound SO2 contains two oxygen atoms for every sulfur atom. The atomic weight of sulfur is 32, whereas the atomic weight of oxygen is 16. If the compound were SO, 16 grams of oxygen would be present with 32 grams of oxygen. But, the formula is SO2, and therefore ......?

Correct
The compound SO2 contains two oxygen atoms for every sulfur atom. The atomic weight of sulfur is 32, whereas the atomic weight of oxygen is 16. Thus, if the compound contains 32 g sulfur, it must contain 2 x 16 = 32 g of oxygen.

Incorrect
This answer would be correct if the question were how many grams of sulfur are present with 16 g of oxygen in the compound S2O.

Problem Twenty One

What is the ratio of the number of atoms in 16 ounces of sulfur to the number in 1 ounce of hydrogen?

press for answer

0.5 to 1
The atomic weight of sulfur is 32 and the atomic weight of hydrogen is 1.0 (we are using only two signifcant figures). A sample of 32 ounces of sulur would therefore contain the same number of atoms as 1 ounce of hydrogen. Because we only have 16 ounces, the ratio is 16/32 to 1 or 0.5 to 1.

Problem Twenty Two

If the compound NH3 contains 6 g hydrogen how many grams of nitrogen are also present?

press for answer

28 g
The ratio of atomic weights (nitrogen to hydrogen) is 14 to 1. The compound contains three atoms of hydrogen for every atom of nitrogen. The 6 grams of hydrogen represents three atoms of hydrogen. Thus, each hydrogen contributes 2 grams and the nitrogen must contribute 14 x 2 = 28 g.

Problem Twenty Three

How many grams of hydrogen contain the same number of atoms as 12.01 grams of carbon?

press for answer

1.008 g
The average hydrogen atom weighs 1.008/12.01 times as much as the average atom of carbon. Thus, if we have 1.008 g of hydrogen we will have as many atoms as are present in 12.01 carbon.

It should now be clear that the atomic weights in the periodic chart are numbers that can be used to determine the ratio of the number of atoms in various amounts of the elements. The weights (masses) represent the average mass of an atom of the element in atomic mass units. So far, we have no idea of the value of an atomic mass unit in a unit of mass, such as grams, with which we are familiar.

The Mole

Chemists usually weigh elements and compounds in grams, and it is, therefore, of particular interest to know the actual number of atoms in an amount of an element equal to its atomic mass in grams. In other words, how many atoms are present in 12.01 grams of carbon, in 1.008 grams of hydrogen, or in 32.04 grams of sulfur? We know, of course, that the number of atoms in each of these quantities is the same. But what is the number? This number has been determined experimentally in a number of different ways and turns out to be an extremely large number-- 6.022 x 1023. It is known as Avogadro's number, in honor of the Italian chemist.

Problem Twenty Four
Which of the following statements expresses the significance of the number 6.02 x 1023?

a) There are 6.022 x 1023 amu in a gram            b) There are 6.022 x 1023 atoms in a gram            c) There are 6.022 x 1023 atoms in a mass of an element equal to its atomic mass in grams.

Incorrect
Avogadro's number has nothing to do with the number of amu's in a gram.

Incorrect
This would mean that there are the same number of atoms in 1 g of any element. We already know that atoms of different elements have different masses.

Correct

Problem Twenty Five

How many atoms are present in 24 grams of carbon?

press for answer

1.2 x 1024 atoms
There are 6.0 x 1023 atoms in 12 g of carbon. In 24 grams there are 24 g / 12 g x 6.0 x 1023 atoms = 1.2 x 1024 atoms. [Notice that we will frequently express Avogadro's number to only 1 or 2 significant figures.] You could also solve this by proportion. If there are 6.0 x 1023 atoms in 12 g carbon, how many are present in 24 g? Mathematically, this is expressed:
6.0 x 1023 atoms/ 12 g = x / 24 g
x = 1.2 x 1024 atoms

Avogadros's number, 6.022 x 1023, is also called a mole, in the same way that 500 sheets of paper is called a ream, or 144 is referred to as a gross.

Problem Twenty Six

How many ping-pong balls are there in a mole of ping-pong balls and what would their molar mass be if one ping-pong ball weighs 1.0 g?

press for answer

There are 6.02 x 1023 ping-pong balls in a mole of ping-pong balls. If each ball weighs 1.0 grams, the molar mass will be 6.02 x 1023 g. We would normally express this as 6.02 x 1023 g/mole.

Are you saying that we can have a mole of anything? Sure, can't you have a dozen pencils, a dozen ducks, a dozen bricks? However, we normally use the term mole in dealing with atomic particles--atoms, ions, molecules.

Problem Twenty Seven
How many atoms are present in 31 g of phosphorus?

a) 1 mole            b) 6.022 x 1023            c) Avogardro's number

All three answers are correct.

Let's try some questions that compare terms like ream or gross or dozen with mole.

Problem Twenty Eight

If one ream of paper weighs 1.0 pounds, how many reams are there in 300 pounds of paper?

press for answer

There are 300 reams.
300 lb/1.0 lb per ream = 300 reams

Problem Twenty Nine

How many moles of carbon are there in 500 g of carbon?

press for answer

41.6 mole
Unlike the previous question, we are not told how much a mole of carbon weighs. However, we know that in one gram-atomic weight of an element there is one mole of atoms. (One gram-atomic weight is the atomic mass of an element in grams. It is also called the molar mass.)

Problem Thirty

As the head chick in the hen-house, you have responsibility for 100 eggs, which weigh a total of 1.00 x 104 g. How many dozen eggs do you have under your wings and how much does a dozen weigh?

press for answer

8.33 dozen, and a dozen weighs 1.20 x 103 g

Problem Thirty One

Your Zork particle counter has just informed you that there are 3 x 1020 atoms of radon in your basement. How many moles of radon is this?

press for answer

3 x 1020 particles/ 6 x 1023 particles per mole) = 5 x 10-4 mole.
In case you were tempted to divide 6 x 1023 by 3 x 1020, remember that 1020 is a smaller number than 1023 and therefore the number of particles must be less than one mole. That means that you must divide by the larger number.

Problem Thirty Two

In order to do well in your favorite course (well, chemistry, what else?) you have ordered 20 gross of pencils. How many pencils will you get?

press for answer

2880 pencils
A gross is 144, therefore 20 gross is 144 x 20 = 2880.

Problem Thirty Three

Your lab partner is staring out the window as usual when he suddenly exclaims, "Gadzooks! There are 1 x 10-23 moles of dogs running down the street!" Could he be right for a change?

press for answer

Yes, 1 x 10-23 moles is:
1 x 10-23 moles x 6 x 1023 dogs per mole = 6 dogs.
If you like to solve problems with dimensional analysis you would probably set this up as:
1 x 10-23 moles x 6 x 1023 dogs / 1 mole = 6 dogs.

Problem Thirty Four
How many moles of atoms are there in 24 grams of carbon?

a) 12 x 1023            b) 2            c) 24 x 6 x 1023

Incorrect
A mole is 6.02 x 1023 and in 12 grams of carbon there is one mole 6.02 x 1023 of atoms.

Correct
One mole of atoms is present in the atomic mass of the element in grams. The number of moles in 24 g of carbon is: 24 g /12 g per mole = 2 mole.

Incorrect
See answer to (b)

Problem Thirty Five
How many grams of calcium must be weighed out in order to obtain a sample containing 3 moles of atoms?

a) 40 g            b) 3 x 6 x 1023 g            c) 120 g

Incorrect
One mole of calcium - 6 x 1023 atoms - is contained in the molar mass of calcium (40 g). Thus, one mole weighs 40 g. Three moles weigh 3 x 40 g = 120 g.

Incorrect
You do not have the relationship between moles and Avogadro's number straight. One mole of calcium is 6 x 1023 atoms and has a mass of 40 g (its molar mass).

Correct
The atomic mass of calcium is 40, and, therefore, a mole of calcium is 40 grams. Three moles of calcium is 3 mole x 40 g/mole = 120 g.

Problem Thirty Six

What is the weight of an average atom of calcium?

press for answer

6.7 x 10-23 g
The weight of 6.0 x 1023 atoms of calcium is 40 grams. Therefore, the weight of one average atom of calcium is:
40 g per mole / 6.0 x 1023 atoms per mole = 6.7 x 10-23 g per atom,
or, if you prefer:
(40 g / 1 mole) x (1 mole / 6.0 x 1023 atoms) = 6.7 x 10-23 g/atom

Now that we know how to determine the mass of an atom in grams, we can figure out the relationship between amu's and grams.

Problem Thirty Seven

First, calculate the mass of a 12C atom in grams.

press for answer

1.99 x 10-23 g
A mole of 12C has a mass of 12.00 g. There are 6.022 x 1023 atoms in one mole, and therefore an atom of 12C weighs:
12.00 g / 6.022 x 1023 atoms = 1.99 x 10-23 g per atom

Problem Thirty Eight

An amu is defined as 1/12 of the mass of a 12C atom. Determine the mass of one amu in grams.

press for answer

1.66 x 10-24 g = 1 amu
One-twelfth of the mass of a 12C atom is:
1.99 x 10-23 g / 12 = 1.66 x 10-24 g

The Mole and Compounds

We have already found that the formula of a compound represents the ratio of the atoms of the various elements in the compound. For example, magnesium chloride has the formula MgCl2, which reveals that the compound contains two chlorine atoms for every magnesium atom.

Problem Thirty Nine
In the compound Fe2O3, how many atoms are present for every two iron atoms?

a) 1.5            b) 2            c) 3

Incorrect
The ratio of atoms is 2 (Fe) to 3 (O), or, 1 to 1.5. But the question asks how many oxygen atoms are present for every two iron atoms.

Incorrect
If this were the case, the formula would be Fe2O2.

Correct

The subscripts in formulas can also be interpreted in terms of moles. If there are two atoms of chlorine for every atom of magnesium in MgCl2, then there must also be two moles of chlorine atoms for every one mole of magnesium. Or, if we prefer, we can say that for every 12 x 1023 atoms of chlorine there are 6 x 1023 atoms of magnesium.

Problem Forty

The formula for barium hydroxide is Ba(OH)2. The parentheses indicate that for every barium ion there are two hydroxide ions (OH- is the hydroxide ion). How many moles of oxygen are present with one mole of barium?

press for answer

Two
The parentheses can be removed and the formula rewritten as BaO2H2, which should make the mole-mole relationship clear.

We can now calculate the molar mass (sometimes also called the formula weight) of a compound. This is the mass of one mole of the compound and can be determined by simply adding the atomic masses of each of the elements (taking into account the number of each) in the compound. For barium hydroxide, we need to add the atomic weight of barium, two times the atomic weight of oxgyen, and two times the atomic weight of hydrogen. So, this would be 137 + (2 x 16) + (2 x 1) = 171 g. Thus, one mole of Ba(OH)2 has a mass of 171 g.

Problem Forty One
How much barium hydroxide must be weighed out in order to obtain 0.10 mole?

a) 17 g            b) 34 g            c) 171 g

Correct
If one mole of barium hydroxide is 171 g, then a tenth of a mole must be:
0.1 mole x 171 g/mole = 17 g, or if you prefer to do this by proportion:
1 mole/171 g = 0.1 mole/x g, x = 17 g

Incorrect
See the answer to (a)

Problem Forty Two

A student weighs out a sample of 20 g of barium hydroxide. How many moles of barium hydroxide is this?

press for answer

0.12 mole
20 g/ 171 g per mole = 0.12 mole, or by proportion:
1 mole/171 g = x mole/20 g, x = 0.12 mole

Problem Forty Three

Calculate the molar mass of Cu2O.

press for answer

143 g
2 x 63.5 + 16 = 143 g

Problem Forty Four

A 5 x 10-3 mole sample of Cu2O is required. What weight is required?

press for answer

5 x 10-3 mole x 143 g/mole = 0.7 g

In addition to the formulas that we have used above, which are called empirical formulas, there are several formulas that are used for molecular compounds. Before we define the molecular formula let's be sure we are clear about the meaning of the empirical formula. The empirical formula gives the simplest whole number ratio of the number of atoms (moles) of the each of the elements in the compound. The empirical formula is the only formula used for ionic compounds - compounds that contain charged atoms (ions). There are also compounds that do not contain ions. The atoms in these compounds are held together by covalent bonds. Many of these compounds contain molecules, the building blocks of a molecular compound. Benzene, an interesting molecular compound, contains molecules that have six carbon atoms attached to one another in a hexagonal ring. Each carbon has one hydrogen attached to it. The structural formula (the name tells you that the formula gives the structure of the molecule) is shown below:

benzene

[Note: this is not an electron-dot formula (yes, yet another kind of formula), which you will learn about in your discussion of bonding. This structural formula simply shows which atoms are connected.] As you can see, each molecule of benzene contains six carbons and six hydrogens. The molecular formula of benzene therefore is written C6H6.

The molecular weight or molar mass of benzene is entirely analogous to the formula weight, which is derived from the empirical formula.

Problem Forty Five

Calculate the molecular weight of benzene.

press for answer

78 g
6 x 12 + 6 x 1 = 78 g

Problem Forty Six
A 78 g sample of benzene contains how many moles of benzene molecules?

a) 1 mole            b) 6 moles            c) 6 x 1023

Correct
One mole of benzene has a mass of 78 g.

Incorrect
You may be thinking about the number of moles of carbon atoms in 78 g.

Incorrect
You are still having problems with the difference between a mole and the number of objects that it represents (just like the difference between a dozen and 12).

It is important to understand that the empirical formula of benzene is CH, which indicates the one to one mole ratio of carbon to hydrogen. The molecular formula, C6H6, gives the number of each atom in one molecule of the compound.

We sometimes need to convert a formula, be it empirical or molecular, to the weight percentageThe use of percentages is very common in science. The word comes from the Latin, per centum, meaning by the hundreds, and means per hundred parts. For example, if we have a class that is 40% women, we know that out of every hundred students, 40 of them are women. If a sample is 40% carbon then for every hundred parts (grams, pounds, tons, whatever), 40 of them are carbon. In order to determine the amount of an element from its percentage, you must first convert the percentage to its decimal equivalent (40% = 0.40) and then multiply that number by the weight. The amount of carbon in 50 grams of benzene (92% carbon) is 0.92 x 50 g = 46 g. of the elements. For example, benzene is 92% carbon and 8% hydrogen. In order to determine the percentage from the formula we first obtain the molar mass from the empirical formula. For benzene, the molar mass is 13 g. The percent of carbon in that 13 g is:

12 g/13 g x 100 = 92 %

Problem Forty Seven

Determine the percentage of oxygen in acetic acid, which has the empirical formula CH2O.

press for answer

The molar mass is 30 g/mole. There are 16 g of oxygen in this 30 g, and, therefore, the percentage of oxygen in that 30 g is:
16 g/30 g x 100 = 53%

Calculation of Empirical Formula from Percent Composition

The conversion of experimental composition data to empirical formula is an important step in determining the identity of a substance. Suppose that we have a sample of a heavy pale yellow crystalline material that has the experimentally determined composition: 59.0% Ba, 13.5% S, and 27.4% O. We need to convert this percent composition data to an empirical formula, which will tell us the relative number of moles of barium, sulfur and oxygen in the substance. The key to this conversion lies in the understanding of the phrase "relative number of moles of barium, sulfur and oxygen in the substance." If we want the number of moles of each we must know the mass of each element in some amount of the substance and then we must convert each of thoses masses to moles.

In fact, we can choose any amount of the substance and convert it to the mass of each of the elements. If we choose 100 grams of the substance it makes the problem a little easier because we can do the first part of the math in our heads.

Problem Forty Eight

For example, if we have a 100 g sample of the substance and it contains 59.0% barium, what mass of barium is present?

press for answer

59.0 g
100 g x 0.590 = 59.0 g

Likewise, in this same sample we have 13.5 g sulfur and 27.4 g oxygen. In order to convert these masses to moles we must divide each by their atomic mass.

Ba 59.0 g/ 137.33 g/mol = 0.430 mol
S 13.5 g/ 32.07 g/mol = 0.421 mol
O 27.4 g/16.00 g/mol = 1.71 mol

We could at this point write the empirical formula as Ba0.430S0.421O1.71, but we know that the empirical formula should be expressed in whole numbers. Consequently, we divide each of the subscripts by the smallest one (0.421), which gives us subscripts of 1.02, 1.00, and 4.06. These still are not whole numbers, but because these numbers have been derived from experimental data, which contain errors, we can not expect the numbers to come out to exact whole numbers. The numbers appear close enough to 1, 1, and 4 to allow us to safely round them off. Hence, the empirical formula of the substance is BaSO4.

Problem Forty Nine

Determine the empirical formula of a compound that contains 10.0% C, 0.84% H, and 89.1% Cl.

press for answer

CHCl3
Assuming a 100 g sample we have 10.0 g C, 0.84 g H, and 89.1 g Cl. The number of moles of each of the elements is:
10.0 g C/ 12.01 g/mole = 0.833 mole C
0.84 g H/ 1.008 g/mole = 0.833 mole H
89.1 g Cl/ 35.45 g/mole = 2.51 mole Cl
If we divide each of the mole amounts by 0.833 we get 1, 1, and 3.02. The 3.02 can be safely rounded to 3 and the formula therefore is CHCl3.

In some cases the numbers will not come close to whole numbers and we must then multiply each by some number that will produce a set of whole numbers. The following is such a case.

Problem Fifty

Determine the empirical formula of a compound with percent composition data of Fe, 69.9%, O 30.1%.

press for answer

Fe2O3
Assuming a 100 g sample, we will have 69.9 g Fe, and 30.1 g O. The number of moles of each is:
Fe 69.9 g/ 55.85 g/mol = 1.25 mole Fe
O 30.1 g/16.0 g/mol = 1.875 mole O
When we divide both numbers by the smaller (1.25), we get 1.00 and 1.50. Certainly, 1.50 is nowhere close to a whole number and so we must multiply it by the smallest possible number to bring it to a whole number. Obviously, if we multiply all of the mole amounts by 2, we will get Fe2O3.

Problem Fifty One

Determine the empirical formula of a compound with the percent composition of 50.04% C, 5.59% H, and 44.37% O.

press for answer

C3H4O2
Using the procedure above, we get 4.167 mole C, 5.546 mole H, and 2.773 mole O. Dividing by the smallest number gives 1.503 mole C, 2.000 mole H, and 1.000 mole O. Because the number of moles of carbon is not a whole number we must multiply each number by some number (the same number) to make each a whole number. Multiplication by 2 will give the formula C3H4O2.

The compound in the drill question has an empirical formula of C3H4O2. If this were also the molecular formula the compound would have a molecular weight of 3 x 12 + 4 x 1 + 2 x 16 = 72 g/mole. If, in fact, the compound has a molecular weight of 144 g/mole, then the molecular formula must be some multiple of the empirical formula; that is, the subscripts must be multiplied by some whole number. The number, in this case, is 2, and the molecular formula is C6H8O4.

Problem Fifty Two

How many moles of the molecular compound C6H8O4 are there in a 10 g sample of the compound?

press for answer

0.069 moles. 10 g /144 g/mole = 0.069 moles

Problem Fifty Three

How many molecules are there in this sample?

press for answer

4.2 x 1022 molecules
(6.02 x 1023 molecules / 1 mole) x 0.069 moles = 4.2 x 1022 molecules

Problem Fifty Four

How many moles of carbon are present in this sample?

press for answer

0.41 moles carbon
There are six moles of carbon for every mole of the sample and there are 0.069 moles of the sample. Thus, (6 moles C/1 mole C6H8O4) x 0.069 moles C6H8O4 = 0.41 moles carbon

Problem Fifty Five

The empiricial formula of chloroform is CHCl3 and its molar mass is 119. What is the molecular formula of chloroform?

Problem Fifty Six

What is the weight of one molecule of chloroform?

press for answer

1.98 x 10-22 g/molecule
119 g/1 mole x (1 mole/ 6.02 x 1023 molecules) =
1.98 x 10-22 g/molecule

Problem Fifty Seven

How many moles of chloroform are present in 1 mg of the sample [mg = milligram = 10-3 g]?

press for answer

8 x 10-6 mole
1 x 10-3 g /119 g/mole = 8 x 10-6 mole

Problem Fifty Eight

How many moles of chlorine are present in this 1 mg sample of CHCl3?

press for answer

2 x 10-5 mole chlorine
Each mole of chloroform contains 3 moles of chlorine. Thus:
8 x 10-6 mole x 3 = 2 x 10-5 mole chlorine

Problem Fifty Nine

What weight of chlorine is present in this 1 mg sample?

press for answer

7 x 10-4 g Cl
2 x 10-5 mole chlorine x 35.45 g/mole = 7 x 10-4 g Cl

Problem Sixty

What mass of chloroform is required to provide 1 x 1025 atoms of hydrogen?

press for answer

2 x 103 g CHCl3
1 x 1025 atoms H x (1 mole H / 6 x 1023 atoms H) x
(1 mole CHCl3 / 1 mole H) x (119 g CHCl3/ 1 mole CHCl3 ) =
2 x 103 g CHCl3

Moles in Solution

Solutions are homogeneous mixtures of two or more substances. We will restrict our discussion to solutions that contain only two substances: a solute, the substance present in the smallest amount, and the solvent, the substance present in the greatest quantity. One of the important characteristics of a solution is its concentration: the relative amount of solute in the solution. There are a number of ways to express concentration--weight percent, molarity, molality, and mole fraction--but we will concentrate (no pun intended) on molarity.

Molarity is defined as the number of moles of solute per liter of solution. If we dissolve 1.0 mole of NaCl in enough water to make a total of one liter of solution, we will have a 1.0 molar solution of NaCl in water. Molarity is usually designated with a capital M.

Problem Sixty One
Which of the following describes how you would make up a 1.0 M NaCl solution in water?

Incorrect
If you add 58 g NaCl (1.0 mole) to a liter of water you will not have a total of one liter of solution. The presence of the NaCl affects the volume of solution. Thus, you must add just enough water to reach a total of 1.0 liter after the solution has been thoroughly mixed. The vessel that you choose for this depends upon how many significant figures you want in your molarity. If you want only one, a beaker with a 1.0 liter mark is good enough; if you want 2 or 3 significant figures, a graduated cylinder should be used; if you want 4 significant figures you must use a volumetric flask.

Correct
The vessel that you choose for this depends upon how many significant figures you want in your molarity. If you want only one, a beaker with a 1.0 liter mark is good enough; if you want 2 or 3 significant figures, a graduated cylinder should be used; if you want 4 significant figures you must use a volumetric flask. See also the answer to (c).

Correct
Notice that the dissolution of 0.1 mole of NaCl in a total of 100 mL (0.1 L) of solution will provide a 1.0 M solution, as will the dissolution of 0.025 mole in 25 mL (0.025 L), and so on. The vessel that you choose for this depends upon how many significant figures you want in your molarity. If you want only one, a beaker with a 1.0 liter mark is good enough; if you want 2 or 3 significant figures, a graduated cylinder should be used; if you want 4 significant figures you must use a volumetric flask. See also the answer to (b).

Problem Sixty Two
If 50 mL of a solution contains 5.8 g NaCl, what is the molarity of the solution?

a) 0.10 M            b) 2.0 M            c) 0.050 M

Incorrect
The solution contains 0.10 mole of NaCl. If the molarity were 0.10 M the total volume of the solution would have to be 1.0 L.

Correct
The 0.10 mole of solute is present in 0.050 L of solution, thereby making the molarity:
0.10 mole/0.050 L = 2.0 M

Incorrect
Reread and write down the definition of molarity.

Problem Sixty Three

George prepares one liter of a 0.05 M NaCl solution. Jack, a good friend of Mary, came to lab late and needs 50 mL of a 0.05 M NaCl solution for a titration. Mary suggests that Jack ask George for 50 mL of his solution. Jack hesitates to use George's solution because he says that if he takes 50 mL he will not have a 0.05 M solution. He will have 0.05 mole of NaCl in 0.050 L, says Jack, which will make the solution a 1 M solution. What should Mary say?

press for answer

The 50 mL of George's solution will contain 0.05 x 0.05 moles of NaCl, or 0.0025 moles of NaCl. Thus, the concentration will still be 0.05 M:
0.0025 mole/0.05 L = 0.05 M

The previous question emphasizes the fact that no matter how much of a solution you have, the concentration remains the same. Let's go through Mary's reasoning (answer b) again with a different amount of the 0.05 M solution. If we start with 1 liter of the solution and take from it 100 mL, we have taken a tenth of the solution. That tenth will contain a tenth of the original amount of solute (0.05 moles); that is, the 100 mL will contain 0.005 moles of solute. But, that 0.005 moles will be present in a total of 100 ml (0.1 L). Thus, the molarity, which is just moles per liter, will be:

0.005 moles/0.1 L = 0.05 M

Problem Sixty Four

How many moles of solute will be present in 310 mL of the 0.05 M NaCl solution?

press for answer

0.0155 moles
Here's a couple of different ways to solve this one:
1. By proportion:
0.05 mole/1.0 L = x mole/0.310 L x = 0.0155
2. By thinking about the fraction of 1.0 liter in which you know there would be 0.050 mole NaCl. 310 mL is 0.310/1 = 0.31 of a liter, so there should be only 31% of the 0.05 moles:
0.31 x 0.05 moles = 0.0155
3. By recognizing that moles/volume = M, and, therefore, moles = volume x M:
moles = 0.310 L x 0.05 M = 0.0155 moles

Problem Sixty Five

What volume of a 0.05 M NaCl solution will be required to obtain 2 x 10-3 moles of NaCl?

press for answer

0.05 mole/ 1 liter = 0.002 mole x liter x = 0.040 L, or 40 mL

Problem Sixty Six

How many moles of solute will be present in 310 mL of the 0.05 M NaCl solution?

press for answer

20 mL
First, we must calculate the number of moles of NaCl in 50 mL of a 0.02 M solution. This is 0.05 L x 0.02 moles/L = 0.001 moles NaCl. Thus, we need enough 0.05 M NaCl solution to provide 0.001 moles of NaCl. This can be obtained by determining the proportion of 1 liter needed:
0.001 mole/0.05 mole/L = 0.02 L or 20 mL
We could also have solved the last part by proportion; 0.05 mole is to one liter as 0.001 mole is to x liters.
0.31 x 0.05 moles = 0.0155
0.05 mole/1 L = (0.001 mole/ x L) x = 0.02 L
Or, we could have solved the problem by recognizing that a 0.02 M solution is two-fifths as concentrated as a 0.05 M solution. Thus, if we want 50 mL of the 0.02 M solution, we use two-fifth of this volume - 2/5 x 50 = 20 - and then dilute to a total of 50 mL.
In any event, we must measure out 20 mL of 0.05 M NaCl solution and then add to it, with mixing, enough water to make a total of 50 mL of solution. The resulting solution will have a molarity of 0.02 M.

Problem Sixty Seven

How many moles of acetic acid are there in 100 mL of a 0.60 M solution?

press for answer

0.060 moles 0.10 L x 0.60 moles/L = 0.060 moles

Problem Sixty Eight

If 10 mL of a 0.60 M acetic acid solution is diluted to 55 mL, what is the molarity of the resulting solution?

press for answer

0.11 M
The 10 mL of 0.60 M solution contains:
0.01 L x 0.60 mole/L = 0.0060 moles
This 0.006 moles, after dilution, will be in 0.055 L of solution. Therefore, the molarity will be:
0.0060 moles/ 0.055 L = 0.11 M

Problem Sixty Nine

How many mL of a 12 M stock solution of HCl must be used to prepare 100 mL of a 0.10 M HCl solution?

press for answer

0.83 mL
We need 0.10 L x 0.10 mole/L = 0.01 moles HCl. The stock solution contains 12 moles per liter so, we must use:
0.01 mole/ 12 mole/L = 8.3 x 10-4 L or 0.83 mL of the stock solution. This amount of the stock solution must be diluted to a total of 100 mL.

The Mole and Chemical Reactions

Although the quantitative aspects of chemical reactions will be studied in the module on Stoichiometry, we will introduce the idea that substances react in certain mole ratios. For every reaction there is an equation that expresses the mole ratio between the reactants and products. We have looked at a few simple equations early in this module; let's look at a few more from our new vantage point of the mole.

Aluminum reacts with sulfur to form aluminum sulfide as shown below:

2 Al + 3 S → Al2S3

The substances on the left side of the arrow are the reactants, those on the right side are the products. Every equation must obey the Law of Conservation of Mass, which means that all of the aluminum atoms on the left must be converted to aluminum atoms in some form on the right if the reaction goes to completion. The same is true, of course, of the sulfur. In order to balance an equation, we must, therefore, simply make certain that the number of atoms of each element is the same on both sides. In this sense, a chemical equation is like a mathematical equation; there must be an equality between the number of atoms.

Problem Seventy
Methane (CH4) reacts with elemental oxygen, which exists as O2 molecules, to form CO2 and water. Which of the following is a balanced equation for this reaction?

a) CH4 + 4 O → CO2 + 2 H2O            b) CH4 + 2 O2 → CO2 + 2 H2O            c) CH4 + 3 O2 → CO2 + 4 H2O

Incorrect
Elemental oxygen, does not exist as oxygen atoms in nature. It exists in the air we breathe as O2 molecules. Thus, we must use O2 molecules, not O atoms in the equation.

Correct
Always be sure to count up the atoms of each element and make sure you have the same numbers on both sides of the arrow.

Incorrect
The carbon and oxygen are balanced, but the hydrogen is not.

Problem Seventy One

Aluminum metal reacts with HCl to form elemental hydrogen (H2) and AlCl3. Write a balanced equation for this reaction.

press for answer

2 Al + 6 HCl → 3 H2 + 2 AlCl3
Notice that aluminum metal is written as Al.

Problem Seventy Two

Balance the following "equation."
Na2CO3 + HCl → H2CO3 + NaCl

press for answer

Na2CO3 + 2 HCl → H2CO3 + 2 NaCl

What do the coefficients in from of the substances in the balanced equation mean? For example, what do the 2, the 3 and the 1 (there is an implied 1 in front of Al2S3) mean in the following equation?

2 Al + 3 S → Al2S3

We have already encountered one interpretation: Two atoms of aluminum combine with 3 atoms of aluminum to make aluminum sulfide. Notice that in this sentence we did not specify how much aluminum sufide was produced because we can not say one molecule of aluminum sulfide - the compound does not exist in molecular form. It is very convenient to use the mole to interpret equations. Thus, for this equation we can say: Two moles of aluminum combine with three moles of sulfur to make one mole of aluminum sulfide.

Problem Seventy Three
Verbalize the following equation:
H2 + Cl2 → 2 HCl

All three answers are correct.
Make sure that you compare them and understand why each interpretation is correct.

The language in our statements above should make it clear that the coefficients in front of each substance in a chemical equation represent the relative number of moles of each involved in the reaction. For example, in the equation:

H2 + Cl2 → 2 HCl

The coefficients, which are 1, 1, and 2, tell us that if we start the reaction with one mole of H2 and one mole of Cl2, we will get two moles of HCl when the reaction is complete. Or, for the reaction:

Fe2(CO3)3 → Fe2O3 + 3 CO2

we know that if we start the reaction with one mole of Fe2(CO3)3 we will get one mole of Fe2O3 and three moles of CO2 when the reaction is complete.

But, what if we start with some other amount, say 0.15 mole of Fe2(CO3)3 ? Certainly, the same mole ratio applies no matter how much we start with. That is, we will get three times as many moles of CO2 as the number of moles of Fe2(CO3)3 that we start with. Thus, if we start with 0.15 mole of Fe2(CO3)3, we will get:

0.15 mol Fe2(CO3)3 x 3 mole CO2 /1 mole Fe2(CO3)3 = 0.45 mole CO2

Problem Seventy Four
How many moles of Fe2(CO3)3 must we start the reaction with if we want 1.8 mol of CO2 when the reaction has gone to completion?

a) 1 mole            b) 1.8/3 = 0.60 mol            c) 3 x 1.8 = 5.4 mol

Incorrect
The mole ratio of CO2 to Fe2(CO3)3 is 3 to 1 (the coefficients tell us that), but if we want 1.8 mole CO2 we need start with only a third of that number of moles of Fe2(CO3)3

Correct
1.8 mol CO2 x (1 mole Fe2(CO3)3/3 mole CO2) = 0.60 mol

Incorrect
Think about the equation. If you start with 5.4 mole of Fe2(CO3)3 won't you get 3 x 5.4 mole of CO2 ?

Elemental nitrogen and hydrogen, both of which exist as diatomic molecules (N2 and H2), react to form ammonia, according to the equation:

N2 + 3 H2 → 2 NH3

Problem Seventy Five
If 1.0 mole of H2 is consumed during the reaction, how many moles of ammonia are formed?

a) 2 moles            b) 0.67 mole            c) 1.5 moles

Incorrect
2 is simply the coefficient in front of ammonia and we know that three moles of H2 would produce two moles of NH3.

Correct
mole H2 x (2 mole NH3/3 moles H2) = 0.67 mole NH3

Incorrect
Notice that 3 moles of H2 are converted to 2 moles NH3. Thus, we get fewer moles of ammonia than the number of moles of H2 that we start with.

In the problem above, and in many other such problems, it is convenient to divide all of the coefficients by the coefficient of the substance in question. In the above case, we could divide all coefficients by the 3 in front of H2 to get the equation:

1/3 N2 + H2 → 2/3 NH3

This equation makes it clear that every mole of H2 will be converted to two-thirds mole of NH3.

Try the same approach with the reaction of ethane with O2 as shown below:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Problem Seventy Six

How many moles of CO2 are formed as a result of the reaction of 2 moles of O2?

press for answer

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
1.1 mole. First notice that the coefficients tell you that fewer moles of CO2 will be formed. Second, divide all coefficients by the coefficient of O2 to get:
2/7 C2H6 + O2 → 4/7 CO2 + 6/7 H2O
Each mole of O2 that is consumed will produce four-seventh mole of CO2. So, if we have 2 mole O2, we should get:
2 mole O2 x 4/7 = 1.1 mole CO2
or in more strict dimensional analysis methodology:
2 mole O2 x (4/7 mole CO2/1 mole O2) = 1.1 mole CO2
or, slightly differently:
2 mole O2 x (4 mole CO2/ 7 mole O2) = 1.1 mole CO2

Problem Seventy Seven

If 1.0 mole ethane is mixed with 5.0 mole O2 and the 1.0 mole ethane is completely consumed during the reaction, how many moles of O2 will be left at the end of the reaction?

press for answer

The equation
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
tells us that:
1 mole C2H6 x (7 mole O2 / 2 mole C2H6 ) = 3.5 mole O2
will be required to react with the 1.0 mole ethane. Since we started with 5.0 mole O2, 5.0 - 3.5 = 1.5 mole O2 will be left at the end of the reaction.