Introductory Chemistry: A Foundation - Zumdahl S.S., DeCoste D.J. 2019

Chemical Composition
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.

In 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 g. What mass of copper contains exactly the same number of atoms as this sample of aluminum?

An arrow labeled ’contains the same number of atoms’ points from “26.98 g aluminum” to “? grams copper.”

To answer this question, we need to know the average atomic masses for aluminum ( amu) and copper ( amu). Which atom has the greater atomic mass, aluminum or copper? The answer is copper. If we have g of aluminum, do we need more or less than g of copper to have the same number of copper atoms as aluminum atoms? We need more than 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 amu and amu, respectively, g of aluminum and g of copper contain exactly the same number of atoms. So we need g of copper. As we saw in the first section 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

Therefore, g of aluminum contains the same number of aluminum atoms as g of copper contains copper atoms.

Now compare carbon (average atomic mass, amu) and helium (average atomic mass, amu). A sample of g of carbon contains the same number of atoms as 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 (Fig. 8.1). This number (the number of atoms present 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 (mol) can be defined as the number equal to the number of carbon atoms in grams of carbon. Techniques for counting atoms very precisely have been used to determine this number to be . This number is called Avogadro’s number . One mole of something consists of units of that substance. Just as a dozen eggs is eggs, a mole of eggs is eggs. And a mole of water contains molecules.

Figure 8.1.A set of three photos are shown. The first photo shows lead bar weighs 207.2g. The second photo shows silver bars weighs 107.9g. The third photo shows 63.55g of copper placed on a watch glass.

Ken O’Donoghue © Cengage Learning

All these samples of pure elements contain the same number (a mole) of atoms: atoms.

The magnitude of the number is very difficult to imagine. To give you some idea, mole of seconds represents a span of time 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 miles! However, because atoms are so tiny, a mole of atoms or molecules is a perfectly manageable quantity to use in a reaction (Fig. 8.2).

Figure 8.2.A photo show iodine crystals placed on watchglass, powdered sulfur placed on a watchglass, conical flask filled with liquid mercury, and iron nails.

Ken O’Donoghue © Cengage Learning

One-mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur.

Critical Thinking

· What if you were offered million to count from to at a rate of one number each second? Determine your hourly wage. Would you do it? Could you do it?

How do we use the mole in chemical calculations? Recall that Avogadro’s number is defined such that a -g sample of carbon contains atoms. By the same token, because the average atomic mass of hydrogen is amu (see Table 8.1), g of hydrogen contains hydrogen atoms. Similarly, g of aluminum contains 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 atoms ( mole) of that element.

Table 8.2 shows the masses of several elements that contain mole of atoms.

Table 8.2. Comparison of -Mole Samples of Various Elements

Element

Number of Atoms Present

Mass of Sample (g)

Aluminum

Gold

Iron

Sulfur

Boron

Xenon

In summary, a sample of an element with a mass equal to that element’s average atomic mass expressed in grams contains mole 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. However, before we do any calculations, let’s be sure that the process of counting by weighing is clear.

Consider the following “bag” of atoms (symbolized by dots), which contains mole of atoms and has a mass of g. Assume the bag itself has no mass.

An illustration shows a spherical cloud representing sample A that contains densely packed 1 mole of hydrogen atoms (6.022 times 10 to the power of 23 atoms), represented by black dots. The mass of hydrogen atoms equals 1.008 gram.

Now consider another “bag” of hydrogen atoms in which the number of hydrogen atoms is unknown.

An illustration shows a spherical cloud representing sample B that contains sparsely packed hydrogen atoms labeled “Contains an unknown number of H atoms.”

We want to find out how many 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 g.

How does this measured mass help us determine the number of atoms in sample B? We know that mole of atoms has a mass of g. Sample B has a mass of g, which is approximately half the mass of a mole of atoms.

An illustration shows “sample A mass equals 1.008g” (contains 1 mole of H atoms) and sample B mass equals 0.0500g” (must contain about 1 over 2 mole of H atoms). An arrow labeled “because the mass of B is about half the mass of A” points from sample A to sample B.

We carry out the actual calculation by using the equivalence statement

to construct the conversion factor we need:

Let’s summarize. We know the mass of mole of atoms, so we can determine the number of moles of atoms in any other sample of pure hydrogen by weighing the sample and comparing its mass to g (the mass of mole of atoms). We can follow this same process for any element because we know the mass of mole for each of the elements.

Also, because we know that mole is units, once we know the moles of atoms present, we can easily determine the number of atoms present. In the case just considered, we have approximately mole of atoms in sample B. This means that about of , or , atoms is present. We carry out the actual calculation by using the equivalence statement

to determine the conversion factor we need:

These procedures are illustrated in Example 8.3.

Critical Thinking

· What if you discovered Avogadro’s number was not but ? Would this affect the relative masses given on the periodic table? If so, how? If not, why not?

Interactive Example 8.3. Calculating Moles and Number of Atoms

Aluminum , 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 -g sample of aluminum.

Solution

In this case we want to change from mass to moles of atoms:

An arrow points from “10.0 g Al” to “? moles of Al atoms”, from which a second arrow points to “number of Al atoms.”

The mass of mole ( atoms) of aluminum is g. The sample we are considering has a mass of g. Its mass is less than g, so this sample contains less than mole of aluminum atoms. We calculate the number of moles of aluminum atoms in g by using the equivalence statement

to construct the appropriate conversion factor:

Next we convert from moles of atoms to the number of atoms, using the equivalence statement

We have

We can summarize this calculation as follows:

An illustration shows two equations. The first equation reads 10.0 g Al times 1 mole Al over 26.98 g Al gives 0.371 mol Al. The second equation reads 0.371 mol Al times 6.022 times 10 to the twenty-third power Al atoms over mol Al gives 2.23 times 10 to the twenty-third power Al atoms.

Interactive Example 8.4. Calculating the Number of Atoms

A silicon chip used in an integrated circuit of a microcomputer has a mass of mg. How many silicon atoms are present in this chip? The average atomic mass for silicon is amu.

Solution

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:

A flowchart shows the conversion of milligrams of silicon to atoms of silicon as follows: milligrams of silicon atoms are converted to Grams of Si atoms, which are converted to moles of Si atoms, which are then converted to number of Si atoms.

where each arrow in the schematic represents a conversion factor. Because , we have

Next, because the average mass of silicon is amu, we know that mole of atoms weighs g. This leads to the equivalence statement

Thus,

Using the definition of a mole , we have

We can summarize this calculation as follows:

An illustration shows three equations. The first equation reads 5.68mg Si times 1g over 1000 mg gives 5.68 times 10 to the negative 3 power g Si. The second equation reads 5.68 times 10 to the negative 3 power g Si times 1 mol over 28.09g gives 2.02 times 10 to the negative 4 power mol Si. The third equation reads 2.02 times 10 to the negative 4 power mol Si times 6.022 times 10 to the twenty-third power Si atoms over mol gives 1.22 times 10 to the twentieth power Si atoms.

Problem Solving: Does the Answer Make Sense?

When you finish a problem, always think about the “reasonableness” of your answers. In Example 8.4, mg of silicon is clearly much less than mole of silicon (which has a mass of g), so the final answer of atoms (compared to atoms in a mole) at least lies in the right direction. That is, atoms is a smaller number than . 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.

As you can see, the problems are getting more complicated to solve. In the next section we will discuss strategies that will help you become a better problem solver.

Self-Check: Exercise 8.3

· Chromium 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 atoms and the mass of the sample.

See Problems 8.19, 8.20, 8.21, 8.22, 8.23, and 8.24.