CH150: Chapter 6 – Quantities in Chemistry

Chapter 6 – Quantities in Chemical Reactions

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6.5: Mole-Mole Relationships in Chemical Reactions

6.6: Mole-Mass and Mass-Mass Problems

6.7 Limiting Reagent and Percent Yield

6.1: Chapter Introduction

So far, we have talked about chemical reactions in terms of individual atoms and molecules. Although this works, most of the reactions occurring around us involve much larger amounts of chemicals. Even a tiny sample of a substance will contain millions, billions, or a hundred billion billions of atoms and molecules. How do we compare amounts of substances to each other in chemical terms when it is so difficult to count to a hundred billion billion?

Actually, there are ways to do this, which we will explore in this chapter. In doing so, we will increase our understanding of stoichiometry, which is the study of the numerical relationships between the reactants and the products in a balanced chemical reaction.

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6.2: 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.

Figure 6.1 Water Molecules: The ratio of hydrogen atoms to oxygen atoms used to make water molecules is always 2:1, no matter how many water molecules are being made.

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.

A mole is defined as 6.02 × 10²³ items and thus, it is used by chemists to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (abbreviated mol) represents 6.02 × 10²³ things. The number 6.02 × 10²³, called Avogadro’s number, 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 6.02 × 10²³ Oxygen atoms, we say we have 1 mole of Oxygen atoms. If we have 2 mol of Na atoms, we have 2 × (6.02 × 10²³) Na atoms, or 1.2044 × 10²⁴ Na atoms. Similarly, if we have 0.5 mol of benzene (C₆H₆) molecules, we have 0.5 × (6.02 × 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—6.02 × 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 (6.02 × 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”):

Table 6.1: Molecular Relationships

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

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6.3: Atomic and Molar Mass

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 amu, 1 mol of H atoms has a mass of approximately 1 gram. And whereas one sodium atom has an approximate mass of 23 amu, 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 atomic mass units, but also tell us the mass of 1 mole of atoms in grams! This is because all atoms are made up of the same parts (protons, neutrons, and electrons) with the protons and neutrons having nearly identical masses. Electrons, since they are so light, are negligent in their contribution to atomic mass, even in the largest atoms. Thus, an atomic or molecular mass is indicative of how many atoms or molecules are present. Thus, an important three way relationship is formed: 1 mol = atomic or molecular mass in grams = 6.02 X 10²³ atoms or molecules. This effectively gives us a way to count molecules in the laboratory using a common balance! Note that in chemical equations and calculations that mole concentrations are abbreviated as mol.

Recall that, the mass of an ionic compound (referred to as the formula mass) or a covalent molecule (referred to as the molecular mass)—is simply the sum of the masses of its atoms. To calculate formula or molecular masses, it is important that you keep track of the number of atoms of each element in the molecular formula to obtain the correct molecular mass.

For Example:

A molecule of NaCl contains 1 Na⁺ and 1 Cl^–. Thus, we can the formula mass of this compound by adding together the atomic masses of sodium and chlorine, as found on the periodic table (Figure 6.1).

Figure 6.1 Periodic Table of the Elements

For a larger molecule, like glucose (C₆H₁₂O₆), that has multiple atoms of the same type, simply multiply the atomic mass of each atom by the number of atoms present, and then add up all the atomic masses to get the final molecular mass.

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 (the sum of 2 oxygen atoms), 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 mass, whether the substance is an element, an ionic compound, or a covalent compound.

Figure 6.2: The Amazing Mole. The major mole conversion factors are shown.

The relationship of the mole quantities to gram conversion factors listed above in Figure 6.2 are notably some of the most useful equations in all of chemistry. They make it possible to set up chemical reactions in a safe and efficient manner and they have tremendous impact on the economics of many industrial and manufacturing processes and the production of medicine. If you are a serious student of chemistry, I would recommend printing out table 6.2 and keeping a copy in your notebook. It will be extremely useful in setting up a multitude of word problems and is functionally useful in the laboratory.

So why is the relationship between the mole and compound mass so important? In the microscopic world, chemical equations are set up on the scale of molecules. Figure 6.3 depicts a typical chemical reaction. As we learned in Chapter 5, the coefficients in front of each compound represent the number of molecules needed for the reaction to proceed. In this combustion reaction, 2 molecules of butane (C₄H₁₀) react with 13 molecules of oxygen to produce 8 molecules of carbon dioxide and 10 molecules of water. In the lab, however, chemists are unable to count out molecules and place them in a reaction flask. Molecules are way too small to be seen by the naked eye and there is no equipment available that is capable of sorting and counting molecules in this way. Mass, on the other hand, can easily be measured using a balance. Thus, the relationship of mass to the number of molecules present becomes a very important conversion. Since the mole represents a fixed number of molecules (6.02 X 10²³), the coefficients used in chemical reactions can also be thought of as molar ratios. Instead of reading our equation in terms of molecules, we can read it in terms of moles. At the macroscopic level, the reaction below reads: 2 moles of butane (C₄H₁₀) react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

Figure 6.3 Combustion Reaction of Butane. The example shows the molecular ratios of the substrates and products of the reaction.

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6.4 Mole-Mass Conversions

The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion), as shown in Figure 6.2. 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 6.3). 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:

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.

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 6.4 is a chart for determining what conversion factor is needed, and Figure 6.5 is a flow diagram for the steps needed to perform a conversion. Note that it takes one mathematical step to convert from moles to mass or from mass to moles.

Figure 6.4 A Simple Flowchart for Converting between Mass and Moles of a Substance.

Figure 6.5 A Flowchart Illustrating the Steps in Performing a Unit Conversion.

To Your Health: Minerals

For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the non-carbon elements, our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission.

The US Department of Agriculture has established some recommendations for the daily intake (RDIs) of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.

Table 6.2: Recommended Daily Intake (RDIs) of Mineral Supplements

Table 6.2 illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women.

Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly.

Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich).

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6.5 Mole-Mole Relationships in Chemical Reactions

In this section you will learn how to use a balanced chemical reaction to determine molar relationships between the substances.

In Chapter 5, 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, 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:

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.

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6.6 Mole-Mass and Mass-Mass Problems

In this section you will learn to convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.

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:

Figure 6.6: Flowchart for Calculating Mole to Mass Conversions using Chemical Equations.

Collectively, these conversions are called mole-mass calculations.

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. Graphically, it is represented in these two steps:

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:

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:

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 calculations. 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:

Figure 6.7 Flow chart for conducting conversions using chemical equations.

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.

To Your Health: The Synthesis of Taxol

Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of C₄₇H₅₁NO₁₄. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.

After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.

Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.

Figure 6.8 The Structure of the Cancer Drug Taxol. Because of the complexity of the molecule, hydrogen atoms are not shown, but they are present on every atom to give the atom the correct number of covalent bonds (four bonds for each carbon atom).

6.7 Limiting Reagent and Percent Yield

By the end of this section, you will be able to:

Explain the concepts of theoretical yield and limiting reactants/reagents.
Derive the theoretical yield for a reaction under specified conditions.
Calculate the percent yield for a reaction.

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

Limiting Reagent

Consider another food analogy, making grilled cheese sandwiches (Fig. 6.9)

$1 slice of cheese + 2 slices of bread ⟶ 1 sandwich$

Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

This figure has three rows showing the ingredients needed to make a sandwich. The first row reads, “1 sandwich = 2 slices of bread + 1 slice of cheese.” Two slices of bread and one slice of cheese are shown. The second row reads, “Provided with: 28 slices of bread + 11 slices of cheese.” There are 28 slices of bread and 11 slices of cheese shown. The third row reads, “We can make: 11 sandwiches + 6 slices of bread left over.” 11 sandwiches are shown with six extra slices of bread.

Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

$H_{2} (s) + {Cl}_{2} (g) ⟶ 2 HCl (g)$

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reagent, and the other substance is the excess reagent. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H₂ and 2 moles of Cl₂. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.

An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example:

As seen above, the chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Fig. 6.10).

Figure 6.10. Finding the Limiting Reagent. When H₂ and Cl₂ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.

Figure from: Wacowich-Sgarbi, S. and the Langara Chemistry Dept. (2018) BCCampus

View this interactive simulation illustrating the concepts of limiting and excess reactants.

Example Problem:

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

$3 Si (s) + 2 N_{2} (g) ⟶ {Si}_{3} N_{4} (s)$

Which is the limiting reactant when 2.00 g of Si and 1.50 g of N₂ react?

Solution:

Test Yourself:

Which is the limiting reactant when 5.00 g of H₂ and 10.0 g of O₂ react and form water?

Answer

O₂

Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete. Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

Example:

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

${CuSO}_{4} (a q) + Zn (s) ⟶ Cu (s) + {ZnSO}_{4} (a q)$

What is the percent yield?

Solution:

For this problem, we have been given the actual yield of copper, which is 0.392 g, and we have been giving the starting concentration of the copper sulfate reactant at 1.274 g. Thus, we need to calculate the theoretical yield before we can access the percent yield.

Test Yourself:

What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF₂Cl₂ from 32.9 g of CCl₄ and excess HF?

${CCl}_{4} + 2 HF ⟶ {CF}_{2} {Cl}_{2} + 2 HCl$

Answer

48.3%

Green Chemistry and Atom Economy

The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at this website). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Fig 6.11). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%.

Figure 6.11 Green Chemistry and Ibuprofen Production. (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)

Figure from: Wacowich-Sgarbi, S. and the Langara Chemistry Dept. (2018) BCCampus

In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.

Key Takeaway

A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.

Exercises

Given the following unbalanced chemical equation,

H₃PO₄ + NaOH → H₂O + Na₃PO₄

what mass of H₂O is produced by the reaction of 2.35 mol of H₃PO₄?
Given the following unbalanced chemical equation,

C₂H₆ + Br₂ → C₂H₄Br₂ + HBr

what mass of HBr is produced if 0.884 mol of C₂H₆ is reacted?
Certain fats are used to make soap, the first step being to react the fat with water to make glycerol (also known as glycerin) and compounds called fatty acids. One example is as follows:
$C 3 H 5 (OOC(CH 2) 14 CH 3) 3 a fat + 3H 2 O \to C 3 H 5 (OH) 3 glycerol + 3CH 3 (CH 2) 14 COOH fatty acid$

How many moles of glycerol can be made from the reaction of 1,000.0 g of C₃H₅(OOC(CH₂)₁₄CH₃)₃?
Photosynthesis in plants leads to the general overall reaction for producing glucose (C₆H₁₂O₆):

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

How many moles of glucose can be made from the reaction of 544 g of CO₂?
Precipitation reactions, in which a solid (called a precipitate) is a product, are commonly used to remove certain ions from solution. One such reaction is as follows:

Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaNO₃(aq)

How many grams of Na₂SO₄ are needed to precipitate all the barium ions produced by 43.9 g of Ba(NO₃)₂?
Nitroglycerin [C₃H₅(ONO₂)₃] is made by reacting nitric acid (HNO₃) with glycerol [C₃H₅(OH)₃] according to this reaction:

C₃H₅(OH)₃ + 3HNO₃ → C₃H₅(ONO₂)₃ + 3H₂O

If 87.4 g of HNO₃ are reacted with excess glycerol, what mass of nitroglycerin can be made?
Antacids are bases that neutralize acids in the digestive tract. Magnesium hydroxide [Mg(OH)₂] is one such antacid. It reacts with hydrochloric acid in the stomach according to the following reaction:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

How many grams of HCl can a 200 mg dose of Mg(OH)₂ neutralize?
Acid rain is caused by the reaction of nonmetal oxides with water in the atmosphere. One such reaction involves nitrogen dioxide (NO₂) and produces nitric acid (HNO₃):

3NO₂ + H₂O → 2HNO₃ + NO

If 1.82 × 10¹³ g of NO₂ enter the atmosphere every year due to human activities, potentially how many grams of HNO₃ can be produced annually?
A simplified version of the processing of iron ore into iron metal is as follows:

2Fe₂O₃ + 3C → 4Fe + 3CO₂

How many grams of C are needed to produce 1.00 × 10⁹ g of Fe?
The SS Hindenburg contained about 5.33 × 10⁵ g of H₂ gas when it burned at Lakehurst, New Jersey, in 1937. The chemical reaction is as follows:

2H₂ + O₂ → 2H₂O

How many grams of H₂O were produced?

Answers

127 g
1.236 mol
23.9 g
0.251 g
1.61 × 10⁸ g

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6.8 Chapter Summary

To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.

Chemical reactions relate quantities of reactants and products. Chemists use the mole unit to represent 6.022 × 10²³ things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro’s number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. Molar masses of substances can be determined by summing the appropriate masses from the periodic table; the final molar mass will have units of grams.

Because one mole of a substance will have a certain mass, we can use that relationship to construct conversion factors that will convert a mole amount into a mass amount, or vice versa. Such mole-mass conversions typically take one algebraic step.

Chemical reactions list reactants and products in molar amounts, not just molecular amounts. We can use the coefficients of a balanced chemical equation to relate moles of one substance in the reaction to moles of other substances (stoichiometry). In a mole-mass calculation, we relate the number of moles of one substance to the mass of another substance. In a mass-mass calculation, we relate the mass of one substance to the mass of another substance.

Additional Exercises

If the average male has a body mass of 70 kg, of which 60% is water, how many moles of water are in an average male?
If the average female is 60.0 kg and contains 0.00174% iron, how many moles of iron are in an average female?
How many moles of each element are present in 2.67 mol of each compound?
1. HCl
2. H₂SO₄
3. Al(NO₃)₃
4. Ga₂(SO₄)₃
How many moles of each element are present in 0.00445 mol of each compound?
1. HCl
2. H₂SO₄
3. Al₂(CO₃)₃
4. Ga₂(SO₄)₃
What is the mass of one hydrogen atom in grams? What is the mass of one oxygen atom in grams? Do these masses have a 1:16 ratio, as expected?
What is the mass of one sodium atom in grams?
If 6.63 × 10⁻⁶ mol of a compound has a mass of 2.151 mg, what is the molar mass of the compound?
Hemoglobin (molar mass is approximately 64,000 g/mol) is the major component of red blood cells that transports oxygen and carbon dioxide in the body. How many moles are in 0.034 g of hemoglobin?

Answers

2,330 mol
1. 2.67 mol of H and 2.67 mol of Cl
2. 5.34 mol of H, 2.67 mol of S, and 10.68 mol of O
3. 2.67 mol of Al, 8.01 mol of N, and 24.03 mol of O
4. 5.34 mol of Ga, 8.01 mol of S, and 32.04 mol of O
H = 1.66 × 10⁻²⁴ g and O = 2.66 × 10⁻²³ g; yes, they are in a 1:16 ratio.
324 g/mol

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6.9 References:

Chapter 6 materials have been adapted and modified from the following creative commons resources unless otherwise noted:
1. Anonymous. (2012) Introduction to Chemistry: General, Organic, and Biological (V1.0). Published under Creative Commons by-nc-sa 3.0. Available at: http://2012books.lardbucket.org/books/introduction-to-chemistry-general-organic-and-biological/index.html
2. Poulsen, T. (2010) Introduction to Chemistry. Published under Creative Commons by-nc-sa 3.0. Available at: http://openedgroup.org/books/Chemistry.pdf
3. OpenStax (2015) Atoms, Isotopes, Ions, and Molecules: The Building Blocks. OpenStax CNX.Available at: http://cnx.org/contents/be8818d0-2dba-4bf3-859a-737c25fb2c99@12.

4. Wacowich-Sgarbi, S. and the Langara Chemistry Department. (2018) CHEM1114 – Introduction to Chemistry. BCcampus. Available at: https://pressbooks.bccampus.ca/chem1114langaracollege/

CH150: Chapter 6 – Quantities in Chemistry

Chapter 6 – Quantities in Chemical Reactions

6.1: Chapter Introduction

6.2: The Mole

6.3: Atomic and Molar Mass

6.4 Mole-Mass Conversions

To Your Health: Minerals

6.5 Mole-Mole Relationships in Chemical Reactions

2H2 + O2 → 2H2O

4H2 + 2O2 → 4H2O

22H2 + 11O2 → 22H2O

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

2 mol H2 + 1 mol O2 → 2 mol H2O

2H2 + O2 → 2H2O

6.6 Mole-Mass and Mass-Mass Problems

Fe2O3 + 3SO3 → Fe2(SO4)3

To Your Health: The Synthesis of Taxol

6.7 Limiting Reagent and Percent Yield

Limiting Reagent

Example Problem:

Solution:

Test Yourself:

Percent Yield

Example:

Solution:

Test Yourself:

Green Chemistry and Atom Economy

Key Takeaway

Exercises

Answers

6.8 Chapter Summary

Additional Exercises

Answers

6.9 References:

Tools

Resources

2H₂ + O₂ → 2H₂O

4H₂ + 2O₂ → 4H₂O

22H₂ + 11O₂ → 22H₂O

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

2 mol H₂ + 1 mol O₂ → 2 mol H₂O

2H₂ + O₂ → 2H₂O

Fe₂O₃ + 3SO₃ → Fe₂(SO₄)₃