Ch. 2 Components of Matter, atomic theory, periodic table, chemical bonding, formulas, names PDF

Chapter 2: The Components of Matter 2.1 Elements, Compounds, and Mixtures: An Atomic Overview 2.2 The Law of Mass Conservation 2.5 The Atomic Theory Today 2.6 Elements: A First Look at the Periodic Table 2.7 Compounds: Introduction to Bonding 2.8 Formulas, Names, and Masses of Compounds ©McGraw-Hill...

Chapter 2: The Components of Matter 2.1 Elements, Compounds, and Mixtures: An Atomic Overview 2.2 The Law of Mass Conservation 2.5 The Atomic Theory Today 2.6 Elements: A First Look at the Periodic Table 2.7 Compounds: Introduction to Bonding 2.8 Formulas, Names, and Masses of Compounds ©McGraw-Hill Education. Definitions for Components of Matter Element - the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means. Molecule - a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit. ©McGraw-Hill Education. Definitions for Components of Matter (2) Compound – a pure substance composed of molecules (two or more elements that are chemically combined). Mixture –two or more substances (elements and/or compounds) that are physically mixed but not chemically bonded. Types of mixtures: – Homogeneous (solutions): uniform mixing at molecular scale. Examples: liquid solutions, solid solutions (alloys, glasses, minerals), gas mixtures – Heterogeneous: nonuniform mixing at molecular scale. Nonuniformity and boundaries may be detected. Examples: colloids (sol, gel, aerosol, foam), rocks, most materials (plastic, wood, paper, etc.) ©McGraw-Hill Education. Table 2.1 Some Properties of Sodium, Chlorine, and Sodium Chloride Property Sodium Melting point Chlorine Sodium Chloride 97.8°C -101°C 801°C Boiling point 881.4°C -34°C 1413°C Color Silvery Yellow-green Colorless (white) Density 0.97 g/cm3 0.0032 g/cm3 2.16 g/cm3 Behavior in water Reacts Dissolves slightly Dissolves freely ©McGraw-Hill Education. + Source:(Sodium, Chlorine, Sodium chloride) © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 2.1: Problem and Plan PROBLEM: The following scenes represent an atomic-scale view of three samples of matter. Describe each sample as an element, compound, or mixture. PLAN: A sample that contains only one type of particle is either an element or a compound. The particles of an element consist of only one type of atom, whereas the particles of a compound have two or more types of atom bonded together. ©McGraw-Hill Education. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Sample Problem 2.1: Solution SOLUTION: Sample (a) contains three different types of particles and is therefore a mixture. Sample (b) contains only one type of particle and each particle has only one atom. This is an element. Sample (c) contains only one type of particle, each of which contains two different types of atoms. This is a compound. ©McGraw-Hill Education. Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Figure 2.2: The law of mass conservation. The total mass of substances does not change during a chemical reaction. ©McGraw-Hill Education. Source: © McGraw-Hill Education/Stephen Frisch, photographer Law of Mass Conservation The total mass of substances present does not change during a chemical reaction. reactant 1 = total mass calcium oxide + carbon dioxide CaO + CO2 56.08 g ©McGraw-Hill Education. product + reactant 2 + 44.00 g total mass calcium carbonate CaCO3 100.08 g Figure 2.3 Law of Definite (or Constant) Composition ©McGraw-Hill Education. Source: (top): © Punchstock RF; (bottom): © Alexander Cherednichenko/Shutterstock.com Calcium carbonate ©McGraw-Hill Education. Analysis by Mass (grams/20.0 g) Mass Fraction (parts/1.00 part) Percent by Mass (parts/100 parts) 8.0 g calcium 0.40 calcium 40% calcium 2.4 g carbon 0.12 carbon 12% carbon 9.6 g oxygen 0.48 oxygen 48% oxygen 20.0 g 1.00 part by mass 100% by mass Law of Multiple Proportions If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. Example: Carbon Oxides A & B Carbon Oxide I : 57.1% oxygen and 42.9% carbon Carbon Oxide II : 72.7% oxygen and 27.3% carbon ©McGraw-Hill Education. Multiple Proportions Demonstration Assume that you have 100 g of each compound. Carbon Oxide I Carbon Oxide II g oxygen/100 g compound 57.1 72.7 g carbon/100 g compound 42.9 27.3 g oxygen/g carbon 57.1 = 1.33 42.9 72.7 = 2.66 27.3 2.66 g O / g C in II 2 = 1.33 g O / g C in I 1 ©McGraw-Hill Education. 2.5 Structure of the Atom The atom is an electrically neutral, spherical entity composed of a positively charged central nucleus surrounded by one or more negatively charged electrons. The atomic nucleus consists of protons and neutrons. Figure 2.7 ©McGraw-Hill Education. Properties of the Three Key Subatomic Particles Table 2.2 Charge Name Relative Absolute (C)* (Symbol) Mass Relative (amu)† Absolute (g) Location in Atom Proton (p+) 1+ +1.60218x10-19 1.00727 1.67262x10-24 Nucleus Neutron (n0) 0 0 1.67493x10-24 Nucleus Electron (e-) 1- -1.60218x10-19 0.00054858 9.10939x10-28 Outside nucleus * The † 1.00866 coulomb (C) is the SI unit of charge: 1 C = 1 A x 1 s The atomic mass unit (amu) equals 1.66054x10-24 g. ©McGraw-Hill Education. Atomic Symbol, Number and Mass X = Atomic symbol of the element A = mass number; A = Z + N Z = atomic number (the number of protons in the nucleus; an atom is electrically neutral because it has the same number of electrons, Z) N = number of neutrons in the nucleus Figure 2.8 ©McGraw-Hill Education. Isotopes Isotopes are atoms of an element with the same number of protons, but a different number of neutrons. Isotopes have the same atomic number Z, but a different mass number A. Therefore, isotopes have different atomic masses. Since chemical properties of an element are given by the number of electrons, isotopes of an element have nearly identical chemical behavior. Figure 2.8 ©McGraw-Hill Education. Determining the Number of Subatomic Particles in the Isotopes of an Element Sample Problem 2.4: Problem and Plan PROBLEM: Silicon (Si) has three naturally occurring isotopes: 28Si, 29Si, and 30Si. Determine the number of protons, neutrons, and electrons in each silicon isotope. PLAN: The mass number (A; left superscript) of each of the three isotopes is given, which is the sum of protons and neutrons. From the List of Elements, we find the atomic number (Z, number of protons), which equals the number of electrons. We obtain the number of neutrons by subtracting Z from A. ©McGraw-Hill Education. Sample Problem 2.4: Solution SOLUTION: The atomic number of silicon is Z = 14; since N = A – Z, therefore 28Si has 14p+, 14e–, and 14n0 (28 – 14 = 14) 29Si has 14p+, 14e–, and 15n0 (29 – 14 = 15) 30Si has 14p+, 14e–, and 16n0 (30 – 14) = 16 ©McGraw-Hill Education. Electron Impact Ionization in a Mass Spectrometer + Formation of a positively charged neon particle (Ne ) A mass spectrometer can only detect charged particles (ions). Figure 2.9A ©McGraw-Hill Education. A Magnet Analyzer Mass Spectrometer and the output signal (mass spectrum). ©McGraw-Hill Education. Source: © James King Holmes/Oxford Centre for Molecular Sciences/Science Source. Atomic Masses of the Isotopes The mass of an atom is measured and reported relative to the mass of an atomic standard. The atomic standard today is the 126𝐶 isotope. The name of the standard is the atomic mass unit (amu). It is equal to 1/12 of the mass of 12C isotope: 1 1 𝑎𝑚𝑢 = 𝑚 126𝐶 = 1.66054 × 10−24 𝑔 12 Another name for the amu is the dalton (Da) © McGraw Hill Atomic Masses of the Elements Source: https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl © McGraw Hill Calculating Atomic Mass of an Element from Isotopic Composition From isotopic mass and relative abundance data, we can calculate the atomic mass of an element, the average of the masses of its naturally occurring isotopes, weighted according to their abundances: Atomic mass =  ( isotopic mass )( fractional abundance of isotope ) Example: what is the atomic mass of Neon? 𝐴𝑡𝑜𝑚𝑖𝑐 𝑀𝑎𝑠𝑠 𝑁𝑒 = 𝑚 20𝑁𝑒 × 𝐹𝐴 20𝑁𝑒 + 𝑚 21𝑁𝑒 × 𝐹𝐴 21𝑁𝑒 + 𝑚 22𝑁𝑒 × 𝐹𝐴 22𝑁𝑒 = 19.992 × 0.9048 + 20.993 × 0.0027 + 21.991 × 0.0925 = 𝟐𝟎. 𝟏𝟖 © McGraw Hill Calculating the Atomic Mass of an Element Sample Problem 2.5: Problem and Plan PROBLEM: Silver (Ag, Z = 47) has two naturally occurring isotopes, 107Ag and 109Ag. From the mass spectrometric data provided, calculate the atomic mass of Ag. Isotope 107Ag Mass (amu) 106.90509 108.90476 109Ag PLAN: Find the weighted average of the isotopic masses. ©McGraw-Hill Education. Abundance (%) 51.84 48.16 Sample Problem 2.5: Solution mass portion from 107Ag = 106.90509 amu x 0.5184 = 55.42 amu mass portion from 109Ag = 108.90476 amu x 0.4816 = 52.45 amu atomic mass of Ag = 55.42 amu + 52.45 amu = 107.87amu ©McGraw-Hill Education. The Modern Periodic Table Figure 2.10 ©McGraw-Hill Education. Some Metals, Metalloids, and Nonmetals Figure 2.11 ©McGraw-Hill Education. Source: © McGraw-Hill Education/Stephen Frisch, photographer Identifying an Element from its Z value Sample Problem 2.6: Problem and Plan PROBLEM: For each of the following Z values, give the name, symbol, and group and period numbers of the element and classify it as a main group metal, transition metal, inner-transition metal, nonmetal, or metalloid: (a) Z = 38; (b) Z = 17; (c) Z = 27. PLAN: The Z value is the atomic number of an element. The list of elements inside the front cover is alphabetical, so we look up the name and symbol of the element. Then we use the periodic table to find the group number (top of the column) and the period number (left end of the row) in which the element is located. We classify the element from the color coding in the periodic table. ©McGraw-Hill Education. Sample Problem 2.6: Solution SOLUTION: (a) Strontium, Sr, is in Group 2A(2) and Period 5, and it is a main group metal. (b) Chlorine, Cl, is in Group 7A(17) and Period 3, and it is a nonmetal. (c) Cobalt, Co, is in Group 8B(9) and Period 4, and it is a transition metal. ©McGraw-Hill Education. 2.7 Compounds: Introduction to Chemical Bonding Elements are rarely found in nature Elements combine among themselves in chemical reactions to form compounds. Atoms of elements combine their electrons to form chemical bonds. Most common chemical bonds are: 1) Covalent Bond: sharing of electrons over all the atoms in a molecule. This is the prevailing bonding in Chemistry. 2) Ionic Bond: transferring of 1 – 3 electrons from one atom to another. ©McGraw-Hill Education. Molecules and Ions Molecule – the basic unit of a molecular element or covalent compound, consisting of two or more atoms bonded by the sharing of electrons. Most covalent substances consist of molecules. Ion – a single atom or covalently bonded group of atoms that has an overall electrical charge. There are no molecules in an ionic compound but a pair of ions of opposite charges. ©McGraw-Hill Education. The formation of an ionic compound Transferring electrons from the atoms of one element to those of another results in an ionic compound. Ion = Traveler (Greek). It travels in an electric field. Cation = positively charged ion (+) Anion = negative (-) ion Figure 2.11 ©McGraw-Hill Education. Source: A(1−2), E: © McGraw-Hill Education/Stephen Frisch, photographer Factors that influence the strength of ionic bonding Coulomb’s Law for electric charges: 𝑐ℎ𝑎𝑟𝑔𝑒1 × 𝑐ℎ𝑎𝑟𝑔𝑒2 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 ∝ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 ©McGraw-Hill Education. The relationship between ions formed and the nearest noble gas Metals in groups 1A, 2A, and Al (3A) lose electrons to form positive ions. The electric charge of the ion equals the # of boxes to the previous noble gas. Nonmetals in groups 7A, 6A, and N (5A) gain electrons to form negative ions. The electric charge of the ion equals the # of boxes to the next noble gas. ©McGraw-Hill Education. Sample Problem 2.7: Problem and Plan PROBLEM: What monatomic ions would you expect the following elements to form? (a) Iodine (Z = 53) (b) Calcium (Z = 20) (c) Aluminum (Z = 13) PLAN: We use the given Z value to find the element in the periodic table and see where its group lies relative to the noble gases. Elements in Groups 1A, 2A, and 3A lose electrons to attain the same number as the nearest noble gas and become positive ions; those in Groups 5A, 6A, and 7A gain electrons and become negative ions. ©McGraw-Hill Education. Sample Problem 2.7: Solution SOLUTION: a) Iodine is in Group 7A, the halogens. Like any member of this group, it gains 1 electron to attain the same number as the nearest Group 8A member, in this case, Xe. b) Calcium is in Group 2A, the alkaline earth metals. Like any Group 2A member, it loses 2 electrons to attain the same number as the nearest noble gas, Ar. c) Aluminum is a metal in the boron family [Group 3A] and thus loses 3 electrons to attain the same number as its nearest noble gas, Ne. ©McGraw-Hill Education. Formation of a covalent bond between two H atoms Covalent bonds form when elements share electrons, which usually occurs between nonmetals. Figure 2.15 ©McGraw-Hill Education. Elements that occur as molecules Figure 2.16 ©McGraw-Hill Education. Polyatomic Ions: Covalent Bonds within Ions The case of carbonate ion in calcium carbonate A polyatomic ion consists of two of more atoms covalently bonded together and has an overall charge. In many reactions the polyatomic ion will remain together as a unit. The electric charge is dispersed over the entire polyatomic ion. ©McGraw-Hill Education. Source: (calcium carbonate crystals) © papa1266/Shutterstock.com 2.8 Chemical Formulas A chemical formula consists of element symbols with numerical subscripts. Example: Na2SO4 The chemical formula indicates the type and number of each atom present in the smallest unit of a substance. Smallest unit of substance: either a molecule or an ionic pair. The subscript refers to the chemical symbol preceding it. Polyatomic anion must be enclosed in curved brackets if a subscript follows it. Example: Ca(NO3)2 Subscript “1” is never shown. It is implied when a subscript is not shown. Example: NaCl ©McGraw-Hill Education. Naming Binary Ionic Compounds A binary ionic compound is made of one type of cation (+ ion) and one type of anion (- ion). In a binary ionic compound, both the cation and the anion are monatomic. For all ionic compounds, the name and formula lists the cation first and the anion second. The name of the cation is the same as the name of the metal. Many metal names end in -ium. The anion is named by adding the suffix -ide to the root of the nonmetal name. Zinc and bromine form zinc bromide. ©McGraw-Hill Education. Symbols for Common Monatomic Ions* Cations Formula Name Charge +1 H+ Li+ Na+ K+ Cs+ Ag+ hydrogen lithium sodium potassium cesium silver –1 H– F– Cl– Br– I– hydride fluoride chloride bromide iodide +2 Mg2+ Ca2+ Sr2+ Ba2+ Zn2+ Cd2+ magnesium calcium strontium barium zinc cadmium –2 O2– S2– oxide sulfide +3 Al3+ aluminum –3 N3– nitride Charge *Listed ©McGraw-Hill Education. Anions Formula Name by charge; those in boldface are most common. Naming Binary Ionic Compounds Sample Problem 2.8: Problem and Plan PROBLEM: Name the ionic compound formed from each of the following pairs of elements: a) magnesium and nitrogen b) iodine and cadmium c) strontium and fluorine d) sulfur and cesium PLAN: Use the periodic table to decide which element is the metal and which the nonmetal. The metal (cation) is named first and the suffix –ide is added to the root of the non-metal name. ©McGraw-Hill Education. Sample Problem 2.8: Solution SOLUTION: a) magnesium nitride b) cadmium iodide c) strontium fluoride d) cesium sulfide ©McGraw-Hill Education. Some common monatomic ions of the elements Figure 2.19 Most main-group elements form one type of monatomic ion. Most transition elements form two types of monatomic ions. ©McGraw-Hill Education. Formula Unit of Binary Ionic Compounds A formula unit of a binary ionic compound represents the relative number of cations and anions. In any ionic compound, the total electric charge of all cations and anions must be 0 (electrically neutral). The subscript refers to the element before it. Subscripts 1 and 0 are never used. A missing subscript is interpreted as 1. ©McGraw-Hill Education. Determining Formulas of Binary Ionic Compounds Sample Problem 2.9: Problem and Plan PROBLEM: Write formula units for each of the compounds named in Sample Problem 2.8: a) magnesium nitride b) cadmium iodide c) strontium fluoride d) cesium sulfide PLAN: We find the smallest number of each ion that will produce a neutral compound. These numbers appear as right subscripts to the relevant element symbol. ©McGraw-Hill Education. Sample Problem 2.9: Solution SOLUTION: a) Mg2+ and N3–; three Mg2+(6+) and two N3– (6-); Mg3N2 b) Cd2+ and I–; one Cd2+(2+) and two I– (2-); CdI2 strontium fluoride c) Sr2+ and F–; one Sr2+(2+) and two F– (2-); SrF2 d) Cs+ and S2–; two Cs+(2+) and one S2– (2-); Cs2S ©McGraw-Hill Education. Compounds of Metals That Form More Than One Ion Many metals can form more than one type of ion. Ionic compounds of these elements are named like those of elements that form only one monoatomic ion with the addition of a roman numeral within parentheses immediately following the metal ion’s name to indicate its charge. © McGraw Hill Some Metals That Form More Than One Monatomic Ion* Element Ion Formula Systematic Name Common Name Chromium Cr2+ Cr3+ Co2+ Co3+ Cu+ Cu2+ Fe2+ Fe3+ Pb2+ Pb4+ Hg22+ Hg2+ Sn2+ Sn4+ chromium(II) chromium(III) cobalt(II) cobalt(III) copper(I) copper(II) iron(II) iron(III) lead(II) lead(IV) mercury (I) mercury (II) tin(II) tin(IV) chromous chromic Cobalt Copper Iron Lead Mercury Tin *Listed ©McGraw-Hill Education. cuprous cupric ferrous ferric mercurous mercuric stannous stannic alphabetically by metal name; the ions in boldface are most common. Sample Problem 2.10: Problem and Plan Determining Formulas of Binary Ionic Compounds PROBLEM: Give the systematic name for each formula or the formula for each name for the following compounds: a) tin(II) fluoride b) CrI3 c) ferric oxide d) CoS PLAN: Find the smallest number of each ion that will produce a neutral formula. ©McGraw-Hill Education. Sample Problem 2.10: Solution SOLUTION: a) Tin(II) is Sn2+; fluoride is F–; so the formula is SnF2. b) The anion I– is iodide; 3I– means that Cr (chromium) is +3. CrI3 is chromium(III) iodide. c) Ferric is a common name for Fe3+; oxide is O2–; therefore the formula is Fe2O3. d) Co is cobalt; the anion S2– is sulfide, which requires the cation to be Co2+, and the compound is cobalt(II) sulfide. ©McGraw-Hill Education. Naming Compounds Containing Polyatomic Ions Ionic compounds containing polyatomic ions are named like binary ionic compounds except that the name of the polyatomic ion is used without adding the suffix –ide, unless the name of the polyatomic ion already contains that suffix (hydroxide). Suffixes for polyatomic anions are: – -ite for fewer O atoms – -ate for more O atoms – Example: NO2- (nitrite), NO3- (nitrate) When suffixes are not sufficient, prefixes hypo- or per- are added. © McGraw Hill Naming oxoanions # oxygens highest higher lower lowest ©McGraw-Hill Education. Prefix Root Per- Hypo- root root root root Suffix -ate -ate -ite -ite Examples ClO4ClO3ClO2ClO- perchlorate chlorate chlorite hypochlorite Some Common Polyatomic Ions* Formula Name Formula Name Cations NH4+ H 3 O+ ammonium hydronium Common Anions CH3COOacetate CNcyanide OHhydroxide ClOhypochlorite ClO2chlorite ClO3chlorate NO2nitrite NO3nitrate MnO4permanganate * Bold face ions are most common. ©McGraw-Hill Education. CO32HCO3CrO42Cr2O72O22PO43H2PO42SO32SO42- carbonate hydrogen carbonate chromate dichromate peroxide phosphate dihydrogen phosphate sulfite sulfate Hydrates (hydrated ionic compounds) Have an integer number of water molecules in a formula unit: Number Prefix 1 mono- – Epsom salt: MgSO4·7 H2O is called magnesium sulfate heptahydrate 2 di- 3 tri- – Washing soda: Na2CO3·10 H2O is called sodium carbonate decahydrate 4 tetra- 5 penta- 6 hexa- 7 hepta- 8 octa- 9 nona- 10 deca- Water is typically weakly bound and can be removed by heating: © McGraw Hill Sample Problem 2.11: Problem and Plan Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions PROBLEM: Give the systematic name for each formula or the formula for each name for the following compounds : a) Fe(ClO4)2 b) sodium sulfite c) Ba(OH)2·8H2O PLAN: Remember to use parentheses when more than one unit of a particular polyatomic ion is present in the compound. ©McGraw-Hill Education. Sample Problem 2.11: Solution SOLUTION: a) ClO4– is perchlorate; Fe must have a 2+ charge since there are 2 ClO4– ions. This is iron(II) perchlorate. b) The anion sulfite is SO32–; therefore 2 Na+ ions are needed for each sulfite. The formula is Na2SO3. c) ©McGraw-Hill Education. The ionic compound is barium hydroxide. When water is included in the formula, we use the term “hydrate” and a prefix that indicates the number of molecules of H2O. This compound is barium hydroxide octahydrate. Sample Problem 2.12: Problem Recognizing Incorrect Names and Formulas of Ionic Compounds PROBLEM: Explain what is wrong with the name or formula at the end of each statement, and correct it: a) Ba(C2H3O2)2 is called barium diacetate. b) Sodium sulfide has the formula (Na)2SO3. c) Iron(II) sulfate has the formula Fe2(SO4)3. d) Cesium carbonate has the formula Cs2(CO3). ©McGraw-Hill Education. Sample Problem 2.12: Solution SOLUTION: a) The charge of Ba2+ must be balanced by two C2H3O2– ions. The prefix “di” is not required and is not used in this way when naming ionic compounds. The correct name is simply barium acetate. b) An ion of a single element does not need parentheses, and sulfide is S2–, not SO32–. The correct formula is Na2S. c) Sulfate or SO42– has a 2- charge, and only one Fe2+ is needed to form a neutral compound. The formula should be FeSO4. d) The parentheses are unnecessary, since only one CO32– ion is present. The correct formula is Cs2CO3. ©McGraw-Hill Education. Naming Binary Covalent Compounds For a binary covalent compound, the element with the lower group number in the periodic table is first in the name and formula. Its name remains unchanged. If both elements are in the same group, the heavier element is listed first: SO2. The element that is second is named using the root with the suffix –ide. Numerical prefixes indicate the number of atoms of each element present. Greek numerical prefixes indicate the number of atoms of each element present: mono, di, tri, tetra, penta, etc. No Roman numerals are used since there are no electric charges. Prefix mono is optional (implied) and may be omitted. © McGraw Hill Naming Binary Covalent Compounds Some of these compounds are very common and have trivial (old) names: – H2O is water – NH3 is ammonia – CH4 is methane. Two non-metals always form a covalent compound. Metals in groups 1(A), 2(A) except for Be, and 3(B) always form ionic compounds with group 6(A) and 7(A) nonmetals. Beryllium forms covalent compounds only. All other metals form some ionic compounds and some covalent compounds. Example: PbCl2 is ionic = lead(II) chloride; PbCl4 is covalent = lead tetrachloride. ©McGraw-Hill Education. Numerical Prefixes for Hydrates and Binary Covalent Compounds Table 2.6 ©McGraw-Hill Education. Naming Rules for Ionic and Covalent Compounds © McGraw Hill Sample Problem 2.14: Problem Determining Names and Formulas of Binary Covalent Compounds PROBLEM: a) What is the formula of carbon disulfide? b) What is the name of PCl5? c) ©McGraw-Hill Education. Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms. Sample Problem 2.14: Solution SOLUTION: a) The prefix di- means “two.” The formula is CS2. b) P is the symbol for phosphorus; there are five chlorine atoms, which is indicated by the prefix penta-. The name is phosphorus pentachloride. c) ©McGraw-Hill Education. Nitrogen (N) comes first in the name (lower group number). The compound is dinitrogen tetroxide, N2O4. Sample Problem 2.15: Problem Recognizing Incorrect Names and Formulas of Binary Covalent Compounds PROBLEM: Explain what is wrong with the name or formula at the end of each statement, and correct it: a) SF4 is monosulfur pentafluoride. b) Dichlorine heptoxide is Cl2O6. c) ©McGraw-Hill Education. N2O3 is dinitrotrioxide. Sample Problem 2.15: Solution SOLUTION: a) There are two mistakes. Mono- is not needed if there is only one atom of the first element, and the prefix for four is tetra-, not penta-. The correct name is sulfur tetrafluoride. b) The prefix hepta- indicates seven, not six. The correct formula is Cl2O7. c) ©McGraw-Hill Education. The full name of the first element is needed, and a space separates the two element names. The correct name is dinitrogen trioxide. Naming Binary Acids Binary acid solutions form when certain covalent compounds of Hydrogen dissolve in water: HF, HCl, HBr, HI, H2S. The name for the acid solution changes: – Pure, gaseous HCl is called hydrogen chloride – Dissolved in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid - hydro + chlor + ic + acid H2S in water changes the name from dihydrogen sulfide to hydrosulfuric acid. HCN in water changes name from hydrogen cyanide to hydrocyanic acid. This is technically a ternary acid. ©McGraw-Hill Education. Naming Oxoacids Oxoacids are compounds containing hydrogen and an oxoanion. Oxoacid names are similar to those of the oxoanions, except for two suffix changes: -ate in the anion becomes –ic in the acid. -ite in the anion becomes –ous in the acid. To name oxoacids: Oxyanion root + suffix -ic or -ous + the word acid The oxoanion prefixes hypo- and per- are retained. Thus, BrO4- is perbromate, and HBrO4 is perbromic acid; BrO2− is bromite, and HBrO2 is bromous acid. © McGraw Hill Sample Problem 2.13: Problem Determining Names and Formulas of Anions and Acids PROBLEM: Name the following anions and give the name and formula of the acid derived from each: a) Br – b) IO3– c) CN– d) HSO4– ©McGraw-Hill Education. Sample Problem 2.13: Solution SOLUTION: a) The anion is bromide; the acid is hydrobromic acid, HBr. b) The anion is iodate; the acid is iodic acid, HIO3. c) The anion is cyanide; the acid is hydrocyanic acid, HCN. d) The anion is hydrogen sulfate; the acid is sulfuric acid, H2SO4. (In this case, the suffix is added to the element name sulfur, not to the root, sulf-.) ©McGraw-Hill Education. Molecular Masses from Chemical Formulas Molecular mass = sum of atomic masses For the H2O molecule: molecular mass = (2 x atomic mass of H) + (1 x atomic mass of O) = (2 x 1.008 amu) + (1 x 16.00 amu) = 18.02 amu – In this course we limit atomic masses to 4 significant figures. For ionic compounds we refer to a formula mass since ionic compounds do not consist of molecules. ©McGraw-Hill Education. Sample Problem 2.16: Problem and Plan Calculating the Molecular Mass of a Compound PROBLEM: Using the periodic table, calculate the molecular (or formula) mass of: a) tetraphosphorous trisulfide b) ammonium nitrate PLAN: Write the formula and then multiply the number of atoms by the respective atomic masses. Add the masses for each compound. ©McGraw-Hill Education. Sample Problem 2.16: Solution SOLUTION: a) P4S3: Molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S) = (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu b) NH4NO3: Formula mass = (2 x atomic mass of N) + (4 x atomic mass of H) + (3 x atomic mass of O) = (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu) = 80.05 amu ©McGraw-Hill Education. Sample Problem 2.17: Problem and Plan Using Molecular Depictions to determine Formula, Name, and Mass for a compound PROBLEM: Each scene represents a binary compound. Determine its formula, name, and molecular (or formula) mass. PLAN: Each compound contains only two elements. Find the simplest whole number ratio of atoms in each compound and use this formula to determine the name and the molecular (or formula) mass. ©McGraw-Hill Education. Sample Problem 2.17: Solution SOLUTION: a) There is 1 brown Na+ for every green F-, so the formula is NaF, an ionic compound, which is named sodium fluoride. Formula mass = (1 x atomic mass of Na) + (1 x atomic mass of F) = 22.99 amu + 19.00 amu = 41.99 amu b) There are 3 green F for every blue N, so the formula is NF3, a covalent compound, which is named nitrogen trifluoride. Molecular mass = (1 x atomic mass of N) + (3 x atomic mass of F) = 14.01 amu + (3 x 19.00) = 71.01 amu ©McGraw-Hill Education.