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The Mole
Objectives: To understand the mole concept and Avogadro’s number.
To learn to convert among moles, mass, and number of atoms in a
given sample.
I
n the previous section we used atomic mass units for mass, but these are
extremely small units. In the laboratory a much larger unit, the gram, is
the convenient unit for mass. In this section we will learn to count atoms in
samples with masses given in grams.
Let’s assume we have a sample of aluminum that has a mass of 26.98 g.
What mass of copper contains exactly the same number of atoms as this sample of aluminum?
26.98 g
aluminum
Contains the same
number of atoms
? grams
copper
To answer this question, we need to know the average atomic masses for aluminum (26.98 amu) and copper (63.55 amu). Which atom has the greater
atomic mass, aluminum or copper? The answer is copper. If we have 26.98 g
of aluminum, do we need more or less than 26.98 g of copper to have the same
number of copper atoms as aluminum atoms? We need more than 26.98 g of
copper because each copper atom has a greater mass than each aluminum atom.
Therefore, a given number of copper atoms will weigh more than an equal
number of aluminum atoms. How much copper do we need? Because the average masses of aluminum and copper atoms are 26.98 amu and 63.55 amu,
respectively, 26.98 g of aluminum and 63.55 g of copper contain exactly the
same number of atoms. So we need 63.55 g of copper. As we saw in Section 6.1
when we were discussing candy, samples in which the ratio of the masses is the
same as the ratio of the masses of the individual atoms always contain the same
number of atoms. In the case just considered, the ratios are
26.98 amu
26.98g
63.55g
63.55 amu
Ratio of
sample
masses
W H AT I F ?
What if you were offered
$1 million to count from 1 to
6 1023 at a rate of one
number each second?
Determine your hourly wage.
Would you do it? Could you do
it?
Ratio of
atomic
masses
Therefore, 26.98 g of aluminum contains the same number of aluminum
atoms as 63.55 g of copper contains copper atoms.
Now compare carbon (average atomic mass, 12.01 amu) and helium (average atomic mass, 4.003 amu). A sample of 12.01 g of carbon contains the
same number of atoms as 4.003 g of helium. In fact, if we weigh out samples of all the elements such that each sample has a mass equal to that element’s average atomic mass in grams, these samples all contain the same
number of atoms (Figure 6.1). This number (the number of atoms present
Figure 6.1
All these samples of pure elements contain the same number
(a mole) of atoms: 6.022 1023
atoms.
6.3 The Mole
159
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CHEMISTRY
Avogadro’s number (to four
significant figures) is
6.022 1023. One moles of
anything is 6.022 1023 units
of that substance.
CHEMISTRY
The mass of 1 mol of an
element is equal to its average
atomic mass in grams.
in all of these samples) assumes special importance in chemistry. It is called
the mole, the unit all chemists use in describing numbers of atoms. The
mole (abbreviated mol) can be defined as the number equal to the number of
carbon atoms in 12.01 grams of carbon. Techniques for counting atoms very
precisely have been used to determine this number to be 6.022 1023. This
number is called Avogadro’s number. One mole of something consists of
6.022 10 23 units of that substance.
Just as a dozen eggs is 12 eggs, a mole
of eggs is 6.022 1023 eggs. And a
mole of water contains 6.022 1023
H2O molecules.
The magnitude of the number
6.022 1023 is very difficult to imagine. To give you some idea, 1 mol of
seconds represents a span of time 4
million times as long as the earth has
already existed! One mole of marbles
is enough to cover the entire earth to
a depth of 50 miles! However, because
atoms are so tiny, a mole of atoms or Figure 6.2
molecules is a perfectly manageable One-mole samples of iron (nails), iodine
crystals, liquid mercury, and powdered
quantity to use in a reaction (Figure
sulfur.
6.2).
How do we use the mole in chemical calculations? Recall that Avogadro’s
number is defined such that a 12.01-g sample of carbon contains 6.022 1023 atoms. By the same token, because the average atomic mass of hydrogen is 1.008 amu (Table 6.1), 1.008 g of hydrogen contains 6.022 1023
hydrogen atoms. Similarly, 26.98 g of aluminum contains 6.022 1023 aluminum atoms. The point is that a sample of any element that weighs a number of grams equal to the average atomic mass of that element contains
6.022 1023 atoms (1 mol) of that element.
Table 6.2 shows the masses of several elements that contain 1 mol of
atoms.
In summary, a sample of an element with a mass equal to that element’s average atomic mass expressed in grams contains 1 mol of atoms.
To do chemical calculations, you must understand what the mole means
and how to determine the number of moles in a given mass of a substance.
TABLE 6.2
Comparison of 1-Mol Samples of Various Elements
160
Element
Number of Atoms Present
Mass of Sample (g)
Aluminum
6.022 1023
26.98
Gold
6.022 10
23
196.97
Iron
6.022 10
23
55.85
Sulfur
6.022 1023
32.07
Boron
6.022 10
23
10.81
Xenon
6.022 10
23
Chapter 6 Chemical Composition
131.3
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However, before we do any calculations, let’s be sure that the process of counting by weighing is clear. Consider the following “bag” of H atoms (symbolized by dots), which contains 1 mol (6.022 1023) of H atoms and has a mass
of 1.008 g. Assume the bag itself has no mass.
Contains 1 mol H atoms
(6.022 × 1023 atoms)
Sample A
Mass = 1.008 g
Now consider another “bag” of hydrogen atoms in which the number of hydrogen atoms is unknown.
Contains an unknown
number of H atoms
A 1-mol sample of graphite (a
form of carbon) weighs 12.01 g.
Sample B
We want to find out how many H atoms are present in sample (“bag”) B.
How can we do that? We can do it by weighing the sample. We find the mass
of sample B to be 0.500 g.
How does this measured mass help us determine the number of atoms in
sample B? We know that 1 mol of H atoms has a mass of 1.008 g. Sample B
has a mass of 0.500 g, which is approximately half the mass of a mole of H
atoms.
Sample A
Mass = 1.008 g
Sample B
Mass = 0.500 g
Contains 1 mol
of H atoms
Must contain
about 1/2 mol
of H atoms
Because the mass of B
is about half the
mass of A
We carry out the actual calculation by using the equivalence statement
1 mol H atoms 1.008 g H
to construct the conversion factor we need:
1 mol H
0.500 g H 0.496 mol H in sample B
1.008 g H
Let’s summarize. We know the mass of 1 mol of H atoms, so we can determine the number of moles of H atoms in any other sample of pure hydrogen by weighing the sample and comparing its mass to 1.008 g (the mass of
6.3 The Mole
161
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1 mol of H atoms). We can follow this same process for any element, because
we know the mass of 1 mol for each of the elements.
Also, because we know that 1 mol is 6.022 1023 units, once we know
the moles of atoms present, we can easily determine the number of atoms present. In the case considered above, we have approximately 0.5 mol of H atoms
in sample B. This means that about 1/2 of 6 1023, or 3 1023, H atoms are
present. We carry out the actual calculation by using the equivalence statement
1 mol 6.022 1023
to determine the conversion factor we need:
6.022 1023 H atoms
0.496 mol H atoms 1 mol H atoms
2.99 1023 H atoms in sample B
These procedures are illustrated in Example 6.3.
Example 6.3
Calculating Moles and Number of Atoms
Aluminum (Al), a metal with a high strength-to-weight ratio and a
high resistance to corrosion, is often used for structures such as
high-quality bicycle frames. Compute both the number of moles of
atoms and the number of atoms in a 10.0-g sample of aluminum.
Solution
In this case we want to change from mass to moles of atoms:
A bicycle with an aluminum
frame.
10.0 g
Al
? moles
of Al
atoms
The mass of 1 mol (6.022 1023 atoms) of aluminum is 26.98 g. The sample we are considering has a mass of 10.0 g. Its mass is less than 26.98 g,
so this sample contains less than 1 mol of aluminum atoms. We calculate
the number of moles of aluminum atoms in 10.0 g by using the equivalence statement
1 mol Al 26.98 g Al
to construct the appropriate conversion factor:
1 mol Al
10.0 g Al 0.371 mol Al
26.98 g Al
Next we convert from moles of atoms to the number of atoms, using the
equivalence statement
6.022 1023 Al atoms 1 mol Al atoms
We have
6.022 1023 Al atoms
0.371 mol Al 2.23 1023 Al atoms
1 mol Al
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Chapter 6 Chemical Composition
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We can summarize this calculation as follows:
10.0 g Al 0.371 mol Al atoms 1 mol
26.98 g
0.371 mol Al
6.022 1023 Al atoms
mol
2.23 1023 Al atoms
Example 6.4
Calculating the Number of Atoms
A silicon chip used in an integrated circuit of a microcomputer has a mass
of 5.68 mg. How many silicon (Si) atoms are present in this chip? The average atomic mass for silicon is 28.09 amu.
Solution
A silicon chip of the type
used in electronic equipment.
Our strategy for doing this problem is to convert from milligrams of silicon to grams of silicon, then to moles of silicon, and finally to atoms of
silicon:
Milligrams of
Si atoms
Grams of
Si atoms
Moles of
Si atoms
Number of
Si atoms
Each arrow in the schematic represents a conversion factor. Because
1 g 1000 mg, we have
1 g Si
5.68 mg Si 5.68 103 g Si
1000 mg Si
Next, because the average mass of silicon is 28.09 amu, we know that
1 mol of Si atoms weighs 28.09 g. This leads to the equivalence statement
1 mol Si atoms 28.09 g Si
W H AT I F ?
What if you discovered
Avogadro’s number was not
6.02 1023 but 3.01 1023?
Would your discovery affect the
relative masses given on the
periodic table? If so, how? If
not, why not?
Thus,
1 mol Si
5.68 103 g Si 2.02 104 mol Si
28.09 g Si
Using the definition of a mole (1 mol 6.022 1023), we have
6.022 1023 atoms
2.02 104 mol Si 1.22 1020 Si atoms
1 mol Si
We can summarize this calculation as follows:
5.68 mg Si 1g
1000 mg
5.68 103 g Si 1 mol
28.09 g
23
2.02 104 mol Si 6.022 10 Si atoms
mol
5.68 103 g Si
2.02 104 mol Si
1.22 1020 Si atoms
(continued)
6.3 The Mole
163
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(continued)
Self-Check Exercise 6.3
Chromium (Cr) is a metal that is added to steel to improve its resistance
to corrosion (for example, to make stainless steel). Calculate both the number of moles in a sample of chromium containing 5.00 1020 atoms and
the mass of the sample.
Problem Solving: Does the Answer Make Sense?
CHEMISTRY
The values for the average
masses of the atoms of the
elements are listed inside the
front cover of this book.
Focus Questions
When you finish a problem, always think about the “reasonableness” of your
answers. In Example 6.4, 5.68 mg of silicon is clearly much less than 1 mol
of silicon (which has a mass of 28.09 g), so the final answer of 1.22 1020
atoms (compared to 6.022 1023 atoms in a mole) at least lies in the right
direction. That is, 1.22 1020 atoms is a smaller number than 6.022 1023.
Also, always include the units as you perform calculations and make sure the
correct units are obtained at the end. Paying careful attention to units and
making this type of general check can help you detect errors such as an inverted conversion factor or a number that was incorrectly entered into your
calculator.
Sections 6.1–6.3
1. Suppose you work in a hardware store. The manager asks you to get an
order for a major customer who is waiting impatiently. You need 1200
matched sets of nuts and bolts. Unfortunately the nuts and bolts are not
boxed—they are loose in big buckets. How can you earn a bonus from
your boss and make the customer happy by giving him the nuts and
bolts in less than 2 minutes?
2. Why isn’t the average atomic mass of any element a whole number
(for example, C, 12 amu; H, 1 amu; N, 14 amu)?
3. A mole of any substance contains Avogadro’s number of units.
a. Write an equivalence statement for this definition.
b. Write the possible conversion factors from this relationship.
c. To determine the following tell which conversion factor you would
need to use:
1. Moles Al from atoms Al
2. Atoms Au from mol Au
4. Table 6.2 gives the mass (g) of a mole of some elements. Use this information and your conversion factors to find the mass of:
a. 0.25 mol Al
b. 0.500 mol Au
c. 5.0 mol Fe
d. 25 mol S
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Chapter 6 Chemical Composition