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Advanced Chemistry Chapter Three Stoichiometry. 3.1 Counting by Weighing Average Mass = total mass/ number of objects For purposes of counting, objects.

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Presentation on theme: "Advanced Chemistry Chapter Three Stoichiometry. 3.1 Counting by Weighing Average Mass = total mass/ number of objects For purposes of counting, objects."— Presentation transcript:

1 Advanced Chemistry Chapter Three Stoichiometry

2 3.1 Counting by Weighing Average Mass = total mass/ number of objects For purposes of counting, objects behave as though they are identical

3 3.2 Atomic Masses

4 Atomic Masses The modern system of atomic masses is based on 12 C, as the standard. Developed in 1961

5 Mass Spectrometer An instrument that passes atoms or molecules through a beam of high- speed electrons, which in turn knock electrons off the atoms or molecules being analyzed and change them into positive ions.

6 Mass Spectrometer An applied electric field accelerates the ions into a magnetic field. The amount of deflection that occurs with each ion depends upon its mass.

7 Mass Spectrometer The most massive ions are deflected the smallest amount. A comparison of the positions where the ions hit the deflector plate provides accurate values of relative masses.

8 A Scientist Injecting a Sample into a Mass Spectrometer. (right) Schematic Diagram of a Mass Spectrometer

9 Atomic masses Naturally occurring isotopes are averaged to reflect the percent of abundance of those isotopes. Counting by averaging the mass of atoms allows for an accurate atomic mass for chemical calculations.

10 Mass spec Peaks

11 Mass Spec Bar Graph

12 Mass Spectrum of Natural Copper What is the average mass of natural copper?

13 Mass Spectrum of Natural Copper What is the average mass of natural copper? 63.55 amu/atom

14 3.3 The mole

15 Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 6.022 x 10 23 A sample of a natural element with a mass equal to the element’s atomic mass expressed in grams.

16 Question? What is the mass, in grams, of 12 atoms of Aluminum?

17 Answer

18 Answer

19 Answer

20 Question? How many moles and number of atoms are in a 10.0g sample of Aluminum?

21 Answer

22 3.4 Molar Mass

23 Molar Mass Is the mass in grams of one mole of the compound. “molecular weight”

24 Question? The formula for juglone, a dye, is C 10 H 6 O 3. What is the molar mass?

25 Answer The formula for juglone, a dye, is C 10 H 6 O 3. What is the molar mass?

26 Question? If the molar mass of juglone is 174.1g, How many moles of Juglone are in a 1.56 x 10 -2 g sample?

27 Answer

28 3.5 Learning to Solve Problems

29 Conceptual Problem Solving 1.Read the problem and decide on final goal. Gather facts and state the problem as simply as possible. Where are we going? Where are we going?

30 Conceptual Problem Solving 2.Work backwards from the final goal to decide where to start. How do we get there? How do we get there?

31 Conceptual Problem Solving 3.Once a solution is obtained, check to see in answer is reasonable. Does it make sense? Does it make sense?

32 3.6 Percent Composition of Compounds

33 Mass Percentage Compare the mass of each element in one mole to the molar mass of the compound.

34 Question? What is the mass percentage of C, H and O in the following molecule: C 10 H 14 O?

35 Answer

36 3.7 Determining the Formula of a Compound

37 Empirical vs. Molecular Formula Empirical formula is the formula of a molecule in its smallest whole number ratio. Molecular formula is the exact formula of the molecule as it exists. Example: Empirical = CH 5 N Molecular = (CH 5 N) n

38 Substances Whose Empirical and Molecular Formulas Differ

39 Determining the Empirical Formula Use percent composition as a 100g molecule sample. Divide the element mass sample by the molar mass of each element to determine the molar ratios between elements. Divide the molar ratios by the smallest ratio. If needed, multiply all ratios by the same number to obtain low whole numbers.

40 Determining the Molecular Formula Determine the empirical formula by using mole ratios between elements. Divide the molar mass of the molecular formula by the molar mass of the empirical formula. (n) Multiply every element in the formula by (n).

41 Question? Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents): 71.65% Cl24.27% C4.07% H 71.65% Cl24.27% C4.07% H The molar mass is 98.96 g/mol

42 Answer 71.65% Cl24.27% C4.07% H molar mass = 98.96 g/mol Empirical: ClCH 2 Molecular: Cl 2 C 2 H 4

43 3.8 Chemical Equations

44 Chemical Equations Representation of the chemical reaction process Reactants – left side Products – right side

45 Chemical Equations Atoms are reorganized. Bonds have been broken, and new ones have been formed. Atoms are neither created nor destroyed therefore all atoms present in the reactants must be accounted for among the products.

46 Chemical Equations Subscripts apply to an atom or atoms in parenthesis. Coefficients apply to entire molecule/compound.

47 Chemical Equations Physical states should be given. Solid – (s) Liquid – (l) Gas – (g) Dissolved in water – (aq)

48 3.9 Balancing Chemical Equations

49 Writing and Balancing Equations Determine what reaction is occurring. What are the reactants, the products, and the physical states involved. Write the unbalanced equation that summarizes the reaction.

50 Writing and Balancing Equations Balance the equation by inspection, starting with the most complicated molecules. Determine what coefficients are necessary so that the same number of each type of atom appears on both reactant and product sides. Do not change the identities (formulas) of any of the reactants or products.

51 Question? (NH 4 ) 2 Cr 2 O 7(s) Cr 2 O 3(s) + N 2(g) + H 2 O (g)

52 Answer 1 1 + 1 + 4

53 Question? NH 3(g) + O 2(g) NO (g) + H 2 O (g)

54 Answer 4 + 5 4 + 6

55 3.10 Stoichiometric Calculations: Amounts of Reactants and Products

56 Stoichiometry Write a balanced equation. Coefficients in the balanced equation provide the mole ratios used in the conversion of mass and other quantities from one molecule/compound to another.

57 Calculating Mass of Reactants and Products

58 Stoichiometry

59 Question? NaHCO 3(s) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2(aq) Mg(OH) 2(s) + 2HCl (aq) 2H 2 O (l) + MgCl 2(aq) Which one of these antacids neutralizes more acid with a 1.0g sample?

60 Answer NaHCO 3(s) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2(aq) Mg(OH) 2(s) + 2HCl (aq) 2H 2 O (l) + MgCl 2(aq) Which one of these antacids neutralizes more acid with a 1.0g sample? NaHCO 3 : 1.19 x 10 -2 mol HCl neutralized Mg(OH) 2 : 3.42 x 10 -2 mol HCl neutralized

61 3.11 The Concept of Limiting Reagent

62 Stoichiometric Mixtures A stoichiometric mixture is one that contains the relative amounts of reactants that match the numbers in the balanced equation. Assuming the reaction goes to completion, all reactants will be consumed to form products.

63 Limiting Reactant The reactant that runs out first and therefore limits the amount of product that can form. To determine how much product can be formed from a given mixture of reactants, the limiting reactant must first be determined.

64 Solving a Stoichiome try Problem Involving Masses of Reactants and Products

65

66 Isopentyl Acetate

67 Carvon e

68 Carbon Dioxide

69 Water

70 Figure 3.5 A Schematic Diagram of the Combustion Device Used to Analyze Substances for Carbon and Hydrogen

71 Figure 3.9 Three Different Stoichiometric Mixtures of Methane and Water, which React One-to-One

72 Figure 3.10 A Mixture of CH4 and H20 Molecules

73 Figure 3.11 Methane and Water Have Reacted to Form Products

74 Figure 3.12 Hydrogen and Nitrogen React to Form Ammonia

75 Jellybeans Can be Counted by Weighing

76 Weighing Hex Nuts

77 Copper Nugget

78 Figure 3.4 Samples Containing One Mole Each of Copper, Aluminum, Iron, Sulfur, Iodine, and Mercury

79 Pure Aluminum

80 Bee Stings Cause the Release of Isopentyl Acetate

81 Penicillin is Isolated from a Mold that Can be Grown in Large Quantities in Fermentation Tanks

82 Figure 3.7 The Two Forms of Dichloroenthane

83 Figure 3.8 The Structure of P4O10.

84 Computer-Generated Molecule of Caffeine

85 Methan e Reacts with Oxygen to Produce Flame

86 Hydrochl oric Acid Reacts with Solid Sodium Hydroge n Carbona te to Produce Gaseous Carbon Dioxide

87 Decomposition of Ammonium Dichromate

88

89 Astrona ut Sidney M. Gutierr ez

90 Milk of Magnes ia

91 Race Cars use Methanol as a Fuel

92 Table 3.1 Comparison of 1 Mole Samples of Various Elements

93 Table 3.2 Information Conveyed by the Balanced Equation for the Combustion of Methane


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