Wired Chemist

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.

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."

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

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

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).

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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:

[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.

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 %

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.

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_.421O_1.71, 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.

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.

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.

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.

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

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.

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.

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

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

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