Chapter 1: Introductory Ideas

CH1010-1040 homepage

1-1:The Evolution of the Science of Chemistry
1-2:Atomic Masses
1-3:Formulas
1-4:Ions
1-5:Types of Compounds
1-6:The Mole
1-7:Stoichiometric Calculations for Compounds
1-8:Chemical Reactions
Applications

Major Concept Area: The Atomic/Molecular View of Matter

Specific Concepts in this Chapter:

Macroscopic, Microscopic, and Molecular Viewpoints. The scope of scientific inquiry is magnificent. It ranges from the very small (the size of an elementary particle, 10-16 meters) to the very large (the size of the universe, 1026 meters); the very brief (the time required for a photon of light to traverse an atomic nucleus, about 10-24 seconds) to the very lengthy (the age of the universe, about 15 billion years, or 1018 seconds); the very light (the mass of an electron, 10-27 grams) to the very heavy (the mass of all of the matter in the universe, presently unknown, but certainly larger than 1058 g). Thus, scientists must be adept at discussing and comprehending different scales of size, time, mass, energy, electric charge, and so on. Chemists and physicists in particular use words that give some indication of the scale in which they are thinking. Several such words are macroscopic, microscopic, and atomic (molecular). A macroscopic phenomenon is an event or property that is characteristic of an ensemble (i.e., a large collection) of atoms or molecules. It can be observed in a sample of matter of large size, where large implies that the sample can be weighed. The best available analytical balances are capable of detecting a mass of 1 microgram (10-6 g). A microscopic phenomenon is one that can be viewed with an optical microscope, an instrument that allows magnification factors as high as 1000. An atomic (or molecular) phenomenon is an event or property that is characteristic of an individual atom or molecule, or of a small collection of atoms or molecules. Such small samples of matter are too small to be weighed, and too small to be seen with an optical microscopic. Macroscopic properties represent averages or cumulative totals of atomic/molecular properties. For example, the speed of a gas molecule is a molecular property. The macroscopic manifestation of the molecular speed is the temperature of the gas. Thus temperature, which we can observe on the macroscopic scale and can in fact discuss without any reference to molecules, is a reflection of average behavior on the molecular scale. Similarly, the momentum of a gas molecule is a molecular property. The cumulative result of huge numbers of molecules transferring momentum to the container walls during collisions is the macroscopic property called pressure. Again, we may detect and discuss pressure for large samples of gas without any reference to molecules, even though it is a direct consequence of their existence. As you progress through this text, you will find yourself becoming conversant with both the macroscopic and atomic/molecular viewpoints. It is important that you realize, though, that the molecular interpretation is very often inferred from our observations of macroscopic behavior. Only recently have we become technologically capable of observing the atomic realm. In doing so, we have found that our inferences about it, made from many observations over many many years, are largely correct.

The subject matter of this textbook is the science of chemistry. We may define chemistry as the branch of science concerned with the molecular structure of matter, the relation between this structure and the behavior of matter, and the practical applications of this behavior. This definition takes us into the realm of atoms and molecules. As you progress through your study of chemistry, you will become more comfortable in this realm. Throughout the journey, we attempt to keep you focussed on the questions, "What do the molecules look like, what are they doing, and how is this related to the properties and behavior of the material in which we are interested?" By the end, you should have a good sense of the manner in which chemists look at the world.

1-1 The Evolution of the Science of Chemistry. It is assumed from the outset that you know a bit about matter, so we will spend minimal time on concepts that you have probably learned before. We will devote just a few lines to the concept of a pure substance, which may be either an element or a compound. A sample of matter is considered to be a pure substance if it cannot be separated into two or more distinctly different components by physical methods. A physical method can be something as simple as using a pair of forceps (tweezers) to separate two clearly different types of particles from a sample, to something as sophisticated as instrumental chromatography. A pure substance has exactly the same composition throughout its bulk; this composition remains unchanged no matter what physical operations we carry out on the substance. An element is a fundamental pure substance. It cannot be separated into other pure substances by chemical means (such as decomposing by heat or light). In modern terms, it consists of just one kind of atom. A compound is a combination of two or more elements with a specific, reproducible composition, which is the same throughout the bulk of any sample of the compound. In modern terms, a compound consists of combinations of atoms of two or more elements in very specific arrangements. We will have much more to say about this later. Appendices A-D should be consulted for discussion of the basic introductory subject material that we have briefly considered here.

Almost all of what we have to say in this book pertains to pure substances. It is only rarely possible to do meaningful experiments on matter that is not pure. Consequently, the concern of the chemist is with pure substances. The chemist is a manipulator of matter; s/he must start from a known point, with a known and pure material, to manipulate successfully.

Modern science began when people began to do experiments. Experiments consist of making observations of various types on well-controlled, well-defined situations and processes. From careful experiments it is possible to notice regularities in nature that are not evident in the erratic and chaotic course of ordinary life. Having done experiments, one can proclaim the regularities with confidence, and attempt to find explanations for them. One of the first individuals to take the experimental approach was Antoine Lavoisier, a native of France, who at the end of the 18th century carried out a series of well-controlled experiments involving the burning of substances. Lavoisier found that, to within the fairly large margin of error that his equipment required, the total mass of the products of a chemical reaction is the same as the total mass of the reactants; mass is conserved in chemical processes.

Law of Conservation of Mass: In any chemical reaction, the total mass of products is the same as the total mass of reactants.

Thus when a steel garden tool rusts as the iron of which it is composed reacts with oxygen from the air, the mass of rust obtained is equal to the sum of the masses of iron and oxygen used to form it.

Only a few years after Lavoisier did his experiments, Joseph Proust discovered via careful experiments of his own that simple chemical compounds such as sodium chloride (modern name) and water always contain the same elements in the same relative amounts, no matter what the source of the particular sample analyzed. Thus samples of water from a lake, from a distant stream, and even from the ocean (once impurities are removed) all contain only hydrogen and oxygen, and always contain 8 times more oxygen than hydrogen, by mass. The compound ammonia is formed by reaction of the elements nitrogen (N) and hydrogen (H). No matter what its source or how it is made, ammonia always contains 82.2% nitrogen and 17.8% hydrogen by mass. The relative amounts of N and H in an ammonia sample are always the same. Observations like these led Proust to propose the

Law of Definite Proportions: the relative masses of the elements in a compound is always the same, regardless of the source of the sample of compound tested.

Finally, an interesting corollary to the Law of Definite Proportions became known at about this same time. This corollary, called the Law of Multiple Proportions, is stated (rather clumsily) as follows.

Law of Multiple Proportions: If two elements, A and B, combine to form two different compounds, the masses of B that combine with a particular mass of A give a whole number ratio.

For example, in addition to ammonia, nitrogen and hydrogen form a compound called diazene that contains 93.3% nitrogen and 6.7% hydrogen by mass. The mass of hydrogen that would combine with 1.00g of nitrogen to form diazene is therefore

1.00 g nitrogen * (6.7 g hydrogen/93.3 g nitrogen) = 0.0718 g hydrogen

Similarly, the mass of hydrogen combining with 1.00 g of nitrogen to form ammonia is

1.00 g nitrogen * (17.8 g hydrogen/82.2 g nitrogen) = 0.217 g hydrogen

The ratio of these two masses of hydrogen is 0.217/0.0718 = 3, in agreement with the Law of Multiple Proportions.

These three laws are the foundation of modern chemistry. They provided the basis for the first credible theory of atoms, proposed by John Dalton.

The Atomic Theory. In 1803, John Dalton proposed that all matter is composed of atoms. Today we know this to be true, because we have seen atoms using advanced microscopic techniques. The postulates that Dalton proposed, although in some cases not altogether true, remain valid for the most part today. They are given below, with the untrue parts enclosed in square brackets:

  1. Each element is composed of extremely small, [indivisible] particles called atoms.
  2. [All atoms of an element are identical.]
  3. Atoms of different elements have different properties (including different mass).
  4. Atoms of an element are not changed into different types of atoms by chemical reactions; atoms are neither created nor destroyed in such reactions.
  5. Compounds are formed when atoms of more than one element combine.
  6. In a particular compound, the relative number and kind of atoms are constant.

The atomic theory is the single most important idea in science. Its essential statement merits repeating, with due emphasis:

All matter is composed of atoms, either single atoms or well-defined combinations of atoms.

Dalton's theory explains the Laws of Conservation of Mass and Definite Proportions, both well established before Dalton proposed his atom theory. The 4th postulate above recognizes the validity of the law of mass conservation. The Law of definite proportions is recognized in postulates 2, 5, and 6. With Dalton's theory in hand, the interpretation of the Law of Multiple Proportions is clear: there are 3 times as many hydrogen atoms per nitrogen atom in ammonia as there are in diazene. The integral mass ratios strongly imply the existence of atoms with characteristic masses.

Since the late 1800's, we have learned that atoms are not indivisible. They are composed of still smaller particles, arranged in an interesting way.

The Nuclear Atom. Atoms are composed of three smaller particles, called subatomic (or elementary) particles: the electron (e), the proton (p), and the neutron (n). These have the properties given in Table 1-1.


ParticleMass in gramsMass in amuCharge in electronic charge units
electron9.11 * 10-280.0055-1
proton1.673 * 10-241.0073+1
neutron1.675 * 10-241.00870
Table 1-1: Subatomic Particles



The electron and the proton carry electrical charge. The electron is negatively charged, the proton positively charged, with the magnitudes of the charges being exactly the same at 1.602 * 10-19 coulombs (C). This amount of charge is, for convenience, defined as 1 electronic charge unit. When we speak of the charge of an electron, we will call it -1, but understand that this actually means -1.602 * 10-19 coulombs. An atom is electrically neutral; it contains the same number of protons and electrons. The protons and neutrons are concentrated together in a small region of space called the nucleus. The diameter of the nucleus is about 1 * 10-15 m. These two types of particles are bound together in the nucleus by the so-called nuclear strong force, which is much more powerful than the repulsive electrical force that the protons exert on each other (particles of like electrical charge repel each other). The nuclear strong force is still incompletely understood, primarily because we are not yet able to create energies in the laboratory sufficient to overcome it. The electrons are found in a relatively large volume (diameter about 1 * 10-10 m) surrounding the nucleus, and are held to it by the attractive electrical force between positive protons and negative electrons. The magnitude of this attractive force is given by Coulomb's Law, which states that the force between two electrical charges, q1 and q2, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them:

(1-1-1): F = kq1q2/r2

The electrons are often said to "orbit" the nucleus, but this planetary view of the situation is not correct, because it implies that the electrons are behaving like well-defined little balls of matter. When they are confined in atoms, electrons actually behave more like waves than like particles. They are distributed about the nucleus like standing waves on a string with both ends tied down. It may therefore be best to think of the atom as a tiny massive nucleus surrounded by a somewhat diffuse cloud of electron density. This is pictured in Figure 1-1.

The size of the nucleus is exaggerated in this picture so that you can see it. Drawn to scale, it would be too small to see. Chemistry is concerned with the behavior of the electrons in the atom, and seldom ventures into the nuclear regime. However, the electrical pull of the nucleus on the electrons, particularly the outer ones, is of tremendous importance in chemical behavior. Be mindful of the presence of the nucleus at the center.

Atoms of different elements are distinguished by the number of protons in the nucleus. Hydrogen atoms have 1 proton in the nucleus, carbon atoms 6, iron atoms 26, etc. The number of protons in the nucleus of the atom of an element is called the atomic number (symbol Z) of the element. The total number of protons and neutrons in the nucleus is the mass number, A. Although all atoms of a particular element have the same number of protons, they may have different numbers of neutrons. Atoms of an element with different numbers of neutrons are called isotopes. Carbon has 3 isotopes, each with 6 protons and with 6, 7, and 8 neutrons respectively. These have mass numbers 12, 13, and 14 and are distinguished as follows:

126C
136C
146C

An arrangement of the known elements in increasing order of atomic number Z is called the Periodic Table of the Elements. Figure 1-2 shows the so-called long form of the Periodic Table. The rows of the table are called periods; the columns, numbered 1, 2, 3f through 16f, and 3 through 18, are called groups or families, because all elements in a column exhibit similar chemistry. (For example, the elements in group 1 (except hydrogen) are all metals; react with water to produce hydrogen gas; react vigorously with chlorine to produce salts; and have relatively low melting points.) The periodic (i.e., repeating) behavior of the properties of the elements is readily understandable in terms of the quantum mechanical theory of the atom, which we will take up in Chapter 2.

1-2 Atomic Masses. The concept of atomic mass is crucial to Dalton's atomic theory. With the advent of this theory, chemists became concerned with the masses of atoms and how to measure them. Direct measurement of the masses of single atoms was recognized to be impossible. However, chemists realized that determination of the relative masses of the atoms of different elements would allow great progress to be made in understanding the atomic compositions of molecules, even if the absolute masses could not be determined. For example, suppose one knew that a carbon atom was 12 times heavier than a hydrogen atom. Then the fact that a compound contains 12 g of carbon for each 4 g of hydrogen allows the conclusion that the compound contains 4 atoms of hydrogen per atom of carbon.

When Dalton made his proposals, there was no compound in which the numbers of atoms of each type were known. However, combining masses of the elements were known for a number of compounds. Thus water was known to consist of 89% O and 11% H, indicating that in forming water, 8 g of O combine with each 1 g of H. Although much combining mass data was available, atomic masses could not be assigned to the elements until at least one formula was known. This presented a thorny problem, that was eventually solved by the efforts of Gay Lussac, who worked on the combining volumes of gases; Cannizzaro, and Avogadro.

The Modern Atomic Mass Scale/Mass Spectrometry. We have available to us now a set of atomic masses that are crucial to all quantitative work in chemistry. The masses are based on the arbitrary assignment of a mass of exactly 12.000... atomic mass units to an atom of the 126C isotope. For convenience, this definition gives each element an atomic mass that is numerically close to (but not exactly equal to) its mass number, A (the number of protons and neutrons in the nucleus). The atomic mass unit, symbolized m, is then 1/12 the mass of an atom of 12C. Today we know that 1 m = 1.660 x 10-24 g. The units of atomic mass are m/atom. Atomic masses are included with the symbols of the elements on the inside front cover of the text.

The very precise atomic masses that you find in the periodic table, or in listings of atomic masses, were for the most part not determined from combining mass data. Instead, they were determined by Mass Spectrometry. The operation of a mass spectrometer is illustrated schematically in Figure 1-3a.

A sample of the substance whose mass is to be determined is vaporized in the ion source, and the gaseous atoms or molecules are bombarded with a beam of high energy particles. These cause the ejection of an electron from several of the atoms/molecules, which become positively charged. The positive ions are accelerated through an electric field of known strength and are then passed through a slit, which focuses them to a narrow beam. The beam of positive ions is passed through a magnetic field of adjustable strength, which causes the beam to follow a curved path. The extent of curvature of the path depends on the mass of the ion, which can be determined from the position at which the beam strikes the detector. A modern mass spectrometer can distinguish between ions that differ in mass by as little as 0.0005 m. The mass spectrum of mercury, Hg, is shown in Figure 1-3b. Seven isotopes are evident.

1-3 Formulas. Many chemical compounds consist of molecules. We may define a molecule as a grouping of atoms bound tightly enough together to act as a single entity. For example, a molecule of ammonia consists of one atom of nitrogen and three atoms of hydrogen bound together in a very specific way. A formula is the shorthand representation for a compound, based on the symbols for the constituent elements. Each element symbol is followed by a subscript to indicate the number of atoms of the element in a molecule of the compound. If only one atom occurs, no subscript is used. The formulas for several common compounds are indicated in Table 1-2.


CompoundFormula
waterH2O
ammoniaNH3
methaneCH4
carbon dioxideCO2
sulfur trioxideSO3
glucose (sugar)C6H12O6
Table 1-2

Chemists use several types of formulas, which convey different levels of information. In order of increasing information content, these types are as follows.

a. The empirical formula. It is known that 1.00 g of hydrogen peroxide contains 94.1% oxygen and 5.9% hydrogen by mass. To find the relative numbers of atoms of H and O in a molecule of hydrogen peroxide, we find the ratio of masses of hydrogen and oxygen in the compound and compare this with the ratio of masses of H and O atoms:

mass H in hydrogen peroxide/mass O in hydrogen peroxide = (# atoms H x atomic mass H)/(# atoms O x atomic mass O)

Solving this for the ratio of numbers of atoms, we obtain 1.00. For every atom of O in a molecule of hydrogen peroxide, there is one atom of H. The simplest formula that conveys this information is HO, which is called the empirical formula for hydrogen peroxide.

b. Molecular formula. The molecular formula for a compound shows the actual numbers of atoms of each element present in a molecule. The molecular formula for hydrogen peroxide is H2O2. The formulas given in Table 1-2 are all molecular formulas.

c. Structural formula. The structural formula for a molecule indicates not only how many of each type of atom is present, but also how the atoms are bonded to each other in the molecule. For example, the structural formulas for hydrogen peroxide and ammonia are shown here.

The lines between atoms represent the chemical bonds holding atoms together. The structural formula gives more information than the molecular formula, which in turn is more informative than the empirical formula. The primary goal of a chemist who synthesizes a new compound is to determine the structural formula.

Chemists follow rules when writing formulas for compounds. Unfortunately, different sets of rules are used for compounds of different types. For now, we state two rules that are particularly useful:

For carbon-based compounds, the elements are written in the order, C, H, N, O, others.
For binary (2-element) compounds, the more positive element is written first.
In short order, we will see that in many binary compounds, the more positive element is easily discerned.

Formula Mass and Molecular Mass. Given the formula for a substance, we may define a quantity called the formula mass as the sum of the atomic masses of the atoms in the formula. This places a formula unit of a compound on the same mass scale as the atoms of the elements. Example 1-1 shows a formula mass calculation.


Example 1-1. Calculate the formula mass of magnesium chloride, MgCl2.
Solution. Add the masses of the atoms in the formula: FW = atomic mass(Mg) + 2*atomic mass(Cl) = 24.305 + 2(35.453) = 95.211 m

For compounds such as those in Table 1-2, consisting of discrete molecules (i.e., individual units that contain a characteristic number of atoms of each constituent element), the formula mass is called the molecular mass. The molecular mass is the actual mass of a molecule of the compound, in m. Thus the molecular mass of a molecule of ammonia, NH3, is the sum of the masses of 1 N atom and 3 H atoms: 17.031 atomic mass units. For compounds that do not consist of discrete molecules, such as sodium chloride (NaCl, table salt), we must be content with the formula mass. The formula mass of NaCl is 22.990 + 34.453 = 58.443 m. This is the mass of one formula unit of NaCl. A formula unit consists of one Na atom and one Cl atom, as written in the formula.

The Quantitative Implications of Formulas. A chemical formula carries a lot of quantitative information. This is readily illustrated by example.


Example 1-2. The molecular formula for acetaminophen, the active ingredient in the pain reliever, tylenol, is C8H9NO2. What quantitative relationships are implied by this formula?
Solution. There are a great many such relationships:

There are many others. Reasoning similar to that in the last two statements was used in discussing the Law of Multiple Proportions earlier.

There are some other important quantitative relationships implied in formulas that we will return to shortly.

1-4 Ions. Chemical reactions involve the electrons in the atom, and leave the nuclei untouched. This is because it is relatively easy to either remove electrons from or add them to an atom. Doing so destroys the charge balance of the atom, and produces an electrically charged particle called an ion. Adding one or more electrons to an atom gives a negative entity called a negative ion, or anion. Removing one or more electrons gives a positive entity called a positive ion, or cation. In both cases, the algebraic difference between the numbers of protons and electrons in the ion is the electrical charge of the ion, in electronic charge units. This is indicated as a right superscript.


Example 1-3. Describe the formation of the Na+, Cl-, Ca2+, and S2- ions from the atoms.

Solution. Descriptions are in the table.

atomionp - echarge
Na atom Na+ by removal of 1 e11-10 = 1
Cl atomCl- by addition of 1 e17-18 =-1
Ca atomCa2+ by removal of 2 e20-18 = 2
S atomS2- by addition of 2 e16-18 = -2

Having introduced ions, we ask an obvious question: Why does sodium, Na, prefer to lose an electron to give Na+, while chlorine, Cl, prefers to gain one to form Cl-? The answer is rooted in the detailed structure of the atom, which we address in Chapter 2. Nonetheless, we can still predict easily which elements tend to form cations and which anions, and further, how many electrons a particular atom is likely to gain or lose, based on the position of the atom in the Periodic Table (Figure 1-2). At the right side of the table is a zig-zag line bordering boron on the left and ending between astatine (At) and polonium (Po). This is the boundary between metals (to the left of and below the line) and non-metals (above and to the right). Metals have the following properties:

  1. a shiny, reflective appearance;
  2. ability to conduct heat and electricity;
  3. malleability (capable of being hammered into sheets) and ductility (drawn into wires);
  4. High melting and boiling points (with many exceptions, for example, mercury!)
Non-metals are very different:

  1. poor conductivity of heat and electricity;
  2. brittleness -- cannot be drawn or hammered without fracturing or crumbling.
  3. Many are gases at room temperature, and those that are solids are dull in appearance.
Elements adjacent to the boundary line are called semimetals or metalloids, because they have some properties of both under certain conditions. Based on this classification, we now state some rules about ion formation.
  1. Metals usually form cations, non-metals anions.
  2. Metals in groups 1,2, and 13 form cations having 1, 2, and 3 + charges, respectively. Thus the expected charge is given by the last digit of the group number. The following are examples:
    ElementGroupPredicted Cation
    Li1Li+
    Be2Be2+
    Al13Al3+
    Ba2Ba2+
    Cs1Cs+
    Tl13Tl3+ (also Tl+)
  3. Non-metals tend to form anions with the same number of electrons as the element in group 18 that follows them in the periodic table. The charge on the anion is obtained by subtracting 8 from the last digit of the group number.
    ElementGroupGroup # -8Predicted anion
    N 15 5-8 = -3 N3-
    O166-8 = -2O2-
    Cl177-8 = -1Cl-
    Se166-8 = -2Se2-
  4. Hydrogen forms the positive ion, H+. This ion plays a unique and important role in chemistry. Note that, since the hydrogen atom has only one electron, the hydrogen ion is a bare nucleus consisting of a single proton! We will frequently refer to the H+ ion as the proton.

These rules apply to simple ions, called monatomic ions, formed by adding electrons to or removing them from a single atom. It is also possible to have complex, or polyatomic, ions containing a group of 2 or more atoms, the group as a whole carrying a + or - charge.

Polyatomic ions. A polyatomic ion is a group of atoms bound together as in a molecule, which carries a net + or - charge due to a deficiency or surplus of electrons. The ammonium ion, NH4+, is a polyatomic cation. The sulfate ion, SO42-, is a polyatomic anion. We cannot now state a simple rule for predicting formulas and charges for polyatomic ions. This requires more extensive knowledge of the electronic structure of atoms and the principles of chemical bonding. However, it is important that you be familiar with the formulas and charges of at least the common polyatomic ions. These are listed in Table 1-3, which should be committed to memory.

Compounds containing polyatomic ions are common and numerous. Simple examples are sulfuric acid, H2SO4; phosphoric acid, H3PO4; ammonium perchlorate, NH4ClO4; magnesium nitrate, Mg(NO3)2. The last example shows that when the formula for a polyatomic ion occurs more than once in the formula for a compound, the ion formula is enclosed by parentheses and an appropriate subscript is written after it.


Example 1-4. Aluminum sulfate has 2 aluminum and 3 sulfate ions per formula unit. What is its formula?
Solution. Remembering to enclose the polyatomic ion in parentheses, we obtain

Al2(SO4)3

Aluminum is written first in the formula because it is positive; the sulfate anion is negative.
Table 1-3: Common Polyatomic Ions
NameFormula
ammoniumNH4+
acetateC2H3O2-
azideN3-
carbonateCO32-
hydrogen carbonateHCO3-
hypochloriteClO-
chloriteClO2-
chlorateClO3-
perchlorateClO4-
chromateCrO42-
dichromateCr2O72-
cyanateOCN-
cyanideCN-
hydroxideOH-
nitriteNO2-
nitrateNO3-
permanganateMnO4-
peroxideO22-
phosphatePO43-
hydrogen phosphateHPO42-
dihydrogen phosphateH2PO4-
phosphitePO32-
sulfiteSO32-
hydrogen sulfiteHSO3-
sulfateSO42-
hydrogen sulfateHSO4-
thiosulfateS2O32-

1-5 Types of Compounds. Many compounds consist of discrete molecules, in which atoms are tightly bonded together, but which act as independent units. Examples are water (H2O), methane (CH4), ethanol (C2H6O), and ammonia (NH3). However, many compounds are not molecular. An example is sodium chloride, NaCl, which consists of Na+ and Cl- ions arranged in a regular, essentially infinite, 3-dimensional array called a crystal lattice. A portion of this lattice is shown in Figure 1-4.

Each Na+ ion is surrounded by 6 Cl-, spaced at equal distances from it; similarly, each Cl- is surrounded by 6 Na+. It is not meaningful to associate any one Na+ with any one Cl-, so no discrete molecules exist. It is more proper to consider the entire NaCl crystal as a single giant molecule containing equal huge numbers of Na+ and Cl- ions. NaCl and other compounds with lattice structures consisting of ions are called ionic compounds. The formulas for such compounds are empirical (not molecular), showing the smallest relative numbers of ions necessary to give electrical neutrality. For these compounds, we calculate a formula mass rather than a molecular mass, and we deal in formula units rather than molecules.


Example 1-5. Predict the formula for the ionic compound formed between aluminum and oxygen.
Solution. First predict ion charges: Al3+ and O2-, based on periodic table position. Now write an electrically-neutral formula. This requires 2 Al3+ for each 3 O2-. Thus

Al2O3

Aluminum is written first in the formula because it is the positive species; oxygen is negative.

Ionic compounds such as NaCl, KBr, and Al2O3 are formed by transfer of electrons from one atom to another to form ions. The resulting ions exert strong electrical attractive forces on one another called ionic bonds, which hold them together in the orderly lattice. In contrast, compounds such as water, consisting of discrete molecules, are called molecular (covalent) compounds. Bonds between atoms in these molecules result not from electron transfer, but from electron sharing. The bonds are called covalent bonds.

Given 2 elements, will they form an ionic or a covalent compound? There are 3 simple guidelines that allow such predictions to be made.

  1. If one element is a metal, the other a non-metal, the resulting compound is ionic. Formulas of the compounds can be predicted as in Example 1-5 above. Examples:
    SubstanceMetalNon-metal
    NaCl Na Cl
    CaF2 Ca F
    CsI Cs I
    Al2S3 AlS
  2. Compounds between metals and complex anions are ionic (however, bonding within the anion is covalent).
  3. If both elements are non-metals, the compound is covalent.

We have seen that formulas for ionic compounds are easy to obtain. Those for covalent compounds are as well, with an understanding of valence. The valence of an atom is the number of bonds that the atom can form to other atoms. The primary valence for an atom is predictable, just as is ionic charge. We can see this by looking at the formulas for the hydrogen and fluorine compounds of elements in periods 2 and 3 of the periodic table:

Group:121314151617
Period 2 cpd with H:LiH BeH2 BH3 CH4 NH3 OH2 FH
with F:LiF BeF2 BF3 CF4 NF3 OF2 FF

The number of bonds formed to hydrogen and fluorine varies in a systematic way from left to right across a row of the periodic table, from 1 at the left, to a maximum of 4 in the middle, and back to 1 at the right. These observations suggest two rules of valence.

  1. For atoms in groups 1, 2, 13, and 14, the valence is the last digit of the group number. Thus
    ElementGroupValence
    H 1 1
    Mg 2 2
    Ga 13 3
    C 14 4
  2. For atoms in groups 15 to 18, valence is equal to (8 - last digit of group #).
    element GroupValence
    O 168-6 = 2
    Cl 17 8-7 = 1
    N 15 8-5 = 3

Note that the way valence is defined requires it to be a positive number. The rules are stated so as to guarantee this.

With these rules, the formula for water is sensible. Oxygen (valence 2) makes 2 attachments; hydrogen (valence 1) makes 1 attachment. One oxygen binds with 2 hydrogens, as follows:

H-O-H

The formula HO satisfies the valence of hydrogen, but not that of oxygen. It is therefore incorrect. Similarly the formula for CO2 is sensible:

O=C=O

Carbon has 4 attachments to satisfy a valence of 4; each oxygen has 2 attachments to satisfy a valence of 2.

1-6 The Mole. The mole is the unit of amount in chemistry. It provides a bridge between the atom and the macroscopic amounts of material that we work with in the laboratory. It allows the chemist to weigh amounts of two substances, say iron and sulfur, such that equal numbers of atoms of iron and sulfur are obtained. A mole of a substance is awkwardly defined as the mass of substance containing the same number of fundamental units as there are atoms in exactly 12.000...g of 12C. Fundamental units may be atoms, molecules, or formula units, depending on the substance concerned. At present, our best estimate of the number of atoms in 12.000...g of 12C is 6.022 * 1023, a huge number of atoms. This is obviously a very important quantity. For historical reasons, it is called Avogadro's Number, and is given the symbol No.

Unfortunately, the clumsy definition of the mole obscures its utility. It is nearly analogous to defining a dozen as the mass of a substance that contains the same number of fundamental units as are contained in 733 g of Grade A large eggs. This definition completely obscures the utility of the dozen: that it is 12 things! Similarly, a mole is No things. The mole is the same kind of unit as the dozen -- a certain number of things. But it differs from the dozen in a couple of ways. First, the number of things in a mole is so huge that we cannot identify with it in the way that we can identify with 12 things. Second, 12 is an important number in the English system of weights and measures, so the definition of a dozen as 12 things makes sense. However, the choice of the unusual number, 6.022 * 1023, as the number of things in a mole seems odd. Why is this number chosen? Would it not make more sense to define a mole as 1.0 * 1023 things, a nice (albeit large) integer that everyone can easily remember? To understand why the particular number, 6.022 * 1023 is used, it is necessary to resurrect an older, in some ways more sensible and useful, definition of the mole, which is grounded in the atomic mass scale addressed above.

The atomic mass scale defines the masses of atoms relative to the mass of an atom of 12C, which is assigned a mass of exactly 12.000.. atomic mass units (m). The number 12 is chosen so that the least massive atom, hydrogen, has a mass of about 1 (actually 1.008) on the scale. The atomic mass unit is a very tiny unit of mass appropriate to the scale of single atoms. Originally, of course, chemists had no idea of its value in laboratory-sized units like the gram. The early versions of the atomic mass scale were established by scientists who had no knowledge of the electron, proton, or neutron. When these were discovered in the late 19th and early 20th centuries, it turned out that the mass of an atom on the atomic mass scale was very nearly the same as the total number of protons and neutrons in its nucleus. This is a very useful correpondence, but it was discovered only after the mass scale had been in use for a long time.

In their desire to be able to count atoms by weighing, chemists gradually developed the concept of the "gram-atomic mass", which was defined in exact correspondence with the atomic mass scale:

1 atom of 12C has a mass of 12.000 amu
1 gram-atomic mass of 12C has a mass of 12.000 g

Thus the gram-atomic mass of an element was defined as the atomic mass of the element, expressed in grams. Because the atomic mass scale is numerically preserved in the definition of gram atomic mass, 1 gram-atomic mass of any element could be immediately determined as the atomic mass in grams. Thus 1 gram-atomic mass of sulfur is 32.06 g; 1 gram-atomic mass of hydrogen is 1.008 g, and so on. Analogous terms, such as gram-molecular mass for the molecular mass of a compound expressed in grams, were similarly used. However, having to use a different term depending on whether elements or compounds were being discussed was awkward and inconvenient. For this reason, the term "molar mass (abbrev MM)" (the mass of 1 mole) was adopted to signify the atomic, molecular, or formula mass of a pure substance expressed in grams.

Alternative definition of the mole: a mole of a pure substance is the atomic, molecular, or formula mass of the substance, expressed in grams.

Thus one mole of ethyl alcohol, C2H6O, is 46.069g. One mole of water is 18.015 g. If we mix 46.069 g of ethyl alcohol with 18.015 g of water, we can be assured that the mixture contains 1 molecule of ethyl alcohol per molecule of water. Further, we will know that there are 2 atoms of C and 8 atoms of H per each 2 atoms of O. Thus the mole allows us to measure convenient amounts of material containing known numbers of atoms; i.e., it allows us to count atoms.

The mole enables us to count atoms in the laboratory.

The mole is useful whether or not we know how many atoms of carbon-12 there are in 12.000 g of carbon-12. If we measure one mole of iron and one mole of sulfur, we know that these two samples contain the same number of atoms. This is the important aspect of the mole. How many atoms there are in a mole is of subsidiary importance. Nonetheless, it has become possible to determine this number. It is, of course, 6.022 * 1023 atoms per mole. We thus see that this number is simply a consequence of the choice that 1 mole be the formula mass in grams. It is very nice that we know it; but we do not need to know it for the mole to be useful.

I would even go so far as to say that the modern definition of the mole in terms of a certain number of atoms of 12C is unfortunate, in that it suggests that the number, 6.022 * 1023 things/mole, must be used in any and every calculation involving moles! In practice, we seldom need to know how many atoms or molecules we are working with, so in mole calculations the number 6.022 * 1023 is rarely used. What is invariably used (except for sample calculations in chemistry textbooks; see below!) is the fact that 1 mole of substance is its atomic, formula, or molecular mass in grams.

The dual definitions of the mole can be used to find the mass of 1 m expressed in g. Exactly 12 g of carbon contains No atoms, each with a mass of exactly 12 m. In equation form,

No atoms/mole * 12 m /atom = 12 g/mole

Simplifying, we obtain No m = 1 g. It follows that 1 m = 1/No g = 1.660 * 10-24 g. Avogadro's number is an experimentally measured quantity. Although we are confident that we know its value quite well, some future experiment may cause us to make a small revision in the number. By necessity, the mass of 1 m in grams will change accordingly. This is not worrisome, because neither number is crucial to the utility of the mole.

The importance of the mole concept can be summed up as follows: any statement that can be made about the number of atoms of an element in a molecule or formula unit of a substance can also be made about the number of moles of an element in a mole of the substance. This is true because 1 mole of substance contains No atoms, molecules, or formula units of substance. Based on the formula for glucose, C6H12O6, we can make the following statements:

You might now look back at Example 1-2 and apply these ideas to the formula for acetaminophen. Some examples will familiarize you with the mole.

Example 1-6. How many moles of Fe are in 5.6 g Fe? How many Fe atoms are contained in the sample?
Solution. By definition, 1 mole of Fe is 56.0 g. 5.6 g Fe is therefore 0.1 mole of Fe. The number of Fe atoms in the sample is

0.1 mole * 6.022 x 1023 atoms/mole = 6.022 * 1022 atoms.

5.6 g of iron is not much iron. However, even this small amount contains a huge number of iron atoms.

Example 1-7. How many sulfur atoms are in 1.56 g sulfur?
Solution. We can calculate the number of moles of sulfur from the molar mass and the given mass. Once we have this, the number of atoms is obtained from Avogadro's Number, No.

moles S = 1.56 g/(32.06 g/mole) = 0.0487 mole
atoms S = 0.0487 moles S * 6.022 * 1023 atoms/mole = 2.93 * 1022


Example 1-8. What is the mass in g of 1 atom of sodium?
Solution. If we know the mass of 1 mole of Na, and how many atoms are in a mole, the mass of a single atom should be easy to obtain:

mass Na atom = 22.99 g Na/mole * 1 mole/6.022 * 1023 atoms
= 3.817 * 10-23 g.


Example 1-9. What mass of sulfur contains the same number of moles as are in 10.0 g Fe?
Solution. Figure the number of moles of Fe. This is the desired number of moles of S. Convert moles of S to mass of S using the molar mass.

moles Fe = 10.0 g Fe/(55.85 g Fe/mole) = 0.1791 moles
moles S = moles Fe = 0.1791
g S = 0.1791 moles S * 32.06 g S/mole = 5.71 g S

Example 1-10. Hemoglobin is the oxygen-carrying protein of most mammals. Each molecule of hemoglobin contains 4 atoms of iron. The molar mass of hemoglobin is about 64000 g/mole. How many moles of iron are contained in 0.50 moles of hemoglobin? Calculate the number of iron atoms in 0.128 g of hemoglobin.
Solution. Based on its molar mass, hemoglobin is clearly a large molecule containing many atoms. We are not told what the atoms are, nor how many of each there are. However, we are told that each molecule of hemoglobin contains 4 atoms of iron. We can write the formula for a molecule of hemoglobin as follows:

Fe4X

where X represents the collection of all other atoms present. What we can say about molecules and atoms, we can say about moles. Thus 1 mole of hemoglobin contains 4 moles of iron. Similarly, 0.50 moles of hemoglobin contains 4 * 0.50 = 2.00 moles of iron.

To obtain the second required answer, we convert mass of hemoglobin to moles hemoglobin using the molar mass:

moles Fe4X = 0.128 g Fe4X/64000 g/mole = 2.00 * 10-6 moles

This contains 4 * 2.00 * 10-6 moles of iron. The number of iron atoms is obtained using Avogadro's number:

Number Fe atoms = 4 * 2.00 * 10-6 * 6.023 * 1023 = 4.82 * 1018 Fe atoms

It is interesting that we do not need to know what the other atoms are in the hemoglobin molecule, much less how many of them there are. Knowledge of the number of iron atoms per molecule of hemoglobin is enough.

For oxygen, which exists in nature as diatomic molecules, O2, the statement "a mole of oxygen" is ambiguous. Does it mean a mole of oxygen molecules or a mole of oxygen atoms? These are different things. A mole of oxygen molecules contains 2 moles of oxygen atoms. For elements that exist as molecules, it is best to explicitly state whether molecules or atoms are meant. Thus "1 mole of oxygen molecules" means 6.022 * 1023 O2 molecules, or 2 x 6.022 * 1023 O atoms; "1 mole of oxygen atoms" means 6.022 * 1023 O atoms.

1-7 Stoichiometric Calculations for Compounds. When a chemist speaks of "stoichiometry", s/he means the quantitative mass relationships that govern the formulas of and the reactions between chemical substances. Stoichiometry rests on the validity of the Law of Conservation of Mass. A fundamental property of a compound is that it contains its constituent elements in definite proportions by mass. The percent composition of a compound is a statement of the mass percentages of the elements present in the compound. From the formula for a compound, percent composition is readily calculated. More importantly, knowledge of percent composition leads to the formula. The following examples illustrate this two-way street.


Example 1-11. What is the percent composition of nitrogen dioxide, NO2?
Solution. From the formula, we calculate the mass of one molecule of NO2. This mass contains the mass of 1 atom of N. The mass percent N is thus

%N = AW(N)/MW(NO2) * 100 = (14.0 m/46.0 m) * 100 = 30.4%
%O = 2 * AW(O)/MW(NO2) * 100 = (32.0/46.0) * 100 = 69.6%

The sum of the mass percentages of N and O is 100%, as it must be. In general, the percentage by mass of an element in a compound is obtained by multiplying the molar mass of the element by the number of atoms of the element in 1 molecule of compound, dividing by the molar mass of the compound, and multiplying by 100.

Example 1-12. The percent composition of a sample of the mineral, bustamite, is determined by elemental analysis to be
%Ca = 16.21
%Mn = 22.23
%Si = 22.72
%O = the remainder
What is the empirical formula of bustamite?
Solution. The approach to this problem is less obvious than in the previous example. The general strategy is as follows. Knowledge of percent composition allows calculation of the masses of calcium, manganese, silicon, and oxygen in any specified mass of compound. Since this mass is not important, we specify a convenient value -- 100 g. Once the mass of each element in this amount of bustamite is known, the masses can be converted to moles by division by the molar masses. The result is the number of moles of each element in the given mass of compound. The relative numbers of moles are then converted to whole numbers by dividing by the smallest number of moles. The flow is as follows:

%comp --> mass of each element in 100 g cpd --> moles each element in 100g cpd --> relative numbers of moles in integer form --> empirical formula

mass Ca in 100g cpd = 100 g cpd * 0.1621 g Ca/1 g cpd = 16.21 g Ca
mass Mn = 100 g cpd * 0.2223 g Mn/1 g cpd = 22.23 g Mn
mass Si = 22.72 g Si
mass O = 38.84 g O

(The percent O is determined by subtracting the sum of the other percentages from 100.)

Now convert to moles:

moles Ca = 16.21 g C/(40.08 g Ca/mole) = 0.4044 mole Ca
moles Mn = 22.23 g Mn/(54.938 g Mn/mole) = 0.4046 mole Mn
moles Si = 22.72 g Si/(28.086 g/mole) = 0.8089 mole Si
moles O = 38.84 g O/(15.999 g/mole) = 2.428 mole O

Divide each of these numbers by the smallest of them, 0.4044:

relative moles Ca = 0.4044/0.4044 = 1
relative moles Mn = 0.4046/0.4044 = 1.00
relative moles Si = 0.8089/0.4044 = 2.00
relative moles O = 2.428/0.4044 = 6.00

The empirical formula of bustamite is CaMnSi2O6.

Note: Calculation of the relative numbers of moles rarely gives exactly integers, because the percent composition is determined experimentally and is therefore subject to error. However, since we know that atoms must occur in formulas in integer numbers, it is often obvious how to round off the results.

Using the percent composition to determine the formula is one of the first things that a chemist does with a newly synthesized chemical compound. It is a very important initial step in determining the structural details of the new substance.

1-8 Chemical Reactions. A chemical reaction is a process involving changes in atomic arrangements, accompanied by changes in energy. In a chemical reaction, substances are converted into new substances. Some examples of chemical reactions follow:

These processes involve rearrangements of atoms (for example, in c), the carbon atoms that begin in CO2 end up in glucose molecules!) and all produce or require heat or light energy. In two of these examples we can tell visually that reaction has occurred. In a), yellow S disappears and is replaced by white ZnS; in b) we can see the light and feel the heat; however, in c), nothing happens visually. We cannot see photosynthesis occurring. Thus although a visible change often accompanies a reaction, it need not.

To describe a reaction more concisely and completely than the word descriptions above, chemists use chemical equations. The chemical equation for reaction a) is

Zn + S---->ZnS

Note that 1) chemical formulas are used to represent substances that react and form; 2) Substances that react (the reactants) are written to the left of the arrow; those formed (the products) are written to the right; 3) reactants (and products) are separated by + signs, which conveys that reactants combine or react together; 4) reactants are separated from products by an arrow pointing from the former to the latter. The arrow is read "yield" or "produce"; 5) the number of atoms of each element in the products is the same as the number of atoms of the element in the reactants (the Law of Conservation of Mass).

The chemical reaction for photosynthesis is written below.

CO2 + H2O--->C6H12O6 + O2

However, the equation is incomplete, because the numbers of atoms on left and right are not the same; the equation is not balanced. Balancing is accomplished by placing numerical coefficients (called stoichiometric coefficients) in front of the formulas to equalize the numbers of atoms of all types on both sides of the arrow. The stepwise approach for the photosynthesis reaction is illustrated below.

  1. As written, the reaction shows 1 C atom on the left, 6 on the right. To equalize the number of C atoms, coefficient 6 is placed in front of CO2: 6CO2 + H2O ---> C6H12O6 + O2
  2. There are 2 H atoms left, 12 right. Balance H by placing coefficient 6 before H2O: 6CO2 + 6H2O ---> C6H12O6 + O2
  3. There are 18 O atoms left, only 8 right. Place coefficient 6 before O2 on the right: 6CO2 + 6H2O ---> C6H12O6 + 6O2

Balancing is now complete. Note that it is never permissible to balance a chemical equation by altering the subscripts in the chemical formulas, because the subscripts are characteristic of the chemical species involved. Altering them changes the identity of the species! Balancing can be done only by altering the stoichiometric coefficients. Our reading of the balanced equation is "6 molecules of CO2 react with 6 molecules of H2O to produce 1 molecule of glucose and 6 molecules of O2." Since one mole of any substance contains No molecules, we may also read it as "6 moles of CO2 react with 6 moles of H2O to produce 1 mole of glucose and 6 moles of O2."

Finally, it is sometimes useful to indicate the physical states (solid, liquid, or gas) of reactants and products in the equation, and/or the reaction conditions. We indicate physical state using symbols in parentheses following the formulas. Common symbols are (g), (l), (s), and (aq) for, respectively, gas, liquid, solid, and aqueous (water) solution. Reaction conditions are usually written either above or below the arrow. Photosynthesis occurs at ambient temperature and pressure, in the presence of sunlight. We indicate this as shown below.

(1-8-1): 6CO2(g) +6H2O(l)T=298 K, P=1 atm
------------------>
light
C6H12O6(s) + 6O2(g)

The Chemical Equation as a Recipe. Frequently, people who are first learning about chemical equations interpret the meaning of the equation incorrectly. In this section, we will attempt to concretize chemical equations, which admittedly relate symbols of unseeable atoms and molecules and are therefore abstract, by drawing an analogy between a chemical equation and a recipe. The analogy is very close; all of the important quantitative aspects of chemical reactions apply equally to recipes. A recipe is a prescription for a process by which specified relative amounts of ingredients are transformed into a desired product. It is a written representation of the process of producing the product. Similarly, a chemical equation is a description of a process by which appropriate relative amounts of reactants are transformed into products. It is a written representation of the process of chemical change. A recipe for a simple fruit salad is given below:

(1-8-2): 1 apple + 2 oranges + 10 grapes --> 1 fruit salad

The recipe specifies the relative numbers of ingredients needed per fruit salad to be made. It tells us the relative numbers of apples, oranges, and grapes to be used. However many fruit salads we plan to make, we must use twice as many oranges as apples, and 10 times as many grapes as apples. The number of fruit salads that result will be the same as the number of apples that we started with. The equation very definitely does NOT say anything about the numbers of apples, oranges, and grapes that we actually have on hand. Thus we might purchase a bag of 10 apples, a bag of 10 oranges, and a bag of 200 grapes. The ratios of these numbers do not correspond to the ratios of the coefficients in equation 1-8-2. The coefficients tell us that in making fruit salads, we must use 2 oranges and 10 grapes for each apple that we use. How many fruit salads we can make from our purchased amounts depends on which ingredient we run out of first. For this particular situation, we can make only 5 fruit salads, because we will run out of oranges first. 5 apples and 150 grapes will be left over. If we want more fruit salads, we must buy more oranges!

Chemical equations are exactly like recipes. Thus equation 1-8-1 says that for every 6 molecules of CO2 we use, we must also use 6 molecules of water. We will produce 1 molecule of glucose and 6 molecules of O2. Note that after reaction, we no longer have the water and the CO2; we now have glucose and O2. If we happen to have on hand 20 molecules of CO2 and 12 molecules of H2O, then we can produce only two molecules of glucose using our recipe, because we will run out of H2O. We expect to have 8 molecules of CO2 left over. The equation describes a conversion process.

Chemical equations, like recipes, are usually written according to certain agreed-upon rules (this is probably less true for recipes than for equations!) Thus, a recipe for devils food cake specifies the amounts of ingredients needed for 1 cake. Similarly, chemists agree that they will write equations such that all the coefficients are the smallest possible whole numbers consistent with atom balance. However, just as we can scale the amounts in the recipe to make any number of cakes desired, we can scale the amounts of reactants used in a chemical reaction. Thus in carrying out a chemical process we are not restricted to the specific amounts indicated in the equation (or in cooking, to the specific amounts in the recipe); we can scale the amounts by a factor that is convenient for us. Green plants carry out photosynthesis on a scale of tons; however, the process proceeds according to 1-8-1. That is, for each mole of CO2 used, one mole of H2O is used, and 1/6 mole of glucose and 1 mole of O2 are formed.

Most people find this type of reasoning easy to understand in terms of a recipe, because recipes are familiar, and concrete. The reasoning works the same way for chemical reactions; but it is complicated by the abstract aspect of the formulas. If you find yourself bogged down by the abstraction, ground yourself by thinking of the recipe analogy.

Classification of Chemical Reactions. There are literally millions of known chemical reactions. To learn them all individually is a hopeless task. Instead, we attempt to classify reactions into a manageably small number of categories, and learn the fundamental characteristics of each category. We can then make statements about a reaction category that apply to all individual reactions in the category. In this text, we will be concerned with three major reaction types that we will now define and describe. We are not yet equipped to fully understand certain aspects of these reactions; however, it is best to begin practicing the recognition of reaction type as early as possible, even though it may seem difficult at first. It will become easier. The following three classes of chemical reactions will be discussed in this text:

We will define each type, and briefly discuss its characteristics.

Electron Transfer Reactions. These are reactions in which electrons are transferred from one atom, molecule, or ion to another atom, molecule, or ion. The substance that provides the electrons becomes more positive as a result of the process; the substance that receives the electrons becomes more negative. Some electron transfer reactions are very easy to recognize. For example, the reaction of sodium and chlorine in 1-8-3 is an electron transfer process, because sodium becomes more positive and chlorine more negative in the process:

(1-8-3): 2Na(s) + Cl2(g) ---> 2NaCl(s) Sodium is in the elemental (uncharged) form prior to reaction; after reaction, it exists as Na+ ions in NaCl. Clearly, each sodium atom has lost one electron. At the same time, each Cl atom of Cl2 has gained an electron to become Cl-. The reaction of magnesium and oxygen is also an electron transfer process:

(1-8-4): 2Mg(s) + O2(g) ---> 2MgO(s) Each magnesium atom loses two electrons; each oxygen atom accepts two electrons; electrons are therefore transferred from Mg to O.

Many reactions involve electron transfer. In this chapter, we recognize only two subcategories, which are exemplified by 1-8-3 and 1-8-4: the reaction of a metal with a non-metal; and the reaction of an element with oxygen (combustion). We will discuss combustion below.

Proton Transfer Reactions. These are processes in which a proton, H+, is transferred from one species to another. The species losing the proton is called an acid; the species gaining the proton is called a base. An example of a proton transfer reaction is shown in 1-8-5.

(1-8-5): HCl(aq) + Mg(OH)2(ag) ---> MgCl2(aq) + HOH (H2O) Here a proton, H+, is transferred from HCl to the OH- portion of magnesium hydroxide. Water is produced by combination of H+ and OH-; magnesium chloride is also produced. Similarly, 1-8-6 is a proton transfer process:

(1-8-6): H2SO4(aq) + Na2CO3(aq) ---> Na2SO4(aq) + H2CO3(aq) Here H2SO4 transfers two protons to CO32- to produce H2CO3; sodium sulfate is also produced. If protons are transferred in a reaction, the reaction is an acid-base process.

Double Displacement Reactions. In a double displacement process, the positive and negative parts of two reacting substances exchange: the positive part of one substance ends up with the negative part of the other. For example, the reactions below are double displacement processes:

(1-8-7): NaCl(aq) + AgNO3(aq) ---> NaNO3(aq) + AgCl(s)
(1-8-8): BaCl2(aq) + K2SO4(aq) ---> KCl + BaSO4(s)
You may notice that the proton transfer reactions in 1-8-5 and 1-8-6 are also double displacement reactions. This is certainly true. But because acid-base processes play such a central role in chemistry, chemists choose to classify them separately. We will adhere to this convention.

As we encounter reactions in this text, we will point out the reaction type whenever possible. Acid base processes will be discussed thoroughly in Chapter 13; electron transfer reactions are the subject of Chapter 14. We will now spend some time with combustion reactions, which constitute an important subclass of electron transfer reactions.

Combustion Reactions. In this section, we introduce and discuss an important category of chemical reactions: combustion reactions. A combustion reaction involves the reaction of a substance, element or compound, with oxygen, usually with the accompanying production of heat or light, or sound energy. The balanced equations representing the combustion reactions of elemental phosphorus and of methane are shown in equation form below:

4P(s) + 5 O2(g) ---> P4O10(s)
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g)

P4O10 is called the oxide of phosphorus; similarly, CO2 is an oxide of carbon, and H2O is the oxide of hydrogen. Using this terminology, we generalize from these typical combustion processes as follows:

Element + O2 ---> oxide of element
Compound + O2 ---> oxides of all elements in the compound

These generalizations give us tremendous predictive power, as the next example shows.


Example 1-13. Predict the products of the following combustion reactions, and balance the resulting equations.

S + O2 --->
Fe + O2 --->
H2 + O2 --->
C8H18 + O2 --->
C6H12O6 + O2 --->
NH3 + O2 --->

Solution. The first 3 reactions involve combustion of an element; the product is the oxide of the element:

(1-8-9): S + O2 ---> SO2 Predicting SO3 as the product of combustion of sulfur is also reasonable, although it is SO2 that is actually formed. Further oxidation to SO3 occurs only very slowly unless a suitable catalyst (a substance that speeds a reaction) is present.

(1-8-10): 4Fe + 3 O2 ---> 2Fe2O3
(1-8-11): 2H2 + O2 ---> 2H2O
The remaining three reactions are those of compounds with O2. We expect to produce the oxides of all of the elements in the compound:

(1-8-12): 2C8H18 + 2O2 ---> 16CO2 + 18H2O
(1-8-13): C6H12O6 + 6O2 ---> 6CO2 + 6H2O
The oxide of oxygen is O2. We could place it on the products side, but would have to remove it again in simplifying the equation. The simplest approach is to ignore it.

(1-8-14): 4NH3 + 5O2 ---> 4NO + 6H2O There are a number of oxides of nitrogen, including N2O, NO, NO2, N2O3, and N2O5. NO is the product actually formed. Any of the others would be a reasonable prediction, however.

Each of the reactions in Example 1-13 is of great industrial, technological, or biological importance. The reaction of sulfur with oxygen is the first step in the industrial process for the manufacture of sulfuric acid. Sulfuric acid is produced in larger amount than any other chemical substance, and is used in most other industrial chemical processes. The reaction of sulfur with oxygen also occurs when coal is burned to power the generators in electrical power plants. Although coal contains only a small (and variable) amount of sulfur, coal is burned on such a large scale that significant amounts of SO2 are released into the atmosphere. Conversion of SO2 to SO3 is catalyzed by particulate matter suspended in the air; and SO3 then combines with water to give sulfuric acid:

(1-8-15): SO3 + H2O ---> H2SO4

Sulfuric acid is the major component of acid rain, which destroys forests, aquatic life in lakes and rivers, and man-made objects such as statues and buildings. The acid rain problem remains to be solved.

The reaction of iron with oxygen is better known as rusting. Although rusting is slow, it leads to the inevitable destruction of iron- and steel-based construction. The rusting of iron and steel is an example of corrosion. Corrosion is responsible for many millions of dollars in damage to the infrastructure of the United States each year.

Reaction of hydrogen with oxygen is used to power the Space Shuttle, the centerpiece of the space program of the National Aeronautics and Space Administration. Hydrogen is used as the fuel. Unlike other combustion fuels, hydrogen is very desirable because the only product, other than the desired energy output, is water, which of course does no environmental damage. Currently, although it is possible to produce hydrogen gas from water, it cannot be done economically. When an economical process is developed, hydrogen may be used for less esoteric purposes, such as heating homes and powering automobiles.

The reactant, C8H18, in Reaction 1-8-12 is a major constituent of gasoline, which is used to power the internal combustion engines of automobiles, lawn mowers, snow blowers, jet skis, and the like. Combustion of gasoline releases a large amount of energy, which, when controlled, can be converted effectively to useful work. Gasoline is an example of a hydrocarbon, a compound containing only the elements carbon and hydrogen. There are many known hydrocarbons, the simplest being CH4 (methane), which is a gas under ordinary conditions. Gasoline, with 8 carbons, is a liquid; and hydrocarbons with even bigger molecules (e.g., C20H42) are solids. Methane is used as heating fuel in homes, under the name "natural gas." It is so-called because it is found trapped in huge underground deposits, from which it can be released and piped over long distances.

Glucose, C6H12O6, and its close relatives are the primary biological fuels. Controlled combustion of glucose provides the energy to power the living cell. Almost all living organisms, plant and animal, carry out some version of equation 1-8-13. The reverse of 1-8-13, in which carbon dioxide and water are converted to glucose and oxygen, is photosynthesis. This reaction, driven by the energy from sun light, is performed by green plants. Thus these organisms produce their own food via photosynthesis, then consume it via 1-8-13.

Finally, the combustion of ammonia, NH3, is carried out as the first step in the industrial synthesis of nitric acid, HNO3. Nitric acid is a commercially important substance, and is made in very large quantities each year. A currently very important chemical problem is to develop inexpensive methods for making NH3, the reactant in 1-8-14. The currently used process has been in place since about 1915, and is very expensive.

Introduction to Stoichiometry of Chemical Reactions. Before leaving this topic, we explore some of the quantitative ideas from the previous section. Recall that the important ideas there were


Example 1-14. For the photosynthesis reaction, which is the reverse of 1-8-13, make as many quantitative statements as possible about the numbers of molecules and moles of reactants and products.

(1-8-1): 6CO2(g) + 6H2O(g) ---> C6H12O6(s) + 6O2(g) Solution. There are so many statements that can be made that we will cover only a small fraction of them. However, let's try:

This is sufficient to illustrate the types of things we can say. It is important to be comfortable with the information obtainable from a chemical equation.


Example 1-15. The combustion of ethane, C2H6, is represented in 1-8-16:

(1-8-16): 2C2H6 + 7O2 ---> 4CO2 + 6H2O Suppose that we have on hand 352 g of ethane. How many moles of ethane do we have? How many moles of oxygen will be required to convert it to CO2 and H2O according to the recipe above? How many moles of CO2 will enter the atmosphere?
Solution. The moles of ethane can be calculated from the given mass and the molar mass, which is 2(12.011) + 6(1.008) = 30.070 g/mole.

moles ethane = 352 g/30.070 g/mole = 11.71 moles

According to the recipe, 7 moles of O2 are required for each 2 moles of ethane consumed. Thus

moles O2 required = 7/2 x moles ethane = 40.97 moles oxygen

Again according to the equation, 4 moles of CO2 are produced for each 2 moles of ethane burned:

moles CO2 produced = 4/2 x moles ethane = 23.42 moles carbon dioxide

The type of calculation illustrated in Example 1-15, based on the numerical relationships in the chemical equation, is extremely important in chemistry. We will practice these continually as we progress.

Supplement: An Alternative Approach to Reaction Stoichiometry.

Calculations of the type in Example 1-15 are extremely important in chemistry, yet they seem difficult for many students. Here I present an alternative approach to doing this type of calculation that explicitly shows the amounts of substances present before reaction takes place; the amounts of substances that react and form; and the amounts present after reaction is complete. You may find this system easier to use than that outlined above. In addition, we will find this approach useful in our discussions of chemical equilibrium.

To illustrate the method, let us do a problem similar to that in Example 1-15:

The combustion of ethane, C2H6, is represented in 1-8-16:

(1-8-16): 2C2H6 + 7O2 ---> 4CO2 + 6H2O

Suppose that we have on hand 352 g of ethane and 450 g of oxygen. How many moles of ethane do we have? How many moles of oxygen do we have? How many moles of CO2 and H2O can be produced according to the recipe above?

We begin as before with a balanced equation for the process:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)

Directly under the equation, we write the initial amount (the amount we start with) of each substance:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000

Note that we explicitly acknowledge that before reaction takes place, we have no CO2 or H2O.

Knowing that the coefficients in the chemical equation represent the numbers of molecules or moles of substances involved, we first convert masses to moles, and portray the information in a second line labelled initial, moles:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000
initial, moles11.7114.0600

Now we must decide which of the two reactants, ethane or oxygen, will be used up first. The equation helps us with this, because it indicates that for every 2 moles of ethane used, 7 moles of O2 are needed. Thus to completely use up 11.71 moles of ethane would require 7/2 * 11.71 = 40.99 moles of O2. Clearly there is not this much available initially, so we conclude oxygen is used up first; i.e., it is the limiting reagent. We acknowledge this by writing lim under the initial moles for oxygen:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000
initial, moles11.7114.06
lim
00

Next we write the amounts of reactants that react and the amounts of products that form. These amounts are determined from the amount of the limiting reagent and the stoichiometric coefficients in the chemical equation:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000
initial, moles11.7114.06
lim
00
reacts/forms, moles-(2/7)*14.06
= -4.02
-14.06
lim
(4/7)*14.06
= 8.03
(6/7)*14.06
= 12.05

Note a couple of important things about the reacts/forms line. First, because reactants are used up as reaction takes place, the "reacts" amounts are entered as negative. Second, the limiting reagent is assumed to completely react; i.e., to be used up. Third, the amount of ethane used up and the amounts of CO2 and H2O formed are calculated by applying the stoichiometric coefficients to the number of moles of limiting reagent. Finally, the amounts of all substances present when reaction is over are obtained by adding the initial amounts to the amounts used up or formed:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000
initial, moles11.7114.06
lim
00
reacts/forms, moles-(2/7)*14.06
= -4.02
-14.06
lim
(4/7)*14.06
= 8.03
(6/7)*14.06
= 12.05
final, moles7.6908.0312.05

If the final amounts are desired in grams, they may be easily calculated using molar masses:

2 C2H6 +7 O2(g) ---> 4 CO2(g) +6 H2O(l)
initial,g35245000
initial, moles11.7114.06
lim
00
reacts/forms, moles-(2/7)*14.06
= -4.02
-14.06
lim
(4/7)*14.06
= 8.03
(6/7)*14.06
= 12.05
final, moles7.6908.0312.05
final, grams30.1 g0353 g217 g

Supplement: Mass Spectrometry.

Applications

1-1. Write chemical formulas for the indicated compounds based on the descriptions:
A molecule of hydrazine contains 2 atoms of nitrogen and 4 atoms of hydrogen
A molecule of ethane contains 2 carbon and 6 hydrogen atoms
A molecule of carbon tetrachloride contains 4 chlorine atoms per carbon atom
A molecule of carbonic acid contains 3 oxygen and 2 hydrogen atoms per carbon atom
1-2. Calculate the molar mass for each compound:
NH3
B2H6
C4H10
H2CO3
C6H12O6
1-3. The amino acid, alanine, has the formula, C3H7O2N. Calculate the molar mass of alanine.

1-4. What is the formula of the common ion of each element?
Na
P
Br
Al
Mg
1-5. Predict the formula for the binary compound of each element pair, and indicate whether the compound is ionic or covalent.
hydrogen and oxygen
carbon and fluorine
nitrogen and chlorine
magnesium and bromine
aluminum and sulfur
1-6. For each stated mass, calculate the moles and molecules of substance
3.25 g of carbon
4.67 g of O2
16.56 g of C6H14
8.29 g of C6H12O6

1-7. How many moles of carbon atoms are in 8.29 g of C6H12O6? How many moles of hydrogen atoms? How many moles of oxygen atoms?

1-8. How many moles of hydrogen atoms are in 1.25 g of H3PO4? How many moles of phosphorus atoms? Of oxygen atoms?

1-9. How many atoms of carbon are in 52 molecules of acetaminophen (C8H9NO2)? How many atoms of nitrogen? Of oxygen? Of hydrogen?

1-10. Calculate percent composition of each compound: CH4, P4O10, C6H12O6, CaCl2.

1-11. What is the percent composition of acetaminophen, for which the formula is C8H9NO2?

1-12. An oxide of chlorine analyzes at 38.8%Cl and 61.2%O by mass. What is the empirical formula?

1-13. Caffeine has molar mass 194.2 and percent composition by mass, 49.5%C, 5.2%H, 28.8% N, and the rest O. What is the molecular formula?

1-14. Balance each equation.

  1. ZnS + HCl ---> ZnCl2 + H2S
  2. HCl + Cr ---> CrCl3 + H2
  3. Al + Fe3O4 ---> Al2O3 + Fe
  4. H2 + Br2 ---> HBr
  5. Na2S2O3 + I2 ---> NaI + Na2S4O6
  6. FeCl3 + Na2CO3 ---> Fe2(CO3)3 + NaCl
  7. Ca(OH)2 + H3PO4 ---> Ca3(PO4)2 + H2O
  8. Na2O + (NH4)2SO4 ---> Na2SO4 + H2O + NH3
  9. FeS2 + O2 ---> Fe2O2 + SO2
  10. C7H6O2 + O2 ---> CO2 + H2O
  11. NH3 + O2 ---> NO + H2O
1-15. Make at least 5 quantitative statements based on each formula.

  1. Urea, CH4N2O
  2. Naproxen, C14H14O3
  3. Cinnamaldehyde, C9H8O
  4. Dinitrogen pentoxide, N2O5
  5. Oxalic acid, H2C2O4 1-16. Predict the products of the following combustion reactions.

    1-17. Nitrogen and hydrogen react according to

    N2 + 3H2 ---> 2NH3 When 56.02 g of N2 is mixed with 9.00 moles of H2 and reaction occurs,

    1. 6.00 moles NH3 is produced
    2. 18.00 moles NH3 is produced with 1.00 mole N2 left over
    3. 2.00 moles NH3 is produced with 6.00 moles H2 left over
    4. 4.00 moles NH3 is produced with 3.00 moles H2 left over
    5. 4.00 moles NH3 is produced with 6 1/3 moles H2 left over 1-18. One type of sugar has molecular formula C12H22O11. An amount of sugar equal to 0.10 moles contains

      1. 342.297 g of sugar
      2. 12 moles of carbon atoms
      3. 2.2 * 6.022 * 1023 H atoms
      4. 11 oxygen atoms
      5. 12 * 6.022 * 1023 C atoms
      1-19. A compound of sulfur, oxygen, and fluorine contains 25.84%S and 61.26%F. The empirical formula for the compound is

      1. SO2F8
      2. S2OF4
      3. SOF2
      4. SOF4
      5. SOF 1-20. The equation for a chemical reaction is

        2S + 3O2 ---> 2SO3 A mixture of S and O2 in a closed container is shown.
        Circle the diagram below that best represents the product mixture.

        1-21. The amino acid, histidine, is instrumental in the binding of oxygen to the blood protein, hemoglobin; in other words, it helps us to breathe. A molecule of histidine contains

        6 carbon atoms
        7 hydrogen atoms
        one-third as many oxygen atoms as carbon atoms
        1.5 times as many nitrogen atoms as oxygen atoms
        What is the formula for histidine?
        What is the molar mass of histidine?

        1-22. Each molecule of hemoglobin contains 4 atoms of iron. The molar mass of hemoglobin is about 64000 g/mole. How many moles of iron are contained in 0.50 * 10-4 moles of hemoglobin?
        Calculate the number of iron atoms in 0.128 g of hemoglobin.
        Spreadsheet Applications


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