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1 Molecules and Compounds Read Chapter 3. Study all examples and complete all exercises. Complete all bold numbered problems.

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Presentation on theme: "1 Molecules and Compounds Read Chapter 3. Study all examples and complete all exercises. Complete all bold numbered problems."— Presentation transcript:

1 1 Molecules and Compounds Read Chapter 3. Study all examples and complete all exercises. Complete all bold numbered problems.

2 2 Chapter 3 Outline Molecular FormulaMolecular Formula Molar MassMolar Mass Empirical and Molecular FormulaEmpirical and Molecular Formula NomenclatureNomenclature

3 3 Compounds & Molecules NaCl, salt Buckyball, C 60

4 4 Compounds & Molecules COMPOUNDSCOMPOUNDS are a combination of 2 or more elements in definite ratios by mass. The character of each element is lost when forming a compound. MOLECULES smallest unit of a compound that retains the characteristics of the compound.MOLECULES are the smallest unit of a compound that retains the characteristics of the compound.

5 5 MOLECULAR FORMULAS Formula for glycine is C 2 H 5 NO 2Formula for glycine is C 2 H 5 NO 2 In one molecule there areIn one molecule there are –2 C atoms –5 H atoms –1 N atom –2 O atoms

6 6 WRITING FORMULAS FormulaFormula HOCH 2 CH 2 OH to show atom ordering or in the form of a structural formulaor in the form of a structural formula

7 7 Molecular Modeling

8 8 Ball & stick Space-filling Drawing of glycine CHHCHHOOHN

9 9 Resources for Molecular Modeling Oxford Molecular/CAChe Scientific software on Saunders General Chemistry CD-ROMOxford Molecular/CAChe Scientific software on Saunders General Chemistry CD-ROM Rasmol and Chime on the InternetRasmol and Chime on the Internet See http://www.saundercollege.comSee http://www.saundercollege.com

10 10 ELEMENTS THAT EXIST AS MOLECULES Allotrope of C Buckyball C 60 Buckyball, C 60

11 11

12 12 IONS AND IONIC COMPOUNDS IONS are atoms or groups of atoms with a positive or negative charge.IONS are atoms or groups of atoms with a positive or negative charge. Taking away an electron from an atom gives a CATION with a positive charge.Taking away an electron from an atom gives a CATION with a positive charge. Adding an electron to an atom gives an ANION with a negative charge.Adding an electron to an atom gives an ANION with a negative charge.

13 13 Formation of Cations & Anions

14 14 Formation of Cations & Anions A cation forms when an atom loses one or more electrons. An anion forms when an atom gains one or more electrons Mg --> Mg 2+ + 2 e- F + e - --> F -

15 15 PREDICTING ION CHARGES In general metals (Mg) lose electrons ---> cationsmetals (Mg) lose electrons ---> cations nonmetals (F) gain electrons ---> anionsnonmetals (F) gain electrons ---> anions Li 1+ F 1- LiF MONATOMIC IONS

16 16 Figure 3.7 Sn 4+ Pb 4+ Also

17 17 METALS M ---> n e- + M n+ where n = periodic group Na + Mg 2+ Al 3+ Transition metals --> M 2+ or M 3+ are most common

18 18 NONMETALS NONMETAL + n e- ------> X n- where n = 8 - Group number C 4- carbide N 3- nitride O 2- oxide F - fluoride Bromine

19 19 Groups of atoms with a charge. (See back of periodic chart) POLYATOMIC IONS

20 20 Some Common Polyatomic Ions HNO 3 nitric acid NO 3 - nitrate ion

21 21 Some Common Polyatomic Ions NH4+ NH4+ NH4+ NH4+ ammonium ion ammonium ion One of the few common polyatomic cations One of the few common polyatomic cations

22 22 Some Common Polyatomic Ions CO 3 2- carbonate ion HCO 3 - bicarbonate ion hydrogen carbonate

23 23 Some Common Polyatomic Ions PO 4 3- phosphate ion CH 3 CO 2 - acetate ion

24 24 SO 4 2- Sulfate ion SO 3 2- SO 3 2- Sulfite ion Some Common Polyatomic Ions

25 25 NO 3 - NO 3 - Nitrate ion NO 2 - Nitrite ion Some Common Polyatomic Ions

26 26 COMPOUNDS FORMED FROM IONS CATION + ANION COMPOUND CATION + ANION COMPOUND Na + + Cl - NaCl Na + + Cl - NaCl A neutral compound requires equal number of + and - charges.

27 27 IONIC COMPOUNDS NH4+NH4+NH4+NH4+ Cl - ammonium chloride, NH 4 Cl

28 28 Some Ionic Compounds Ca 2+ + 2 F - CaF 2 Calcium fluoride Mg 2+ + 2 NO 3 - Mg(NO 3 ) 2 Magnesium nitrate 3 Fe 2+ + 2 PO 4 3- Fe 3 (PO 4 ) 2 Iron(II) phosphate Calcium Fluoride

29 29 Sample Questions Predict the charges for the ions formed from: SeGa PSr Give the formula for each ion in Al 2 (SO 4 ) 3 Give the formula for the ionic compound that forms between Na and S Ga and O Ba and N Answers

30 30 Properties of Ionic Compounds Forming NaCl from Na and Cl 2 A metal atom can transfer an electron to a nonmetal. electrostatic forces.The resulting cation and anion are attracted to each other by electrostatic forces. 30

31 31 Electrostatic Forces ELECTROSTATIC FORCES. The oppositely charged ions in ionic compounds are attracted to one another by ELECTROSTATIC FORCES. These forces are governed by COULOMB’S LAW.

32 32 Electrostatic Forces COULOMB’S LAW As ion charge increases, the attractive force _______________. As the distance between ions increases, the attractive force ________________. This idea is important and will come up many times in future discussions! Force of attraction == (charge on + )(charge on - ) (distance between ions) 2

33 33 Importance of Coulomb’s Law NaCl, Na + and Cl -, m.p. 801 o C MgO, Mg 2+ and O 2- m.p. 2800 o C AlN, Al 3+ and N 3- m.p. 2900 o C

34 34 Names of Compounds Rules for nomenclature are found in section 3.5Rules for nomenclature are found in section 3.5 STUDY them carefully!!STUDY them carefully!! We will be studying nomenclature in the laboratory in Experiment MA.We will be studying nomenclature in the laboratory in Experiment MA.

35 35 Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO.Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg?How can we figure out how much oxide is produced from a given mass of Mg?

36 36 Counting Atoms Chemistry is a quantitative science — we need a “counting unit.” The MOLEThe MOLE 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12 C.

37 37 Particles in a Mole 6.02 x 10 23 Avogadro’s Number Amedeo Avogadro 1776-1856 There is Avogadro’s number of particles in a mole of any substance.

38 38 Mole in Chemistry is NOT: An informer / spyAn informer / spy Dark spot on Cindy Crawford’s upper lipDark spot on Cindy Crawford’s upper lip Rodent that burrows in the groundRodent that burrows in the ground A tunneling machineA tunneling machine Wave breakWave break Spicy Mexican sauceSpicy Mexican sauce An informer / spyAn informer / spy Dark spot on Cindy Crawford’s upper lipDark spot on Cindy Crawford’s upper lip Rodent that burrows in the groundRodent that burrows in the ground A tunneling machineA tunneling machine Wave breakWave break Spicy Mexican sauceSpicy Mexican sauce

39 39 A mole is a convenient measuring tool. PairDozenReam Baseballs2 baseballs12 baseballs13 baseballs 500 baseballs Pineapples 2 Pineapples 12 Pineapples 13 Pineapples 500 Pineapples Calculators 2 Calculators 12 Calculators 13 Calculators 500 Calculators Planets2 Planets12 Planets13 Planets500 Planets Baker’s Dozen

40 40 Mole = mol = 6.022  10 23 “particles” Mole Baseballs 6.022  10 23 baseballs Pineapples 6.022  10 23 Pineapples Calculators 6.022  10 23 Calculators Planets 6.022  10 23 Planets

41 41 Just as 1 doz eggs = 12 eggs or 1 mol of eggs = 6.022  10 23 eggs A mole is a “Number” 1 mol of H = 6.022  10 23 atoms of H 1 mol of O = 6.022  10 23 atoms of O 1 mol of Al = 6.022  10 23 atoms of Al 1 mol of Cr = 6.022  10 23 atoms of Cr

42 42 In chemistry, the mol is more then just a number 1 mol  amu = 1.00 g

43 43 1 mol  amu = 1.00 g Proof: 1 mol = 6.022  10 23 amu = 1.661  10 -24 g 1 mol amu = (6.022  10 23 )  (1.661  10 -24 g) = 1.00 g

44 44 Molar Mass 1 mol of 12 C = 12.00 g of C = 6.02 x 10 23 atoms of C 12.00 g of 12 C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol

45 45 Molar Mass 1 mol of 12 C = 12.00 g of C = 6.02 x 10 23 atoms of C 12.00 g of 12 C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol Find molar mass from periodic table 13 Al 26.9815 atomic number symbol atomic weight

46 46 PROBLEM: How many moles are represented by 0.200 g of Mg? How many atoms in this piece of Mg? 0.200 g = 1 mole 24.3 g 0.00823 mole = 6.02 x 10 23 atom 1 mole 4.95 x 10 21 atom

47 47 MOLECULAR WEIGHT AND MOLAR MASS Molecular weight Molecular weight is the sum of the atomic weights of all atoms in the molecule. Molar mass Molar mass = molecular weight in grams

48 48 What is the molar mass of ethanol, C 2 H 6 O? 1 mol contains 2 mol C (12.0 g C/1 mol) = 24.0 g C 6 mol H (1.0 g H/1 mol) = 6.0 g H 1 mol O (16.0 g O/1 mol) = 16.0 g O TOTAL = molar mass = 46.0 g/mol

49 49 Formula = C 8 H 9 NO 2 Molar mass = 151.0 g/mol Tylenol

50 50 How many moles of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? 21.3 g = 1 mole 46.0 g 0.463 mole

51 51 How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? 0.463 mole = 6.02 x 10 23 molecule 1 mole 2.79 x 10 23 molecule

52 52 How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? 2.79 x 10 23 molecule = 2 atom C 1 molecule 5.58 x 10 23 atom C Sample problems

53 53 Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C 2 H 6 O 52.2% C, 13% H, 34.8% O

54 54 Percent Composition Consider some of the family of nitrogen-oxygen compounds: NO 2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) Structure of NO 2 Chemistry of NO, nitrogen monoxide (nitric oxide)

55 55 Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? Wt.% O = 2 (16.0 g O per mole) 46.0 g x100% = 69.6% Wt. % N= 14.0 g N 46.0 g NO 2 100%= 30.4 %

56 56 Percent Composition What are the weight percentages of N and O in NO? Wt.% O = 16.0 g O 30.0 g NO 100% = 53.3% Wt. % N= 14.0 g N 30.0 g NO 100%= 46.7 %

57 57 Percent Composition Samples Problems 1. Calculate the percent composition of H 2 O. Wt.% O = 16.0 g O 18.0 g H 2 O 100% = 88.9% Wt. % H= 2.0 g H 18.0 g H 2 O 100%= 11 %

58 58 Percent Composition Samples Problems 2. Calculate the percent O in NaOH. Wt.% O  16.0 g O 40.0 g NaOH 100%  40.0%

59 59 Percent Composition Samples Problems 3. Calculate the percent O and the percent water in CuSO 4. 5H 2 O. Wt.% O = 144.0 g O 249.6 g CuSO 4 5H 2 O 100% = 57.69 % Wt.% H 2 O = 90.0 g H 2 O 249.6 g CuSO 4. 5H 2 O 100% = 36.1%

60 60 Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.

61 61 A compound of B and H is 81.10% B. What is its empirical formula? 81.10 g 1 mole 10.8 g 18.90 g 1 mole 1.0 g B H B H 1.00 mole B 2.5 mole H 7.51mole B 19 mole H 7.51 mole B 7.51 mole B 2.00 mole B 5.0 mole H B2H5B2H5

62 62 A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? Is the molecular formula B 2 H 5, B 4 H 10, B 6 H 15, B 8 H 20, etc.? B 2 H 6 is one example of this class of compounds. B2H6B2H6B2H6B2H6

63 63 A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula? EXPERIMENT We need to do an EXPERIMENT to find the MOLAR MASS. Here experiment gives 53.3 g/mol. Compare with the mass of B 2 H 5, 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

64 64 Determine the formula of a compound of Sn and I using the following data. Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 g Mass of iodine (I 2 ) used = 1.947 g Mass of Sn remaining = 0.601 g

65 65 Find the mass of Sn that combined with 1.947 g I 2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Tin and Iodine Compound

66 66 0.455 g 1 mole 118.7 g 1.947 g 1 mole 126.9 g Sn I Sn I 1.00 mole Sn 4.01 mole I 0.0383 mole Sn 0.1534 mole I 0.0383 mole Sn SnI 4 Tin and Iodine Compound

67 67 More Problems 2. Calculate the formula for the iron sulfide that forms when 53.73g Fe react with 46.27 g of sulfur. 53.73 g 1 mole 55.8 g 46.27 g 1 mole 32.1 g Fe S Fe S 1.00 1.50 0.963 1.44 0.963 0.963 Fe 2 S 3 2.00 3.00

68 68 More Problems Calculate the empirical formula a compound containing 90.7% Pb and 9.33% O. 90.7 g 1 mole 207.2 g 9.33 g 1 mole 16.0 g Pb O Pb O 1.00 1.33 0.438 0.583 0.438 0.438 Pb 3 O 4 3.00 3.99

69 69 More Problems Calculate the empirical formula for a compound containing 36.5% Na, 25.4% S and 38.1% O. 36.5 g 1 mole 23.0 g 25.4 g 1 mole 32.1 g Na S O Na S O 2.01 1.00 3.01 Na 2 SO 3 38.1 g 1 mole 16.0 g 1.59 0.791 2.38 0.791 0.791 0.791

70 70 More Problems Calculate the empirical and molecular formulas for nicotine, 74.0% C, 8.7% H and 17.3% N, with a molar mass of 160 g/mole. 74.0 g 1 mole 12.0 g 8.7 g 1 mole 1.0 g C H N 4.98 7.0 1.00 17.3 g 1 mole 14.0 g 6.17 8.7 1.24 1.24 1.24 1.24 Empirical formula C 5 H 7 N

71 71 More Problems 7. Calculate the empirical and molecular formulas for nicotine, 74.0% C, 8.7% H and 17.3% N, with a molar mass of 160 g/mole. Empirical formula C 5 H 7 N Molecular formula C 10 H 14 N 2 160 = 2 81

72 72 Practice Problems Names/Formulas FeO Pb(C 2 H 3 O 2 ) 2 magnesium bromide sodium chromate calcium phosphate ammonium carbonate

73 73 Practice Problems ( NH 4 ) 2 S As 2 O 3 SO 2 silicon disulfide As 2 S 5 dinitrogen monoxide

74 74 Practice Problems As 2 S 3 dinitrogen pentoxide silicon tetrabromide diphosphorus pentoxide HBrO 3 H 3 PO 4 (aq) H 2 CO 3

75 75 Practice Problems Calculations 1. 26 g H 2 is how many moles H 2 ? 2. 4.25 x 10 21 molecules NH 3 is how many grams NH 3 ? 3. 1.5 x 10 2 formula units KClO 3 is how many moles KClO 3 4. 0.0042 mole Fe 2 O 3 is how many formula units Fe 2 O 3 ? 5. 2.15 moles MgSO 4. 7H 2 O is how many g MgSO 4. 7H 2 O?

76 76 Practice Problems 6. 25 molecules HBr is how many mole HBr? 7..00002 g Sn is how many atoms Sn? 8. 7.25 mole H 2 S is how many g H 2 S? 9. 5.2 g Sr(OH) 2 is how many formula units Sr(OH) 2 ? 10. How many moles of CCl 4 will contain 2.4 g of chlorine?

77 77 Practice Problems 11. 19 g of HNO 3 contains how many a) molecules of HNO 3 ? b) grams of O? 12. Calculate the percent composition of NaCl. 13. Calculate the empirical and molecular formulas for a compound containing 43.7g P and 56.3g O, with a molar mass of 140 g/mole.

78 78 Practice Problems Answers Iron(II) oxide, ferrous oxide lead(II) acetate, plumbous acetate MgBr 2 Na 2 CrO 4 Ca 3 (PO 4 ) 2 (NH 4 ) 2 CO 3 ammonium sulfide diarsenic trioxide sulfur dioxide SiS 2 As 2 S 5 N 2 O diarsenic trisulfide N 2 O 5 SiBr 4 P 2 O 5 bromic acid phosphoric acid carbonic acid

79 79 Practice Problems Answers 1. 13 mole2. 0.120 g 3. 2.5 x 10 -22 mole4. 2.5 x 10 21 atom 5. 530. g6. 4.2 x 10 -23 mole 7. 1 x 10 17 atom8. 247 g 9. 2.6 x 10 22 formula units 10. 0.017 mole 11. a) 1.8 x 10 23 molecule b) 14 g 12. 39.3%, 60.7%13. C 5 H 7 N 14. P 2 O 5

80 80 Sample Questions Predict the charges for the ions formed from: Se -2Ga +3 Al 3+ P -3Sr +2 SO 4 2- Give the formula for each ion in Al 2 (SO 4 ) 3 Give the formula for the ionic compound that forms between Na and S Na 2 S Ga and O Ga 2 O 3 Ba and N Ba 3 N 2

81 81 Sample Problems 1. 2.5 mole S = ? atom S 2.5 mole = 6.02 x 10 23 atom 1 mole 1.5 x 10 24 atom S

82 82 Sample Problems 2. 2.1 mole Zn = ? g Zn 2.1 mole = 65.4 g 1 mole 140 g Zn 140 g Zn

83 83 Sample Problems 3. 1.42 g Mg = ? atom Mg 1.42 g = 1 mole 24.3 g 3.55 x 10 22 atom 6.02 x 10 23 atom 1 mole

84 84 Sample Problems 4. 125.2 g O 2 = ? mole O 2 125.2 g = 1 mole 32.0 g 3.91 mole O 2

85 85 Sample Problems 1. 1.5 mole H 2 O = ? g H 2 O 1.5 mole = 18.0 g 1 mole 27 g H 2 0 27 g H 2 0

86 86 Sample Problems 2. 1.50 mole CCl 4 = ? molecules CCl 4 1.5 mole = 6.02 x 10 23 molecule 1 mole 9.03 x 10 23 molecule

87 87 Sample Problems 3. 1.25 mole CaCl 2 = ? formula units CaCl 2 1.25 mole = 6.02 x 10 23 formula unit 1 mole 7.52 x 10 23 formula unit

88 88 Sample Problems 4. 2.5 g K = ? g KOH 2.5 g K = 1 mole K 39.1 g K 3.6 g KOH 3.6 g KOH 1 mole KOH 1 mole K 56.1 g KOH 1 mole KOH

89 89 Sample Problems 5. 34.5 g CaCO 3 = ? g O 34.5gCaCO 3 = 1 moleCaCO 3 100.1gCaCO 3 16.5 g O 16.5 g O 3 moleO 1 moleCaCO 3 16.0 gO 1 moleO


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