Chapter

Chapter

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 3 Molecules, Compounds, and Chemical Equations Roy Kennedy Massachusetts Bay Community College 2008, Prentice Hall MA Wellesley Hills, Elements and Compounds elements combine together to make an almost limitless number of compounds the properties of the compound are totally different from the constituent elements

Tro, Chemistry: A Molecular Approach 2 Formation of Water from Its Elements Tro, Chemistry: A Molecular Approach 3 Chemical Bonds compounds are made of atoms held together by chemical bonds bonds are forces of attraction between atoms the bonding attraction comes from attractions between protons and electrons Tro, Chemistry: A Molecular Approach 4

Bond Types two general types of bonding between atoms found in compounds, ionic and covalent ionic bonds result when electrons have been transferred between atoms, resulting in oppositely charged ions that attract each other generally found when metal atoms bonded to nonmetal atoms covalent bonds result when two atoms share some of their electrons generally found when nonmetal atoms bonded together Tro, Chemistry: A Molecular Approach 5

Tro, Chemistry: A Molecular Approach 6 Representing Compounds with Chemical Formula compounds are generally represented with a chemical formula the amount of information about the structure of the compound varies with the type of formula all formula and models convey a limited amount of information none are perfect representations all chemical formulas tell what elements are in the compound use the letter symbol of the element

7 Types of Formula Empirical Formula Empirical Formula describe the kinds of elements found in the compound and the ratio of their atoms they do not describe how many atoms, the order of attachment, or the shape the formulas for ionic compounds are empirical Tro, Chemistry: A Molecular Approach 8 Types of Formula Molecular Formula Molecular Formula describe the kinds of elements found in the compound and the numbers of their atoms

they do not describe the order of attachment, or the shape Tro, Chemistry: A Molecular Approach 9 Types of Formula Structural Formula Structural Formula describe the kinds of elements found in the compound, the numbers of their atoms, order of atom attachment, and the kind of attachment they do not directly describe the 3-dimensional shape, but an experienced chemist can make a good guess at it use lines to represent covalent bonds each line describes the number of electrons shared by the bonded atoms single line = 2 shared electrons, a single covalent bond double line = 4 shared electrons, a double covalent bond triple line = 6 shared electrons, a triple covalent bond

Tro, Chemistry: A Molecular Approach 10 Representing Compounds Molecular Models Models show the 3-dimensional structure along with all the other information given in structural formula Ball-and-Stick Models use balls to represent the atoms and sticks to represent the attachments between them Space-Filling Models use interconnected spheres to show the electron clouds of atoms connecting together Tro, Chemistry: A Molecular Approach 11

Chemical Formulas Hydrogen Peroxide Molecular Formula = H2O2 Empirical Formula = HO Benzene Molecular Formula = C6H6 Empirical Formula = CH Glucose Molecular Formula = C6H12O6 Empirical Formula = CH2O Tro, Chemistry: A Molecular Approach 12 Types of Formula Tro, Chemistry: A Molecular Approach 13 Molecular View of

Elements and Compounds Tro, Chemistry: A Molecular Approach 14 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 Tro, Chemistry: A Molecular Approach 15 Molecular Elements

Certain elements occur as 2 atom molecules Rule of 7s Other elements occur as polyatomic molecules P4, S8, Se8 7A H2 7 N2 O2 F2 Cl2 Br2 I2

Tro, Chemistry: A Molecular Approach 16 Molecular Elements Tro, Chemistry: A Molecular Approach 17 Ionic vs. Molecular Compounds Propane contains individual C3H8 molecules Tro, Chemistry: A Molecular Approach Table salt contains an array of Na+ ions and Cl- ions 18

Ionic Compounds metals + nonmetals no individual molecule units, instead have a 3-dimensional array of cations and anions made of formula units many contain polyatomic ions several atoms attached together in one ion Tro, Chemistry: A Molecular Approach 19 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 Tro, Chemistry: A Molecular Approach 20 Writing Formulas for Ionic Compounds 1. 2. 3. 4. 5. 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

Tro, Chemistry: A Molecular Approach 21 Write the formula of a compound made from aluminum ions and oxide ions 1. Write the symbol for the metal 2. 3. 4. 5. 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 Tro, Chemistry: A Molecular Approach Al+3 column 3A O2- column 6A Al+3 O2Al2 O3 Al = (2)(+3) = +6 O = (3)(-2) = -6 22 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 Tro, Chemistry: A Molecular Approach 23

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 Tro, Chemistry: A Molecular Approach 24 Formula-to-Name

Rules for Ionic Compounds made of cation and anion some have one or more nicknames that are only learned by experience NaCl = table salt, NaHCO3 = baking soda write systematic name by simply naming the ions If cation is: metal with invariant charge = metal name metal with variable charge = metal name(charge) polyatomic ion = name of polyatomic ion If anion is: nonmetal = stem of nonmetal name + ide polyatomic ion = name of polyatomic ion Tro, Chemistry: A Molecular Approach 25 Metal Cations

Metals withInvariant Variable Metals with Charge Charges metals ionsions can can only metalswhose whose have onehave possible charge more than Groups 1A+1 & 2A+2, Al+3, Ag+1, one Zn+2, Sc+3 possible charge

cation name = metal name determine charge by charge on anion cation name = metal name with Roman numeral charge in parentheses Tro, Chemistry: A Molecular Approach 26 Naming 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 S = sulfide Cl = chloride Si = silicide P = phosphide Tro, Chemistry: A Molecular Approach 27

Naming Binary Ionic Compounds for Metals with Invariant Charge Contain Metal Cation + Nonmetal Anion Metal listed first in formula and name 1. name metal cation first, name nonmetal anion second 2. cation name is the metal name 3. nonmetal anion named by changing the ending on the nonmetal name to -ide Tro, Chemistry: A Molecular Approach 28 Example Naming Binary Ionic with Invariant Charge Metal CsF

1. Identify cation and anion Cs = Cs+ because it is Group 1A F = F- because it is Group 7A 2. Name the cation Cs+ = cesium 3. Name the anion F- = fluoride 4. Write the cation name first, then the anion name cesium fluoride Tro, Chemistry: A Molecular Approach 29 Name the following compounds 1. KCl 2. MgBr2 3. Al2S3

Tro, Chemistry: A Molecular Approach 30 Name the following compounds 1. KCl potassium chloride 2. MgBr2 magnesium bromide 3. Al2S3 aluminum sulfide Tro, Chemistry: A Molecular Approach 31

Naming Binary Ionic Compounds for Metals with Variable Charge Contain Metal Cation + Nonmetal Anion Metal listed first in formula and name 1. name metal cation first, name nonmetal anion 2. 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 ions Table 3.4 3. nonmetal anion named by changing the ending on the nonmetal name to -ide Tro, Chemistry: A Molecular Approach

32 Determining the Charge on a Cation with Variable Charge Au2S3 1. determine the charge on the anion Au2S3 - the anion is S, since it is in Group 6A, its charge is -2 2. determine the total negative charge since there are 3 S in the formula, the total negative charge is 6 3. determine the total positive charge since the total negative charge is -6, the total positive charge is +6 4. divide by the number of cations since there are 2 Au in the formula and the total positive charge is +6, each Au has a +3 charge

Tro, Chemistry: A Molecular Approach 33 Example Naming Binary Ionic with Variable Charge Metal CuF2 1. Identify cation and anion F = F- because it is Group 7 Cu = Cu2+ to balance the two (-) charges from 2 F- 2. Name the cation Cu2+ = copper(II) 3. Name the anion F- = fluoride 4. Write the cation name first, then the anion name

copper(II) fluoride Tro, Chemistry: A Molecular Approach 34 Name the following compounds 1. TiCl4 2. PbBr2 3. Fe2S3 Tro, Chemistry: A Molecular Approach 35 Name the following compounds 1. TiCl4 titanium(IV) chloride 2. PbBr2

lead(II) bromide 3. Fe2S3 iron(III) sulfide Tro, Chemistry: A Molecular Approach 36 Example Writing Formula for Binary Ionic Compounds Containing Variable Charge Metal manganese(IV) sulfide 1. Write the symbol for the cation 2. 3. 4. 5. 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 Tro, Chemistry: A Molecular Approach Mn+4 S2Mn+4 S2- Mn2S4 MnS2 Mn = (1)(+4) = +4 S = (2)(-2) = -4 37

Practice - What are the formulas for compounds made from the following ions? 1. copper(II) ion with a nitride ion 2. iron(III) ion with a bromide ion Tro, Chemistry: A Molecular Approach 38 Practice - What are the formulas for compounds made from the following ions? 1. Cu2+ with N3- Cu3N2 2. Fe+3 with Br- FeBr3

Tro, Chemistry: A Molecular Approach 39 Compounds Containing Polyatomic Ions Polyatomic ions are single ions that contain more than one atom Often identified by (ion) in formula Name and charge of polyatomic ion do not change Name any ionic compound by naming cation first and then anion Tro, Chemistry: A Molecular Approach 40 Some Common Polyatomic Ions Name

Formula Name Formula acetate C2H3O2 hypochlorite ClO carbonate CO32 chlorite

ClO2 hydrogen carbonate HCO3 (aka bicarbonate) chlorate ClO3 perchlorate hydroxide OH ClO4 nitrate NO3

sulfate SO42 nitrite NO2 sulfite SO32 chromate CrO hydrogen sulfate (aka bisulfate) HSO4

dichromate Cr2O72 hydrogen sulfite (aka bisulfite) HSO3 2 4 ammonium NH4+ Tro, Chemistry: A Molecular Approach 41 Patterns for Polyatomic Ions 1. elements in the same column form similar

polyatomic ions same number of Os and same charge ClO3- = chlorate BrO3- = bromate 2. if the polyatomic ion starts with H, add hydrogen- prefix before name and add 1 to the charge CO32- = carbonate HCO3-1 = hydrogen carbonate Tro, Chemistry: A Molecular Approach 42 Periodic Pattern of Polyatomic Ions -ate groups 3A -3 BO3

4A 5A 6A -2 CO3 -1 NO3 -2 SiO3 -3 PO4 -3

AsO4 -2 SO4 -1 ClO3 -2 SeO4 -1 BrO3 -2 TeO4 Tro, Chemistry: A Molecular Approach 7A

-1 IO3 43 Patterns for Polyatomic Ions -ate ion chlorate = ClO3-1 -ate ion + 1 O same charge, per- prefix perchlorate = ClO4-1 -ate ion 1 O same charge, -ite suffix chlorite = ClO2-1 -ate ion 2 O same charge, hypo- prefix, -ite suffix hypochlorite = ClO-1 Tro, Chemistry: A Molecular Approach

44 Example Naming Ionic Compounds Containing a Polyatomic Ion 1. Identify the ions Na2SO4 Na = Na+ because in Group 1A SO4 = SO42- a polyatomic ion 2. Name the cation Na+ = sodium, metal with invariant charge 3. Name the anion SO42- = sulfate 4. Write the name of the cation followed by the name of the anion

sodium sulfate Tro, Chemistry: A Molecular Approach 45 Example Naming Ionic Compounds Containing a Polyatomic Ion 1. Identify the ions Fe(NO3)3 NO3 = NO3- a polyatomic ion Fe = Fe+3 to balance the charge of the 3 NO3-1 2. Name the cation Fe+3 = iron(III), metal with variable charge 3. Name the anion NO3- = nitrate

4. Write the name of the cation followed by the name of the anion iron(III) nitrate Tro, Chemistry: A Molecular Approach 46 Name the following 1. NH4Cl 2. Ca(C2H3O2)2 3. Cu(NO3)2 Tro, Chemistry: A Molecular Approach 47 Name the following 1. NH4Cl

ammonium chloride 2. Ca(C2H3O2)2 calcium acetate 3. Cu(NO3)2 copper(II) nitrate Tro, Chemistry: A Molecular Approach 48 Example Writing Formula for Ionic Compounds Containing Polyatomic Ion Iron(III) phosphate 1. Write the symbol for the cation 2. 3. 4.

5. 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 Tro, Chemistry: A Molecular Approach Fe+3 PO43Fe+3 PO43- Fe3(PO4)3 FePO4 Fe = (1)(+3) = +3 PO4 = (1)(-3) = -3 49

Practice - What are the formulas for compounds made from the following ions? 1. aluminum ion with a sulfate ion 2. chromium(II) with hydrogen carbonate Tro, Chemistry: A Molecular Approach 50 Practice - What are the formulas for compounds made from the following ions? 1. Al+3 with SO42- Al2(SO4)3 2. Cr+2 with HCO3 Cr(HCO3)2

Tro, Chemistry: A Molecular Approach 51 Hydrates hydrates are ionic compounds containing a specific number of waters for each formula unit water of hydration often driven off by heating in formula, attached waters follow Prefix No. of Waters hemi

mono 1 di 2 tri 3 -hydrate after name of ionic compound tetra 4 CoCl26H2O = cobalt(II) chloride hexahydrate

CaSO4H2O = calcium sulfate hemihydrate penta 5 hexa 6 hepta 7 octa 8 CoCl26H2O in name attached waters indicated by suffix

Hydrate CoCl26H2O Tro, Chemistry: A Molecular Approach Anhydrous CoCl2 52 Practice 1. What is the formula of magnesium sulfate heptahydrate? 2. What is the name of NiCl26H2O? Tro, Chemistry: A Molecular Approach 53

Practice 1. What is the formula of magnesium sulfate heptahydrate? MgSO47H2O 2. What is the name of NiCl26H2O? nickel(II) chloride hexahydrate Tro, Chemistry: A Molecular Approach 54 Writing Names of Binary Molecular Compounds of 2 Nonmetals 1. Write name of first element in formula element furthest left and down on the Periodic Table use the full name of the element 2. Writes name the second element in the formula with

an -ide suffix as if it were an anion, however, remember these compounds do not contain ions! 3. Use a prefix in front of each name to indicate the number of atoms a) Never use the prefix mono- on the first element Tro, Chemistry: A Molecular Approach 55 Subscript - Prefixes 1 = mono not used on first nonmetal

2 = di3 = tri4 = tetra5 = penta- 6 = hexa7 = hepta8 = octa9 = nona10 = deca- drop last a if name begins with vowel Tro, Chemistry: A Molecular Approach 56 Example Naming Binary Molecular BF3 1. Name the first element

boron 2. Name the second element with an ide fluorine fluoride 3. Add a prefix to each name to indicate the subscript monoboron, trifluoride 4. Write the first element with prefix, then the second element with prefix Drop prefix mono from first element boron trifluoride Tro, Chemistry: A Molecular Approach 57 Name the following 1. NO2 2. PCl5 3. I2F7 Tro, Chemistry: A Molecular Approach

58 Name the following 1. NO2 nitrogen dioxide 2. PCl5 phosphorus pentachloride 3. I2F7 diiodine heptafluoride Tro, Chemistry: A Molecular Approach 59

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 Tro, Chemistry: A Molecular Approach 60 Write formulas for the following 1. dinitrogen tetroxide 2. sulfur hexafluoride 3. diarsenic trisulfide Tro, Chemistry: A Molecular Approach 61

Write formulas for the following 1. dinitrogen tetroxide N2O4 2. sulfur hexafluoride SF6 3. diarsenic trisulfide As2S3 Tro, Chemistry: A Molecular Approach 62 Acids

acids are molecular compounds that form H+ when dissolved in water to indicate the compound is dissolved in water (aq) is written after the formula not named as acid if not dissolved in water sour taste dissolve many metals like Zn, Fe, Mg; but not Au, Ag, Pt formula generally starts with H e.g., HCl, H2SO4 Tro, Chemistry: A Molecular Approach 63 Reaction of Acids with Metals H2 gas Tro, Chemistry: A Molecular Approach

64 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 Tro, Chemistry: A Molecular Approach

65 Naming Binary Acids write a hydro prefix follow with the nonmetal name change ending on nonmetal name to ic write the word acid at the end of the name Tro, Chemistry: A Molecular Approach 66 Example - Naming Binary Acids HCl(aq) 1. Identify the anion Cl = Cl-, chloride because Group 7A

2. Name the anion with an ic suffix Cl- = chloride chloric 3. Add a hydro- prefix to the anion name hydrochloric 4. Add the word acid to the end hydrochloric acid Tro, Chemistry: A Molecular Approach 67 Naming Oxyacids if polyatomic ion name ends in ate, then change ending to ic suffix if polyatomic ion name ends in ite, then change ending to ous suffix write word acid at end of all names Tro, Chemistry: A Molecular Approach

68 Example Naming Oxyacids H2SO4(aq) 1. Identify the anion SO4 = SO42- = sulfate 2. If the anion has ate suffix, change it to ic. If the anion has ite suffix, change it to -ous SO42- = sulfate sulfuric 3. Write the name of the anion followed by the word acid sulfuric acid (kind of an exception, to make it sound nicer!) Tro, Chemistry: A Molecular Approach 69

Example Naming Oxyacids H2SO3(aq) 1. Identify the anion SO3 = SO32- = sulfite 2. If the anion has ate suffix, change it to ic. If the anion has ite suffix, change it to -ous SO32- = sulfite sulfurous 3. Write the name of the anion followed by the word acid sulfurous acid Tro, Chemistry: A Molecular Approach 70 Name the following 1. H2S 2. HClO3

3. HNO2 Tro, Chemistry: A Molecular Approach 71 Name the following 1. H2S hydrosulfuric acid 2. HClO3 chloric acid 3. HNO2 nitrous acid Tro, Chemistry: A Molecular Approach

72 Writing Formulas for Acids when name ends in acid, formulas starts with H write formulas as if ionic, even though it is molecular hydro prefix means it is binary acid, no prefix means it is an oxyacid for oxyacid, if ending is ic, polyatomic ion ends in ate; if ending is ous, polyatomic ion ends in ous Tro, Chemistry: A Molecular Approach 73 Example Binary Acids hydrosulfuric acid 1. Write the symbol for the cation 2. 3. 4.

5. and its charge Write the symbol for the anion and its charge Charge (without sign) becomes subscript for other ion Add (aq) to indicate dissolved in water Check that the total charge of the cations cancels the total charge of the anions Tro, Chemistry: A Molecular Approach H+ S2- in all acids the cation is H+ hydro means

binary H+ S2- H2S H2S(aq) H = (2)(+1) = +2 S = (1)(-2) = -2 74 Example Oxyacids carbonic acid 1. Write the symbol for the cation 2. 3. 4. 5. and its charge Write the symbol for the anion

and its charge Charge (without sign) becomes subscript for other ion Add (aq) to indicate dissolved in water Check that the total charge of the cations cancels the total charge of the anions Tro, Chemistry: A Molecular Approach H+ CO32- in all acids the cation is H+ no hydro means polyatomic ion -ic means -ate ion H+ CO32-

H2CO3 H2CO3(aq) H = (2)(+1) = +2 CO3 = (1)(-2) = -2 75 Example Oxyacids sulfurous acid 1. Write the symbol for the 2. 3. 4. 5. cation and its charge Write the symbol for the anion and its charge Charge (without sign)

becomes subscript for other ion Add (aq) to indicate dissolved in water Check that the total charge of the cations cancels the total charge of the anions Tro, Chemistry: A Molecular Approach H+ SO32- in all acids the cation is H+ no hydro means polyatomic ion -ous means -ite ion H+ SO32-

H2SO3 H2SO3(aq) H = (2)(+1) = +2 SO3 = (1)(-2) = -2 76 Practice - What are the formulas for the following acids? 1. chlorous acid 2. phosphoric acid 3. hydrobromic acid Tro, Chemistry: A Molecular Approach 77 Practice - What are the formulas for the following acids? 1. H+ with ClO2

HClO2 2. H+ with PO43 H3PO4 3. H+ with Br HBr Tro, Chemistry: A Molecular Approach 78 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 Tro, Chemistry: A Molecular Approach 79 Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g

so the Molar Mass of H2O is 18.02 g/mole Tro, Chemistry: A Molecular Approach 80 Example Find the number of CO2 molecules in 10.8 g of dry ice Given: Find: Concept Plan: Relationships: Solution: 10.8 g CO2 molecules CO2 g CO2 1 mol 44.01 g mol CO2

1 mol CO2 = 44.01 g, 1 mol = 6.022 x 1023 molec CO2 6.022 1023 molecules 1 mol 1 mol CO 2 6.022 10 23 molecules 10.8 g CO 2 44.01 g CO 2 1 mol 1.48 10 23 molecules CO 2 Check: since the given amount is much less than 1 mol CO2, the number makes sense

Practice - Converting Grams to Molecules How many molecules are in 50.0 g of PbO2? (PbO2 = 239.2) Tro, Chemistry: A Molecular Approach 82 Practice - Converting Grams to Molecules How many molecules are in 50.0 g of PbO2? Given: 50.0 g PbO2 Find: molecules PbO2 Relationships: 1 mole PbO2 239.2 g; 1 mol 6.022 x 1023 molec Concept Plan: PbO22 ggPbO 1 mole PbO 2 239.2 g

molPbO PbO22 mol molecPbO PbO22 molec 6.022 1023 molec 1 mole PbO 2 Apply Solution Map: 1 mole PbO 2 6.022 1023 molec 50.0 g PbO 2 1.26 1023 molec PbO 2 239.2 g 1 mole PbO 2 Check Answer: Units are correct. Number makes sense because given amount less

than 1 mole 83 Percent Composition Percentage of each element in a compound By mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding part Percentage 100% whole

Tro, Chemistry: A Molecular Approach 84 Example 3.13 Find the mass percent of Cl in C2Cl4F2 Given: Find: Concept Plan: Relationships: C2Cl4F2 % Cl by mass 4 molar mass Cl Mass % Cl 100% molar mass C 2Cl 4 F2 mass element X in 1 mol Mass % element X 100%

mass 1 mol of compound Solution: 4 molar mass Cl 4(35.45 g/mol) 141.8 g/mol molar mass C 2Cl 4 F2 2(12.01) 4(35.45) 2(19.00) 203.8 g/mol 141.8 g/mol Mass % Cl 100% 69.58% 203.8 g/mol Check: since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense Practice - Determine the Percent Composition of the following CaCl2 Mass % Ca molar mass Ca

100% molar mass CaCl2 2 molar mass Cl Mass % Cl 100% molar mass CaCl 2 2 molar mass Cl 2(35.45 g/mol) 70.90 g/mol molar mass CaCl 2 1(40.08) 2(35.45) 110.98 g/mol 40.08 g/mol Mass % Ca 100% 36.11% 110.98 g/mol 70.90 g/mol Mass % Cl 100% 63.88% 110.98 g/mol Tro, Chemistry: A Molecular Approach 87

Mass Percent as a Conversion Factor the mass percent tells you the mass of a constituent element in 100 g of the compound the fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains 58.64 g Cl this can be used as a conversion factor 100 g CCl2F2 : 58.64 g Cl 58.64 g Cl g CCl 2 F2 g Cl 100 g CCl 2 F2 Tro, Chemistry: A Molecular Approach 100 g CCl 2 F2 g Cl g CCl 2 F2 58.64 g Cl 88

Example 3.14 Find the mass of table salt containing 2.4 g of Na Given: Find: Concept Plan: Relationships: 2.4 g Na, 39% Na g NaCl g Na g NaCl 100 g NaCl 39 g Na 100. g NaCl : 39 g Na Solution: 100 g NaCl

2.4 g Na 6.2 g NaCl 39 g Na Check: since the mass of NaCl is more than 2x the mass of Na, the number makes sense Practice Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? Tro, Chemistry: A Molecular Approach 90 Practice Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? Given: Find: Concept Plan:

Relationships: 19.8 g C, 79.2% C g benzaldehyde gC g benzaldehyde 100 g benzaldehyde 79.2 g C 100. g benzaldehyde : 79.2 g C Solution: 100 g benzaldehyde 19.8 g C 25.0 g benzaldehyde 79.2 g C Check: since the mass of benzaldehyde is more than the mass of C, the number makes sense

Conversion Factors in Chemical Formulas chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules like percent composition Tro, Chemistry: A Molecular Approach 92 Example 3.15 Find the mass of hydrogen in 1.00 gal of water Given: Find: 1.00 gal H2O, dH2O = 1.00 g/ml

gH Concept Plan: gal H2O L H2O g H2O mol H2O mL H2O g H2O moL H gH Relationships:

3.785 L = 1 gal, 1 L = 1000 mL, 1.00 g H2O = 1 mL, 1 mol H2O = 18.02 g, 1 mol H = 1.008 g, 2 mol H : 1 mol H2O Solution: 1.00 gal H 2O 3.785 L 1000 mL 1.00 g 3.785 103 g H 2O 1 gal 1L 1 mL 1 mol H 2 O 2 mol H 1.008 g H 3.785 10 g H 2O 4.23 102 g H 18.02 g 1 mol H 2 O 1 mol H 3 Check: since 1 gallon weighs about 3800 g, and H is light, the number makes sense Practice - How many grams of sodium are in 6.2 g

of NaCl? (Na = 22.99; Cl = 35.45) Tro, Chemistry: A Molecular Approach 94 How many grams of sodium are in 6.2 g of NaCl? Given: 6.2 g NaCl Find: g Na Rel: 1 mole NaCl 58.45 g; 1 mol Na 1 mol NaCl; 1 mol Na 22.99 g Na Concept Plan: NaCl ggNaCl 1 mole NaCl 58.45 g molNaCl NaCl mol

1 mole Na 1 mol NaCl molNa Na mol 22.99 g Na 1 mol Na Na ggNa Apply Concept Plan: 1 mole NaCl 1 mole Na 22.99 g Na 6.2 g NaCl 2.4 g Na 58.45 g

1 mole NaCl 1 mole Na Check Answer: Units are correct. Number makes sense because given amount less than 1 mole NaCl. 95 Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analysis masses of elements formed when decompose or react compound combustion analysis percent composition Tro, Chemistry: A Molecular Approach 96

Finding an Empirical Formula 1) convert the percentages to grams a) b) assume you start with 100 g of the compound skip if already grams a) use molar mass of each element a) if result is within 0.1 of whole number, round to whole number 2) convert grams to moles 3) write a pseudoformula using moles as subscripts 4) divide all by smallest number of moles

5) multiply all mole ratios by number to make all whole numbers a) b) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc. skip if already whole numbers Tro, Chemistry: A Molecular Approach 97 Example 3.17 Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53%

Tro, Chemistry: A Molecular Approach 98 Example: Find the empirical formula of aspirin with the given mass percent composition. Write down the given quantity and its units. Given: C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O Tro, Chemistry: A Molecular Approach 99 Example:

Find the empirical formula of aspirin with the given mass percent composition. Information Given: 60.00 g C, 4.48 g H, 35.53 g O Write down the quantity to find and/or its units. Find: empirical formula, CxHyOz Tro, Chemistry: A Molecular Approach 100 Example: Find the empirical formula of aspirin with

the given mass percent composition. Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: Empirical Formula, CxHyOz Write a Concept Plan: g C, H, O mol C, H, O Tro, Chemistry: A Molecular Approach mol ratio empirical formula

101 Example: Find the empirical formula of aspirin with the given mass percent composition. Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: Empirical Formula, CxHyOz CP: g C,H,O mol C,H,O mol ratio empirical formula Collect Needed Relationships: 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O

Tro, Chemistry: A Molecular Approach 102 Example: Find the empirical formula of aspirin with the given mass percent composition. Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: Empirical Formula, CxHyOz CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Apply the Concept Plan:

calculate the moles of each element 1 mol C 60.00 g C 4.996 mol C 12.01 g C 1 mol H 4.48 g H 4.44 mol H 1.008 g H 1 mol O 35.53 g O 2.220 mol O 16.00 g O Tro, Chemistry: A Molecular Approach 103

Example: Find the empirical formula of aspirin with the given mass percent composition. Information Given: 4.996 mol C, 4.44 mol H, 2.220 mol O Find: Empirical Formula, CxHyOz CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Apply the Concept Plan: write a pseudoformula

C4.996H4.44O2.220 Tro, Chemistry: A Molecular Approach 104 Information Given: C4.996H4.44O2.220 Find: Empirical Formula, CxHyOz CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Example: Find the empirical formula of aspirin with the given mass percent composition. Apply the Concept Plan:

find the mole ratio by dividing by the smallest number of moles C 4.996 H 4.44 O 2.220 2.220 2.220 2.220 C 2.25 H 2O1 Tro, Chemistry: A Molecular Approach 105 Example: Find the empirical formula of aspirin with the given mass percent

composition. Information Given: C2.25H2O1 Find: Empirical Formula, CxHyOz CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Apply the Concept Plan: multiply subscripts by factor to give whole number {C2.25H2O1} x 4 C9H8O4 Tro, Chemistry: A Molecular Approach

106 Practice Determine the empirical formula of hematite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Tro, Chemistry: A Molecular Approach 110 Practice Determine the empirical formula of hematite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, (100 72.4) = 27.6% O in 100 g hematite there are 72.4 g Fe and 27.6 g O Find: FexOy Rel: 1 mol Fe = 55.85 g; 1 mol O = 16.00 g Concept Plan: whole mole number Fe

molFe Fe ggFe mol pseudo- ratio ratio pseudoformula formula g O mol O gO mol O Tro, Chemistry: A Molecular Approach empirical empirical formula formula

111 Practice Determine the empirical formula of hematite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Apply the Concept Plan: 1 mol Fe 72.4 g Fe 1.30 mol Fe 55.85 g 1 mol O 26.7 g O 1.73 mol O 16.00 g Fe1.30O1.73 Fe1.30 O 1.73 Fe1O1.33 1.30

1.30 Fe1O1.33 3 Fe3O 4 Tro, Chemistry: A Molecular Approach 112 Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound Molar Mass molecular formula multiplying factor, n Empirical Formula Mass Tro, Chemistry: A Molecular Approach

113 Example 3.18 Find the molecular formula of butanedione Given: Find: emp. form. = C2H3O; MM = 86.03 g/mol molecular formula Molecular Form. Emp. Form.n Concept Plan: and Relationships: n Molar Mass Emp. Form. Molar Mass

Solution: Molar Mass Emp. Form. 2(12.01g/mol) 3(1.008 g/mol) 1(16.00 g/mol) 43.04 g/mol 86.09 g/mol n 2 43.04 g/mol Check: Molecular Formula C 2 H 3O 2 C 4 H 6 O 2 the molar mass of the calculated formula is in agreement with the given molar mass Practice Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C 5H3. What is its molecular formula? (C = 12.01, H=1.01)

Tro, Chemistry: A Molecular Approach 115 Practice Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01) C5 = 5(12.01 g) = 60.05 g H3 = 3(1.01 g) = 3.03 g C5H3 = 63.08 g 252 g/mol n 4 63.08 g/mol Molecular Formula = {C5H3} x 4 = C20H12 Tro, Chemistry: A Molecular Approach

116 Combustion Analysis a common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O by knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined all the original C forms CO2, the original H forms H2O, the original mass of O is found by subtraction once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found Tro, Chemistry: A Molecular Approach

117 Combustion Analysis Tro, Chemistry: A Molecular Approach 118 Example 3.20 Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO2 = 2.445 g H2O = 0.6003 g Determine the empirical formula of the compound Tro, Chemistry: A Molecular Approach 119 Example 3.20:

Find the empirical formula of compound with the given amounts of combustion products Write down the given quantity and its units. Given: compound = 0.8233 g CO2 = 2.445 g H2O = 0.6003 g Tro, Chemistry: A Molecular Approach 120 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information

Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H Write down the quantity to find and/or its units. Find: empirical formula, CxHyOz Tro, Chemistry: A Molecular Approach 121 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound,

2.445 g CO2, 0.6003 g H Find: Empirical Formula, CxHyOz Write a Concept Plan: g CO2, H2O mol CO2, H2O mol C, H, O Tro, Chemistry: A Molecular Approach mol C, H mol ratio g

C, H g O mol O empirical formula 122 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O

Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formula Collect Needed Relationships: 1 mole CO2 = 44.01 g CO2 1 mole H2O = 18.02 g H2O 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H Tro, Chemistry: A Molecular Approach 123 Example 3.20: Find the empirical formula of compound with the

given amounts of combustion products Information Given: 0.8233 g compound 2.445 g CO2, 0.6003 g H2O Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: calculate the moles of C and H 1 mol CO 2 1 mol C 2.445 g CO 2 0.05556 mol C

44.01 g CO 2 1 mol CO 2 1 mol H 2O 2 mol H 0.6003 g H 2O 0.06662 mol H 18.02 g H 2O 1 mol H 2O Tro, Chemistry: A Molecular Approach 124 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O,

0.05556 mol C, 0.06662 mol H Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: calculate the grams of C and H 12.01 g 0.05556 mol C 0.6673 g C 1 mol C 1.008 g 0.06662 mol H 0.06715 g H 1 mol H Tro, Chemistry: A Molecular Approach 125

Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O, 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: calculate the grams and moles of O 0.8233 g compound - (0.6673 g C 0.06715 g H) 0.0889 g O

1 mol O 0.0889 g O 0.00556 mol O 16.00 g Tro, Chemistry: A Molecular Approach 126 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O, 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O

Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: write a pseudoformula C0.05556H0.06662O0.00556 Tro, Chemistry: A Molecular Approach 127 Example 3.20: Find the empirical formula of compound with the given amounts of combustion

products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O, 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: find the mole ratio by dividing by the smallest number of moles C 0.05556 H 0.06662 O 0.00556 0.00556 0.00556

0.00556 C10 H12O1 Tro, Chemistry: A Molecular Approach 128 Example 3.20: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O, 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: Empirical Formula, CxHyOz CP: g CO2 & H2O mol CO2 & H2O mol C & H

g C & H g O mol O mol ratio emp. formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: multiply subscripts by factor to give whole number, if necessary write the empirical formula C10 H12O1 Tro, Chemistry: A Molecular Approach 129 The smell of dirty gym socks is caused by the compound caproic acid. Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid? (MM C = 12.01, H = 1.008, O =

16.00) Tro, Chemistry: A Molecular Approach 130 Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid? 1 mol CO 2 1 mol C 1.92 g CO 2 0.0436 mol C 44.01 g CO 2 1 mol CO 2 1 mol H 2 O 2 mol H 0.784 g H 2 O

0.0870 mol H 18.02 g H 2 O 1 mol H 2 O 12.01 g 0.0436 mol C 0.524 g C 1 mol C 1.008 g 0.0870 mol H 0.0877 g H 1 mol H Tro, Chemistry: A Molecular Approach 131 g moles C 0.524 0.0436

H 0.0877 0.0870 O 0.232 0.0145 0.844 g compound - (0.524 g C 0.0877 g H) 0.232 g O 1 mol O 0.232 g O 0.0145 mol O 16.00 g C0.0436H0.0870O0.0145 C 0.0436 H 0.0870 O 0.0145 0.0145 0.0145

0.0145 C3H 6O1 Tro, Chemistry: A Molecular Approach 132 C3 312.01 g 36.03 g H 6 61.008 g 6.048 g O1 116.00 g 16.00 g C3 H 6 O 58.08 g 116.2 g/mol n 2 58.08 g/mol

Molecular Formula = {C3H6O} x 2 = C6H12O2 Tro, Chemistry: A Molecular Approach 133 Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules Elements are not transmuted during a reaction Reactants Products 134

Chemical Equations Shorthand way of describing a reaction Provides information about the reaction Formulas of reactants and products States of reactants and products Relative numbers of reactant and product molecules that are required Can be used to determine weights of reactants used and products that can be made Tro, Chemistry: A Molecular Approach 135 Combustion of Methane methane gas burns to produce carbon dioxide gas and gaseous water whenever something burns it combines with O2(g) CH4(g) + O2(g) CO2(g) + H2O(g) O H

H C H H + O O C + H O

H O 1C+4H + 2O Tro, Chemistry: A Molecular Approach 1C+2O +2H+O 1C+2H+3O 136 Combustion of Methane Balanced to show the reaction obeys the Law of Conservation of Mass, it must be balanced CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H

H C H H O + O + O C O 1C + 4H + 4O Tro, Chemistry: A Molecular Approach

O O + H H O + O H H 1C + 4H + 4O 137

Chemical Equations CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) CH4 and O2 are the reactants, and CO2 and H2O are the products the (g) after the formulas tells us the state of the chemical the number in front of each substance tells us the numbers of those molecules in the reaction called the coefficients Tro, Chemistry: A Molecular Approach 138 Chemical Equations CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

this equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides to obtain the number of atoms of an element, multiply the subscript by the coefficient 1C1 4H4 4O2+2 Tro, Chemistry: A Molecular Approach 139 Symbols Used in Equations symbols used to indicate state after chemical (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water energy symbols used above the arrow for

decomposition reactions = heat h = light shock = mechanical elec = electrical Tro, Chemistry: A Molecular Approach 140 Example 3.22 Write a balanced equation for the combustion of butane, C4H10 Write a skeletal equation C4H10(l) + O2(g) CO2(g) + H2O(g) Balance atoms in complex substances first 4C1x4 C4H10(l) + O2(g) 4 CO2(g) + H2O(g) 10 H 2 x 5

C4H10(l) + O2(g) 4 CO2(g) + 5 H2O(g) Balance free elements by adjusting coefficient in front of free element 13/2 x 2 O 13 C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) If fractional coefficients, multiply thru by denominator {C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g)}x 2 Check 2 C4H10(l) + 13 O2(g) 8 CO2(g) + 10 H2O(g) 8 C 8; 20 H 20; 26 O 26 Practice

when aluminum metal reacts with air, it produces a white, powdery compound aluminum oxide reacting with air means reacting with O2 aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O2(g) Al2O3(s) Tro, Chemistry: A Molecular Approach 142 Practice when aluminum metal reacts with air, it produces a white, powdery compound aluminum oxide reacting with air means reacting with O2 aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O2(g) Al2O3(s) 4 Al(s) + 3 O2(g) 2 Al2O3(s)

Tro, Chemistry: A Molecular Approach 143 Practice Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen acids are always aqueous metals are solid except for mercury Tro, Chemistry: A Molecular Approach 144 Practice Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen

acids are always aqueous metals are solid except for mercury Al(s) + HC2H3O2(aq) Al(C2H3O2)3(aq) + H2(g) 2 Al(s) + 6 HC2H3O2(aq) 2 Al(C2H3O2)3(aq) + 3 H2(g) Tro, Chemistry: A Molecular Approach 145 Classifying Compounds Organic vs. Inorganic in the18th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic organic compounds easily decomposed and could not be made in 18th century lab inorganic compounds very difficult to decompose, but able to be synthesized Tro, Chemistry: A Molecular Approach

146 Modern Classifying Compounds Organic vs. Inorganic today we commonly make organic compounds in the lab and find them all around us organic compounds are mainly made of C and H, sometimes with O, N, P, S, and trace amounts of other elements the main element that is the focus of organic chemistry is carbon Tro, Chemistry: A Molecular Approach 147 Carbon Bonding carbon atoms bond almost exclusively covalently compounds with ionic bonding C are generally inorganic when C bonds, it forms 4 covalent bonds

4 single bonds, 2 double bonds, 1 triple + 1 single, etc. carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms Tro, Chemistry: A Molecular Approach 148 Carbon Bonding Tro, Chemistry: A Molecular Approach 149 Classifying Organic Compounds there are two main categories of organic compounds, hydrocarbons and functionalized

hydrocarbons hydrocarbons contain only C and H most fuels are mixtures of hydrocarbons Tro, Chemistry: A Molecular Approach 150 Classifying Hydrocarbons hydrocarbons containing only single bonds are called alkanes hydrocarbons containing one or more C=C are called alkenes hydrocarbons containing one or more CC are called alkynes hydrocarbons containing C6 benzene ring are called aromatic Tro, Chemistry: A Molecular Approach 151

Tro, Chemistry: A Molecular Approach 152 Naming Straight Chain Hydrocarbons consists of a base name to indicate the number of carbons in the chain, with a suffix to indicate the class and position of multiple bonds suffix ane for alkane, ene for alkene, yne for alkyne Base Name No. of C Base Name No. of C meth- 1

hex- 6 eth- 2 hept- 7 prop- 3 oct- 8

but- 4 non- 9 pent- 5 dec- 10 Tro, Chemistry: A Molecular Approach 153 Functionalized Hydrocarbons

functional groups are non-carbon groups that are on the molecule substitute one or more functional groups replacing Hs on the hydrocarbon chain generally, the chemical reactions of the compound are determined by the kinds of functional groups on the molecule Tro, Chemistry: A Molecular Approach 154 Functional Groups Tro, Chemistry: A Molecular Approach 155

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