Fondamenti di Chimica & Educazione Ambientale Lecture 3

Transcription

1 & Educazione Ambientale Lecture 3

2

3 Compounds and Molecules Pure substances composed of two or more different elements are called compounds. Compounds have definite percent composition of their combining elements.

4 Molecular compounds

5 Standard color scheme

6 Some molecules H 2 O 2 CH 3 CH 2 Cl P 4 O 10 CH 3 CH(OH)CH 3 HCO 2 H

7 Ionic compounds z Atoms of almost all elements can gain or lose electrons to form charged species called ions. z Compounds composed of ions are known as ionic compounds. u Metals tend to lose electrons to form positively charged ions called cations. u Non-metals tend to gain electrons to form negatively charged ions called anions.

8 Sodium chloride Extended array of Na + and Cl - ions

9 Simplest formula unit is NaCl

10 Inorganic molecules S 8 P 4

11 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 = ( ) ( ) = g/mol

12 Calculating the Mass Percent Composition of a Compound Calculate the molecular mass M(C 2 HBrClF 3 ) = g/mol For one mole of compound, formulate the mass ratio and convert to percent: %C = ( )g g 100% = 12.17%

13 Calculating the Mass Percent Composition of a Compound %C = ( )g g %H = 1.01g g %Br = 79.90g g %Cl = 35.45g g %F = ( )g g 100% = 12.17% 100% = 0.51% 100% = 40.48% 100% = 17.96% 100% = 28.88%

14 Empirical formula 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.

15 Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?

16 Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Step 1: Determine the mass of each element in a 100g sample. C g H 9.63 g O g

17 Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Step 2: Convert masses to amounts in moles. n C = 62.58gC Step 3: Write a tentative formula. 1molC gC = 5.210molC n H = 9.63gH 1molH 1.008gH = 9.55molH n O = 27.79gO 1molO gO = 1.737molO C 5.21 H 9.55 O 1.74 Step 4: Convert to small whole numbers. C 2.99 H 5.49 O

18 Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Step 5: Convert to a small whole number ratio. Multiply 2 to get C 5.98 H O 2 The empirical formula is C 6 H 11 O 2 Step 6: Determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. The molecular formula is C 12 H 22 O 4

19 Combustion analysis

20 Oxidation States Metals tend to lose electrons. Na D Na + + e - Non-metals tend to gain electrons. Cl + e - D 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.

21 Rules for Oxidation States 1. The oxidation state (OS) of an individual atom in a free element is 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 5. In compounds, the OS of hydrogen is usually In compounds, the OS of oxygen is usually 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.

22 Anions (- Q):Groups 15, 16, 17 (VA, VIA, VIIA) Cations (+ Q):Groups 1, 2, 3 (IA, IIA, IIIA)

23 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.

24 Periodic Trends In Oxidation States

25 Naming 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

26 Inorganic Nomenclature Binary Compounds of Metals and Nonmetals NaCl = sodium chloride electrically neutral name is unchanged ide ending MgI 2 = magnesium iodide Al 2 O 3 = aluminum oxide Na 2 S = sodium sulfide

27

28

29

30 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

31

32 Naming Binary Compounds of the Non-metals Named in order of increasing group number N 2 O 4 SF 6 Dinitrogen tetroxide Sulfur hexafluoride N 2 O Dinitrogen oxide

33 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 HF hydrogen chloride hydrochloric acid hydrogen fluoride hydrofluoric acid

34 Polyatomic Ions Polyatomic ions are very common. Here are a few: ammonium ion NH + 4 acetate ion C 2 H 3 O - 2 carbonate ion CO 2-3 hydrogen carbonate HCO - 3 hypochlorite ClO - phosphate PO 3-4 chlorite ClO - 2 hydrogen phosphate HPO 2-4 chlorate ClO - 3 sulfate SO 2-4 perchlorate ClO - 4 hydrogensulfate HSO - 4

35

36 Ionic Bonding: Gaining (anion) and Losing (cation) Electrons to attain Inert gas electronic structure

37 When an atom gains electrons it becomes an anion. Anion are larger than the neutral atom. When an atom loses electrons it becomes a cation. Cations are smaller than the neutral atom.

38 The attractive force between + and - charges is called a coulombic attraction Force of attraction = K Q 1 Q 2 d 2

39 Predict the ionic bond strength of the following ionic bonds.

40 Consider a cation and an anion NaCl Na Cl MgCl 2 Mg Cl Cl

41 How to write formulas for ionic compounds 1. Identify ions 2. Cross charges 3. Write formula 1 2 Charge is Conserved: The + = -!

42 Naming Ionic Compounds Cations are named as the metal + ion Na + = sodium ion Multicharged ions Stock System Metal (Charge in Roman Numerals) Copper(I) ion ion Cu 1+ Cu 2+ Copper(II) ion

43 Naming Ionic Compounds MgBr 2 Fe 2 O 3 Magnesium bromide Iron(III) oxide Ca 3 (PO 4 ) 2 Calcium phosphate

44 More Examples Mg(NO 3 ) 2 FeO CuBr 2 NaHCO 3

45 Compounds and Molecules Some compounds are composed of molecules.

46 E Covalent Bonding: Sharing electrons to attain inert gas electron configurations. d

47 Naming Organic Compounds Organic compounds abound in nature Fats, carbohydrates and proteins are foods. Propane, gasoline, kerosene, oil. Drugs and plastics Carbon atoms form chains and rings and act as the framework of molecules.

48

49 Molecular Structure The arrangement of atoms in a molecule Isomers: Same number of constituent atoms BUT different arrangements.

50 Visualizations of some hydrocarbons

51 Functional Groups Carboxylic Acid

52 Functional Groups Alcohol

53 Positional Structural Chain Functional Isomerism Geometrical Stereo Optical

54 Chain Isomers C 5 H 12 n-pentane 2-methylbutane 2,2-dimethylpropane

55 Positional Isomers C 3 H 7 OH Propan-1-olo Propan-2-olo

56 Positional Isomers C 7 H 7 OH 2-Methylphenol 3-Methylphenol 4-Methylphenol

57 Functional Isomers C 2 H 6 O Ethanol Methoxymethane Alcohols have the hydroxyl group, OH Ethers have the functional group R O R'

58 Functional Isomers C 3 H 6 O Propanal Propanone Aldehydes and ketones have the carbonyl group C=O. In aldehydes C is attached either to 1 or 2 H atoms; in ketones C is attached to two carbon atoms.

59 Positional Structural Chain Functional Isomerism Geometrical Stereo Optical

60 Geometrical Isomers Geometrical isomerism is an example of stereoisomerism. This occurs when substances have the same molecular formula, but a different arrangement of their atoms in space. Geometrical isomerism can happen if: i) There is a C=C bond in the molecule ii) The molecule has a ring iii) The molecule has a >C=N bond

61 Geometrical Isomers C 4 H 8 Cis-But-2-ene Trans-But-2-ene

62 Geometrical Isomers trans- Pt(NH 3 ) 2 Cl 2 cis-pt(nh 3 ) 2 Cl 2

63 Geometrical Isomers Pt(NH 3 )Cl 3 No isomers

64 Geometrical Isomers cis-pt(nh 3 )Cl 2 Br trans-pt(nh 3 )Cl 2 Br Two isomers

65 Geometrical Isomers N Cl N Cl Pt Cl Cl N Cl Pt Cl N Cl cis Two isomers Cl trans

66 Geometrical Isomers Cl N Cl N Cl N Pt Pt N N Cl N Cl Cl mer Two isomers fac

67 Optical Isomers Optical isomerism is an example of stereo-isomerism. It occurs when substances have the same molecular and structural formula, but one cannot be superimposed on the other. The different forms are called enantioners. optical isomerism non-superimposable mirror images

68 Optical Isomers Optical isomers can occur when there is an asymmetric carbon atom; i.e. a C atom bonded to four different groups.

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93 Lecture 3 End