General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 3: Chemical Compounds Dr. Burak Esat Fatih University Chem 107Fall 2013 1 Compound: A combination of two or more different elements. A substance composed of two or more elements chemically combined in fixed ratios by mass. Compounds are represented by chemical formulas indicating the fixed ratios of each elements chemically combined Water - H 2 O Carbon dioxide - CO 2 Sodium Chloride - NaCl Iron(II) sulfide - FeS Molecule: The smallest entity having the same elemental combination as the compound. Compounds are made of individual molecules 2 1
Chemical Formulas Empirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule. 3 Molecular compounds Structural Formula shows the order in which atoms are bonded together 4 2
Standard color scheme 5 Some molecules H 2 O 2 CH 3 CH 2 Cl P 4 O 10 CH 3 CH(OH)CH 3 HCO 2 H 6 3
Chemical Formulas Empirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule. 7 Some Examples of Compounds with the Same Elemental Ratio s Empirical Formula Molecular Formula CH 2 (unsaturated Hydrocarbons) C 2 H 4, C 3 H 6, C 4 H 8 OH or HO H 2 O 2 S S 8 P P 4 Cl Cl 2 CH 2 O (carbohydrates) C 6 H 12 O 6 8 4
Some Compounds with Empirical Formula CH 2 O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH 2 O 30.03 Formaldehyde Disinfectant; Biological preservative C 2 H 4 O 2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C 3 H 6 O 3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C 4 H 8 O 4 120.10 Erythrose Forms during sugar metabolism C 5 H 10 O 5 150.13 Ribose Component of many nucleic acids and vitamin B 2 C 6 H 12 O 6 180.16 Glucose Major nutrient for energy in cells 9 Mole The Mole is based upon the definition: The amount of substance that contains as many elementary parts (atoms, molecules, or other) as there are atoms in exactly 12 grams of carbon 12. 1 Mole = 6.022045 x 10 23 particles 10 5
One Mole of Common Substances CaCO 3 100.09 g Oxygen Gas (O 2 ) 32.00 g Copper (Cu) 63.55 g Water (H 2 O) 18.02 g 11 Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 10 23 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 10 23 atoms 1 molecule of O 2 = 32.00 amu 1 mole of O 2 = 32.00 g = 6.022 x 10 23 molecule 1 molecule of S 8 = 205.952 amu 1 mole of S 8 = 205.952 g = 6.022 x 10 23 molecules 12 6
Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound Mass/mole of compound Carbon (C) Hydrogen (H) Oxygen (O) 6 atoms 12 atoms 6 atoms 6 moles of 12 moles of 6 moles of atoms atoms atoms 6(6.022 x 10 23 ) 12(6.022 x 10 23 ) 6(6.022 x 10 23 ) atoms atoms atoms 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) =72.06 amu =12.10 amu =96.00 amu 72.06 g 12.10 g 96.00 g 13 Calculate the Molecular Mass of Glucose: C 6 H 12 O 6 Carbon 6 x 12.011 g/mol = 72.066 g Hydrogen 12 x 1.008 g/mol = 12.096 g Oxygen 6 x 15.999 g/mol = 95.994 g 180.156 g 14 7
% Composition H OH HO HO H H H O OH OH H Glucose Molecular formula C 6 H 12 O 6 Empirical formula CH 2 O Molecular Mass: Use the naturally occurring mixture of isotopes, 6 x 12.01 + 12 x 1.01 + 6 x 16.00 = 180.18 Exact Mass: Use the most abundant isotopes, 6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915 = 180.06339 15 Chemical Composition Halothane C 2 HBrClF 3 Mole ratio Mass ratio n C /n halothane m C /m halothane M(C 2 HBrClF ) 3 = 2M C + M H + M Br + M Cl + 3M F = (2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 x 19.00) = 197.38 g/mol 16 8
Example 3.4 Calculating the Mass Percent Composition of a Compound Calculate the molecular mass M(C 2 HBrClF 3 ) = 197.38 g/mol For one mole of compound, formulate the mass ratio and convert to percent: (2 12.01) g % C = 100% = 12.17% 197.38g 17 Example 3-4 (2 12.01) g % C = 100% = 12.17% 197.38g % H 1.01g = 100% = 0.51% 197.38g 79.90g % Br = 100% = 40.48% 197.38g 35.45g % Cl = 100% = 17.96% 197.38g (3 19.00) g % F = 100% = 28.88% 197.38g 18 9
Mass Percent Composition of Na 2 SO 4 Na 2 SO 4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses 2 x Na = 2 x 22.99 = 45.98 1 x S = 1 x 32.07 = 32.07 4 x O = 4 x 16.00 = 64.00 Atomic Masses from Periodic Table 142.05 Percent of each Element % Na = Mass Na / Total mass x 100% % Na = (45.98 / 142.05) x 100% =32.37% % S = Mass S / Total mass x 100% % S = (32.07 / 142.05) x 100% = 22.58% % O = Mass O / Total mass x 100% % O = (64.00 / 142.05) x 100% = 45.05% Check % Na + % S + % O = 100% 32.37% + 22.58% + 45.05% = 100.00% Steps to Determine Empirical Formulas Masses (g) of Individual Elements xa+yb A x B y Molar Mass (g/mol ) Moles of Each Element use no. of moles as subscripts Preliminary Formula change to integer subscripts Empirical Formula 20 10
Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x Moles of Cr = 6.420 g Cr x Moles of O = 7.902 g O x 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O = 0.2469 mol Na = 0.12347 mol Cr = 0.4939 mol O 21 Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na 0.2469 Cr 0.1235 O 0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na 1.99 Cr 1.00 O 4.02 Rounding off to whole numbers: Na 2 CrO 4 Sodium Chromate 22 11
Establishing Formulas from Experimentally Determined Percent Composition 5 Step approach: 1. Choose an arbitrary sample size (100g). 2. Convert masses to amounts in moles. 3. Write a formula. 4. Convert formula to small whole numbers. 5. Multiply all subscripts by a small whole number to make the subscripts integral. 23 Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g Cmpd 24 12
Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x Moles of H = Mass of H x Moles of O = Mass of O x 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O = 3.3306 moles C = 6.6657 moles H = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O 3.33 / 3.33 = CH 2 O 25 Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 Whole-number multiple = M of Glucose = empirical formula mass 180.16 = = 6.00 = 6 30.03 Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 = C 6 H 12 O 6 26 13
Adrenaline Is a Very Important Compound in the Body - I Analysis gives : C = 56.8 % H = 6.50 % O = 28.4 % N = 8.28 % Calculate the Empirical Formula 27 Adrenaline - II Assume 100g! C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N Divide by 0.591 = C = 8.00 mol C = 8.0 mol C or H = 10.9 mol H = 11.0 mol H O = 3.01 mol O = 3.0 mol O C 8 H 11 O 3 N N = 1.00 mol N = 1.0 mol N 28 14
Combustion Train for the Determination of the Chemical Composition of Organic Compounds. C n H m + (n+ m ) O 2 = n CO 2(g) + m H 2 O 2 2 (g) Fig. 3.4 29 Combustion Analysis 30 15
Ascorbic Acid ( Vitamin C ) - I Contains C, H, and O Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO 2 and 2.64 mg H 2 O Calculate it s Empirical formula! C: 9.74 x10-3 g CO 2 x(12.01 g C/44.01 g CO 2 ) = 2.65 x 10-3 g C H: 2.64 x10-3 g H 2 O x (2.016 g H 2 /18.02 gh 2 O) = 2.92 x 10-4 g H Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O 31 Vitamin C Combustion - II C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O Divide each by 2.21 x 10-4 C = 1.00 Multiply each by 3 = 3.00 = 3.0 H = 1.32 = 3.96 = 4.0 O = 1.00 = 3.00 = 3.0 C 3 H 4 O 3 32 16
Determining a Chemical Formula from Combustion Analysis - I Problem: Erythrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO 2 and 0.4194 g H 2 O. Plan: 1) We find the masses of Hydrogen and Carbon using the mass fractions of H in H 2 O, and C in CO 2. 2) The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. 3) We then calculate moles, and 4) Construct the empirical formula, and 5) From the given molar mass we can calculate the molecular 33 formula. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: mol C x M of C Mass fraction of C in CO 2 = = mass of 1 mol CO 2 = 1 mol C x 12.01 g C/ 1 mol C = 0.2729 g C / 1 g CO 2 44.01 g CO 2 Mass fraction of H in H 2 O = mol H x M of H = mass of 1 mol H 2 O 2 mol H x 1.008 g H / 1 mol H = = 0.1119 g H / 1 g H 2 O 18.02 g H 2 O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element 34 17
Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C Mass (g) of C = 1.027 g CO 2 x 1 g CO 2 0.1119 g H Mass (g) of H = 0.4194 g H 2 O x 1 g H 2 O = 0.2803 g C = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C 0.02334 H 0.04656 O 0.02330 = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C 4 H 8 O 4 35 Oxidation States Metals tend to lose electrons. Na -> Na + + e - Non-metals tend to gain electrons. Cl + e - -> Cl - Reducing agents Oxidizing agents We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element. 36 18
Ionic Bonding - An ionic bond is a chemical bond that results from an electrostatic attraction among oppositely charged ions in a compound. They form when electrons are transferred from one atom to another to form ions. Mono-atomic ions form binary ionic compounds Na [Ne] 3s 1 + Cl [Ne] 3s 2 3p 5 Na [Ne] + + Cl [Ar] - Na + Cl - Cations - Metal atoms lose electrons to form + ions. Anions - Nonmetal atoms gain electrons to form - ions. 37 Fig. 2.18 38 19
Rules for Oxidation States 1. The oxidation state (OS) of an individual atom in a free element is 0. 2. The total of the OS in all atoms in: i. Neutral species is 0. ii. Ionic species is equal to the charge on the ion. 3. In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively. 4. In compounds the OS of fluorine is always 1 39 Rules for Oxidation States 5. In compounds, the OS of hydrogen is usually +1 6. In compounds, the OS of oxygen is usually 2. 7. In binary (two-element) compounds with metals: i. Halogens have OS of 1, ii. Group 16 have OS of 2 and iii. Group 15 have OS of 3. 40 20
Example 3-7 Assigning Oxidation States. What is the oxidation state of the underlined element in each of the following? a) P 4 ; b) Al 2 O 3 ; c) MnO 4- ; d) NaH a) P 4 is an element. P OS = 0 b) Al 2 O 3 : O is 2. O 3 is 6. Since (+6)/2=(+3), Al OS = +3. c) MnO 4- : net OS = -1, O 4 is 8. Mn OS = +7. d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and H OS = -1. 41 Fig. 2.23 42 21
Fig. 2.20 43 Predicting the Ion an Element Will Form in Chemical Reactions Problem: What mono-atomic ions will each of the elements form? (a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9) Plan: We use the z value to find the element in the periodic table and the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons. Solution: (a) Ba +2, Barium is an alkaline earth element, Group 2A, and is expected to loose two electrons to attain the same number of electrons as the noble gas Xenon! (b) S -2, Sulfur is in the Oxygen family, Group 6A, and is expected to gain two electrons to attain the same number of electrons as the noble gas Argon! (c) Ti +4, Titanium is in Group 4B, and is expected to loose 4 electrons to attain the same number of electrons as the noble gas Argon! (d) F -, Fluorine is in a halogen, Group 7A, and is expected to gain one electron, to attain the same number of electrons as the noble gas Neon! 44 22
3-5 Naming Compounds: Organic and Inorganic Compounds Trivial names are used for common compounds. A systematic method of naming compounds is known as a system of nomenclature. Organic compounds Inorganic compounds Lead (IV) oxide Lead (II) oxide 45 3-6 Names and Formulas of Inorganic Compounds Binary Compounds of Metals and Nonmetals 46 23
Table 2.3 (p. 67) Common Mono-atomic Ions Cations Anions Charge Formula Name Charge Formula Name 1+ H + hydrogen 1- H - hydride Li + lithium F - fluoride Na + sodium Cl - chloride K + potassium Br - bromide Cs + cesium I - iodide Ag + silver 2+ Mg 2+ magnesium 2- O 2 - oxide Ca 2+ calcium S 2 - sulfide Sr 2+ strontium Ba 2+ barium Zn 2+ zinc Cd 2+ cadmium 3+ Al 3+ aluminum 3- N 3 - nitride Listed by charge; those in boldface are most common 47 Give the Name and Chemical Formulas of the Compounds Formed from the Following Pairs of Elements a) Sodium and Oxygen Na 2 O Sodium Oxide b) Zinc and Chlorine ZnCl 2 Zinc Chloride c) Calcium and Fluorine CaF 2 Calcium Fluoride d) Strontium and Nitrogen Sr 3 N 2 Strontium Nitride e) Hydrogen and Iodine HI Hydrogen Iodide f) Scandium and Sulfur Sc 2 S 3 Scandium Sulfide 48 24
Some Metals That Form More than One Oxidation State Element Ion Formula Systematic Name Common Name Chromium Cr +2 Chromium (II) Chromous Cr +3 Chromium (III) Chromic Cobalt Co +2 Cobalt (II) Co +3 Cobalt (III) Copper Cu +1 Copper (I) Cuprous Cu +2 Copper (II) Cupric Iron Fe +2 Iron (II) Ferrous Fe +3 Iron (III) Ferric Lead Pb +2 Lead (II) Pb +4 Lead (IV) Manganese Mn +2 Manganese (II) Mn +3 Manganese (III) Mercury Hg +2 2 Mercury (I) Mercurous Hg +2 Mercury (II) Mercuric Tin Sn +2 Tin (II) Stannous Sn +4 Tin (IV) Stannic Table 2.4 (p. 69) 49 Determining Names and Formulas of Ionic Compounds of Elements That Form More than One Ion Give the systematic names for the formulas or the formulas for the names of the following compounds. a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is: Fe 2 S 3 b) CoF 2 - the anion is Fluoride (F -1 ) and there are two F -1, the cation is Cobalt and it must be Co +2 therefore the compound is: Cobalt (II) Fluoride c) Stannic Oxide - Stannic is the common name for Tin (IV), Sn +4, the Oxide ion is O -2, therefore the formula of the compound is: SnO 2 d) NiCl 3 - The anion is chloride (Cl -1 ), there are three anions, so the Nickel cation is Ni +3, therefore the name of the compound is: 50 Nickel (III) Chloride 25
51 Rules for Families of Oxoanions Families with Two Oxoanions The ion with more O atoms takes the nonmetal root and the suffix -ate. The ion with fewer O atoms takes the nonmetal root and the suffix -ite. Families with Four Oxoanions (usually a Halogen) The ion with most O atoms has the prefix per-, the nonmetal root and the suffix -ate. The ion with one less O atom has just the suffix -ate. The ion with two less O atoms has the just the suffix -ite. The ion with three less O atoms has the prefix hypo- and the 52 suffix -ite. 26
Examples of Names and Formulas of Oxoanions and Their Compounds - I KNO 2 Potassium Nitrite BaSO 3 Barium Sulfite Mg(NO 3 ) 2 Magnesium Nitrate Na 2 SO 4 Sodium Sulfate LiClO 4 Lithium Perchlorate Ca(BrO) 2 Calcium Hypobromite NaClO 3 Sodium Chlorate Al(IO 2 ) 3 Aluminum Iodite RbClO 2 Rubidium Chlorite KBrO 3 Potassium Bromate CsClO Cesium Hypochlorite LiIO 4 Lithium Periodate 53 Examples of Names and Formulas of Oxoanions and Their Compounds - II Calcium Nitrate Ca(NO 3 ) 2 Ammonium Sulfite (NH 4 ) 2 SO 3 Strontium Sulfate SrSO 4 Lithium Nitrite LiNO 2 Potassium Hypochlorite KClO Lithium Perbromate LiBrO 4 Rubidium Chlorate RbClO 3 Calcium Iodite Ca(IO 2 ) 2 Ammonium Chlorite NH 4 ClO 2 Boron Bromate B(BrO 3 ) 3 Sodium Perchlorate NaClO 4 Magnesium Hypoiodite Mg(IO) 2 54 27
55 Naming Oxoanions - Examples Prefixes Root Suffixes Chlorine Bromine Iodine per ate perchlorate perbromate periodate [ ClO 4- ] [ BrO 4- ] [ IO 4- ] ate chlorate bromate iodate [ ClO 3- ] [BrO 3- ] [ IO 3- ] No. of O atoms ite chlorite bromite iodite [ ClO 2- ] [ BrO 2- ] [ IO 2- ] hypo ite hypochlorite hypobromite hypoiodite [ ClO - ] [ BrO - ] [ IO - ] 56 28
Naming Acids 1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride (HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid hydrochloric acid 2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion -ate suffix becomes an -ic suffix in the acid. Anion -ite suffix becomes an -ous suffix in the acid. The oxoanion prefixes hypo- and per- are retained. Thus, BrO 4 - is perbromate, and HBrO 4 is perbromic acid; IO 2- is iodite, and HIO 2 is iodous acid. 57 Binary Acids Acids produce H + when dissolved in water. They are compounds that ionize in water. Emphasize the fact that a molecule is an acid by altering the name. HCl hydrogen chloride hydrochloric acid HF hydrogen fluoride hydrofluoric acid 58 29
Determining Names and Formulas of Anions and Acids Problem: Name the following anions and give the names and formulas of the acid solutions derived from them: a) I - b) BrO - c) SO -2 3 d) NO - 3 e) CN - Solution: a) The anion is Iodide; and the acid is Hydroiodic acid, HI b) The anion is hypobromite; and the acid is hypobromous acid, HBrO c) The anion is Sulfite; and the acid is Sulfurous acid, H 2 SO 3 d) The anion is Nitrate; and the acid is Nitric acid, HNO 3 e) The anion is Cyanide; and the acid is Hydrocyanic acid, HCN 59 Some Compounds of Greater Complexity Effect of Moisture Blue anhydrous CoCl 2 Pink hexahydrate CoCl 2 6 H 2 O %H 2 O = 6 mol H 2 O 18.02 g H 2O 1 mol H 2 O 237.9 g CoCl 2 6 H 2 O 100% = 45.45% H 2 O 60 30
Hydrates Compounds Containing Water Molecules MgSO 4 7H 2 O Magnesium Sulfate heptahydrate CaSO 4 2H 2 O Calcium Sulfate dihydrate Ba(OH) 2 8H 2 O Barium Hydroxide octahydrate CuSO 4 5H 2 O Copper II Sulfate pentahydrate Na 2 CO 3 10H 2 O Sodium Carbonate decahydrate 61 Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions a) BaCl 2 5 H 2 O Ba +2 is the cation Barium, Cl - is the Chloride anion. There are five water molecules therefore the name is: Barium Chloride Pentahydrate b) Magnesium Perchlorate Magnesium is the Mg +2 cation, and perchlorate is the ClO 4- anion, therefore we need two perchlorate anions for each Mg cation therefore the formula is: Mg( ClO 4 ) 2 c) (NH 4 ) 2 SO 3 NH 4+ is the ammonium ion, and SO 3-2 is the sulfite anion, therefore the name is: Ammonium Sulfite d) Calcium Nitrate Calcium is the Ca +2 cation, and nitrate is the NO 3- anion, therefore the formula is: Ca(NO 3 ) 2 62 31
Binary Compounds of Two Non-Metals Molecular compounds usually write the positive OS element first. HCl hydrogen chloride Some pairs form more than one compound mono 1 penta 5 di 2 hexa 6 tri 3 hepta 7 tetra 4 octa 8 63 Names and Formulas of Binary Molecular (Covalent) Compounds 1) The element with the lower group number in the periodic table is the first word in the name; the element with the higher group number is the second word. (Important exception: When the compound contains oxygen and a halogen, the halogen is named first.) 2) If both elements are in the same group, the one with the higher period number is named first. 3) The second element is named with its root and the suffix -ide. 4) Covalent compounds have Greek numerical prefixes to indicate the number of atoms of each element in the compound. The first word has a prefix only when more than one atom of the element is present; the second word always has a numerical prefix. 64 32
65 66 33
67 68 34
Naming Alkanes Alkanes are hydrocarbons that are called saturated hydrocarbons, they contain only single bonds, no multiple bonds! Alkanes have the general formula --- C n H 2n+2 Each carbon atom has four bonds to others atoms! The names for alkanes all end in -ane Alkanes are found in three distinct groups: a) Straight chain hydrocarbons b) Branched chain hydrocarbons c) Cyclic hydrocarbons 69 70 35
Structural Isomers Isomers have the same molecular formula but have different arrangements of atoms in space. Are the following pairs isomers? (c) H 71 The First 10 Straight-Chain Alkanes Name Formula Structural Formulas Methane CH 4 Ethane C 2 H 6 Propane C 3 H 8 Butane C 4 H 10 H H H H C H H C C H H H H CH 3 -CH 2 -CH 3 CH3 -(CH 2 ) 2 -CH 3 Pentane C 5 H 12 Hexane C 6 H 14 CH 3 -(CH 2 ) 3 -CH 3 CH 3 -(CH 2 ) 4 -CH 3 Heptane C 7 H 16 Octane C 8 H 18 CH 3 -(CH 2 ) 5 -CH 3 CH 3 -(CH 2 ) 6 -CH 3 Nonane C 9 H 20 Decane C 10 H 22 Table 2.7 CH 3 -(CH 2 ) 7 -CH 3 CH 3 -(CH 2 ) 8 -CH 3 72 36
Functional Groups Alcohols 73 Functional Groups Carboxylic Acid 74 37
Two Compounds with Molecular Formula C 2 H 6 O Property Ethanol Dimethyl Ether M (g/mol) 46.07 46.07 Color Colorless Colorless Melting point - 117 o C - 138.5 o C Boiling point 78.5 o C - 25 o C Density (at 20 o C) 0.789 g/ml 0.00195 g/ml Use Intoxicant in In refrigeration H H alcoholic beverages H H H C C O H H C O C H Table 3.4 H H H H 75 76 38
77 78 39