MathConnections
3.4 Interpreting Equations and the Mole
The chemical equations that we have been learning to write are symbolic descriptions of chemical reactions. But to interpret these equations, we must think about them from another point of view, in terms of the actual substances and processes they represent.
As often happens in chemistry, we can do this at either the microscopic or the mac- roscopic level. The microscopic interpretation visualizes reactions between indi- vidual molecules, and that interpretation is the one we have used thus far. The macroscopic interpretation pictures reactions between bulk quantities. Neither perspective is inherently better than the other; we simply use the point of view that is best suited to whatever circumstances we are considering. To make the connection between the two worlds, we need to look at the way chemists count the molecules in a large sample.
Interpreting Chemical Equations
In describing how to write and balance a chemical equation, we adopted a way of looking at reactions that seems natural. We spoke of one molecule reacting with another mol- ecule to form some new compound that we called the product. Real chemical reactions rarely involve just one or two molecules though, so we will need to think of chemical equations in terms of larger quantities, too. Let’s return to our fi rst example, the explo- sive reaction between hydrogen and oxygen to produce water:
2 H2(g) + O2(g) : 2 H2O(g)
We have read this chemical equation as saying that “two H2 molecules react with one O2 molecule to form two H2O molecules.” But suppose we carried out the same reaction starting with 20 H2 molecules rather than two. The balanced equation says that we need one oxygen molecule for every two H2 molecules. So in this case, we could say that 20 H2 molecules react with 10 O2 molecules to form 20 H2O molecules.
Two key features of these statements must be emphasized. First, both statements are correct because they use the ratios established by the stoichiometric coeffi cients.
There are twice as many hydrogen molecules as there are oxygen molecules in each case. Second, both statements refer to numbers of particles. When we interpret a chemi- cal reaction, we must remember that the stoichiometric coeffi cients always refer to numbers of particles. In the example we just considered, the numbers were 1 and 2 or 10 and 20. We could just as easily have said 1,000,000 and 2,000,000 or 45,600,000 and 91,200,000. Provided we use numbers that satisfy the correct stoichiometric ratio, we will always have a reasonable description of an actual chemical reaction.
Avogadro’s Number and the Mole
When we begin to consider chemical reactions with macroscopic amounts of materials, the numbers of molecules involved become staggeringly large. But people generally don’t like to work with huge numbers, because it is very diffi cult to develop an intuitive feeling for them. So inevitably, when confronted with the fact that any real sample con- tains an almost preposterous number of molecules, scientists invented a way to make the numbers manageable. The simple way to do this is just to count the molecules in terms of some large quantity rather than individually. The particular quantity that chemists have chosen to count by is called the mole. One mole is defi ned as the num- ber of atoms in exactly 12 grams of 12C. This number is also referred to as Avogadro’s number, and its value is 6.022 × 1023. When chemists want to count molecules, they count by moles, so that the resulting numbers will be easier to work with.
There are several reasons why this number is useful in chemistry. First, it is a large number that gives a useful unit for counting particles as small as atoms. In this sense, the mole is to a chemist what a dozen is to a baker. Second, one mole is the same number of particles no matter what substance we are talking about. So a mole of H2 contains the same number of particles as a mole of TNT (trinitrotoluene, C7H5N3O6);
each has 6.022 × 1023 molecules.
You may be thinking that the value of Avogadro’s number seems like a rather pe- culiar choice for a counting unit. Why not just make it an even 1023, for example, and simplify some calculations? The answer is that the chosen value of Avogadro’s number produces a very convenient connection between the mass of an atom or molecule and the mass of a mole of those atoms or molecules. Our defi nition of the mole said that a mole of 12C has a mass of exactly 12 g. Earlier, we said that an atom of 12C has a mass of exactly 12 amu. This is not merely a coincidence, but rather a very deliberate deci- sion. Once we have defi ned the amu and the gram independently, Avogadro’s number is fi xed as the number of amu in one gram. It may help to envision a hypothetical experiment: suppose that we had a balance that could accurately measure the mass of individual atoms and a way of loading atoms onto that balance one at a time. We could then determine Avogadro’s number by loading carbon-12 atoms onto the bal- ance, one by one, until the mass read exactly 12 grams. Of course, no such experiment is possible, so Avogadro’s number has been measured by more indirect methods.
The mass of 6.022 × 1023 atoms of any element is the molar mass of that ele- ment, and its value in grams per mole is given in most periodic tables. The molar mass of silicon, for example, is greater than that of carbon because an atom of silicon is heavier than an atom of carbon. And the molar mass of each element takes into ac- count the natural isotopic abundances. So the molar mass of carbon is 12.011 g/mol, refl ecting the weighted average between the masses of 12C and 13C in the same way as we discussed in Chapter 2.
Because of its fundamental importance, Avogadro’s number has been measured extremely carefully. The internationally accepted value is actually
6.02214179 × 10 23.
Because of its fundamental importance, Avogadro’s number has been measured extremely carefully. The internationally accepted value is actually
6.02214179 × 10 23.
Charles Steele
Figure 3.9 ❚ One mole samples of various elements are shown.
Back row (left to right): bromine, aluminum, mercury, and copper.
Front row (left to right): sulfur, zinc, and iron.
The mole is the key to the macroscopic interpretation of a chemical equation.
Consider the same equation we have been discussing:
2 H2(g) + O2(g) : 2 H2O(g)
If we want to read this in terms of moles, we can say “two moles of H2 and one mole of O2 react to form two moles of H2O.” Because each mole contains the same number of molecules, the 2:1 mole ratio between the reactants is the same as the 2:1 ratio for the numbers of molecules. Chemical equations and their stoichiometric coeffi cients always provide ratios of numbers of particles, not masses.
Determining Molar Mass
Because balanced chemical equations are always expressed in terms of numbers of particles, it will be convenient to have some easy way to determine the number of par- ticles in a given sample of some substance. But there is no simple laboratory instru- ment that can measure the number of moles in a sample directly. Instead, we usually determine the number of moles indirectly from the mass of a sample.
The link between the mass of a sample and the number of moles present is the molar mass of the substance in question. To determine the molar mass of a compound, we can exploit the idea of conservation of mass. Consider one mole of water as an example. We know that one mole of the compound must contain Avogadro’s number of H2O molecules. Furthermore, we know that each of those molecules must con- tain one O atom and two H atoms. Avogadro’s number of oxygen atoms is one mole, and we know from the periodic table that one mole of O atoms has a mass of 16.0 g.
Because each molecule contains two hydrogen atoms, the entire one mole sample will contain two moles of H atoms. Again consulting the periodic table, we see that a mole of H atoms has a mass of 1.0 g, so two moles must have a mass of 2.0 g. The masses of the O and H atoms must be the same as the mass of the mole of H2O, so we can sim- ply add them to get 18.0 g as the mass of one mole of H2O. In other words, the molar mass of H2O is 18.0 g/mol. We can do the same thing for any compound; the sum of the molar masses of all of the atoms is the molar mass of the compound. In Example Problem 3.5, we determine the molar mass of several explosive compounds.
E X A M P L E P RO B L E M 3 . 5
Determine the molar mass of each of the following compounds, all of which have been used as explosives: (a) lead azide, PbN6, (b) nitroglycerin, C3H5N3O9, (c) mercury fulminate, Hg(ONC)2
Strategy We must determine the mass contributed by each element and then add them up to calculate the molar mass. When parentheses appear in the formula, each atom inside the parentheses must be multiplied by its own subscript and by the sub- script appearing after the right-hand parenthesis.
Solution
(a) PbN6: 1 mole Pb: 1 × 207.2 g/mol = 207.2 g/mol 6 moles N: 6 × 14.0067 g/mol = 84.0402 g/mol
Molar mass = 291.2 g/mol
(b) C3H5N3O9: 3 moles C: 3 × 12.011 g/mol = 36.033 g/mol 5 moles H: 5 × 1.0079 g/mol = 5.0395 g/mol 3 moles N: 3 × 14.0067 g/mol = 42.0202 g/mol 9 moles O: 9 × 15.9994 g/mol = 143.995 g/mol
Molar mass = 227.087 g/mol
For simplicity, we use whole numbers for the molar masses of hydrogen and oxygen here. In general, you should use molar masses with at least as many signifi cant fi gures as the actual data involved in the problem you are working.
For simplicity, we use whole numbers for the molar masses of hydrogen and oxygen here. In general, you should use molar masses with at least as many signifi cant fi gures as the actual data involved in the problem you are working.
(c) Hg(ONC)2: 1 mole Hg: 1 × 200.59 g/mol = 200.59 g/mol 2 moles O: 2 × 15.9994 g/mol = 31.9988 g/mol 2 moles N: 2 × 14.0067 g/mol = 28.0134 g/mol 2 moles C: 2 × 12.011 g/mol = 24.022 g/mol Molar mass 284.62 g/mol Analyze Your Answer Looking at these three compounds, we see that nitroglyc- erin has the smallest molar mass even though it clearly contains the most atoms per molecule. Does this make sense? The key factor is that both lead azide and mercury fulminate contain elements with large molar masses, so it is not surprising that nitro- glycerin is the least massive of these three molecules.
Discussion The number of signifi cant fi gures to which the molar masses of the elements are known varies from one element to another, as you can see if you examine the periodic table inside the back cover of this book. So the appropriate number of signifi cant fi gures in the molar masses here varies according to the elements in each molecule.
Check Your Understanding What are the molar masses of the following com- pounds, which are used in the preparation of various explosives? (a) H2SO4, (b) HNO3, (c) (NH2)2HNO3