1. Unit 3 (Chp 1,2,3): Matter, Measurement, & Stoichiometry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Chemistry: The study of matter and the changes it undergoes. Quantitative or Qualitative Ni +HCl nickel hydrochloric acid NiCl 2 nickel(II) chloride H 2 + hydrogen solid aqueous gas solid crystals metal solution

3. Matter Atom: Element: Compound: H H CC CC OO Na simplest particle with properties of element same type of atom (1 or more) different atoms bonded H 2 OCO 2 NaCl H2H2 O2O2 C molecule

4. salt, baking soda, water, sugar oxygen, iron, hydrogen, gold Matter Mixture Pure Substance ElementsCompounds Heterogeneous Mixture Homogeneous Mixture separate physically cannot separate physically differences or unevenly mixed uniform or evenly mixed separate chemically cannot separate Chemical changes Physical changes (solutions) filtering distillation (boiling) (suspensions/colloids) NaCl NaHCO 3 H 2 O C 12 H 22 O 11 O 2 Fe H 2 Au chromatography

5. Physical Separation: Separates heterogeneous mixtures (solids from liquids). Filtration:

6. Physical Separation: Distillation: Separates solution by boiling point differences.

7. Physical Separation: Chromatography: Separates solution by differences in solubility (attractions).

8. Metric Prefixes BASE UNIT:1 m 1 L 1 g 0.01 m 0.001 L 0.000 001 g 0.000 000 001 m PrefixSymbolMultiplier Examples: 1,000,000,000 B 1,000,000 J 1,000 g (atoms) (light wavelength) (nuclei)

9. Precision in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty) % Error = |Experimental – Accepted| x100 Accepted

10. Significant Digits measured digits (using marks on instrument) last estimated digit (one digit past marks) do not overstate the precision 5.23 cm 5.230 cm

11. Significant Zeroes 1.All nonzero digits are significant. 2.Captive Zeroes between two significant figures are significant. 3.Leading Zeroes at the beginning of a number are never significant. 4.Trailing Zeroes: Sig, if there’s a decimal point. NOT, if there is no decimal point. 0.0003700400 grams 0’s

12. Power of 10 is the number of places the decimal has been moved. Examples:42000 = 4.2 x 10 4 0.0508 = 5.08 x 10 –2 positive power: move decimal right  to obtain the original # in standard notation. negative power: move decimal left  to obtain the original # in standard notation. Scientific Notation

13. 1. Convert the numbers to scientific notation. (i) 24500 (ii) 0.000985 (iii) 12002 2. Convert to standard notation. (i) 4.2 x 10 5 (ii) 2.15 x 10 -4 (iii) 3 x 10 -3 2.45 x 10 4 9.85 x 10 –4 1.2002 x 10 4 420,000 0.000215 0.003 Scientific Notation

14. round answers to keep the fewest decimal places round answers to keep the fewest significant digits Sigs Digs in Operations + or – x or ÷ 3.48 + 2.2 = 6.40 x 2.0 = 5.68 5.7 12.8 13

15. 1. How many sig digs are in each number? (i) 250.0 (ii) 4.7 x 10 –5 (iii) 34000000 (iv) 0.03400 2. Round the answer to the correct sig digs. (i) 34.5 x 23.46 (ii) 123/3 (iii) 23.888897 + 11.2 (iv) 2.50 x 2.0 – 3 Sig Digs Practice WS 1s 4 2 2 4 809 40 35.1 2

16. WARM UP (for QUIZ!!!) Review WS 1s #1, 3, 10 Complete WS 1a #1, 2, 8, 9, 10

17. Law of Definite Proportions  elemental formulas (composition) of pure compounds cannot vary. 2 H’s & 1 O is ALWAYS water. Water is ALWAYS 2 H’s & 1 O. 2 H’s & 2 O’s is NOT water. √ H2O√ H2O X H 2 O 2 H O H HH O O

18. Law of Conservation of Mass The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place. __H 2 + __O 2 __H 2 O 2 2 Balancing Equations!!!

19. Symbols of Elements 12 6 C Mass Number = p’s + n’s Atomic Number (Z) = p’s Element Symbol All atoms of the same element have the same number of protons (same Z), but… can have different mass numbers. HOW?

20. 1111 H 2121 H 3131 H protium deuteriumtritium element: mass: why? same or different same # of protons (& electrons), but different # of neutrons Isotopes

21. Average Atomic Mass average atomic mass: calculated as a weighted average of isotopes by their relative abundances. (6.015)(0.0750) + (7.016)(0.925) = 6.94 amu Avg. Mass = (Mass 1 )(%) + (Mass 2 )(%) … lithium-6 (6.015 amu), which has a relative abundance of 7.50%, and lithium-7 (7.016 amu), which has a relative abundance of 92.5%.

22. WS Atomic Structure element sample atomized, ionized magnetic field ~75% ~25% isotopes separated by difference in mass (35)(~0.75) + (37)(~0.25) = ? Cl (avg at. Mass) = Mass Spectrometry

23. Molecular (Covalent) Compounds Covalent compounds contain nonmetals that “share” electrons to form molecules. (molecular compounds)

24. Diatomic Molecules These seven elements occur naturally as molecules containing two atoms. “H-air-ogens” 7

25. Binary Molecular Compounds list less electronegative atom first. (left to right on PT) use prefix for the number of atoms of each element. change ending to –ide. CO 2 : carbon dioxide CCl 4 : carbon tetrachloride N 2 O 5 : ________________ CuSO 4 ∙ 5H 2 O copper(II) sulfate pentahydrate dinitrogen pentoxide (ionic & covalent)

26. Ions Cations metals lose e’s (+) charge (metal) ion Anions nonmetals gain e’s (–) charge (nonmetal)ide

27. Ionic Bonds Attraction between +/– ions formed by metals & nonmetals transferring e – ’s.

28. Formulas of Ionic Compounds Compounds are electrically neutral, so the formulas can be determined by: –Crisscross the charges as subscripts (then erase) –If needed, reduce to lowest whole number ratio. Pb 4+ O 2– Pb 2 O 4 PbO 2

29. Naming Ionic Compounds 1)Cation: Write metal name (ammonium NH 4 + ) For transition metals with multiple charges, write charge as Roman numeral in parentheses. Iron(II) chloride, FeCl 2 Iron(III) chloride, FeCl 3 2)Anion: Write nonmetal name with –ide OR the polyatomic anion name. (–ate, –ite) Iron(II) sulfide, FeS Magnesium sulfate, MgSO 4

30. Common Polyatomic Ions NameSymbolCharge *ammoniumNH 4 + 1+ *acetate (ethanoate) C 2 H 3 O 2 – (CH 3 COO – ) 1– *hydroxideOH – 1– *perchlorateClO 4 – 1– *chlorateClO 3 – 1– chloriteClO 2 – 1– hypochloriteClO – 1– bromateBrO 3 – 1– iodateIO 3 – 1– *nitrateNO 3 – 1– nitriteNO 2 – 1– cyanideCN – 1– *permanganateMnO 4 – 1– *bicarbonate (hydrogen carbonate) HCO 3 – 1– *carbonateCO 3 2– 2– *sulfateSO 4 2– 2– sulfiteSO 3 2– 2– *chromateCrO 4 2– 2– dichromateCr 2 O 7 2– 2– *phosphatePO 4 3– 3– * these 12 will be on Quiz 1 - all 20 Polyatomic Ions will be on Quiz 2 WS 2d

31. perchlorateClO 4 – chlorateClO 3 – chloriteClO 2 – hypochloriteClO – CNOF SiPSCl AsSeBr TeI nitrateNO 3 – nitriteNO 2 – InOutIon Name – – “Oxyanion” Names (elbO’s) 4 3 2 1 4 3 ___-ate ___-ite sulfateSO 4 2– sulfiteSO 3 2– phosphatePO 4 3– per-___-ate hypo-___-ite

32. nitrateNO 3 – nitriteNO 2 – sulfateSO 4 2– sulfiteSO 3 2– perchlorateClO 4 – chlorateClO 3 – chloriteClO 2 – hypochloriteClO – Ion Acid add H + Name Acids from these oxyanions: InOutIon NameAcid Name 4– per-___-ate 34 ___-ate 23 ___-ite 1– hypo-___-ite per-___-ic acid ___-ic acid ___-ous acid hypo-___-ous acid Naming Acids WS 2e

33. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

34. Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

35. Anatomy of a Chemical Equation Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

36. Anatomy of a Chemical Equation States ( s, l, g, aq ) written in parentheses next to each compound CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

37. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Subscripts show how many atoms of each element

38. Anatomy of a Chemical Equation Coefficients show the amount of each particle and are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

39. Reaction Types

40. Combination 2 Mg (s) + O 2 (g)  2 MgO (s) Demo: MgO 2 → 1 A + B → AB

41. 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g) Decomposition 1 → 2 AB → A + B (50 milliseconds!)

42. Replacement Reactions (or “Displacement”) Single Replacement AB + C → A + CB Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) Double Replacement AB + CD → AD + CB video

43. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) Often involve hydrocarbons reacting with oxygen in the air WS 4a C x H y + _O 2  _CO 2 + _H 2 O Combustion C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g)

44. Formula Weights

45. Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula Formula Weight of calcium chloride, CaCl 2, is… Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu Sum of the atomic weights for the atoms in a molecule or compound Molecular Weight of ethane, C 2 H 6, is… Molecular Weight (MW) C: 2(12.01 amu) + H: 6(1.008 amu) 30.07 amu

46. Percent Composition One can find the percent by mass of a compound of each element in the compound by using this equation. % element = (# of atoms)(AW) (FW) x 100

47. Percent Composition So the percentage of carbon in ethane (C 2 H 6 ) is… %C = (2)(12.01) (30.07) 24.02 30.07 = x 100 = 79.88% C

48. Moles

49. Avogadro Constant One mole of particles contains the Avogadro constant of those particles 6.022 x 10 23

50. Mole Relationships One mole of atoms, ions, or molecules contains the Avogadro constant of those particles 6.022 x 10 23 In 1 mol Na 2 CO 3, how many… Na atoms? C atoms? O atoms? How many donuts in 1 mol of donuts? How many boogers in 1 mol of boogers? Which has more atoms, 1 mol CH 3 or 1 mol NH 3 ? How about CH 3 CH 2 OH or H 2 SO 4 ?

51. Molar Mass the mass of 1 mol of a substance (g/mol) –molar mass (in g/mol) of an element is the atomic mass (in amu) on the periodic table –formula weight (amu) of a compound same number as the molar mass (g/mol) of 1 mole of particles of that compound

52. Using Moles Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. Mass (grams) Particles (atoms) (molecules) (units) Moles (groups of 6.022x10 23 particles) molar mass # g 1 mol # g Avogadro constant 6.022x10 23 1 mol 6.022x10 23 macro- bridgemicro-

53. Using Moles 1.What is the mass of 1 mole of copper(II) bromide, CuBr 2 ? 2.How many moles are there in 112 g of copper(II) bromide, CuBr 2 ? 3.How many particles present in each of the questions #1 & #2 above? (63.55) + 2(79.90) = 223.35 g 112 g CuBr 2 x 1 mol CuBr 2 223.35 g CuBr 2 = 0.501 mol CuBr 2 0.501 mol x 6.022 x 10 23 particles 1 mol = 3.02 x 10 23 particles = 6.022 x 10 23 particles

54. Balanced chemical equations show the amount of: Most important are the ratios of reactants and products in moles, or… mol-to-mol ratios Stoichiometry : calculations of quantities in chemical rxns –how much reactant is consumed or –how much product is formed atoms, molecules, moles, and mass

55. g A g B mol B g A 1 mol A g B 1 mol B molar mass A molar mass B mol A Rxn: A (aq) + 2 B (aq)  C (aq) + 2 D (aq) Stoichiometric Calculations 1 mol A g A OR mol-to-mol ratio 2 mol B 1 mol A 2 mol B OR Coefficients of balanced equation ??????

56. Stoichiometric problems have 1-3 Steps: (usually) 1)Convert grams to moles (if necessary) using the molar mass (from PT) 2)Convert moles (given) to moles (wanted) using the mol ratio (from coefficients) 3)Convert moles to grams (if necessary) using the molar mass (from PT) grams A x 1 mol A. grams A = _ mol B mol A x x grams B 1 mol B 1) molar mass2) mole ratio3) molar mass

57. Example : g of A  g of B Solid magnesium is added to an aqueous solution of hydrochloric acid. What mass of H 2 gas will be produced from completely reacting 18.0 g of HCl with magnesium metal? Mg (s) + 2 HCl (aq)  MgCl 2 (aq) + H 2 (g) = ____ g H 2 18.0 g HCl x g HCl mol HCl x mol H 2 g H 2 36.46 x 2.016 1 2 1 1 0.498 g H 2 molar mass A molar mass B g of A Stoichiometric Calculations HW p. 114 #58 mole ratio B/A

58. Finding Empirical Formulas

59. Types of Formulas Empirical formulas: the lowest ratio of atoms of each element in a compound. Molecular formulas: the total number of atoms of each element in a compound. CH 3 C2H6C2H6 C2H4OC2H4O C 6 H 12 O 3 molecular mass = emp. form. empirical mass multiple

60. Percent to Mass Mass to Mole Divide by Small Times ‘til Whole Steps (rhyme) ÷ moles by smallest to get mole ratio of atoms MM from PT assume 100 g x (if necessary) to get whole numbers of atoms Calculating Empirical Formulas 75 % C75 g C6.2 mol C 25 % H25 g H24.8 mol H 1 C 4 H CH 4 from Mass % Composition

61. Butane is 17.34% H and 82.66% C by mass. Determine its empirical formula. If molecular mass is 58 g∙mol –1, what is the Molecular Formula? 82.66 g C 17.34 g H 1) Percent to Mass 2) Mass to Mole 82.66 g C x = 6.883 mol C 17.34 g H x = 17.20 mol H 1 mol C 12.01 g C 1 mol H 1.008 g H 6.883 mol 3) Divide by Small 4) Times ’til Whole = 1  1 C = 2.499  2.5 H x 2 = 2 C C2H5C2H5 x 2 = 5 H C 4 H 10 58 29.06 = 2 2 (C 2 H 5 ) = molecular mass empirical mass HW p. 113 #43a, 48

62. Percent to Mass Mass to Mole Divide by Small Times ‘til Whole Calculating Empirical Formulas

63. Hydrocarbons with C and H are analyzed through combustion with O 2 in a chamber.  g C is from the g CO 2 produced  g H is from the g H 2 O produced  g X is found by subtracting (g C + g H) from g sample Combustion Analysis Step 1 is “combustion to mass”

64. When 4-ketopentenoic acid is analyzed by combustion, a 0.3000 g sample produces 0.579 g of CO 2 and 0.142 g of H 2 O. The acid contains only C, H, and O. What is the empirical formula of the acid? Combustion Analysis Example 1

65. 0.579 g CO 2 x 1 mol CO 2 44.01 g CO 2 1 mol H 2 O 18.02 g H 2 O 1 mol C 1 mol CO 2 x 12.01 g C 1 mol C x 2 mol H 1 mol H 2 O x 1.008 g H 1 mol H x 0.142 g H 2 O x = 0.158 g C = 0.0159 g H 0.3000 g sample – (0.158 g C) – (0.0159 g H) = = 0.126 g O ? g C ? g H ? g O Step 1: “combustion to mass”

66. 0.00788 mol 0.158 g C x 1 mol C 12.01 g C 1 mol H 1.008 g H 0.0132 mol C = 0.0158 mol H = 0.0159 g H x = = 1.67 C 2 H 0.126 g O x 1 mol O 16.00 g O 0.00788 mol O = =1 O x 3 = 5 C x 3 = 6 H x 3 = 3 O C5H6O3C5H6O3

67. A sample of a chlorohydrocarbon with a mass of 4.599 g, containing C, H and Cl, was combusted in excess oxygen to yield 6.274 g of CO 2 and 3.212 g of H 2 O. Calculate the empirical formula of the compound. If the compound has a MW of 193 g∙mol –1, what is the molecular formula? Example 2 Combustion Analysis

68. 6.274 g CO 2 x 1 mol CO 2 44.01 g CO 2 1 mol H 2 O 18.02 g H 2 O 1 mol C 1 mol CO 2 x 12.01 g C 1 mol C x 2 mol H 1 mol H 2 O x 1.008 g H 1 mol H x 3.212 g H 2 O x = 1.712 g C = 0.3593 g H 4.599 g sample – (1.712 g C) – (0.3593 g H) = = 2.528 g Cl ? g C ? g H ? g Cl Step 1: “combustion to mass”

69. 0.07131 mol 1.712 g C x 1 mol C 12.01 g C 1 mol H 1.008 g H 0.1425 mol C = 0.3564 mol H = 0.3593 g H x = = 2 C 5 H 2.528 g Cl x 1 mol Cl 35.45 g Cl 0.07131 mol Cl = =1 Cl C 2 H 5 Cl If the compound has a MW of 193 g∙mol –1, what is the molecular formula? MW EW 193 64.51 = 3 C 6 H 15 Cl 3 HW p. 114 #52b

70. Which ingredient will run out first? If out of sugar, you should stop making cookies. How Many Cookies Can I Make? Sugar is the limiting ingredient, because it will limit the amount of cookies you can make.

71. BeforeAfter limiting 2 H 2 + O 2 2 H 2 O Initial: ? mol ? mol ? mol Change: End: 10 7 0 0 mol 2 mol 10 mol –10 –5+10 H2H2 O2O2 Which is limiting? excess

72. BeforeAfter 2 H 2 + O 2 2 H 2 O Initial: ? mol ? mol ? mol Change: End: 10 7 0 0 mol 2 mol 10 mol –10 –5+10 H2H2 O2O2 O 2 is in smallest amount, but… H 2 is in smallest “stoichiometric” amount Does limiting mean smallest amount of reactant?No!

73. Solid aluminum metal is reacted with aqueous copper(II) chloride in solution 2 Al + 3 CuCl 2  2 AlCl 3 + 3 Cu 54.0 g Al x 1 mol Al 26.98 g Al 3.00 mol CuCl 2 = 3 mol CuCl 2 2 mol Al x Limiting Reactant convert reactant A to reactant B to compare If available < needed (limiting) If available > needed (excess) 54.0 g Al 4.50 mol CuCl 2 (Which is limiting?) (4.50 mol CuCl 2 ) available > needed (3.00 mol CuCl 2 ) CuCl 2 is excess Al is limiting

74. Theoretical Yield theoretical yield: the maximum amount of product that can be formed –calculated by stoichiometry –limited by LR (use LR only to calculate) different from actual yield (or experimental), amount recovered in the experiment 54.0 g Al x 1 mol Al 26.98 g Al = 3 mol Cu 2 mol Al x 63.55 g Cu 1 mol Cu x 191 g produced Cu limiting HW p. 115 #72

75. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make %Yield = x 100 Actual Theoretical (calculate using the LR only) % Error = |Accepted – Experimental| x100 Accepted NOT % Error:

76. Aluminum will react with oxygen gas according to the equation below 4 Al + 3 O 2 2 Al 2 O 3 In one such reaction, 23.4 g of Al are allowed to burn in excess oxygen. 39.3 g of aluminum oxide are formed. What is the percentage yield? Percent Yield

77. 101.96 g Al 2 O 3 1 mol Al 2 O 3 4 Al + 3 O 2 2 Al 2 O 3 Percent Yield HW p. 116 #79, 77 39.3 g of aluminum oxide are formed. What is the percentage yield? 23.4 g Al 1mol Al 26.98 g Al x 2 mol Al 2 O 3 4 mol Al x = 44.2 g Al 2 O 3 x %Yield = x 100 39.3 g 44.2 g 88.9 %