Chem 101 Chapter 3- Molecules, Ions, and their Compounds

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Unformatted text preview: The Basic Tools of Chemistry 3— Molecules, Ions, and Their Compounds DNA: The Most Important Molecule DNA is the substance in every plant and animal that carries the exact blueprint of that plant or animal. The structure of this molecule, the cornerstone of life, was uncovered in 1953, and James D. Watson, Francis Crick, and Maurice Wilkins shared the 1962 Nobel Prize in medicine and physiology for the work. It was one of the most important scientific discoveries of the 20th century, and the story has recently been told by Watson in his book The Double Helix. When Watson was a graduate student at Indiana University, he had an interest in the gene and said he hoped that its biological role might be solved “without my learning any chemistry.” Later, however, he and Crick found out just how useful chemistry can be when they began to unravel the structure of DNA. Solving important problems requires teamwork among scientists of many kinds so Watson went to Cambridge University in England in 1951. There he met Crick, who, Watson said, talked louder and faster than anyone else. Crick shared Watson’s belief in the fundamental importance of DNA, and the pair soon learned that Maurice Wilkins and Rosalind Franklin at King’s College in London were using a technique called x-ray crystallography to learn more about DNA’s structure. Watson and Crick believed that understanding this structure was crucial to understanding genetics. To solve the structural problem, however, they needed experimental data of the type that could come from the experiments at King’s College. The King’s College group was initially reluctant to share their data; and, what is more, they did not seem to share Watson and Crick’s sense of urgency. There was also an ethical dilemma: Could Watson and Crick work on a problem that others had claimed as theirs? “The English sense of fair play would not allow Francis to move in on Maurice’s problem,” said Watson. James D. Watson and Francis Crick. In a photo taken in 1953, Watson (left) and Crick (right) stand by their model of the DNA double helix. Together with Maurice Wilkins, Watson and Crick received the Nobel Prize in medicine and physiology in 1962. 96 A. Barrington Brown/Science Source/Photo Researchers, Inc. Chapter Goals See Chapter Goals Revisited (page 130). Test you knowledge of these goals by taking the exam-prep quiz on the General ChemistryNow CD-ROM or website. 3.1 3.2 3.3 3.4 3.5 3.6 3.7 Chapter Outline Molecules, Compounds, and Formulas Molecular Models Ionic Compounds: Formulas, Names, and Properties Molecular Compounds: Formulas, Names, and Properties Formulas, Compounds, and the Mole Describing Compound Formulas Hydrated Compounds • Interpret, predict, and write formulas for ionic and molecular compounds. • • • • Name compounds. Understand some properties of ionic compounds. Calculate and use molar mass. Calculate percent composition for a compound and derive formulas from experimental data. Rosalind Franklin of King’s College, London. She died in 1958 at the age of 37. Because Nobel Prizes are never awarded posthumously, she did not share in this honor with Watson, Crick, and Wilkins. Watson and Crick approached the problem through a technique chemists now use frequently—model building. They built models of the pieces of the DNA chain, and they tried various chemically reasonable ways of fitting them together. Finally, they discovered that one arrangement was “too pretty not to be true.” Ultimately, the experimental evidence of Wilkins and Franklin confirmed the “pretty structure” to be the real DNA structure. As you will see, chemists often use models to help guide them to experimental evidence that is definitive. The story of how Watson, Crick, Wilkins, and Franklin ultimately came to share information and insight is an interesting human drama and illustrates how scientific progress is often made. For more on this interesting human and scientific drama, read Rosalind Franklin: The Dark Lady of DNA by Brenda Maddox and Watson’s book The Double Helix. Courtesy American Society for Microbiology Structure of DNA: Sugar, Phosphate, and Bases Watson and Crick recognized early on that the overall structure of DNA was a helix; that is, the atomic-level building blocks formed chains that twisted in space like the strands of a grapevine. They also knew which chemical elements it contained and roughly how they were grouped together. What they did not know was the detailed structure of the helix. By the spring of 1953, however, they had the answer. The atomic-level building blocks of DNA form two chains twisted together in a double helix. DNA is a very large molecule that consists of two chains of atoms (P, C, and O) that twist together. The P, C, and O atoms are parts of phosphate ions (P) and sugar molecules. The chains are joined by four different molecules (adenine, thymine, guanine, and cytosine) belonging to a general class of molecules called bases. P S A P S P S P S P G S C P S P G S P T G P S T P A P S C S P ST A S T C S P S P S C H H C O -O P O O T S A P S P S O -O P O O CH2 C H H C O -O P O CH2 O H C H C H N HC N O O C H H C H N HC Four Bases 1. Adenine 2. Thymine H O N C C N H C N H O C N H N C C CH3 H H C H C O O C N CH H C H H C O O C H H C CH2 O P O– O T A C P T P S 3.4 nm P S P S Phosphate Sugar A sample of DNA. © BSIP/Emakoff/Science Source/Photo Researchers, Inc. S C 3. Guanine N N C C C O C N H H 4. Cytosine O C N C H N H N CH CH A C H H C CH2 O P O– O O P S A P 97 98 Chapter 3 Molecules, Ions, and Their Compounds To Review Before You Begin • Know how to calculate and use molar amounts (Section 2.5) I Throughout the chapter this icon introduces a list of resources on the General ChemistryNow CD-ROM or website (http://now .brookscole.com/kotz6e) that will: help you evaluate your knowledge of the material provide homework problems allow you to take an exam-prep quiz provide a personalized Learning Plan targeting resources that address areas you should study n 1953 the structure of the giant molecule DNA, deoxyribonucleic acid, was finally understood (page 96). Chromosomes, which are present in the nuclei of almost all living cells, consist of DNA. Recently discovered knowledge of the human genome, which is the complete structure of the DNA in every one of our 23 chromosomes, is widely expected to revolutionize the practice of medicine. To comprehend modern molecular biology—indeed all of modern chemistry—we have to understand the structures and properties of molecules. This chapter marks the beginning of our attempt to acquaint you with this important subject. 3.1—Molecules, Compounds, and Formulas A molecule is the smallest identifiable unit into which a pure substance like sugar and water can be divided and still retain the composition and chemical properties of the substance. Such substances are composed of identical molecules consisting of atoms of two or more elements bound firmly together. For example, atoms of the element aluminum, Al, combine with molecules of the element bromine, Br2, to produce the compound aluminum bromide, Al2Br6 (Figure 3.1). 2 Al(s) aluminum • • • • 3 Br2( ) ¡ bromine Al2Br6(s) aluminum bromide ¡ (a) (b) (c) Active Figure 3.1 Reaction of the elements aluminum and bromine. (a) Solid aluminum and (in the beaker) liquid bromine. (b) When the aluminum is added to the bromine, a vigorous chemical reaction produces white, solid aluminum bromide, Al2Br6 (c). See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. Photos: Charles D. Winters 3.1 Molecules, Compounds, and Formulas NAME MOLECULAR FORMULA C2H6O CONDENSED FORMULA CH3CH2OH STRUCTURAL FORMULA MOLECULAR MODEL 99 Ethanol H H H C C O H H H Dimethyl ether C2H6O CH3OCH3 H H C H O H C H H Figure 3.2 Four approaches to showing molecular formulas. Here the two molecules have the same molecular formula. However, once they are written as condensed or structural formulas, and illustrated with a molecular model, it is clear that these molecules are different. (See the General ChemistryNow Screen 3.4 Representing Compounds, for a tutorial on identifying molecular representations.) To describe this chemical change (or chemical reaction) on paper, the composition of each element and compound is represented by a symbol or formula. Here one molecule of Al2Br6 is composed of two Al atoms and six Br atoms. How do compounds differ from elements? When a compound is produced from its elements, the characteristics of the constituent elements are lost. Solid, metallic aluminum and red-orange liquid bromine, for example, react to form Al2Br6, a white solid (see Figure 3.1). Active Figure 3.1 ■ Standard Colors for Atoms in Molecular Models The colors listed here are used in this book and are generally used by chemists. The colors of some common atoms are: carbon atoms Formulas For molecules more complicated than water, there is often more than one way to write the formula. For example, the formula of ethanol (also called ethyl alcohol ) can be represented as C2H6O (Figure 3.2). This molecular formula describes the composition of ethanol molecules—two carbon atoms, six hydrogen atoms, and one atom of oxygen occur per molecule—but it gives us no structural information. Structural information—how the atoms are connected and how the molecule fills space—is important, however, because it helps us understand how a molecule can interact with other molecules, which is the essence of chemistry. To provide some structural information, it is useful to write a condensed formula, which indicates how certain atoms are grouped together. For example, the condensed formula of ethanol, CH3CH2OH (see Figure 3.2), informs us that the molecule consists of three “groups”: a CH3 group, a CH2 group, and an OH group. Writing the formula as CH3CH2OH also shows that the compound is not dimethyl ether, CH3OCH3, a compound with the same molecular formula but a different structure and distinctly different properties. That ethanol and dimethyl ether are different molecules is further apparent from their structural formulas (see Figure 3.2). This type of formula gives us an even higher level of structural detail, showing how all of the atoms are attached within a molecule. The lines between atoms represent the chemical bonds that hold atoms together in this molecule [ Chapters 9 and 10]. hydrogen atoms oxygen atoms nitrogen atoms chlorine atoms ■ Isomers Compounds having the same molecular formula but different structures are called isomers. (See Chapter 11 and General ChemistryNow Screen 3.4 Representing Compounds.) 100 Chapter 3 Molecules, Ions, and Their Compounds Example 3.1—Molecular Formulas Problem The acrylonitrile molecule is the building block for acrylic plastics (such as Orlon and Acrilan). Its structural formula is shown here. What is the molecular formula for acrylonitrile? H C H CH2CHCN Condensed formula Molecular model C C H N Structural formula Strategy Count the number of atoms of each type. Solution Acrylonitrile has three C atoms, three H atoms, and one N atom. Therefore, its molecular formula is C3H3N. Comment When writing molecular formulas of organic compounds (compounds with C, H, and other elements) the convention is to write C first, then H, and finally other elements in alphabetical order. Exercise 3.1—Molecular Formulas The styrene molecule is the building block of polystyrene, a material used for drinking cups and building insulation. What is the molecular formula of styrene? H H C C C C C H C C H H H C6H5CHCH2 Condensed formula Molecular model C H H Structural formula 3.2—Molecular Models Molecular structures are often beautiful in the same sense that art is beautiful. For example, there is something intrinsically beautiful about the pattern created by water molecules assembled in ice (Figure 3.3). More important, however, is the fact that the physical and chemical properties of a molecular compound are often closely related to its structure. For example, two well-known features of ice are easily related to its structure. The first is the shape of ice crystals: The sixfold symmetry of macroscopic ice crystals also appears at the particulate level in the form of six-sided rings of hydrogen and oxygen atoms. The second is water’s unique property of being less dense when solid than it is when liquid. The lower density of ice, which has enormous consequences for earth’s climate, results from the fact that molecules of water are not packed together tightly. 3.2 Mehau Kulyk/Science Photo Library/ Photo Researchers, Inc.; model by S. M. Young Molecular Models 101 Figure 3.3 Ice. Snowflakes are six-sided structures, reflecting the underlying structure of ice. Ice consists of six-sided rings formed by water molecules, in which each side of a ring consists of two O atoms and an H atom. Because molecules are three-dimensional, it is often difficult to represent their shapes on paper. Certain conventions have been developed, however, that help represent three-dimensional structures on two-dimensional surfaces. Simple perspective drawings are often used (Figure 3.4).c Wood or plastic models are also a useful way of representing molecular structure. These models can be held in the hand and rotated to view all parts of the molecule. Several kinds of molecular models exist. In the ball-and-stick model, spheres, usually in different colors, represent the atoms, and sticks represent the bonds holding them together. These models make it easy to see how atoms are attached to one another. Molecules can also be represented using space-filling models. These models are more realistic because they offer a better representation of relative sizes of atoms and their proximity to each other when in a molecule. A disadvantage of pictures of space-filling models is that atoms can often be hidden from view. H H Charles D. Winters H C H Simple perspective drawing Plastic model Ball-and-stick model Space-filling model All visualizing techniques represent the same molecule. Active Figure 3.4 Ways of depicting the methane (CH4) molecule. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. 102 Chapter 3 Molecules, Ions, and Their Compounds A Closer Look Computer Resources for Molecular Modeling With the availability of relatively low-cost, high-powered computers, the use of molecular modeling programs has become common. Although the computer screen is two-dimensional, the perspective drawings obtained from molecular-modeling programs are usually quite good. In addition, most programs offer an option to rotate the model on the computer screen to allow the viewer to see the structure from any desired angle. Both ball-and-stick and space-filling representations can be portrayed. Most of the drawings in this book were prepared with the commercial molecular modeling software from CAChe/Fujitsu. The General ChemistryNow CD-ROM includes a program for visualizing molecules and for measuring atom–atom distances and angles. The site on the World Wide Web for this textbook (http://www.brookscole.com) contains a link to RasMol and Chime, molecular visualization software. Models of many of the compounds mentioned in this book are available through the General ChemistryNow CD-ROM and website. You can visualize these molecules using the software on the CD-ROM, or, if you download RasMol or Chime and configure your browser properly, you can download files from the Internet will that allow you to visualize these models on your own computer. A model of caffeine as viewed with RasMol (left) and the CAChe/Fujitsu software (right). Example 3.2—Using Molecular Models Problem A model of uracil, an important biological molecule, is given here. Write its molecular formula. Molecular model Strategy The standard color codes used for the atoms are as follows: carbon atoms hydrogen atoms white; nitrogen atoms blue; and oxygen atoms red. gray; Solution Uracil has four C atoms, four H atoms, two N atoms, and two O atoms, giving a formula of C4H4N2O2. Exercise 3.2—Formulas of Molecules Cysteine, whose molecular model and structural formula are illustrated here, is an important amino acid and a constituent of many living things. What is its molecular formula? See Example 3.2 and page 99 for the color coding of the model. 3.3 Ionic Compounds: Formulas, Names, and Properties 103 O C O NH3 H C H C H S H Molecular model Structural formula 3.3—Ionic Compounds: Formulas, Names, and Properties The compounds you have encountered so far in this chapter are molecular compounds—that is, compounds that consist of discrete molecules at the particulate level. Ionic compounds constitute another major class of compounds. They consist of ions, atoms or groups of atoms that bear a positive or negative electric charge. Many familiar compounds are composed of ions (Figure 3.5). Table salt, or sodium chloride (NaCl ), and lime (CaO) are just two. To recognize ionic compounds, and to be able to write formulas for these compounds, it is important to know the formulas and charges of common ions. You also need to know the names of ions and be able to name the compounds they form. See the General ChemistryNow CD-ROM or website: • Screen 3.5 Ions, for tutorials on determining the number of protons and electrons in an ion and determining ionic charge Hematite, Fe2O3 Calcite, CaCO3 Common Name Calcite Fluorite Gypsum Hematite Orpiment Name Calcium carbonate Calcium fluoride Calcium sulfate dihydrate Iron(III) oxide Arsenic sulfide Formula CaCO3 CaF2 CaSO4.2 H20 Fe2O3 As2S3 Ions Involved Ca2 , CO32 Ca2 , F Ca2 , SO42 Fe3 , O2 As3 , S2 Charles D. Winters Gypsum, CaSO4 2 H2O Fluorite, CaF2 Orpiment, As2S3 Figure 3.5 Some common ionic compounds. 104 Chapter 3 Molecules, Ions, and Their Compounds Ions Atoms of many elements can lose or gain electrons in the course of a chemical reaction. To be able to predict the outcome of chemical reactions [ Section 5.6], you need to know whether an element will likely gain or lose electrons and, if so, how many. Cations If an atom loses an electron (which is transferred to an atom of another element in the course of a reaction), the atom now has one fewer negative electrons than it has positive protons in the nucleus. The result is a positively charged ion called a cation (see Figure 3.6). (The name is pronounced “cat -ion.”) Because it has an excess of one positive charge, we write the cation’s symbol as, for example, Li : Li atom ¡ e (3 protons and 3 electrons) ■ Writing Ion Formulas When writing the formula of an ion, the charge on the ion must be included. Li cation (3 protons and 2 electrons)tive Figu.6 Anions Conversely, if an atom gains one or more electrons, there is now one or more negatively charged electrons than protons. The result is an anion (see Figure 3.6). (The name is pronounced “ann ¿ -ion.”) O atom 2e ¡ O2 anion (8 protons and 10 electrons) (8 protons and 8 electrons) Here the O atom has gained two electrons so we write the anion’s symbol as O2 . 3e e 3p 3n 2e 3p 3n Lithium ion, Li Lithium, Li Li 3p 3n 3e Li 3p 3n 2e 9e 9p 10n Fluorine, F e 10e F 9p 10n 9p 10n 9e F 9p 10n 10e Fluoride ion, F Active Figure 3.6 Ions. A lithium-6 atom is electrically neutral because the number of positive charges (three protons) and negative charges (three electrons) are the same. When it loses one electron, it has one more positive charge than negative charge, so it has a net charge of 1 . We symbolize the resulting lithium cation as Li . A fluorine atom is also electrically neutral, having nine protons and nine electrons. A fluorine atom can acquire an electron to produce a F anion. This anion has one more electron than it has protons, so it has a net charge of 1 . See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. 3.3 Ionic Compounds: Formulas, Names, and Properties 105 How do you know whether an atom is likely to form a cation or an anion? It depends on whether the element is a metal or a nonmetal. • Metals generally lose electrons in the course of their reactions to form cations. • Nonmetals frequently gain one or more electrons to form anions in the course of their reactions. Monatomic Ions Monatomic ions are single atoms that have lost or gained electrons. As indicated in Figure 3.7, metals typically lose electrons to form monatomic cations, and nonmetals typically gain electrons to form monatomic anions. How can you predict the number of electrons gained or lost? Typical charges on such ions are indicated in Figure 3.7. Metals of Groups 1A–3A form positive ions having a charge equal to the group number of the metal. Electron Change Group 1A 2A 3A Metal Atom Na (11 protons, 11 electrons) Ca (20 protons, 20 electrons) Al (13 protons, 13 electrons) Resulting Metal Cation 1 ¡ Na (11 protons, 10 electrons) 2 ¡ Ca2 (20 protons, 18 electrons) 3 ¡ Al3 (13 protons, 10 electrons) Transition metals (B-group elements) also form cations. Unlike the A-group metals, however, no easily predictable pattern of behavior occurs for transition metal cations. In addition, many transition metals form several different ions. An iron-containing compound, for example, may contain either Fe2 or Fe3 ions. Indeed, 2 and 3 ions are typical of many transition metals (see Figure 3.7). Electron Change 2 2 3 Group 7B 8B 8B Metal Atom Mn (25 protons, 25 electrons) Fe (26 protons, 26 electrons) Fe (26 protons, 26 electrons) Resulting Metal Cation ¡ Mn2 (25 protons, 23 electrons) ¡ Fe2 (26 protons, 24 electrons) ¡ Fe3 (26 protons, 23 electrons) 1A H Li Na K Rb Cs Mg2 Ca2 Sr2 Ba2 3B 4B Ti4 2A Metals Transition metals Metalloids Nonmetals 5B 6B 7B Cr2 Mn2 Fe2 Cr3 Fe3 8B Co2 Co3 1B Cu Cu2 Ag 2B Zn2 Cd2 Hg22 Hg2 Sn2 Pb2 Bi3 3A 4A 5A N3 Al3 P3 6A O2 S2 Se2 Te2 7A H F Cl Br I 8A Ni2 Figure 3.7 Charges on some common monatomic cations and anions. Metals usually form cations and nonmetals usually form anions. (The boxed areas show ions of identical charge.) 106 Chapter 3 Molecules, Ions, and Their Compounds Nonmetals often form ions having a negative charge equal to 8 minus the group number of the element. For example, nitrogen is in Group 5A, so it forms an ion having a charge of 3 because a nitrogen atom can gain three electrons. Electron Change 3( 2( 1( 8 8 8 Group 5A 6A 7A Nonmetal Atom N (7 protons, 7 electrons) S (16 protons, 16 electrons) Br (35 protons, 35 electrons) Resulting Nonmetal Anion 5) ¡ N3 (7 protons, 10 electrons) 6) ¡ S2 (16 protons, 18 electrons) 7) ¡ Br (35 protons, 36 electrons) Notice that hydrogen appears at two locations in Figure 3.7. The H atom can either lose or gain electrons, depending on the other atoms it encounters. Electron lost: H (1 proton, 1 electron) ¡ H (1 proton, 0 electrons) e Electron gained: H (1 proton, 1 electron) e ¡ H (1 proton, 2 electrons) Finally, the noble gases do not form monatomic cations or anions in chemical reactions. Ion Charges and the Periodic Table As illustrated in Figure 3.7, the metals of Groups 1A, 2A, and 3A form ions having 1 , 2 , and 3 charges; that is, their atoms lose one, two, or three electrons, respectively. For cations formed from A-group elements, the number of electrons remaining on the ion is the same as the number of electrons in an atom of the noble gas that precedes it in the periodic table. For example, Mg2 has 10 electrons, the same number as in an atom of the noble gas neon (atomic number 10). An atom of a nonmetal near the right side of the periodic table would have to lose a great many electrons to achieve the same number as a noble gas atom of lower atomic number. (For instance, Cl, whose atomic number is 17, would have to lose 7 electrons to have the same number of electrons as Ne.) If a nonmetal atom were to gain just a few electrons, however, it would have the same number as a noble gas atom of higher atomic number. For example, an oxygen atom has eight electrons. By gaining two electrons per atom it forms O2 , which has ten electrons, the same number as neon. Anions having the same number of electrons as the noble gas atom succeeding it in the periodic table are commonly observed in chemical compounds. Exercise 3.3—Predicting Ion Charges Predict formulas for monatomic ions formed from (a) K, (b) Se, (c) Ba, and (d) Cs. In each case indicate the number of electrons gained or lost by an atom of the element in forming the anion or cation, respectively. For each ion, indicate the noble gas atom having the same total number of electrons. ■ Cation Charges and the Periodic Table 1A 2A 3A Group 1A, 2A, 3A metals form Mn cations where n group number. Polyatomic Ions Polyatomic ions are made up of two or more atoms, and the collection has an electric charge (Figure 3.8 and Table 3.1). For example, carbonate ion, CO32 , a common polyatomic anion, consists of one C atom and three O atoms. The ion has two units of negative charge because there are two more electrons (a total of 32) in the ion than there are protons (a total of 30) in the nuclei of one C atom and three O atoms. A common polyatomic cation is NH4 , the ammonium ion. In this case, four H atoms surround an N atom, and the ion has a 1 electric charge. This ion has ten 3.3 Ionic Compounds: Formulas, Names, and Properties 107 Photos: Charles D. Winters CO32 PO43 SO42 Celestite, SrSO4 Strontium sulfate Calcite, CaCO3 Calcium carbonate Apatite, Ca5F(PO4)3 Calcium fluorophosphate Common ionic compounds based on polyatomic ions. Active Figure 3.8 See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. electrons, but there are 11 positively charged protons in the nuclei of the N and H atoms (seven and one each, respectively). Table 3.1 Formula CATION: Positive Ion NH4 Formulas and Names of Some Common Polyatomic Ions Name ammonium ion Based on a Group 7A element ClO hypochlorite ion ClO2 chlorite ion ClO3 chlorate ion ClO4 perchlorate ion Based on a transition metal CrO42 chromate ion Cr2O72 dichromate ion MnO4 permanganate ion Formula Name ANIONS: Negative Ions Based on a Group 4A element CN cyanide ion acetate ion CH3CO2 carbonate ion CO32 hydrogen carbonate ion HCO3 (or bicarbonate ion) Based on a Group 5A element nitrite ion NO2 nitrate ion NO3 phosphate ion PO43 hydrogen phosphate ion HPO42 H2PO4 dihydrogen phosphate ion Based on a Group 6A element OH hydroxide ion sulfite ion SO32 sulfate ion SO42 hydrogen sulfate ion HSO4 (or bisulfate ion) Formulas of Ionic Compounds Ionic compounds are composed of ions. For an ionic compound to be electrically neutral—to have no net charge—the numbers of positive and negative ions must be such that the positive and negative charges balance. In sodium chloride, the sodium ion has a 1 charge (Na ) and the chloride ion has a 1 charge (Cl ). These ions must be present in a 1 : 1 ratio, and the formula is NaCl. 108 ■ Balancing Ion Charges Aluminum, a metal in Group 3A, loses three electrons to form the Al3 cation. Oxygen, a nonmetal in Group 6A, gains two electrons to form an O2 anion. Notice that the charge on the cation is the subscript on the anion, and vice versa. 2 Al3 3 O2 ¡ Al2O3 Chapter 3 Molecules, Ions, and Their Compounds The gem ruby is largely the compound formed from aluminum ions (Al3 ) and oxide ions (O2 ). Here the ions have positive and negative charges that are of different absolute value. To have a compound with the same number of positive and negative charges, two Al3 ions [total charge 2 (3 ) 6 ] must combine with three O2 ions [total charge 3 (2 ) 6 ] to give a formula of Al2O3. Calcium is a Group 2A metal, and it forms a cation having a 2 charge. It can combine with a variety of anions to form ionic compounds such as those in the following table: Compound CaCl2 CaCO3 Ca3(PO4)2 Ion Combination Ca Ca 2 2 This often works well, but be careful. The subscripts in Ti4 O2 are reduced to the simplest ratio (1 Ti to 2 O, rather than, 2 Ti to 4 O). Ti4 2 O2 ¡ TiO2 Overall Charge on Compound (2 ) (2 ) 3 2 (1 ) 0 (3 ) 0 2 0 2 Cl CO3 2 (2 ) 3 Ca2 2 PO43 (2 ) In writing formulas, the convention is that the symbol of the cation is given first, followed by the anion symbol. Also notice the use of parentheses when more than one polyatomic ion is present. Example 3.3—Ionic Compound Formulas Problem For each of the following ionic compounds, write the symbols for the ions present and give the number of each: (a) MgBr2, (b) Li2CO3, and (c) Fe2(SO4)3. Strategy Divide the formula of the compound into the cation and the anion. To accomplish this you will have to recognize, and remember, the composition and charges of common ions. Solution (a) MgBr2 is composed of one Mg2 ion and two Br ions. When a halogen such as bromine is combined only with a metal, you can assume the halogen is an anion with a charge of 1 . Magnesium is a metal in Group 2A and always has a charge of 2 in its compounds. (b) Li2CO3 is composed of two lithium ions, Li , and one carbonate ion, CO32 . Li is a Group 1A element and always has a 1 charge in its compounds. Because the two 1 charges balance the negative charge of the carbonate ion, the latter must be 2 . ˇ (c) Fe2(SO4)3 contains two iron ions, Fe3 , and three sulfate ions, SO42 . The way to recognize this is to recall that sulfate has a 2 charge. Because three sulfate ions are present (with a total charge of 6 ), the two iron cations must have a total charge of 6 . This is possible only if each iron cation has a charge of 3 . Comment Remember that the formula for an ion must include its composition and its charge. Formulas for ionic compounds are always written with the cation first and then the anion. Example 3.4—Ionic Compound Formulas Problem Write formulas for ionic compounds composed of an aluminum cation and each of the following anions: (a) fluoride ion, (b) sulfide ion, and (c) nitrate ion. Strategy First decide on the formula of the Al cation and the formula of each anion. Combine the Al cation with each type of anion to form an electrically neutral compound. Solution An aluminum cation is predicted to have a charge of 3 Group 3A. because Al is a metal in 3.3 Ionic Compounds: Formulas, Names, and Properties 109 (a) Fluorine is a Group 7A element. The charge of the fluoride ion is predicted to be 1 (from 8 7 1). Therefore, we need 3 F ions to combine with one Al3 . The formula of the compound is AlF3. (b) Sulfur is a nonmetal in Group 6A, so it forms a 2 anion. Thus, we need to combine two Al3 ions [total charge is 6 2 (3 )] with three S2 ions [total charge is 6 3 (2 )]. The compound has the formula Al2S3. (c) The nitrate ion has the formula NO3 (see Table 3.1). The answer here is therefore similar to the AlF3 case, and the compound has the formula Al(NO3)3. Here we place parentheses around NO3 to show that three polyatomic NO3 ions are involved. Comment The most common error students make is not knowing the correct charge on an ion. Exercise 3.4—Formulas of Ionic Compounds (a) Give the number and identity of the constituent ions in each of the following ionic compounds: NaF, Cu(NO3)2, and NaCH3CO2. (b) Iron, a transition metal, forms ions having at least two different charges. Write the formulas of the compounds formed between two different iron cations and chloride ions. (c) Write the formulas of all neutral ionic compounds that can be formed by combining the cations Na and Ba2 with the anions S2 and PO43 . Names of Ions Naming Positive Ions (Cations) With a few exceptions (such as NH4 ), the positive ions described in this text are metal ions. Positive ions are named by the following rules: 1. For a monatomic positive ion (that is, a metal cation) the name is that of the metal plus the word “cation.” For example, we have already referred to Al3 as the aluminum cation. 2. Some cases occur, especially in the transition series, in which a metal can form more than one type of positive ion. In these cases the charge of the ion is indicated by a Roman numeral in parentheses immediately following the ion’s name. For example, Co2 is the cobalt(II) cation, and Co3 is the cobalt(III) cation. Finally, you will encounter the ammonium cation, NH4 , many times in this book and in the laboratory. Do not confuse the ammonium cation with the ammonia molecule, NH3, which has no electric charge and one less H atom. ■ “-ous” and “-ic” Endings An older naming system for metal ions uses the ending -ous for the ion of lower charge and -ic for the ion of higher charge. For example, there are cobaltous (Co2 ) and cobaltic (Co3 ) ions, and ferrous (Fe2 ) and ferric (Fe3 ) ions. We do not use this system in this book, but some chemical manufacturers continue to use it. Problem-Solving Tip 3.1 Formulas for Ions and Ionic Compounds Writing formulas for ionic compounds takes practice, and it requires that you know the formulas and charges of the most common ions. The charges on monatomic ions are often evident from the position of the element in the periodic table, but you simply have to remember the formula and charges of polyatomic ions; especially the most common ones such as nitrate, sulfate, carbonate, phosphate, and acetate. If you cannot remember the formula of a polyatomic ion, or if you encounter an ion you have not seen before, you may be able to figure out its formula and the name of one of its compounds. For example, suppose you are told that NaCHO2 is sodium formate. You know that the sodium ion is Na , so the formate ion must be the remaining portion of the compound; it must have a charge of 1 to balance the 1 charge on the sodium ion. Thus, the formate ion must be CHO2 . Finally, when writing the formulas of ions, you must include the charge on the ion (except in an ionic compound formula). Writing Na when you mean sodium ion is incorrect. There is a vast difference in the properties of the element sodium (Na) and those of its ion (Na ). 110 Chapter 3 Molecules, Ions, and Their Compounds 1 H 3 2 hydride ion N 3 O2 oxide ion F fluoride ion nitride ion P3 S2 Se2 Te2 Cl chloride ion phosphide sulfide ion ion Br I iodide ion selenide bromide ion ion Figure 3.9 Names and charges of some common monatomic anions. telluride ion Naming Negative Ions (Anions) There are two types of negative ions: those having only one atom (monatomic) and those having several atoms (polyatomic). 1. A monatomic negative ion is named by adding -ide to the stem of the name of the nonmetal element from which the ion is derived (Figure 3.9). The anions of the Group 7A elements, the halogens, are known as the fluoride, chloride, bromide, and iodide ions and as a group are called halide ions. 2. Polyatomic negative ions are common, especially those containing oxygen (called oxoanions). The names of some of the most common oxoanions are given in Table 3.1. Although most of these names must simply be learned, some guidelines can help. For example, consider the following pairs of ions: NO3 is the nitrate ion; NO2 is the nitrite ion. SO42 is the sulfate ion; SO32 is the sulfite ion. ■ Naming Oxoanions per . . . ate increasing oxygen content . . . ate . . . ite hypo . . . ite The oxoanion having the greater number of oxygen atoms is given the suffix -ate, and the oxoanion having the smaller number of oxygen atoms has the suffix -ite. For a series of oxoanions having more than two members, the ion with the largest number of oxygen atoms has the prefix per- and the suffix -ate. The ion having the smallest number of oxygen atoms has the prefix hypo- and the suffix -ite. The oxoanions containing chlorine are good examples. ClO4 ClO3 ClO2 ClO perchlorate ion chlorate ion chlorite ion hypochlorite ion Oxoanions that contain hydrogen are named by adding the word “hydrogen” before the name of the oxoanion. If two hydrogens are in the compound, we say “dihydrogen.” Many hydrogen-containing oxoanions have common names that are used as well. For example, the hydrogen carbonate ion, HCO3 , is called the bicarbonate ion. 3.3 Ionic Compounds: Formulas, Names, and Properties 111 Ion HPO4 HCO3 HSO4 HSO3 2 Systematic Name hydrogen phosphate ion dihydrogen phosphate ion hydrogen carbonate ion hydrogen sulfate ion hydrogen sulfite ion Common Name H2PO4 bicarbonate ion bisulfate ion bisulfite ion Names of Ionic Compounds The name of an ionic compound is built from the names of the positive and negative ions in the compound. The name of the positive cation is given first, followed by the name of the negative anion. If an element such as titanium can form cations with more than one charge, the charge is indicated by a Roman numeral. Examples of ionic compound names are given below. Ionic Compound CaBr2 NaHSO4 (NH4)2CO3 Mg(OH)2 TiCl2 Co2O3 Ions Involved Ca 2 Name calcium bromide sodium hydrogen sulfate ammonium carbonate magnesium hydroxide titanium(II) chloride cobalt(III) oxide and 2 Br Na and HSO4 2 NH4 and CO32 Mg2 and 2 OH Ti 2 and 2 Cl 2 Co3 and 3 O2 See the General ChemistryNow CD-ROM or website: • Screen 3.6 Polyatomic Ions, for a tutorial on the names of polyatomic ions • Screen 3.9 Naming Ionic Compounds, for a tutorial on naming ionic compounds Exercise 3.5—Names of Ionic Compounds 1. Give the formula for each of the following ionic compounds. Use Table 3.1 and Figure 3.9. (a) ammonium nitrate (b) cobalt(II) sulfate (c) nickel(II) cyanide 2. Name the following ionic compounds: (a) MgBr2 (b) Li2CO3 (c) KHSO3 (d) vanadium(III) oxide (e) barium acetate (f) calcium hypochlorite (d) KMnO4 (e) (NH4)2S (f) CuCl and CuCl2 Properties of Ionic Compounds What is the “glue” that causes ions of opposite electric charge to be held together and to form an orderly arrangement of ions in an ionic compound? As described in 112 Chapter 3 Molecules, Ions, and Their Compounds 1 1 n 1 n 1 d dsmall 2 2 dlarge Force vector Li (a) F LiF As ion charge increases, force of attraction increases (b) Coulomb’s law and electrostatic forces. (a) Ions such as Li and F are held together by an electrostatic force. Here a lithium ion is attracted to a fluoride ion, and the distance between the nuclei of the two ions is d. (b) Forces of attraction between ions of opposite charge increase with increasing ion charge and decrease with increasing distance (d). (The force of attraction is proportional to the length of the arrow in this figure.) See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. As distance increases, force of attraction decreases Active Figure 3.10 Section 2.1, when a substance having a negative electric charge is brought near a substance having a positive electric charge, a force of attraction occurs between them (Figure 3.10). In contrast, a force of repulsion occurs when two substances with the same charge—both positive or both negative—are brought together. These forces are called electrostatic forces, and the force of attraction or repulsion between ions is given by Coulomb’s law (Equation 3.1). charge on and ions charge on electron Force of attraction k (n e)(n e) d2 distance between ions (3.1) proportionality constant where, for example, n is 3 for Al3 and n is 2 for O2 . Based on Coulomb’s law, the force of attraction between oppositely charged ions increases Photo: Charles D. Winters; model, S. M. Young. • As the ion charges (n and n ) increase. Thus, the attraction between ions having charges of 2 and 2 is greater than that between ions having 1 and 1 charges (see Figure 3.10). • As the distance between the ions becomes smaller [Figure 3.10; Chapter 9]. Ionic compounds do not consist of simple pairs or small groups of positive and negative ions. The simplest ratio of cations to anions in an ionic compound is represented by its formula, but an ionic solid consists of millions upon millions of ions arranged in an extended three-dimensional network called a crystal lattice. A portion of the lattice for NaCl, illustrated in Figure 3.11, represents a common way of arranging ions for compounds that have a 1 : 1 ratio of cations to anions. Ionic compounds have characteristic properties that can be understood in terms of the charges of the ions and their arrangement in the lattice. Because each ion is surrounded by oppositely charged nearest neighbors, it is held tightly in its allotted location. At room temperature each ion can move just a bit around its aver- Figure 3.11 Sodium chloride. A crystal of NaCl consists of an extended lattice of sodium ions and chloride ions in a 1:1 ratio. (See General ChemistryNow Screen 3.8 Ionic Compounds, to view an animation on the formation of a sodium chloride crystal lattice.) 3.3 Ionic Compounds: Formulas, Names, and Properties 113 Problem-Solving Tip 3.2 Is a Compound Ionic? Students often ask how to know whether a compound is ionic. No method works all of the time, but here are some useful guidelines. 1. Most metal-containing compounds are ionic. So, if a metal atom appears in the formula of a compound, a good first guess is that it is ionic. (There are interesting exceptions, but few come up in introductory chemistry.) It is helpful in this regard to recall trends in metallic behavior: All elements to the left of a diagonal line running from boron to tellurium in the periodic table are metallic. 2. If there is no metal in the formula, it is likely that the compound is not ionic. The exceptions here are compounds composed of polyatomic ions based on nonmetals (e.g., NH4Cl or NH4NO3). 3. Learn to recognize the formulas of polyatomic ions (see Table 3.1). Chemists write the formula of ammo- nium nitrate as NH4NO3 (not as N2H4O3) to alert others to the fact that it is an ionic compound composed of the common polyatomic ions NH4 and NO3 . As an example of these guidelines, you can be sure that MgBr2 (Mg2 with Br ) and K2S (K with S2 ) are ionic compounds. On the other hand, the compound CCl4, formed from two nonmetals, C and Cl, is not ionic. age position. However, considerable energy must be added before an ion can move fast enough and far enough to escape the attraction of its neighboring ions. Only if enough energy is added will the lattice structure collapse and the substance melt. Greater attractive forces mean that ever more energy—higher and higher temperatures—is required to cause melting. Thus, Al2O3, a solid composed of Al3 and O2 ions, melts at a much higher temperature (2072 °C) than NaCl (801 °C), a solid composed of Na and Cl ions. Most ionic compounds are “hard” solids. That is, the solids are not pliable or soft. The reason for this characteristic is again related to the lattice of ions. The nearest neighbors of a cation in a lattice are anions, and the force of attraction makes the lattice rigid. However, a blow with a hammer can cause the lattice to cleave cleanly along a sharp boundary. The hammer blow displaces layers of ions just enough to cause ions of like charge to become nearest neighbors. The repulsion between like charges then forces the lattice apart (Figure 3.12). Charles D. Winters (a) (b) Figure 3.12 Ionic solids. (a) An ionic solid is normally rigid owing to the forces of attraction between oppositely charged ions. When struck sharply, however, the crystal can cleave cleanly. (b) When a crystal is struck, layers of ions move slightly, and ions of like charge become nearest neighbors. Repulsions between ions of similar charge cause the crystal to cleave. (See the General ChemistryNow Screen 3.10 Properties of Ionic Compounds, to watch a video of cleaving a crystal.) 114 Chapter 3 Molecules, Ions, and Their Compounds See the General ChemistryNow CD-ROM or website: • Screen 3.8 Ionic Compounds, to watch a video of the sodium chlorine reaction and for a simulation on the relationship between cations and anions in ionic compounds Exercise 3.6—Coulomb’s Law Explain why the melting point of MgO (2830 °C), much higher than the melting point of NaCl (801 °C). 3.4—Molecular Compounds: Formulas, Names, and Properties Many familiar compounds are not ionic, they are molecular: the water you drink, the sugar in your coffee or tea, or the aspirin you take for a headache. Ionic compounds are generally solids, whereas molecular compounds can range from gases to liquids to solids at ordinary temperatures (see Figure 3.13). As size and molecular complexity increase, compounds generally exist as solids. We will explore some of the underlying causes of these general observations in Chapter 13. Some molecular compounds have complicated formulas that you cannot, at this stage, predict or even decide if they are correct. However, there are many simple compounds you will encounter often, and you should understand how to name them and, in many cases, know their formulas. Let us look first at molecules formed from combinations of two nonmetals. These “two-element” compounds of nonmetals, often called binary compounds, can be named in a systematic way. Hydrogen forms binary compounds with all of the nonmetals except the noble gases. For compounds of oxygen, sulfur, and the halogens, the H atom is generally Photo: Charles D. Winters Figure 3.13 Molecular compounds. Ionic compounds are generally solids at room temperature. In contrast, molecular compounds can be gases, liquids, or solids. The models are of caffeine (in coffee), water, and citric acid (in lemons). 3.4 Molecular Compounds: Formulas, Names, and Properties 115 written first in the formula and is named first. The other nonmetal is named as if it were a negative ion. Compound HF HCl H2S Name hydrogen fluoride hydrogen chloride hydrogen sulfide Virtually all binary molecular compounds of nonmetals are a combination of elements from Groups 4A–7A with one another or with hydrogen. The formula is generally written by putting the elements in order of increasing group number. When naming the compound, the number of atoms of a given type in the compound is designated with a prefix, such as “di-,” “tri-,” “tetra-,” “penta-,” and so on. Compound NF3 NO NO2 N2O N2O4 PCl3 PCl5 SF6 S2F10 Systematic Name nitrogen trifluoride nitrogen monoxide nitrogen dioxide dinitrogen monoxide dinitrogen tetraoxide phosphorus trichloride phosphorus pentachloride sulfur hexafluoride disulfur decafluoride ■ Formulas of Binary Nonmetal Compounds Containing Hydrogen Simple hydrocarbons (compounds of C and H) such as methane and ethane have formulas written with H following C, and the formulas of ammonia and hydrazine have H following N. Water and the hydrogen halides, however, have the H atom preceding O or the halogen atom. Tradition is the only explanation for such irregularities in writing formulas. Finally, many of the binary compounds of nonmetals were discovered years ago and have common names. Compound CH4 C2H6 C3H8 C4H10 NH3 N2H4 PH3 NO N2O H2O Common Name methane ethane propane butane ammonia hydrazine phosphine nitric oxide nitrous oxide (“laughing gas”) water methane, CH4 propane, C3H8 ■ Hydrocarbons Compounds such as methane, ethane, propane, and butane belong to a class of hydrocarbons called alkanes. (See Chapter 11 and General ChemistryNow Screen 3.13, Alkanes.) See the General ChemistryNow CD-ROM or website: ethane, C2H6 butane, C4H10 • Screen 3.12 Binary Compounds of the Nonmetals, for a tutorial on naming compounds of the nonmetals • Screen 3.13 Alkanes, for a simulation and exercise on naming alkanes 116 Chapter 3 Molecules, Ions, and Their Compounds Exercise 3.7—Naming Compounds of the Nonmetals 1. Give the formula for each of the following binary, nonmetal compounds: (a) carbon dioxide (b) phosphorus triiodide (c) sulfur dichloride (d) boron trifluoride (e) dioxygen difluoride (f) xenon trioxide 2. Name the following binary, nonmetal compounds: (a) N2F4 (b) HBr (c) SF4 (d) BCl3 (e) P4O10 (f) ClF3 3.5—Formulas, Compounds, and the Mole The formula of a compound tells you the type of atoms or ions in the compound and the relative number of each. For example, one molecule of methane, CH4, is made up of one atom of C and four atoms of H. But suppose you have Avogadro’s number of C atoms (6.022 1023) combined with the proper number of H atoms. The compound’s formula tells us that four times as many H atoms are required (4 6.022 1023 H atoms) to give Avogadro’s number of CH4 molecules. What masses of atoms are combined, and what is the mass of this many CH4 molecules? C 6.022 10 C atoms 23 4H 4 6.022 10 H atoms 23 ¡ 6.022 23 CH4 10 CH4 molecules 1.000 mol of CH4 molecules 16.04 g of CH4 molecules 1.000 mol of C 12.01 g of C atoms 4.000 mol of H atoms 4.032 g of H atoms Because we know the number of moles of C and H atoms, we know the masses of carbon and hydrogen that combine to form CH4. It follows that the mass of CH4 is the sum of these masses. That is, 1 mol of CH4 has a mass equivalent to the mass of 1 mol of C atoms (12.01 g) plus 4 mol of H atoms (4.032 g). Thus, the molar mass, M, of CH4 is 16.04 g/mol [ Section 2.5]. ■ Molar Mass or Molecular Weight Although chemists often use the term “molecular weight,” we should more properly cite a compound’s molar mass. The SI unit of molar mass is kg/mol, but chemists worldwide usually express it in units of g/mol. See “NIST Guide to SI Units” at www.NIST.gov Molar and Molecular Masses Element or Compound O2 P4 NH3 H2O CH2Cl2 * Molar Mass, M (g/mol) 32.00 123.9 17.03 18.02 84.93 Average Mass of One Molecule* (g/molecule) 5.314 2.057 2.828 2.992 1.410 10–23 10–22 10–23 10–23 10–22 See text, page 117, for the calculation of the mass of one molecule. Ionic compounds such as NaCl do not exist as individual molecules. Thus, we write the simplest formula that shows the relative number of each kind of atom in a “formula unit” of the compound, and the molar mass is calculated from this formula. To differentiate substances like NaCl that do not contain molecules, chemists sometimes refer to their formula weight instead of their molecular weight. 3.5 Formulas, Compounds, and the Mole 117 Figure 3.14 illustrates 1-mol quantities of several common compounds. To find the molar mass of any compound, you need only add up the atomic masses for each element in one formula unit. As an example, let us find the molar mass of aspirin, C9H8O4. In one mole of aspirin there are 9 mol of carbon atoms, 8 mol of hydrogen atoms, and 4 mol of oxygen atoms, which add up to 180.2 g/mol of aspirin. Mass of C in 1 mol C9H8O4 Mass of H in 1 mol C9H8O4 Mass of O in 1 mol C9H8O4 Total mass of 1 mol of C9H8O4 12.01 g C 1 mol C 1.008 g H 8 mol H 1 mol H 16.00 g O 4 mol O 1 mol O molar mass of C9H8O4 9 mol C 108.1 g C 8.064 g H 64.00 g O 180.2 g H CH3 C O H C C C H O C C C H O C OH As was the case with elements, it is important to be able to convert the mass of a compound to the equivalent number of moles (or moles to mass) [ Section 2.5]. For example, if you take 325 mg (0.325 g) of aspirin in one tablet, what amount of the compound have you ingested? Based on a molar mass of 180.2 g/mol, there are 0.00180 mol of aspirin per tablet. 0.325 g aspirin 1 mol aspirin 180.2 g aspirin 0.00180 mol aspirin ■ Aspirin Formula Aspirin has the molecular formula C9H8O4 and a molar mass of 180.2 g/mol. Aspirin is the common name of the compound acetylsalicylic acid. Using the molar mass of a compound it is possible to determine the number of molecules in any sample from the sample mass and to determine the mass of one molecule. For example, the number of aspirin molecules in one tablet is 0.00180 mol aspirin 6.022 1023 molecules 1 mol aspirin 1.08 1021 molecules and the mass of one molecule is 180.2 g aspirin 1 mol aspirin 1 mol aspirin 6.022 1023 molecules 2.99 10 22 g/molecule Figure 3.14 One-mole quantities of some compounds. H2O 18.02 g/mol Charles D. Winters Aspirin, C9H8O4 180.2 g/mol Copper(II) chloride dihydrate, CuCl2 2 H2O 170.5 g/mol Iron(III) oxide, Fe2O3 159.7 g/mol 118 Chapter 3 Molecules, Ions, and Their Compounds See the General ChemistryNow CD-ROM or website: • Screen 3.14 Compounds, Molecules, and the Mole, for a simulation exploring the relationship between mass, moles, molecules, and atoms, and a tutorial on determining molar mass • Screen 3.15 Using Molar Mass, for a tutorial on determining moles from mass and a second tutorial on determining mass from moles Example 3.5—Molar Mass and Moles Problem You have 16.5 g of oxalic acid, H2C2O4. (a) What amount is represented by 16.5 g of oxalic acid? (b) How many molecules of oxalic acid are in 16.5 g? (c) How many atoms of carbon are in 16.5 g of oxalic acid? (d) What is the mass of one molecule of oxalic acid? Strategy The first step in any problem involving the conversion of mass and moles is to find the molar mass of the compound in question. Then you can perform the other calculations as outlined by the scheme shown here to find the number of molecules from the amount of substance and the number of atoms of a particular kind: mol g molecules mol C atoms molecule Mass, g use molar mass Moles use Avogadro’s number Molecules use formula Number of C atoms (See the General ChemistryNow Screen 3.14 Compounds and Moles, and Screen 3.15 Molar Mass.) Solution (a) Moles represented by 16.5 g Let us first calculate the molar mass of oxalic acid: 12.01 g C 1 mol C 1.008 g H 1 mol H 16.00 g O 1mol O 2 mol C per mol H2C2O4 2 mol H per mol H2C2O4 4 mol O per mol H2C2O4 24.02 g C per mol H2C2O4 2.016 g H per mol H2C2O4 64.00 g O per mol H2C2O4 Molar mass of H2C2O4 90.04 g per mol H2C2O4 Now calculate the amount in moles. The molar mass (expressed in units of 1 mol/90.04 g) is the conversion factor in all mass-to-mole conversions. 1 mol 90.04 g H2C2O4 16.5 g H2C2O4 0.183 mol H2C2O4 3.6 Describing Compound Formulas 119 (b) Number of molecules Use Avogadro’s number to find the number of oxalic acid molecules in 0.183 mol of H2C2O4. 0.183 mol 6.022 1023 molecules 1 mol 1.10 1023 molecules (c) Number of C atoms Because each molecule contains two carbon atoms, the number of carbon atoms in 16.5 g of the acid is 1.10 1023 molecules 2 C atoms 1 molecule 2.20 1023 C atoms (d) Mass of one molecule The units of the desired answer are grams per molecule, which indicates that you should multiply the starting unit of molar mass (grams per mole) by (1/Avogadro’s number) (units are mole/molecule), so that the unit “mol” cancels. 90.04 g 1 mol 1 mol 1023 molecules 1.495 10 22 6.0221 g/molecule Exercise 3.8—Molar Mass and Moles-to-Mass Conversions (a) Calculate the molar mass of citric acid, C6H8O7, and MgCO3. (b) If you have 454 g of citric acid, what amount (moles) does this represent? (c) To have 0.125 mol of MgCO3, what mass (g) must you have? 3.6—Describing Compound Formulas Given a sample of an unknown compound, how can its formula be determined? The answer lies in chemical analysis, a major branch of chemistry that deals with the determination of formulas and structures. ■ Molecular Composition Molecular composition can be expressed as a percent (mass of an element in a 100-g sample). For example, NH3 is 82.27% N. Therefore, it has 82.27 g of N in 100.0 g of compound. Percent Composition Any sample of a pure compound always consists of the same elements combined in the same proportion by mass. This means molecular composition can be expressed in at least three ways: • In terms of the number of atoms of each type per molecule or per formula unit—that is, by giving the formula of the compound • In terms of the mass of each element per mole of compound • In terms of the mass of each element in the compound relative to the total mass of the compound—that is, as a mass percent Suppose you have 1.000 mol of NH3 or 17.03 g. This mass of NH3 is composed of 14.01 g of N (1.000 mol ) and 3.024 g of H (3.000 mol ). If you compare the mass of N to the total mass of compound, 82.27% of the total mass is N (and 17.76% is H). 82.27% of NH3 mass is nitrogen. N H} HH H 17.76% of NH3 mass is hydrogen. Note that the %N and %H do not add up to exactly 100%. This is not unusual and does not mean there is an error. The last digit of the answer is limited by the accuracy of the data used. 120 Chapter 3 Molecules, Ions, and Their Compounds Mass of N per mole of NH3 Mass percent N in NH3 1 mol N 1 mol NH3 14.01 g N 1 mol N 14.01 g N/1 mol NH3 mass of N in 1 mol NH3 mass of 1 mol NH3 14.01 g N 100% 17.03 g NH3 82.27% 1or 82.27 g N in 100.0 g NH3 2 3 mol H 1 mol NH3 1.008 g H 1 mol H 3.024 g H/1 mol NH3 Mass of H per mole of NH3 Mass percent H in NH3 mass of H in 1 mol NH3 100% mass of 1 mol NH3 3.024 g H 100% 17.03 g NH3 17.76% 1or 17.76 g H in 100.0 g NH3 2 ˇ These values represent the mass percent of each element, or percent composition by mass. They tell you that in a 100.0-g sample there are 82.27 g of N and 17.76 g of H. See the General ChemistryNow CD-ROM or website: • Screen 3.16 Percent Composition, for a tutorial on detemining percent composition Example 3.6—Using Percent Composition Problem What is the mass percent of each element in propane, C3H8? What mass of carbon is contained in 454 g of propane? Strategy First find the molar mass of C3H8 and then calculate the mass percent of C and H per mole of C3H8. Using the knowledge of the mass percent of C, calculate the mass of carbon in 454 g of C3H8. Solution (a) The molar mass of C3H8 is 44.10 g/mol. (b) Mass percent of C and H in C3H8: 3 mol C 1 mol C3H8 12.01 g C 1 mol C 36.03 g C/1 mol C3H8 36.03 g C 44.10 g C3H8 100% 81.70% C ˇ Mass percent of C in C3H8 8 mol H 1 mol C3H8 1.008 g H 1 mol H 8.064 g H/1 mol C3H8 8.064 g H 44.10 g C3H8 100% 18.29% H Mass percent of H in C3H8 (c) Mass of C in 454 g of C3H8: 454 g C3H8 81.70 g C 100.0 g C3H8 371 g C 3.6 Describing Compound Formulas 121 Exercise 3.9—Percent Composition (a) Express the composition of ammonium carbonate, (NH4)2CO3, in terms of the mass of each element in 1.00 mol of compound and the mass percent of each element. (b) What is the mass of carbon in 454 g of octane, C8H18? Empirical and Molecular Formulas from Percent Composition Now let us consider the reverse of the procedure just described: using relative mass or percent composition data to find a molecular formula. Suppose you know the identity of the elements in a sample and have determined the mass of each element in a given mass of compound by chemical analysis [ Section 4.6]. You can then calculate the relative amount (moles) of each element and from this the relative number of atoms of each element in the compound. For example, for a compound composed of atoms of A and B, the steps from percent composition to a formula are Convert weight percent to mass %A %B gA gB Convert mass to moles x mol A y mol B Find mole ratio x mol A y mol B Ratio gives formula AxBy Let us derive the formula for hydrazine, a close relative of ammonia and a compound used to remove oxygen from water used for heating and cooling. Step 1: Convert mass percent to mass. The mass percentages in a sample of hydrazine are 87.42% N and 12.58% H. Thus, in a 100.00-g sample of hydrazine, there are 87.42 g of N and 12.58 g of H. Step 2: Convert the mass of each element to moles. The amount of each element in the 100.00-g sample is 87.42 g N 12.58 g H 1 mol N 14.007 g N 1 mol H 1.008 g H 6.241 mol N 12.48 mol H ■ Deriving a Formula Percent composition gives the mass of an element in 100 g of sample. However, any amount of sample is appropriate if you know the mass of an element in that sample mass. See Example 3.8. Step 3: Find the mole ratio of elements. Use the amount (moles) of each element in the 100.00-g of sample to find the amount of one element relative to the other. For hydrazine, this ratio is 2 mol of H to 1 mol of N, 12.48 mol H 6.241 mol N 2.00 mol H ¡ NH2 1.00 mol N ■ Deriving a Formula—Mole Ratios When finding the ratio of moles of one element relative to another, always divide the larger number by the smaller one. showing that there are 2 mol of H atoms for every 1 mol of N atoms in hydrazine. Thus, in one molecule, two atoms of H occur for every atom of N; that is, the formula is NH2. This simplest whole-number atom ratio is called the empirical formula. 122 Chapter 3 Molecules, Ions, and Their Compounds Percent composition data allow us to calculate the atom ratios in a compound. A molecular formula, however, must convey two pieces of information: (1) the relative numbers of atoms of each element in a molecule (the atom ratios) and (2) the total number of atoms in the molecule. For hydrazine there are twice as many H atoms as N atoms, so the molecular formula could be NH2. Recognize, however, that percent composition data give only the simplest possible ratio of atoms in a molecule. The empirical formula of hydrazine is NH2, but the true molecular formula could also be NH2, N2H4, N3H6, N4H8, or any other formula having a 1 : 2 ratio of N to H. To determine the molecular formula from the empirical formula, the molar mass must be obtained from experiment. For example, experiments show that the molar mass of hydrazine is 32.0 g/mol, twice the formula mass of NH2, which is 16.0 g/mol. Thus, the molecular formula of hydrazine is two times the empirical formula of NH2, that is, N2H4. As another example of the usefulness of percent composition data, let us say that you collected the following information in the laboratory for isooctane, the compound used as the standard for determining the octane rating of a fuel: % carbon 84.12; % hydrogen 15.88; molar mass 114.2 g/mol. These data can be used to calculate the empirical and molecular formulas for the compound. The data inform us that 84.12 g of C and 15.88 g of H occur in a 100.0-g sample. From this, we find the amount (moles) of each element in this sample. 84.12 g C 15.88 g H 1 mol C 12.011 g C 1 mol H 1.0079 g H 7.004 mol C 15.76 mol H This means that, in any sample of isooctane, the ratio of moles of H to C is 15.76 mol H 7.004 mol C 2.250 mol H 1.000 mol C Mole ratio Now the task is to turn this decimal fraction into a whole-number ratio of H to C. To do this, recognize that 2.25 is the same as 21 or 9/4. Therefore, the ratio of C to 4 H is 2 1 mol H 4 1 mol C Mole ratio ■ Isooctane and the Octane Rating Isooctane, C8H18, is the standard against which the octane rating of gasoline is determined. Octane numbers are assigned by comparing the burning performance of gasoline with the burning performance of mixtures of isooctane and heptane. Gasoline with an octane rating of 90 matches the burning characteristics of a mixture of 90% isooctane and 10% heptane. 2.25 mol H 1.00 mol C 9/4 mol H 1 mol C 9 mol H 4 mol C You now know that nine H atoms occur for every four C atoms in isooctane. Thus, the simplest or empirical formula is C4H9. If C4H9 were the molecular formula, the molar mass would be 57.12 g/mol. However, we know from experiment that the actual molar mass is 114.2 g/mol, twice the value for the empirical formula. 114.2 g/mol of isooctane 57.12 g/mol of C4H9 2.00 mol C4H9 per mol of isooctane The molecular formula is therefore C8H18. 3.6 Describing Compound Formulas 123 See the General ChemistryNow CD-ROM or website: • Screen 3.17 Determining Empirical Formulas, for a tutorial on determining empirical formulas • Screen 3.18 Determining Molecular Formulas, for a tutorial on determining molecular formulas Charles D. Winters Example 3.7—Calculating a Formula from Percent Composition Problem Eugenol is the major component in oil of cloves. It has a molar mass of 164.2 g/mol and is 73.14% C and 7.37% H; the remainder is oxygen. What are the empirical and molecular formulas of eugenol? Strategy To derive a formula we need to know the mass percent of each element. Because the mass percents of all elements must add up to 100.0%, we find the mass percent of O from the difference between 100.0% and the mass percents of C and H. Next, we assume the mass percent of each element is equivalent to its mass in grams, and convert each mass to moles. Finally, the ratio of moles gives the empirical formula. The mass of a mole of compound having the calculated empirical formula is compared with the actual, experimental molar mass to find the true molecular formula. Solution The mass of O in a 100.0-g sample is 100.0 g 73.14 g C Mass of O The amount of each element is 73.14 g C 7.37 g H 19.49 g O 1 mol C 12.011 g C 1 mol H 1.008 g H 1 mol O 15.999 g O 6.089 mol C 7.31 mol H 1.218 mol O 7.37 g H 19.49 g mass of O Eugenol, C10H12O2, is an important component in oil of cloves. To find the mole ratio, the best approach is to base the ratios on the smallest number of moles present—in this case, oxygen. mol C mol O mol H mol O 6.089 mol C 1.218 mol O 7.31 mol H 1.218 mol O 4.999 mol C 1.000 mol O 6.00 mol H 1.000 mol O 5 mol C 1 mol O 6 mol H 1 mol O Now we know there are 5 mol of C and 6 mol of H per 1 mol of O. Thus, the empirical formula is C5H6O. The experimentally determined molar mass of eugenol is 164.2 g/mol. This is twice the mass of C5H6O (82.1 g/mol). 164.2 g/mol of eugenol 82.10 g/mol of C5H6O The molecular formula is C10H12O2. 2.00 mol C5H6O per mol of eugenol 124 Chapter 3 Molecules, Ions, and Their Compounds Comment There is another approach to finding the molecular formula here. Knowing the percent composition of eugenol and its molar mass, we could calculate that in 164.2 g of eugenol there are 120.1 g of C (10 mol of C), 12.1 g of H (12 mol of H), and 32.00 g of O (2 mol of O). This gives us a molecular formula of C10H12O2. However, you must recognize that this approach can only be used when you know both the percent composition and the molar mass. Exercise 3.10—Empirical and Molecular Formulas (a) What is the empirical formula of naphthalene, C10H8? (b) The empirical formula of acetic acid is CH2O. If its molar mass is 60.05 g/mol, what is the molecular formula of acetic acid? Exercise 3.11—Calculating a Formula from Percent Composition Isoprene is a liquid compound that can be polymerized to form natural rubber. It is composed of 88.17% carbon and 11.83% hydrogen. Its molar mass is 68.11 g/mol. What are its empirical and molecular formulas? Exercise 3.12—Calculating a Formula from Percent Composition Camphor is found in “camphor wood,” much prized for its wonderful odor. It is composed of 78.90% carbon and 10.59% hydrogen. The remainder is oxygen. What is its empirical formula? Determining a Formula from Mass Data The composition of a compound in terms of mass percent gives us the mass of each element in a 100.0-g sample. In the laboratory we often collect information on the composition of compounds slightly differently. We can 1. Combine known masses of elements to give a sample of the compound of known mass. Element masses can be converted to moles, and the ratio of moles gives the combining ratio of atoms—that is, the empirical formula. This approach is described in Example 3.8. 2. Decompose a known mass of an unknown compound into “pieces” of known composition. If the masses of the “pieces” can be determined, the ratio of moles of the “pieces” gives the formula. An example is a decomposition such as Ni1CO2 4 1/2 ¡ Ni1s2 4 CO1g2 Problem-Solving Tip 3.3 Finding Empirical and Molecular Formulas • The experimental data available to find a formula may be in the form of percent composition or the masses of elements combined in some mass of compound. No matter what the starting point, the first step is always to convert masses of elements to moles. • Be sure to use at least three significant figures when calculating empirical formulas. Using fewer significant figures often gives a misleading result. • When finding atom ratios, always divide the larger number of moles by the smaller one. • Empirical and molecular formulas often differ for molecular compounds. In contrast, the formula of an ionic compound is generally the same as its empirical formula. • Determining the molecular formula of a compound after calculating the empirical formula requires knowing the molar mass. • When both the percent composition and the molar mass are known for a compound, the alternative method mentioned in the comment to Example 3.7 could be used. However, you must recognize that this approach can only be used when you know both the percent composition and the molar mass. 3.6 Describing Compound Formulas 125 The masses of Ni and CO can be converted to moles, whose 1 : 4 ratio would reveal the formula of the compound. We will describe this approach in Chapter 4 [ Section 4.6]. Example 3.8—Formula of a Compound from Combining Masses Problem Gallium oxide, GaxOy, forms when gallium is combined with oxygen. Suppose you allow 1.25 g of gallium (Ga) to react with oxygen and obtain 1.68 g of GaxOy. What is the formula of the product? Strategy Calculate the mass of oxygen in 1.68 g of product (which you already know contains 1.25 g of Ga). Next, calculate the amounts of Ga and O (in moles) and find their ratio. Solution The masses of Ga and O combined in 1.68 g of product are 1.68 g product 1.25 g Ga 0.43 g O Next, calculate the amount of each reactant: 1 mol Ga 69.72 g Ga 1 mol O 16.0 g O 1.25 g Ga 0.43 g O 0.0179 mol Ga 0.027 mol O Find the ratio of moles of O to moles of Ga: 0.027 mol O 0.0179 mol Ga 1.5 mol O 1. 0 mol Ga Mole ratio It is 1.5 mol O/1.0 mol Ga, or 3 mol O to 2 mol Ga. Thus, the product is gallium oxide, Ga2O3. Example 3.9—Determining a Formula from Mass Data Problem Tin metal (Sn) and purple iodine (I2) combine to form orange, solid tin iodide with an unknown formula. Sn metal solid I2 ¡ solid SnxIy Weighed quantities of Sn and I2 are combined, where the quantity of Sn is more than is needed to react with all of the iodine. After SnxIy has been formed, it is isolated by filtration. The mass of excess tin is also determined. The following data were collected: Mass of tin (Sn) in original mixture Mass of iodine (I2) in original mixture Mass of tin (Sn) recovered after reaction 1.056 g 1.947 g 0.601 g Strategy The first step is to find the masses of Sn and I that are combined in SnxIy. The masses are then converted to moles, and the ratio of moles reveals the compound’s empirical formula. 126 (a) Weighed samples of tin (left) and iodine (right). Chapter 3 Molecules, Ions, and Their Compounds (c) The hot reaction mixture is filtered to recover unreacted tin. (d) When the solvent cools, solid, orange tin iodide forms and is isolated. (b) The tin and iodine are heated in a solvent. The formula of a compound of tin and iodine can be found by determining the mass of iodine that combines with a given mass of tin. Solution First, let us find the mass of tin that combined with iodine. Mass of Sn in original mixture Mass of Sn recovered Mass of Sn combined with 1.947 g I2 Now convert the mass of tin to the amount of tin. 0.455 g Sn 1 mol Sn 118.7 g Sn 0.00383 mol Sn 1.056 g 0.601 g 0.455 g No I2 was recovered; it all reacted with Sn. Therefore, 0.00383 mol of Sn combined with 1.947 g of I2. Because we want to know the amount of I that combined with 0.00383 mol of Sn, we calculate the amount of I from the mass of I2. 1.947 g I2 1 mol I2 253.81 g I2 2 mol I 1 mol I2 0.01534 mol I Finally, we find the ratio of moles. mol I mol Sn 0.01534 mol I 0.00383 mol Sn 4.01 mol I 1.00 mol Sn 4 mol I 1 mol Sn There are four times as many moles of I as moles of Sn in the sample. Therefore, there are four times as many atoms of I as atoms of Sn per formula unit. The empirical formula is SnI4. Exercise 3.13—Determining a Formula from Combining Masses Analysis shows that 0.586 g of potassium metal combines with 0.480 g of O2 gas to give a white solid having a formula of KxOy. What is the empirical formula of the compound? Determining a Formula by Mass Spectrometry We have described chemical methods of determining a molecular formula, but many instrumental methods are available as well. One of them is mass spectrometry (Figure 3.15). We introduced this technique in Chapter 2, where it was used to show Charles D. Winters 3.6 Describing Compound Formulas 127 A Closer Look Mass Spectrometry, Molar Mass, and Isotopes Bromobenzene, C6H5Br, has a molecular weight of 157.010. Why, then, are there two prominent lines at 156 and 158 in the mass spectrum of the compound? The answer shows us the influence of isotopes on molecular weight. Bromine has two naturally occurring isotopes: 79Br and 81Br. They are 50.7% and 49.3% abundant, respectively. What is the mass of C6H5Br based on each isotope? If we use the most abundant isotopes of C and H (12C and 1H), the mass of the compound having only the 79Br isotope, C6H579Br, is 156. The mass of the compound containing only the 81Br isotope, C6H581Br, is 158. These two lines have the highest mass-to-charge ratio in the spectrum. The calculated molecular weight of bromobenzene is 157.010, a value calculated from the atomic weights of the elements. These atomic weights reflect the abundances of all isotopes. In contrast, the mass spectrum has a line for each possible combination of isotopes. This explains why there are also small lines at the mass-tocharge ratios of 157 and 159. They arise from various combinations of 1H, 12C, 13C, 79 Br, and 81Br atoms. In fact, careful analysis of such patterns can unambiguously identify a molecule. Bromobenzene mass spectrum 100 80 Relative abundance of ions 156 60 158 (12C)6(1H)581Br (12C)6(1H)579Br 40 20 0 0 40 80 Mass-to-charge ratio (m/Z) 120 160 Figure 3.15 Mass spectrum of 100 CH2OH (m/Z 31 u) ethanol, CH3CH2OH. A prominent peak or line in the spectrum is the “parent” ion (CH3CH2OH ) at mass 46. (The “parent” ion is the heaviest ion observed.) The mass designated by the peak for the “parent” ion confirms the formula of the molecule. Other peaks are for “fragment” ions. This pattern of lines can provide further, unambiguous evidence of the formula of the compound. (The horizontal axis is the mass-to-charge ratio of a given ion. Because almost all observed ions have a charge of Z 1, the value observed is the mass of the ion.) (See A Closer Look: Mass Spectrometry, Molar Mass, and Isotopes.) Relative abundance of ions 80 60 C2H5 (m/Z 29 u) CH3 (m/Z 15 u) CH3CH2O (m/Z 45 u) CH3CH2OH (m/Z 46 u) 40 20 0 10 20 30 40 50 Mass-to-charge ratio (m/Z) 128 Chapter 3 Molecules, Ions, and Their Compounds Charles D. Winters the existence of isotopes and to measure their relative abundance [ Figure 2.8]. If a compound can be turned into a vapor, the vapor can be passed through an electron beam in a mass spectrometer where high-energy electrons collide with the gasphase molecules. These high-energy collisions cause the molecule to lose electrons and turn the molecules into positive ions. These ions usually break apart or fragment into smaller pieces. As illustrated in Figure 3.15, the cation created from ethanol (CH3CH2OH ) fragments ( losing an H atom) to give another cation (CH3CH2O ), which further fragments. A mass spectrometer detects and records the masses of the different particles. Analysis of the spectrum can help identify a compound and can give an accurate molar mass. Active Figure 3.17 Figure 3.16 Gypsum wallboard. Gypsum is hydrated calcium sulfate, CaSO4 2 H2O. 3.7—Hydrated Compounds If ionic compounds are prepared in water solution and then isolated as solids, the crystals often have molecules of water trapped within the lattice. Compounds in which molecules of water are associated with the ions of the compound are called hydrated compounds. The beautiful blue copper(II) compound in Figure 3.14, for example, has a formula that is conventionally written as CuCl2 2 H2O. The dot between CuCl2 and 2 H2O indicates that 2 mol of water is associated with every mole of CuCl2; it is equivalent to writing the formula as CuCl2(H2O)2. The name of the compound, copper(II) chloride dihydrate, reflects the presence of 2 mol of water per mole of CuCl2. The molar mass of CuCl2 2 H2O is 134.5 g/mol (for CuCl2) plus 36.0 g/mol (for 2 H2O) giving a total mass of 170.5 g/mol. Hydrated compounds are common. The walls of your home may be covered with wallboard, or “plaster board” (Figure 3.16). These sheets contain hydrated calcium sulfate, or gypsum (CaSO4 2 H2O), as well as unhydrated CaSO4, sandwiched between paper. Gypsum is a mineral that can be mined. Now, however, it is more commonly a byproduct in the manufacture of superphosphate fertilizer from Ca5F(PO4)3 and sulfuric acid. If gypsum is heated between 120 and 180 °C, the water is partly driven off to give CaSO4 1 H2O, a compound commonly called “plaster of Paris.” If you have ever 2 broken an arm or leg and had to have a cast, the cast may have been made of this compound. It is an effective casting material because, when added to water, it forms a thick slurry that can be poured into a mold or spread out over a part of the body. As it takes on more water, the material increases in volume and forms a hard, inflexible solid. These properties also make plaster of Paris a useful material to artists, because the expanding compound fills a mold completely and makes a high-quality reproduction. Hydrated cobalt(II) chloride is the red solid in Figure 3.17. When heated it turns first purple and then deep blue as it loses water to form anhydrous CoCl2; “anhydrous” means a substance without water. On exposure to moist air, anhydrous CoCl2 takes up water and is converted back into the red hydrated compound. It is this property that allows crystals of the blue compound to be used as a humidity indicator. You may have seen them in a small bag packed with a piece of electronic equipment. The compound also makes a good “invisible ink.” A solution of cobalt(II) chloride in water is red, but if you write on paper with the solution it cannot be seen. When the paper is warmed, however, the cobalt compound dehydrates to give the deep blue anhydrous compound, and the writing becomes visible. There is no simple way to predict how much water will be present in a hydrated compound, so it must be determined experimentally. Such an experiment may involve heating the hydrated material so that all the water is released from the solid 3.7 Hydrated Compounds 129 Charles D. Winters Active Figure 3.17 Dehydrating hydrated cobalt(II) chloride, CoCl2 6H2O. (left) Cobalt chloride hexahydrate, CoCl2 6H2O, is a deep red compound. (left and center) When it is heated, the compound loses the water of hydration and forms the deep blue compound CoCl2. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by and exercise. and evaporated. Only the anhydrous compound is left. The formula of hydrated copper(II) sulfate, commonly known as “blue vitriol,” is determined in this manner in Example 3.10. See the General ChemistryNow CD-ROM or website: • Screen 3.19 Hydrated Compounds, for a tutorial on detemining mass and moles of compounds of a hydrated compound and an exercise on analyzing a mixture White CuSO4 Blue CuSO4 5 H2O Example 3.10—Determining the Formula of a Hydrated Compound Problem You want to know the value of x in blue, hydrated copper(II) sulfate, CuSO4 x H2O—that is, the number of water molecules for each unit of CuSO4. In the laboratory you weigh out 1.023 g of the solid. After heating the solid thoroughly in a porcelain crucible (see figure), 0.654 g of nearly white, anhydrous copper(II) sulfate, CuSO4, remains. 1.023 g CuSO4 x H2O ¡ 0.654 g CuSO4 heat Charles D. Winters ? g H2O Strategy To find x we need to know the amount of H2O per mole of CuSO4. Therefore, first we find the mass of water lost by the sample from the difference between the mass of hydrated compound and the anhydrous form. Finally, we find the ratio of amount of water lost (moles) to the amount of anhydrous CuSO4. Heating a Hydrated Compound The formula of a hydrated compound can be determined by heating a weighed sample enough to cause the compound to release its water of hydration. Knowing the mass of the hydrated compound before heating, and the mass of the anhydrous compound after heating, we can determine the mass of water in the original sample. 130 Chapter 3 Molecules, Ions, and Their Compounds Solution Find the mass of water. Mass of hydrated compound Mass of anhydrous compound, CuSO4 Mass of water Next convert the masses of CuSO4 and H2O to moles. 0.369 g H2O 1 mol H2O 18.02 g H2O 1 mol CuSO4 159.6 g CuSO4 0.0205 mol H2O 1.023 g 0.654 g 0.369 g 0.654 g CuSO4 0.00410 mol CuSO4 The value of x is determined from the mole ratio. 0.0205 mol H2O 0.00410 mol CuSO4 5.00 mol H2O 1.00 mol CuSO4 The water-to-CuSO4 ratio is 5 : 1, so the formula of the hydrated compound is CuSO4 5 H2O. Its name is copper1II2 sulfate pentahydrate. Exercise 3.14—Determining the Formula of a Hydrated Compound Hydrated nickel(II) chloride is a beautiful green, crystalline compound. When heated strongly, the compound is dehydrated. If 0.235 g of NiCl2 x H2O gives 0.128 g of NiCl2 on heating, what is the value of x ? Chapter Goals Revisited Now that you have studied this chapter, you should ask whether you have met the chapter goals. In particular, you should be able to Interpret, predict, and write formulas for ionic and molecular compounds a. Recognize and interpret molecular formulas, condensed formulas, and structural formulas (Section 3.1). b. Recognize that metal atoms commonly lose one or more electrons to form positive ions, called cations, and nonmetal atoms often gain electrons to form negative ions, called anions (see Figure 3.7). c. Recognize that the charge on a metal cation in Groups 1A, 2A, and 3A is equal to the group number in which the element is found in the periodic table (Mn , n Group number) (Section 3.3). Charges on transition metal cations are often 2 or 3 , but other charges are observed. General ChemistryNow homework: Study Question(s) 11 • • See the General ChemistryNow CD-ROM or website to: Assess your understanding with homework questions keyed to each goal Check your readiness for an exam by taking the exam-prep quiz and exploring the resources in the personalized Learning Plan it provides d. Recognize that the negative charge on a single-atom or monatomic anion, Xn , is given by n 8 group number (Section 3.3). Key Equations 131 e. Write formulas for ionic compounds by combining ions in the proper ratio to give no overall charge (Section 3.4). Name compounds a. Give the names or formulas of polyatomic ions, knowing their formulas or names, respectively (Table 3.1 and Section 3.3). b. Name ionic compounds and simple binary compounds of the nonmetals (Sections 3.3 and 3.4). General ChemistryNow homework: SQ(s) 7, 19, 21, 27, 29 Understand some properties of ionic compounds a. Understand the importance of Coulomb’s law (Equation 3.1), which describes the electrostatic forces of attraction and repulsion of ions. Coulomb’s law states that the force of attraction between oppositely charged species increases with electric charge and with decreasing distance between the species (Section 3.3). General ChemistryNow homework: SQ(s) 26 Calculate and use molar mass a. Understand that the molar mass of a compound (often called the molecular weight ) is the mass in grams of Avogadro’s number of molecules (or formula units) of a compound (Section 3.5). For ionic compounds, which do not consist of individual molecules, the sum of atomic masses is often called the formula mass (or formula weight ). b. Calculate the molar mass of a compound from its formula and a table of atomic weights (Section 3.5). General ChemistryNow homework: SQ(s) 31, 33 c. Calculate the number of moles of a compound that is represented by a given mass, and vice versa (Section 3.5). General ChemistryNow homework: SQ(s) 35 Calculate percent composition for a compound and derive formulas from experimental data a. Express the composition of a compound in terms of percent composition (Section 3.6). General ChemistryNow homework: SQ(s) 41, 45 b. Use percent composition or other experimental data to determine the empirical formula of a compound (Section 3.6). General ChemistryNow homework: SQ(s) 47, 52, 53, 94 c. Understand how mass spectrometry can be used to find a molar mass (Section 3.6). d. Use experimental data to find the number of water molecules in a hydrated compound (Section 3.7). General ChemistryNow homework: SQ(s) 55, 57. 59 Key Equations Equation 3.1 (page 112) Coulomb’s law describes the dependence of the force of attraction between ions of opposite charge (or the force of repulsion between ions of like charge) on ion charge and the distance between ions. charge on and ions charge on electron Force of attraction k (n e)(n e) d2 distance between ions proportionality constant 132 Chapter 3 Molecules, Ions, and Their Compounds Study Questions ▲ denotes more challenging questions. ■ denotes questions available in the Homework and Goals section of the General ChemistryNow CD-ROM or website. Blue numbered questions have answers in Appendix O and fully worked solutions in the Student Solutions Manual. Structures of many of the compounds used in these questions are found on the General ChemistryNow CD-ROM or website in the Models folder. Assess your understanding of this chapter’s topics with additional quizzing and conceptual questions at http://now.brookscole.com/kotz6e 4. The molecule illustrated here is methanol. Using Figure 3.4 as a guide, decide which atoms are in the plane of the paper, which lie above the plane, and which lie below. Sketch a ball-and-stick model. If available to you, go to the General ChemistryNow CD-ROM or website and find the model of methanol. Practicing Skills Molecular Formulas and Models (See Examples 3.1 and 3.2 and Exercises 3.1 and 3.2.) 1. A ball-and-stick model of sulfuric acid is illustrated here. Write the molecular formula for sulfuric acid and draw the structural formula. Describe the structure of the molecule. Is it flat? That is, are all the atoms in the plane of the paper? (Color code: sulfur atoms are yellow; oxygen atoms are red; and hydrogen atoms are white.) Ions and Ion Charges (See Exercise 3.3, Figure 3.7, Table 3.1, and the General ChemistryNow Screens 3.5 and 3.6.) 5. What charges are most commonly observed for monatomic ions of the following elements? (a) magnesium (c) nickel (b) zinc (d) gallium 6. What charges are most commonly observed for monatomic ions of the following elements? (a) selenium (c) iron (b) fluorine (d) nitrogen 7. ■ Give the symbol, including the correct charge, for each of the following ions: (a) barium ion (b) titanium(IV) ion (c) phosphate ion (d) hydrogen carbonate ion (e) sulfide ion (f ) perchlorate ion (g) cobalt(II) ion (h) sulfate ion 8. Give the symbol, including the correct charge, for each of the following ions: (a) permanganate ion (b) nitrite ion (c) dihydrogen phosphate ion (d) ammonium ion (e) phosphate ion (f ) sulfite ion 2. A ball-and-stick model of toluene is illustrated here. What is its molecular formula? Describe the structure of the molecule. Is it flat or is only a portion of it flat? (Color code: carbon atoms are gray and hydrogen atoms are white.) 3. A model of the cancer chemotherapy agent cisplatin is given here. Write the molecular formula for the compound and draw its structural formula. ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 133 9. When a potassium atom becomes a monatomic ion, how many electrons does it lose or gain? What noble gas atom has the same number of electrons as a potassium ion? 10. When oxygen and sulfur atoms become monatomic ions, how many electrons does each lose or gain? Which noble gas atom has the same number of electrons as an oxygen ion? Which noble gas atom has the same number of electrons as a sulfur ion? Ionic Compounds (See Examples 3.3 and 3.4 and the General ChemistryNow Screen 3.8.) 11. ■ Predict the charges of the ions in an ionic compound containing the elements barium and bromine. Write the formula for the compound. 12. What are the charges of the ions in an ionic compound containing cobalt(III) and fluoride ions? Write the formula for the compound. 13. For each of the following compounds, give the formula, charge, and the number of each ion that makes up the compound: (a) K2S (d) (NH4)3PO4 (b) CoSO4 (e) Ca(ClO)2 (c) KMnO4 14. For each of the following compounds, give the formula, charge, and the number of each ion that makes up the compound: (a) Mg(CH3CO2)2 (d) Ti(SO4)2 (b) Al(OH)3 (e) KH2PO4 (c) CuCO3 15. Cobalt forms Co2 and Co3 ions. Write the formulas for the two cobalt oxides formed by these transition metal ions. 16. Platinum is a transition element and forms Pt2 and Pt4 ions. Write the formulas for the compounds of each of these ions with (a) chloride ions and (b) sulfide ions. 17. Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (a) AlCl2 (c) Ga2O3 (b) KF2 (d) MgS 18. Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (a) Ca2O (c) Fe2O5 (b) SrBr2 (d) Li2O Naming Ionic Compounds (See Exercise 3.5 and the General ChemistryNow Screen 3.9.) 19. ■ Name each of the following ionic compounds: (a) K2S (c) (NH4)3PO4 (b) CoSO4 (d) Ca(ClO)2 20. Name each of the following ionic compounds: (a) Ca(CH3CO2)2 (c) Al(OH)3 (b) Ni3(PO4)2 (d) KH2PO4 21. ■ Give the formula for each of the following ionic compounds: (a) ammonium carbonate (b) calcium iodide (c) copper(II) bromide (d) aluminum phosphate (e) silver(I) acetate 22. Give the formula for each of the following ionic compounds: (a) calcium hydrogen carbonate (b) potassium permanganate (c) magnesium perchlorate (d) potassium hydrogen phosphate (e) sodium sulfite 23. Write the formulas for the four ionic compounds that can be made by combining each of the cations Na and Ba2 with the anions CO32 and I . Name each of the compounds. 24. Write the formulas for the four ionic compounds that can be made by combining the cations Mg2 and Fe3 with the anions PO43 and NO3 . Name each compound formed. Coulomb’s Law (See Equation 3.1, Figure 3.10, and the General ChemistryNow Screen 3.7.) 25. Sodium ion, Na , forms ionic compounds with fluoride, F , and iodide, I . The radii of these ions are as follows: Na 116 pm; F 119 pm; and I 206 pm. In which ionic compound, NaF or NaI, are the forces of attraction between cation and anion stronger? Explain your answer. 26. ■ Consider the two ionic compounds NaCl and CaO. In which compound are the cation–anion attractive forces stronger? Explain your answer. Naming Binary, Nonmetal Compounds (See Exercise 3.6 and the General ChemistryNow Screen 3.12.) 27. ■ Name each of the following binary, nonionic compounds: (a) NF3 (b) HI (c) BI3 (d) PF5 28. Name each of the following binary, nonionic compounds: (a) N2O5 (b) P4S3 (c) OF2 (d) XeF4 29. ■ Give the formula for each of the following compounds: (a) sulfur dichloride (b) dinitrogen pentaoxide (c) silicon tetrachloride (d) diboron trioxide (commonly called boric oxide) ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 134 Chapter 3 Molecules, Ions, and Their Compounds 30. Give the formula for each of the following compounds: (a) bromine trifluoride (b) xenon difluoride (c) hydrazine (d) diphosphorus tetrafluoride (e) butane Molecules, Compounds, and the Mole (See Example 3.5 and the General ChemistryNow Screens 3.14 and 3.15.) 31. ■ Calculate the molar mass of each of the following compounds: (a) Fe2O3, iron(III) oxide (b) BCl3, boron trichloride (c) C6H8O6, ascorbic acid (vitamin C) 32. Calculate the molar mass of each of the following compounds: (a) Fe(C6H11O7)2, iron(II) gluconate, a dietary supplement (b) CH3CH2CH2CH2SH, butanethiol, has a skunk-like odor (c) C20H24N2O2, quinine, used as an antimalarial drug 33. ■ Calculate the molar mass of each hydrated compound. Note that the water of hydration is included in the molar mass. (See Section 3.7.) (a) Ni(NO3)2 6 H2O (b) CuSO4 5 H2O 34. Calculate the molar mass of each hydrated compound. Note that the water of hydration is included in the molar mass. (See Section 3.7.) (a) H2C2O4 2 H2O (b) MgSO4 7 H2O, Epsom salts 35. ■ What mass is represented by 0.0255 mol of each of the following compounds? (a) C3H7OH, propanol, rubbing alcohol (b) C11H16O2, an antioxidant in foods, also known as BHA (butylated hydroxyanisole) (c) C9H8O4, aspirin 36. Assume you have 0.123 mol of each of the following compounds. What mass of each is present? (a) C14H10O4, benzoyl peroxide, used in acne medications (c) Pt(NH3)2Cl2, cisplatin, a cancer chemotherapy agent 37. Acetonitrile, CH3CN, was found in the tail of Comet HaleBopp in 1997. What amount (moles) of acetonitrile is represented by 2.50 kg? 38. Acetone, (CH3)2CO, is an important industrial solvent. If 1260 million kg of this organic compound is produced annually, what amount (moles) is produced? 39. Sulfur trioxide, SO3, is made industrially in enormous quantities by combining oxygen and sulfur dioxide, SO2. What amount (moles) of SO3 is represented by 1.00 kg of sulfur trioxide? How many molecules? How many sulfur atoms? How many oxygen atoms? ▲ More challenging ■ In General ChemistryNow 40. An Alka-Seltzer tablet contains 324 mg of aspirin (C9H8O4), 1904 mg of NaHCO3, and 1000. mg of citric acid (H3C6H5O7). (The last two compounds react with each other to provide the “fizz,” bubbles of CO2, when the tablet is put into water.) (a) Calculate the amount (moles) of each substance in the tablet. (b) If you take one tablet, how many molecules of aspirin are you consuming? Percent Composition (See Exercise 3.6 and the General ChemistryNow Screen 3.16.) 41. ■ Calculate the mass percent of each element in the following compounds. (a) PbS, lead(II) sulfide, galena (b) C3H8, propane (c) C10H14O, carvone, found in caraway seed oil 42. Calculate the mass percent of each element in the following compounds: (a) C8H10N2O2, caffeine (b) C10H20O, menthol (c) CoCl2 6 H2O 43. Calculate the weight percent of lead in PbS, lead(II) sulfide. What mass of lead (in grams) is present in 10.0 g of PbS? 44. Calculate the weight percent of iron in Fe2O3, iron(III) oxide. What mass of iron (in grams) is present in 25.0 g of Fe2O3? 45. ■ Calculate the weight percent of copper in CuS, copper(II) sulfide. If you wish to obtain 10.0 g of copper metal from copper(II) sulfide, what mass of the sulfide (in grams) must you use? 46. Calculate the weight percent of titanium in the mineral ilmenite, FeTiO3. What mass of ilmenite (in grams) is required if you wish to obtain 750 g of titanium? Empirical and Molecular Formulas (See Example 3.7 and the General ChemistryNow Screens 3.16–3.18.) 47. ■ Succinic acid occurs in fungi and lichens. Its empirical formula is C2H3O2 and its molar mass is 118.1 g/mol. What is its molecular formula? 48. An organic compound has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound? 49. Complete the following table: Empirical Formula Molar Mass (g/mol) 26.0 116.1 _______ Molecular Formula _______ _______ C8H16 (a) CH (b) CHO (c) ________ Blue-numbered questions answered in Appendix O Study Questions 135 50. Complete the following table: Empirical Formula Molar Mass (g/mol) 150.0 44.1 _______ Molecular Formula _______ _______ B4H10 61. Zinc metal (2.50 g) combines with 9.70 g of iodine to produce zinc iodide, ZnxIy. What is the formula of this ionic compound? 62. You combine 1.25 g of germanium, Ge, with excess chlorine, Cl2. The mass of product, GexCly, is 3.69 g. What is the formula of the product, GexCly? (a) C2H3O3 (b) C3H8 (c) _______ 51. Acetylene is a colorless gas used as a fuel in welding torches, among other things. It is 92.26% C and 7.74% H. Its molar mass is 26.02 g/mol. What are the empirical and molecular formulas of acetylene? 52. ■ A large family of boron-hydrogen compounds has the general formula Bx H y. One member of this family contains 88.5% B; the remainder is hydrogen. Which of the following is its empirical formula: BH2, BH3, B2H5, B5H7, or B5H11? 53. ■ Cumene is a hydrocarbon, a compound composed only of C and H. It is 89.94% carbon, and its molar mass is 120.2 g/mol. What are the empirical and molecular formulas of cumene? 54. Nitrogen and oxygen form a series of oxides with the general formula NxOy. One of them, a blue solid, contains 36.84% N. What is the empirical formula of this oxide? 55. ■ Mandelic acid is an organic acid composed of carbon (63.15%), hydrogen (5.30%), and oxygen (31.55%). Its molar mass is 152.14 g/mol. Determine the empirical and molecular formulas of the acid. 56. Nicotine, a poisonous compound found in tobacco leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formulas of nicotine? Determining Formulas from Mass Data (See Examples 3.8–3.10 and the General ChemistryNow Screens 3.17–3.19.) 57. ■ If Epsom salt, MgSO4 x H2O, is heated to 250 ° C, all the water of hydration is lost. On heating a 1.687-g sample of the hydrate, 0.824 g of MgSO4 remains. How many molecules of water occur per formula unit of MgSO4? 58. The “alum” used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2 x H2O. To find the value of x, you can heat a sample of the compound to drive off all of the water and leave only KAl(SO4)2. Assume you heat 4.74 g of the hydrated compound and that the sample loses 2.16 g of water. What is the value of x? 59. ■ A new compound containing xenon and fluorine was isolated by shining sunlight on a mixture of Xe (0.526 g) and F2 gas. If you isolate 0.678 g of the new compound, what is its empirical formula? 60. Elemental sulfur (1.256 g) is combined with fluorine, F2, to give a compound with the formula SFx, a very stable, colorless gas. If you have isolated 5.722 g of SFx, what is the value of x? General Questions These questions are not designated as to type or locations in the chapter. They may combine several concepts. More challenging questions are marked with the icon ▲. 63. Write formulas for all of the compounds that can be made by combining the cations NH4 and Ni2 with the anions CO32 and SO42 . 64. Using the General ChemistryNow CD-ROM or website, find a model for each of the following molecules. Write the molecular formula and draw the structural formula. (a) acetic acid (b) methylamine (c) formaldehyde 65. How many electrons are in a strontium atom (Sr)? Does an atom of Sr gain or lose electrons when forming an ion? How many electrons are gained or lost by the atom? When Sr forms an ion, the ion has the same number of electrons as which one of the noble gases? 66. The compound (NH4)2SO4 consists of two polyatomic ions. What are the names and electric charges of these ions? What is the molar mass of this compound? 67. Which of the following compounds has the highest weight percent of chlorine? (a) BCl3 (d) AlCl3 (b) AsCl3 (e) PCl3 (c) GaCl3 68. Which of the following samples has the largest number of ions? (a) 1.0 g of BeCl2 (d) 1.0 g of SrCO3 (b) 1.0 g of MgCl2 (e) 1.0 g of BaSO4 (c) 1.0 g of CaS 69. Which of the following compounds (NO, CO, MgO, or CaO) has the highest weight percent of oxygen? 70. The chemical compound alum has the formula KAl(SO4)2 12 H2O. Give formulas for the ions that make up this ionic compound. 71. Knowing that the formula of sodium borate is Na3BO3, give the formula and charge of the borate ion. Is the borate ion a cation or an anion? 72. What is the difference between an empirical formula and a molecular formula? Use the compound ethane, C2H6, to illustrate your answer. ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 136 Chapter 3 Molecules, Ions, and Their Compounds 73. The structure of one of the bases in DNA, adenine, is shown here. Which represents the greater mass: 40.0 g of adenine or 3.0 1023 molecules of the compound? iron(II) sulfate, FeSO4, and the other with iron(II) gluconate, Fe(C6H11O7)2. If you take 100. mg of each compound, which will deliver more atoms of iron? 81. ▲ Spinach is high in iron (2 mg per 90-g serving). It is also a source of the oxalate ion, C2O42 ; however, oxalate ion combines with iron ions to form iron oxalate, Fex(C2O4)y, a substance that prevents your body from absorbing the iron. Analysis of a 0.109-g sample of iron oxalate shows that it contains 38.82% iron. What is the empirical formula of the compound? 82. A compound composed of iron and carbon monoxide, Fex(CO)y, is 30.70% iron. What is the empirical formula for the compound? 83. Ma huang, an extract from the ephedra species of plants, contains ephedrine. The Chinese have used this herb more than 5000 years to treat asthma. More recently the substance has been used in diet pills that can be purchased over the counter in herbal medicine shops. However, very serious concerns have been raised regarding these pills following reports of serious heart problems with their use. (a) Write the molecular formula for ephedrine, draw its structural formula, and calculate its molar mass. (b) What is the weight percent of carbon in ephedrine? (c) Calculate the amount (moles) of ephedrine in a 0.125-g sample. (d) How many molecules of ephedrine are there in 0.125 g? How many C atoms? 74. Which has the larger mass, 0.5 mol of BaCl2 or 0.5 mol of SiCl4? 75. ■ A drop of water has a volume of about 0.05 mL. How many molecules of water are in a drop of water? (Assume water has a density of 1.00 g/cm3.) 76. Capsaicin, the compound that gives the hot taste to chili peppers, has the formula C18H27NO3. (a) Calculate its molar mass. (b) If you eat 55 mg of capsaicin, what amount (moles) have you consumed? (c) Calculate the mass percent of each element in the compound. (d) What mass of carbon (in milligrams) is there in 55 mg of capsaicin? 77. Calculate the molar mass and the mass percent of each element in the blue solid Cu(NH3)4SO4 H2O. What are the mass (in grams) of copper and the mass of water in 10.5 g of the compound? 78. Write the molecular formula and calculate the molar mass for each of the molecules shown here. Which has the larger percentage of carbon? Of oxygen? (a) Ethylene glycol (used in antifreeze) H H H O C C O H 84. Saccharin is more than 300 times sweeter than sugar. It was first made in 1897, a time when it was common practice for chemists to record the taste of any new substances they synthesized. (a) Write the molecular formula for the compound and draw its structural formula. (S atoms are yellow.) (b) If you ingest 125 mg of saccharin, what amount (moles) of saccharin have you ingested? (c) What mass of sulfur is contained in 125 mg of saccharin? H H (b) Dihydroxyacetone (used in artificial tanning lotions) H H O C H O H C C H O H 79. Malic acid, an organic acid found in apples, contains C, H, and O in the following ratios: C1H1.50O1.25. What is the empirical formula of malic acid? 80. Your doctor has diagnosed you as being anemic—that is, as having too little iron in your blood. At the drugstore you find two iron-containing dietary supplements: one with ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 137 85. Which of the following pairs of elements are likely to form ionic compounds when allowed to react with each other? Write appropriate formulas for the ionic compounds you expect to form, and give the name of each. (a) chlorine and bromine (b) phosphorus and bromine (c) lithium and sulfur (d) indium and oxygen (e) sodium and argon (f ) sulfur and bromine (g) calcium and fluorine 86. Name each of the following compounds, and tell which ones are best described as ionic: (a) ClF3 (f ) OF2 (b) NCl3 (g) KI (c) SrSO4 (h) Al2S3 (d) Ca(NO3)2 (i) PCl3 (e) XeF4 ( j) K3PO4 87. Write the formula for each of the following compounds, and tell which ones are best described as ionic: (a) sodium hypochlorite (b) boron triiodide (c) aluminum perchlorate (d) calcium acetate (e) potassium permanganate (f ) ammonium sulfite (g) potassium dihydrogen phosphate (h) disulfur dichloride (i) chlorine trifluoride ( j) phosphorus trifluoride 88. Complete the table by placing symbols, formulas, and names in the blanks. Cation ______ Ba2 ______ ______ Al 3 90. Fluorocarbonyl hypofluorite is composed of 14.6% C, 39.0% O, and 46.3% F. If the molar mass of the compound is 82 g/mol, determine the empirical and molecular formulas of the compound. 91. Azulene, a beautiful blue hydrocarbon, is 93.71% C and has a molar mass of 128.16 g/mol. What are the empirical and molecular formulas of azulene? 92. Cacodyl, a compound containing arsenic, was reported in 1842 by the German chemist Robert Wilhelm Bunsen. It has an almost intolerable garlic-like odor. Its molar mass is 210 g/mol, and it is 22.88% C, 5.76% H, and 71.36% As. Determine its empirical and molecular formulas. 93. The action of bacteria on meat and fish produces a compound called cadaverine. As its name and origin imply, it stinks! (It is also present in bad breath and adds to the odor of urine.) It is 58.77% C, 13.81% H, and 27.40% N. Its molar mass is 102.2 g/mol. Determine the molecular formula of cadaverine. 94. ■ ▲ Transition metals can combine with carbon monoxide (CO) to form compounds such as Fe(CO)5 (Study Question 3.82). Assume that you combine 0.125 g of nickel with CO and isolate 0.364 g of Ni(CO)x . What is the value of x ? 95. ▲ A major oil company has used a gasoline additive called MMT to boost the octane rating of its gasoline. What is the empirical formula of MMT if it is 49.5% C, 3.2% H, 22.0% O, and 25.2% Mn? 96. ▲ Elemental phosphorus is made by heating calcium phosphate with carbon and sand in an electric furnace. What is the weight percent of phosphorus in calcium phosphate? Use this value to calculate the mass of calcium phosphate (in kilograms) that must be used to produce 15.0 kg of phosphorus. 97. ▲ Chromium is obtained by heating chromium(III) oxide with carbon. Calculate the weight percent of chromium in the oxide and then use this value to calculate the quantity of Cr2O3 required to produce 850 kg of chromium metal. 98. ▲ Stibnite, Sb2S3, is a dark gray mineral from which antimony metal is obtained. What is the weight percent of antimony in the sulfide? If you have 1.00 kg of an ore that contains 10.6% antimony, what mass of Sb2S3 (in grams) is in the ore? 99. ▲ Direct reaction of iodine (I2) and chlorine (Cl2) produces an iodine chloride, IxCly , a bright yellow solid. If you completely used up 0.678 g of iodine and produced 1.246 g of IxCly , what is the empirical formula of the compound? A later experiment showed that the molar mass of IxCly was 467 g/mol. What is the molecular formula of the compound? 100. ▲ In a reaction 2.04 g of vanadium combined with 1.93 g of sulfur to give a pure compound. What is the empirical formula of the product? 101. ▲ Iron pyrite, often called “fool’s gold,” has the formula FeS2. If you could convert 15.8 kg of iron pyrite to iron metal, what mass of the metal would you obtain? ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Anion ______ ______ Cl F CO3 2 Name ammonium bromide __________________ iron(II) chloride __________________ __________________ iron(III) oxide Formula ______ BaS ______ PbF2 ______ ______ ______ ______ 89. Complete the table by placing symbols, formulas, and names in the blanks. Cation ______ ______ ______ ______ Al3 ______ Anion ______ ______ Br ______ ______ ______ Name __________________ aluminum phosphate lithium bromide _________________ aluminum oxide iron(III) carbonate Formula LiClO4 ______ ______ Ba(NO3)2 ______ ______ ▲ More challenging 138 Chapter 3 Molecules, Ions, and Their Compounds 102. Which of the following statements about 57.1 g of octane, C8H18, is (are) not true? (a) 57.1 g is 0.500 mol of octane. (b) The compound is 84.1% C by weight. (c) The empirical formula of the compound is C4H9. (d) 57.1 g of octane contains 28.0 g of hydrogen atoms. 103. The formula of barium molybdate is BaMoO4. Which of the following is the formula of sodium molybdate? (a) Na4MoO (c) Na2MoO3 (e) Na4MoO4 (b) NaMoO (d) Na2MoO4 104. ▲ A metal M forms a compound with the formula MCl4. If the compound is 74.75% chlorine, what is the identity of M? 105. Pepto-Bismol, which helps provide soothing relief for an upset stomach, contains 300. mg of bismuth subsalicylate, C21H15Bi3O12, per tablet. If you take two tablets for your stomach distress, what amount (in moles) of the “active ingredient” are you taking? What mass of Bi are you consuming in two tablets? 106. ▲ The weight percent of oxygen in an oxide that has the formula MO2 is 15.2%. What is the molar mass of this compound? What element or elements are possible for M? 107. The mass of 2.50 mol of a compound with the formula ECl4, in which E is a nonmetallic element, is 385 g. What is the molar mass of ECl4? What is the identity of E? 108. ▲ The elements A and Z combine to produce two different compounds: A2Z3 and AZ2. If 0.15 mol of A2Z3 has a mass of 15.9 g and 0.15 mole of AZ2 has a mass of 9.3 g, what are the atomic masses of A and Z? 109. ▲ Polystyrene can be prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. A sample prepared by this method has the empirical formula Br3C6H3(C8H8)n, where the value of n can vary from sample to sample. If one sample has 10.46% Br, what is the value of n? 110. A sample of hemoglobin is found to be 0.335% iron. If hemoglobin contains one iron atom per molecule, what is the molar mass of hemoglobin? What is the molar mass if there are four iron atoms per molecule? 112. An ionic compound can dissolve in water because the cations and anions are attracted to water molecules. The drawing here shows how a cation and a water molecule, which has a negatively charged O atom and positively charged H atoms, can interact. Which of the following cations should be most strongly attracted to water: Na , Mg2 , or Al3 ? Explain briefly. H2O Mg2 Water molecules interacting with a magnesium ion. 113. ▲ When analyzed, an unknown compound gave these experimental results: C, 54.0%; H, 6.00%; and O, 40.0%. Four different students used these values to calculate the empirical formulas shown here. Which answer is correct? Why did some students not get the correct answer? (a) C4H5O2 (c) C7H10O4 (b) C5H7O3 (d) C9H12O5 114. ▲ Two general chemistry students working together in the lab weigh out 0.832 g of CaCl2 2 H2O into a crucible. After heating the sample for a short time and allowing the crucible to cool, the students determine that the sample has a mass of 0.739 g. They then do a quick calculation. On the basis of this calculation, what should they do next? (a) Congratulate themselves on a job well done. (b) Assume the bottle of CaCl2 2 H2O was mislabeled; it actually contained something different. (c) Heat the crucible again, and then reweigh it. 115. ▲ Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. This question considers some uranium chemistry. (a) A small sample of uranium metal (0.169 g) is heated to between 800 and 900 ° C in air to give 0.199 g of a dark green oxide, UxOy. How many moles of uranium metal were used? What is the empirical formula of the oxide, UxOy? What is the name of the oxide? How many moles of UxOy must have been obtained? (b) The naturally occurring isotopes of uranium are 234U, 235 U, and 238U. Knowing that uranium’s atomic weight is 238.02 g/mol, which isotope must be the most abundant? (c) If the hydrated compound UO2(NO3)2 z H2O is heated gently, the water of hydration is lost. If you have 0.865 g of the hydrated compound and obtain 0.679 g of UO2(NO3)2 on heating, how many molecules of water of hydration are in each formula unit of the original compound? (The oxide UxOy is obtained if the hydrate is heated to temperatures over 800 ° C in the air.) Summary and Conceptual Questions The following questions use concepts from the preceding chapters. 111. A piece of nickel foil, 0.550 mm thick and 1.25 cm square, is allowed to react with fluorine, F2, to give a nickel fluoride. (a) How many moles of nickel foil were used? (The density of nickel is 8.902 g/cm3.) (b) If you isolate 1.261 g of the nickel fluoride, what is its formula? (c) What is its complete name? ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 139 117. The common chemical compound alum has the formula KAl(SO4)2 12 H2O. An interesting characteristic of alum is that it is possible to grow very large crystals of this compound. Suppose you have a crystal of alum in the form of a cube that is 3.00 cm on each side. You want to know how many aluminum atoms are contained in this cube. Outline the steps to determine this value, and indicate the information that you need to carry out each step. 118. Cobalt(II) chloride hexahydrate dissolves readily in water to give a red solution. If we use this solution as an “ink,” we can write secret messages on paper. The writing is not Charles D. Winters 116. The “simulation” section on General ChemistryNow Screen 3.7 Coulomb’s Law, helps you explore Coulomb’s law. You can change the charges on the ions and the distance between them. If the ions experience an attractive force, arrows point from one ion to the other. Repulsion is indicated by arrows pointing in opposite directions. Change the ion charges (from 1 to 2 to 3 ). How does this affect the attractive force? How close can the ions approach before significant repulsive forces set in? How does this distance vary with ion charge? visible when the water evaporates from the paper. When the paper is heated, however, the message can be read. Explain the chemistry behind this observation. (See General ChemistryNow Screen 3.20 Chemical Puzzler.) A solution of CoCl2 6 H2O. Using the secret ink to write on paper. Heating the paper reveals the writing. Do you need a live tutor for homework problems? Access vMentor at General ChemistryNow at http://now.brookscole.com/kotz6e for one-on-one tutoring from a chemistry expert ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O ...
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This note was uploaded on 02/18/2009 for the course CHEM 101 taught by Professor Williamson during the Fall '08 term at Texas A&M.

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