Chapter 6 Quantities in Chemical Reactions

6.1 The Mole

Figure 6.1 "Water Molecules" shows that we need 2 hydrogen atoms and 1 oxygen atom to make 1 water molecule. If we want to make 2 water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make 5 molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom.

One problem we have, however, is that it is extremely difficult, if not impossible, to organize atoms one at a time. As stated in the introduction, we deal with billions of atoms at a time. How can we keep track of so many atoms (and molecules) at a time? We do it by using mass rather than by counting individual atoms.

A hydrogen atom has a mass of approximately 1 u. An oxygen atom has a mass of approximately 16 u. The ratio of the mass of an oxygen atom to the mass of a hydrogen atom is therefore approximately 16:1.

If we have 2 atoms of each element, the ratio of their masses is approximately 32:2, which reduces to 16:1—the same ratio. If we have 12 atoms of each element, the ratio of their total masses is approximately (12 × 16):(12 × 1), or 192:12, which also reduces to 16:1. If we have 100 atoms of each element, the ratio of the masses is approximately 1,600:100, which again reduces to 16:1. As long as we have equal numbers of hydrogen and oxygen atoms, the ratio of the masses will always be 16:1.

The same consistency is seen when ratios of the masses of other elements are compared. For example, the ratio of the masses of silicon atoms to equal numbers of hydrogen atoms is always approximately 28:1, while the ratio of the masses of calcium atoms to equal numbers of lithium atoms is approximately 40:7.

So we have established that the masses of atoms are constant with respect to each other, as long as we have the same number of each type of atom. Consider a more macroscopic example. If a sample contains 40 g of Ca, this sample has the same number of atoms as there are in a sample of 7 g of Li. What we need, then, is a number that represents a convenient quantity of atoms so we can relate macroscopic quantities of substances. Clearly even 12 atoms are too few because atoms themselves are so small. We need a number that represents billions and billions of atoms.

Chemists use the term moleA number of things equal to 6.022 × 10²³ items. to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (mol) represents 6.022 × 10²³ things. The number 6.022 × 10²³, called Avogadro’s numberThe value 6.022 × 10²³. after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules. Thus, if we have 6.022 × 10²³ O atoms, we say we have 1 mol of O atoms. If we have 2 mol of Na atoms, we have 2 × (6.022 × 10²³) Na atoms, or 1.2044 × 10²⁴ Na atoms. Similarly, if we have 0.5 mol of benzene (C₆H₆) molecules, we have 0.5 × (6.022 × 10²³) C₆H₆ molecules, or 3.011 × 10²³ C₆H₆ molecules.

Notice that we are applying the mole unit to different types of chemical entities. In these examples, we cited moles of atoms and moles of molecules. The word mole represents a number of things—6.022 × 10²³ of them—but does not by itself specify what “they” are. They can be atoms, formula units (of ionic compounds), or molecules. That information still needs to be specified.

Because 1 H₂ molecule contains 2 H atoms, 1 mol of H₂ molecules (6.022 × 10²³ molecules) has 2 mol of H atoms. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of ethanol (C₂H₆O), we can construct the following relationships (Table 6.1 "Molecular Relationships"):

The following example illustrates how we can use these relationships as conversion factors.

The fact that 1 mol equals 6.022 × 10²³ items can also be used as a conversion factor.

6.2 Atomic and Molar Masses

Now that we have introduced the mole and practiced using it as a conversion factor, we ask the obvious question: why is the mole that particular number of things? Why is it 6.022 × 10²³ and not 1 × 10²³ or even 1 × 10²⁰?

The number in a mole, Avogadro’s number, is related to the relative sizes of the atomic mass unit and gram mass units. Whereas one hydrogen atom has a mass of approximately 1 u, 1 mol of H atoms has a mass of approximately 1 gram. And whereas one sodium atom has an approximate mass of 23 u, 1 mol of Na atoms has an approximate mass of 23 grams.

One mole of a substance has the same mass in grams that one atom or molecule has in atomic mass units. The numbers in the periodic table that we identified as the atomic masses of the atoms not only tell us the mass of one atom in u but also tell us the mass of 1 mol of atoms in grams.

The mole concept can be extended to masses of formula units and molecules as well. The mass of 1 mol of molecules (or formula units) in grams is numerically equivalent to the mass of one molecule (or formula unit) in atomic mass units. For example, a single molecule of O₂ has a mass of 32.00 u, and 1 mol of O₂ molecules has a mass of 32.00 g. As with atomic mass unit–based masses, to obtain the mass of 1 mol of a substance, we simply sum the masses of the individual atoms in the formula of that substance. The mass of 1 mol of a substance is referred to as its molar massThe mass of 1 mol of atoms or molecules., whether the substance is an element, an ionic compound, or a covalent compound.

Be careful when counting atoms. In formulas with polyatomic ions in parentheses, the subscript outside the parentheses is applied to every atom inside the parentheses. For example, the molar mass of Ba(OH)₂ requires the sum of 1 mass of Ba, 2 masses of O, and 2 masses of H:

Because molar mass is defined as the mass for 1 mol of a substance, we can refer to molar mass as grams per mole (g/mol). The division sign (/) implies “per,” and “1” is implied in the denominator. Thus, the molar mass of bilirubin can be expressed as 584.05 g/mol, which is read as “five hundred eighty four point zero five grams per mole.”

6.3 Mole-Mass Conversions

Example 3 in Section 6.2 "Atomic and Molar Masses" stated that the mass of 2 mol of U is twice the molar mass of uranium. Such a straightforward exercise does not require any formal mathematical treatment. Many questions concerning mass are not so straightforward, however, and require some mathematical manipulations.

The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversionThe conversion from moles of material to the mass of that same material. (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).

We established that 1 mol of Al has a mass of 26.98 g (Example 3 in Section 6.2 "Atomic and Molar Masses"). Stated mathematically,

1 mol Al = 26.98 g Al

We can divide both sides of this expression by either side to get one of two possible conversion factors:

$$\frac{1\text{ mol Al}}{\text{26}\text{.98 g Al}}\text{ and }\frac{\text{26}\text{.98 g Al}}{\text{1 mol Al}}$$

The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”

Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure 6.2 "A Simple Flowchart for Converting between Mass and Moles of a Substance" is a chart for determining what conversion factor is needed, and Figure 6.3 "A Flowchart Illustrating the Steps in Performing a Unit Conversion" is a flow diagram for the steps needed to perform a conversion.

6.4 Mole-Mole Relationships in Chemical Reactions

In Chapter 5 "Introduction to Chemical Reactions", you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.

Consider the simple chemical equation

2H₂ + O₂ → 2H₂O

The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as

4H₂ + 2O₂ → 4H₂O

The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as

22H₂ + 11O₂ → 22H₂O

because 22:11:22 also reduces to 2:1:2.

Suppose we want to use larger numbers. Consider the following coefficients:

12.044 × 10²³ H₂ + 6.022 × 10²³ O₂ → 12.044 × 10²³ H₂O

These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 10²³ is 1 mol, while 12.044 × 10²³ is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as

2 mol H₂ + 1 mol O₂ → 2 mol H₂O

We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is

2H₂ + O₂ → 2H₂O

Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”

By the same token, the ratios we constructed in Chapter 5 "Introduction to Chemical Reactions" can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:

$$\frac{2{\text{ mol H}}_{\text{2}}}{1{\text{ mol O}}_{\text{2}}}\text{ or }\frac{{\text{1 mol O}}_{\text{2}}}{2{\text{ mol H}}_{\text{2}}}$$ $$\frac{2{\text{ mol H}}_{\text{2}}\text{O}}{1{\text{ mol O}}_{\text{2}}}\text{ or }\frac{{\text{1 mol O}}_{\text{2}}}{2{\text{ mol H}}_{\text{2}}\text{O}}$$ $$\frac{2{\text{ mol H}}_{\text{2}}}{2{\text{ mol H}}_{\text{2}}\text{O}}\text{ or }\frac{{\text{2 mol H}}_{\text{2}}\text{O}}{2{\text{ mol H}}_{\text{2}}}$$

We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry.

6.5 Mole-Mass and Mass-Mass Problems

We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:

Collectively, these conversions are called mole-mass calculationsA stoichiometry calculation converting between masses and moles of different substances in a chemical reaction..

As an example, consider the balanced chemical equation

Fe₂O₃ + 3SO₃ → Fe₂(SO₄)₃

If we have 3.59 mol of Fe₂O₃, how many grams of SO₃ can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO₃ in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO₃ needed. Then using the molar mass of SO₃ as a conversion factor, we determine the mass that this number of moles of SO₃ has.

The first step resembles the exercises we did in Section 6.4 "Mole-Mole Relationships in Chemical Reactions". As usual, we start with the quantity we were given:

$$3.59\cancel{{\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\times \frac{{\text{3 mol SO}}_{\text{3}}}{\text{1}\cancel{{\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}=10.77{\text{ mol SO}}_{\text{3}}$$

The mol Fe₂O₃ units cancel, leaving mol SO₃ unit. Now, we take this answer and convert it to grams of SO₃, using the molar mass of SO₃ as the conversion factor:

$$10.77\cancel{{\text{ mol SO}}_{\text{3}}}\times \frac{\text{80}{\text{.06 g SO}}_{\text{3}}}{\text{1}\cancel{{\text{ mol SO}}_{\text{3}}}}=862{\text{ g SO}}_{\text{3}}$$

Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO₃ will react with 3.59 mol of Fe₂O₃. Many problems of this type can be answered in this manner.

The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:

We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.

It is a small step from mole-mass calculations to mass-mass calculationsA stoichiometry calculation converting between the mass of one substance and the mass of a different substance in a chemical reaction.. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:

This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.