Chapter 5 Molecules and Compounds

Molecules and Compounds Salt Sodium – shiny, reactive, poisonous Chlorine – pale yellow gas, reactive, poisonous Sodium chloride – table salt Sugar Carbon – pencil or diamonds Hydrogen – flammable gas Oxygen – a gas in air Combine to form white crystalline sugar

Law of Constant Composition all pure substances have constant composition all samples of a pure substance contain the same elements in the same percentages (ratios) mixtures have variable composition

Compounds Display Constant Composition If we decompose water by electrolysis, we find 16.0 grams of oxygen to every 2.00 grams of hydrogen. Water has a constant Mass Ratio of Oxygen to Hydrogen of 8.0.

Why do Compounds Show Constant Composition smallest piece of a compound is called a molecule every molecule of a compound has the same number and type of atoms as determined by the electronic structures of the atoms (more on that later in the year) since all the molecules of a compound are identical… every sample will have the same ratio of the elements every sample of the compound will have the same properties

EXAMPLE 5.1 Constant Composition of Compounds Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition. To show this, compute the mass ratio of one element to the other by dividing the larger mass by the smaller one. For the first sample: For the second sample: Solution: Since the ratios are the same for the two samples, these results are consistent with the law of constant composition.

What is the carbon-hydrogen mass ratio for methane (CH4)? 1 0.33 3 4 0.25

5.1B If the mass ratio of lead(II) sulfide is 270.0 g lead and 41.8 g sulfur, how much lead is required to completely react with 85.6 g of sulfur? 13.3 185 228 312 553

Formulas Describe Compounds water = H2O \ two atoms of hydrogen and 1 atom of oxygen table sugar = C12H22O11 \12 atoms of C, 22 atoms of H and 11 atoms O

Order of Elements in a Formula metals written first NaCl nonmetals written in order from Table 5.1 CO2 are occasional exceptions for historical or informational reasons H2O, but NaOH Table 5.1 Order of Listing Nonmetals in Chemical Formulas C P N H S I Br Cl O F

Molecules with Polyatomic Ions symbol of the polyatomic ion called nitrate ion called sulfate Mg(NO3)2 compound called magnesium nitrate CaSO4 compound called calcium sulfate parentheses to group two NO3’s no parentheses for one SO4 implied “1” subscript on magnesium on calcium

Molecules with Polyatomic Ions subscript indicating two NO3 groups no subscript indicating one SO4 group Mg(NO3)2 compound called magnesium nitrate CaSO4 compound called calcium sulfate implied “1” subscript on nitrogen, total 2 N on sulfur, total 1 S stated “3” subscript on oxygen, total 6 O stated “4” subscript on oxygen, total 4 O

Classifying Materials atomic elements = elements whose particles are single atoms molecular elements = elements whose particles are multi-atom molecules molecular compounds = compounds whose particles are molecules made of only nonmetals ionic compounds = compounds whose particles are cations and anions

Molecular Elements H2 N2 O2 F2 P4 S8 Cl2 Br2 I2 Certain elements occur as 2 atom molecules Rule of 7’s there are 7 common diatomic elements find the element with atomic number 7, N make a figure 7 by going over to Group 7A, then down don’t forget to include H2 VIIA 7 H2 N2 O2 F2 P4 S8 Cl2 Br2 I2

Molecular Compounds two or more nonmetals smallest unit is a molecule

Ionic Compounds metals + nonmetals no individual molecule units, instead have a 3-dimensional array of cations and anions made of formula units

Molecular View of Elements and Compounds

Classify each of the following as either an atomic element, molecular element, molecular compound or ionic compound aluminum, Al aluminum chloride, AlCl3 chlorine, Cl2 acetone, C3H6O carbon monoxide, CO cobalt, Co

Classify each of the following as either an atomic element, molecular element, molecular compound or ionic compound aluminum, Al = atomic element aluminum chloride, AlCl3 = ionic compound chlorine, Cl2 = molecular element acetone, C3H6O = molecular compound carbon monoxide, CO = molecular compound cobalt, Co = atomic element

Is the compound one of the exceptions to the rules? Formula-to-Name Step 1 Is the compound one of the exceptions to the rules? H2O = water, steam, ice NH3 = ammonia

What major class of compound is it? Ionic or Molecular Formula-to-Name Step 2 What major class of compound is it? Ionic or Molecular

Major Classes Ionic Molecular metal + nonmetal metal first in formula Binary Ionic compounds with polyatomic ions Molecular 2 nonmetals Binary Molecular (or Binary Covalent) Acids – formula starts with H though acids are molecular, they behave as ionic when dissolved in water may be binary or oxyacid

Formula-to-Name Step 3 What major subclass of compound is it? Binary Ionic, Ionic with Polyatomic Ions, Binary Molecular, Binary Acid, Oxyacid

Classifying Compounds Compounds containing a metal and a nonmetal = binary ionic Type I and II Compounds containing a polyatomic ion = ionic with polyatomic ion Compounds containing two nonmetals = binary molecular compounds Compounds containing H and a nonmetal = binary acids Compounds containing H and a polyatomic ion = oxyacids

Apply Rules for the Class and Subclass Formula-to-Name Step 4 Apply Rules for the Class and Subclass

Formula-to-Name Rules for Ionic Made of cation and anion Name by simply naming the ions If cation is: Type I metal = metal name Type II metal = metal name(charge) Polyatomic ion = name of polyatomic ion If anion is: Nonmetal = stem of nonmetal name + ide

Monatomic Nonmetal Anion determine the charge from position on the Periodic Table to name anion, change ending on the element name to –ide 4A = -4 5A = -3 6A = -2 7A = -1 C = carbide N = nitride O = oxide F = fluoride Si = silicide P = phosphide S = sulfide Cl = chloride

How do you know a metal cation is Type II? Metal Cations Type I metals whose ions can only have one possible charge IA, IIA, (Al, Ga, In) determine charge by position on the Periodic Table IA = +1, IIA = +2, (Al, Ga, In = +3) Type II metals whose ions can have more than one possible charge determine charge by charge on anion How do you know a metal cation is Type II? its not Type I !!!

Determine if the following metals are Type I or Type II Determine if the following metals are Type I or Type II. If Type I, determine the charge on the cation it forms. lithium, Li copper, Cu gallium, Ga tin, Sn strontium, Sr

Determine if the following metals are Type I or Type II Determine if the following metals are Type I or Type II. If Type I, determine the charge on the cation it forms. lithium, Li Type I +1 copper, Cu Type II gallium, Ga Type I +3 tin, Sn Type II strontium, Sr Type I +2

Type I Binary Ionic Compounds Contain Metal Cation + Nonmetal Anion Metal listed first in formula & name name metal cation first, name nonmetal anion second cation name is the metal name nonmetal anion named by changing the ending on the nonmetal name to -ide

Type II Binary Ionic Compounds Contain Metal Cation + Nonmetal Anion Metal listed first in formula & name name metal cation first, name nonmetal anion second metal cation name is the metal name followed by a Roman Numeral in parentheses to indicate its charge determine charge from anion charge Common Type II cations in Table 5.5 nonmetal anion named by changing the ending on the nonmetal name to -ide

Examples LiCl = lithium chloride AlCl3 = aluminum chloride PbO = lead(II) oxide PbO2 = lead(IV) oxide Mn2O3 = manganese(III) oxide ZnCl2 = zinc(II) chloride or zinc chloride AgCl = silver(I) chloride or silver chloride

Compounds Containing Polyatomic Ions Polyatomic ions are single ions that contain more than one atom Name any ionic compound by naming cation first and then anion Non-polyatomic cations named like Type I and II Non-polyatomic anions named with -ide

Fixed Charge Metals and Nonmetals IA IIA IIIA VA VIA VIIA Li+1 Be+2 N-3 O-2 F-1 Na+1 Mg+2 Al+3 P-3 S-2 Cl-1 K+1 Ca+2 Zn+2 Ga+3 As-3 Se-2 Br-1 Rb+1 Sr+2 Ag+1 Cd+2 In+3 Te-2 I-1 Cs+1 Ba+2

Some Common Polyatomic Ions Name Formula acetate C2H3O2– carbonate CO32– hydrogen carbonate (aka bicarbonate) HCO3– hydroxide OH– nitrate NO3– nitrite NO2– chromate CrO42– dichromate Cr2O72– ammonium NH4+ Name Formula hypochlorite ClO– chlorite ClO2– chlorate ClO3– perchlorate ClO4– sulfate SO42– sulfite SO32– hydrogensulfate (aka bisulfate) HSO4– hydrogensulfite (aka bisulfite) HSO3–

Patterns for Polyatomic Ions elements in the same column form similar polyatomic ions same number of O’s and same charge ClO3- = chlorate \ BrO3- = bromate if the polyatomic ion starts with H, the name adds hydrogen- prefix before name and add 1 to the charge CO32- = carbonate \ HCO3-1 = hydrogencarbonate

Periodic Pattern of Polyatomic Ions -ate groups IIIA IVA VA VIA VIIA BO 3 -3 CO 3 -2 NO 3 -1 SiO 3 -2 PO 4 -3 SO 4 -2 ClO 3 -1 AsO 4 -3 SeO 4 -2 BrO 3 -1 TeO 4 -2 IO 3 -1

Binary Molecular Compounds of 2 Nonmetals Name first element in formula first use the full name of the element Name the second element in the formula with an -ide as if it were an anion, however, remember these compounds do not contain ions! Use a prefix in front of each name to indicate the number of atoms Never use the prefix mono- on the first element

Subscript - Prefixes 1 = mono - not used on first nonmetal 2 = di- 3 = tri- 4 = tetra- 5 = penta- 6 = hexa- 7 = hepta- 8 = octa- 9 = nona- 10 = deca- drop last “a” if name begins with vowel

Acids Contain H+1 cation and anion in aqueous solution Binary acids have H+1 cation and nonmetal anion Oxyacids have H+1 cation and polyatomic anion

Formula-to-Name Acids acids are molecular compounds that often behave like they are made of ions All names have acid at end Binary Acids = hydro prefix + stem of the name of the nonmetal + ic suffix Oxyacids if polyatomic ion ends in –ate = name of polyatomic ion with –ic suffix if polyatomic ion ends in –ite = name of polyatomic ion with –ous suffix

Example – Naming Binary Acids HCl Is it one of the common exceptions? H2O, NH3, CH4, NaCl, C12H22O11 = No! Identify Major Class first element listed is H,  Acid Identify the Subclass 2 elements,  Binary Acid

Sample - Naming Binary Acids – HCl Identify the anion Cl = Cl-, chloride because Group 7A Name the anion with an –ic suffix Cl- = chloride  chloric Add a hydro- prefix to the anion name hydrochloric Add the word acid to the end hydrochloric acid

Example – Naming Oxyacids H2SO4 Is it one of the common exceptions? H2O, NH3, CH4, NaCl, C12H22O11 = No! Identify Major Class first element listed is H,  Acid Identify the Subclass 3 elements in the formula,  Oxyacid

Example – Naming Oxyacids H2SO4 Identify the anion SO4 = SO42- = sulfate If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous SO42- = sulfate  sulfuric Write the name of the anion followed by the word acid sulfuric acid (kind of an exception, to make it sound nicer!)

Example – Naming Oxyacids H2SO3 Is it one of the common exceptions? H2O, NH3, CH4, NaCl, C12H22O11 = No! Identify Major Class first element listed is H,  Acid Identify the Subclass 3 elements in the formula,  Oxyacid

Example – Naming Oxyacids H2SO3 Identify the anion SO3 = SO32- = sulfite If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous SO32- = sulfite  sulfurous Write the name of the anion followed by the word acid sulfurous acid

Formula-to-Name Flow Chart

Name – to – Formula

Writing the Formulas from the Names For binary molecular compounds, use the prefixes to determine the subscripts For Type I, Type II, Ternary Compounds and Acids Determine the ions present Determine the charges on the cation and anion Balance the charges to get the subscripts

Example – Binary Molecular dinitrogen pentoxide Identify the symbols of the elements nitrogen = N oxide = oxygen = O Write the formula using prefix number for subscript di = 2, penta = 5 N2O5

Compounds that Contain Ions compounds of metals with nonmetals are made of ions metal atoms form cations, nonmetal atoms for anions compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge if Na+ is combined with S2-, you will need 2 Na+ ions for every S2- ion to balance the charges, therefore the formula must be Na2S

Writing Formulas for Ionic Compounds Write the symbol for the metal cation and its charge Write the symbol for the nonmetal anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the sum of the charges of the cation cancels the sum of the anions

Write the formula of a compound made from aluminum ions and oxide ions Write the symbol for the metal cation and its charge Write the symbol for the nonmetal anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions Al+3 column IIIA O2- column VIA Al+3 O2- Al2 O3 Al = (2)∙(+3) = +6 O = (3)∙(-2) = -6

Practice - What are the formulas for compounds made from the following ions? potassium ion with a nitride ion calcium ion with a bromide ion aluminum ion with a sulfide ion

Practice - What are the formulas for compounds made from the following ions? K+ with N3- K3N Ca+2 with Br- CaBr2 Al+3 with S2- Al2S3

Example – Ionic Compounds manganese(IV) sulfide Write the symbol for the cation and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions Mn+4 S2- Mn+4 S2- Mn2S4 MnS2 Mn = (1)∙(+4) = +4 S = (2)∙(-2) = -4

Example – Ionic Compounds Iron(III) phosphate Write the symbol for the cation and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions Fe+3 PO43- Fe+3 PO43- Fe3(PO4)3 FePO4 Fe = (1)∙(+3) = +3 PO4 = (1)∙(-3) = -3

Example – Ionic Compounds ammonium carbonate Write the symbol for the cation and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions NH4+ CO32- NH4+ CO32- (NH4)2CO3 (NH4)2CO3 NH4 = (2)∙(+1) = +2 CO3 = (1)∙(-2) = -2

Practice - What are the formulas for compounds made from the following ions? copper(II) ion with a nitride ion iron(III) ion with a bromide ion aluminum ion with a sulfate ion

Practice - What are the formulas for compounds made from the following ions? Cu2+ with N3- Cu3N2 Fe+3 with Br- FeBr3 Al+3 with SO42- Al2(SO4)3

Example – Binary Acids hydrosulfuric acid in all acids the cation is H+ Write the symbol for the cation and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions H+ hydro means binary S2- H+ S2- H2S H2S H = (2)∙(+1) = +2 S = (1)∙(-2) = -2

Example – Oxyacids carbonic acid in all acids the cation is H+ Write the symbol for the cation and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions H+ no hydro means polyatomic ion CO32- -ic means -ate ion H+ CO32- H2CO3 H2CO3 H = (2)∙(+1) = +2 CO3 = (1)∙(-2) = -2

Practice - What are the formulas for the following acids? chlorous acid phosphoric acid hydrobromic acid

Practice - What are the formulas for the following acids? H+ with ClO2– HClO2 H+ with PO43– H3PO4 H+ with Br– HBr

Formula Mass the mass of an individual molecule or formula unit also known as molecular mass or molecular weight sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts! mass of 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu

EXAMPLE 5.1 Constant Composition of Compounds Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition. To show this, compute the mass ratio of one element to the other by dividing the larger mass by the smaller one. For the first sample: For the second sample: Solution: Since the ratios are the same for the two samples, these results are consistent with the law of constant composition.

EXAMPLE 5.2 Writing Chemical Formulas Write a chemical formula for each of the following: (a) the compound containing two aluminum atoms to every three oxygen atoms (b) the compound containing three oxygen atoms to every sulfur atom (c) the compound containing four chlorine atoms to every carbon atom Since aluminum is the metal, it is listed first. Since sulfur is below oxygen on the periodic table and since it occurs before oxygen in Table 5.1, it is listed first. Since carbon is to the left of chlorine on the periodic table and since it occurs before chlorine in Table 5.1, it is listed first. Solution: (a) Al2O3 (b) SO3 (c) CCl4

EXAMPLE 5.3 Total Number of Each Type of Atom in a Chemical Formula Determine the number of each type of atom in Mg3(PO4)2. Solution: Mg: There are three Mg atoms, as indicated by the subscript 3. P: There are two P atoms, as we see by multiplying the subscript outside the parentheses (2) by the subscript for P inside the parentheses, which is 1 (implied). O: There are eight O atoms, as we see by multiplying the subscript outside the parentheses (2) by the subscript for O inside the parentheses (4).

EXAMPLE 5. 4 Classifying Substances as Atomic Elements, Molecular EXAMPLE 5.4 Classifying Substances as Atomic Elements, Molecular Elements, Molecular Compounds, or Ionic Compounds Classify each of the following substances as an atomic element, molecular element, molecular compound, or ionic compound. (a) krypton (b) CoCl2 (c) nitrogen (d) SO2 (e) KNO3 Solution: (a) Krypton is an element that is not listed as diatomic in Table 5.2; therefore, it is an atomic element. (b) CoCl2 is a compound composed of a metal (left side of periodic table) and nonmetal (right side of the periodic table); therefore, it is an ionic compound. (c) Nitrogen is an element that is listed as diatomic in Table 5.2; therefore, it is a molecular element. (d) SO2 is a compound composed of two nonmetals; therefore, it is a molecular compound. (e) KNO3 is a compound composed of a metal and two nonmetals; therefore, it is an ionic compound.

EXAMPLE 5.5 Writing Formulas for Ionic Compounds Write a formula for the ionic compound that forms from aluminum and oxygen. Solution: Al3+ O2– 1. Write the symbol for the metal and its charge followed by the symbol of the nonmetal and its charge. For many elements, these charges can be determined from their group number in the periodic table (refer to Figure 4.14). 2. Make the magnitude of the charge on each ion (without the sign) become the subscript for the other ion. In this case, the numbers cannot be reduced any further; the correct formula is Al2O3. 3. Reduce the subscripts to give a ratio with the smallest whole numbers. 4. Check that the sum of the charges of the cations exactly cancels the sum of the charges of the anions. Cations: 2(3+) = 6+ Anions: 3(2–) = 6– The charges cancel.

EXAMPLE 5.6 Writing Formulas for Ionic Compounds Write a formula for the ionic compound that forms from magnesium and oxygen. Solution: Mg2+ O2– 1. Write the symbol for the metal and its charge followed by the symbol of the nonmetal and its charge. For many elements, these charges can be determined from their group number in the periodic table (refer to Figure 4.14). 2. Make the magnitude of the charge on each ion (without the sign) become the subscript for the other ion. To reduce the subscripts, divide both subscripts by 2. Mg2O2 ÷ 2 = MgO 3. Reduce the subscripts to give a ratio with the smallest whole numbers. 4. Check that the sum of the charges of the cations exactly cancels the sum of the charges of the anions. Cations: 2+ Anions: 2– The charges cancel.

EXAMPLE 5.7 Writing Formulas for Ionic Compounds Write a formula for the compound that forms from potassium and oxygen. Solution: We first write the symbol for each ion along with its appropriate charge from its group number in the periodic table. K+ O2– We then make the magnitude of each ion’s charge become the subscript for the other ion.   K+ O2–  becomes K2O    No reducing of subscripts is necessary in this case. Finally, we check to see that the sum of the charges of the cations [2(1+) = 2+] exactly cancels the sum of the charges of the anion (2–). The correct formula is K2O.

EXAMPLE 5.8 Naming Type I Ionic Compounds Give the name for the compound MgF2. Solution: The cation is magnesium. The anion is fluorine, which becomes fluoride. The correct name is magnesium fluoride.

EXAMPLE 5.9 Naming Type II Ionic Compounds Give the name for the compound PbCl4. Solution: The name for PbCl4 consists of the name of the cation, lead, followed by the charge of the cation in parenthesis (IV), followed by the base name of the anion, chlor-, with the ending -ide. The full name is lead(IV) chloride. We know the charge on Pb is 4+ because the charge on Cl is 1–. Since there are 4 Cl– anions, the Pb cation must be Pb4+. PbCl4 lead (IV) chloride

EXAMPLE 5.10 Naming Ionic Compounds That Contain a Polyatomic Ion Give the name for the compound K2CrO4. The name for K2CrO4 consists of the name of the cation, potassium, followed by the name of the polyatomic ion, chromate. Solution: K2CrO4 potassium chromate

EXAMPLE 5.11 Naming Molecular Compounds Name each of the following: CCl4 BCl3 SF6. Solution: CCl4 The name of the compound is the name of the first element, carbon, followed by the base name of the second element, chlor, prefixed by tetra- to indicate four, and the suffix -ide. The entire name is carbon tetrachloride. BCl3 The name of the compound is the name of the first element, boron, followed by the base name of the second element, chlor, prefixed by tri- to indicate three, and the suffix -ide. The entire name is boron trichloride. SF4 The name of the compound is the name of the first element, sulfur, followed by the base name of the second element, fluor, prefixed by hexa- to indicate six, and the suffix -ide. The entire name is sulfur hexafluoride.

EXAMPLE 5.12 Naming Binary Acids Give the name of H2S. The base name of S is sulfur, so the name is hydrosulfuric acid. Solution: H2S hydrosulfuric acid

EXAMPLE 5.13 Naming Oxyacids Give the name of HC2H3O2. The oxyanion is acetate, which ends in -ate; therefore, the name of the acid is acetic acid. Solution: HC2H3O2 acetic acid

EXAMPLE 5.14 Nomenclature Using Figure 5.17 Name each of the following: CO, CaF2, HF, Fe(NO3)3, HClO4, H2SO3 Solution: For each compound, the following table shows how to use Figure 5.17 to arrive at a name for the compound. Formula Flow Chart Path Name CO Molecular carbon monoxide CaF2 calcium fluoride HF hydrofluoric acid Fe(NO3)3 iron(III) nitrate HClO4 perchloric acid H2SO3 sulfurous acid Ionic Type I Acid Binary Ionic Type II Acid Oxyacid -ate Acid Oxyacid -ite

EXAMPLE 5.15 Calculating Formula Mass Calculate the formula mass of carbon tetrachloride, CCl4. Solution: To find the formula mass, we sum the atomic masses of each atom in the chemical formula. Formula mass = 1 x (Formula mass C) + 4 x (Formula mass Cl) = 12.01 amu + 4(35.45 amu) = 153.81 amu

EXAMPLE 5.16 Constant Composition of Compounds Two samples said to be carbon disulfide (CS2) are decomposed into their constituent elements. One sample produced 8.08 g S and 1.51 g C, while the other produced 31.3 g S and 3.85 g C. Are these results consistent with the law of constant composition? Solution: Sample 1 These results are not consistent with the law of constant composition and the data given must therefore be in error.

EXAMPLE 5.17 Writing Chemical Formulas Write a chemical formula for the compound containing one nitrogen atom for every two oxygen atoms. Solution: NO2

EXAMPLE 5.18 Total Number of Each Type of Atom in a Chemical Formula Determine the number of each type of atom in PB(ClO3)2. Solution: One Pb atom Two Cl atoms Six O atoms

EXAMPLE 5.19 Classifying Elements as Atomic or Molecular Classify each of the following elements as atomic or molecular: Na, I, N. Solution: Na: atomic I: molecular (I2) N: molecular (N2)

EXAMPLE 5.20 Classifying Compounds as Ionic or Molecular Classify each of the following compounds as ionic or molecular. If they are ionic, classify them as Type I or Type II ionic compounds: FeCl3, K2SO4, CCl4 Solution: FeCl3: ionic, Type II K2SO4: ionic, Type I CCl4: molecular

EXAMPLE 5.21 Writing Formulas for Ionic Compounds Write a formula for the compound that forms from lithium and sulfate ions. Solution: Li+ SO 2– Li2(SO4) In this case, the subscripts cannot be further reduced. Li2SO4 Cations: Anions: 2(1+) = 2+ 2– 4

EXAMPLE 5.22 Naming Type I Binary Ionic Compounds Give the name for the compound Al2O3 Solution: aluminum oxide

EXAMPLE 5.23 Naming Type II Binary Ionic Compounds Give the name for the compound Fe2S3 Solution: 3 sulfide ions x (2–) = 6– 2 iron ions x (?) = 6+ ? = 3+ Charge of each iron ion = 3+ iron(III) sulfide

EXAMPLE 5.24 Naming Compounds Containing a Polyatomic Ion Give the name for the compound Co(ClO4)2. Solution: 3 perchlorate x (1–) = 2– Charge of cobalt ion = 2+ cobalt(II) perchlorate

EXAMPLE 5.25 Naming Molecular Compounds Name the compound NO2. Solution: nitrogen dioxide

EXAMPLE 5.26 Naming Binary Acids Name the acid HI. Solution: hydroiodic acid

EXAMPLE 5.27 Naming Oxyacids with an Oxyanion Ending in -ate Name the acid H2SO4. Solution: The oxyanion is sulfate. The name of the acid is sulfuric acid.

EXAMPLE 5.28 Naming Oxyacids with an Oxyanion Ending in -ite Name the acid HClO2. Solution: The oxyanion is chlorite. The name of the acid is chlorous acid.

Figure: 05-17-12un Title: Problem 95, Ch 5 Caption: A consumer product label.

Figure: 05-17-13un Title: Problem 95, Ch 5, no 2 Caption: A consumer product label.

Figure: 05-17-14un Title: Problem 95, Ch 5, no 3 Caption: A consumer product label.

Figure: 05-17-15un Title: Problem 95, Ch 5, no 4 Caption: A consumer product label.

EXAMPLE 5.29 Calculating Formula Mass Calculate the formula mass of Mg(NO3)2 Solution: Formula mass = 24.31 + 2(14.01) + 6(16.00) = 148.33 amu