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Chapter 3

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  1. Chapter 3

  2. Atomic Mass • amu = Average Atomic Mass Unit • Based on 12C as the standard. • 12C = exactly 12 amu • The average atomic mass (weight) of an element is equal to the sum of the products of each isotope’s mass (in amu) multiplied by it’s relative abundance.

  3. EXAMPLE OF AVERAGE ATOMIC MASS PROBLEM • Naturally occurring chlorine is 75.53% 35Cl which has an atomic mass of 34.969 amu, and 24.47% 37Cl, which has an atomic mass of 36.966 amu. • Calculate the average atomic mass of chlorine.

  4. EXAMPLE OF AVERAGE ATOMIC MASS PROBLEM (CONT) • Average atomic Mass = [ (%/100) (Atomic Mass) ] • Average atomic mass = (0.7553) (34.969 amuCl35) + (0.2447) (36.966 amuCl37) = 26.41 amu + 9.045 amu = 35.46 amu

  5. Formula Weight • AKA: molar mass (g/mol) • Sum of all atomic weights of each atom in its chemical formula Ex: H2S04 1(2H) + 32(1S) + 16(4O) = 98 amu

  6. MOLE Avogadro’s Number • 1 mole of any substance is 6.02 x 1023 particles

  7. Question • How many nitrogen atoms are in 0.25 mol of Ca(NO3)2

  8. Answer 6.02 x 1023 molec Ca(NO3)2 x_2N___ X .25 1 mol molec Ca(NO3)2 = atoms

  9. Law of Conservation of Mass • Mass is never created or destroyed • Reason for balancing chemical equations Lavosier Says : 2 Na + Cl 2  2NaCl 46.0 g 70.9 g 116.9 g

  10. Balancing Equations 1. Write the correct formulas for the reactants and the products 2.Chose the compound that has the greatest number of atoms, then look to the element in that compound that has the greatest number of atoms. 3.Balance this element first by placing a coefficient in front of the corresponding compound on the other side of the equation.

  11. Balancing Equations Cont. 4. Balance H then O 5. Check all coefficients to see that they are in the lowest possible ratio.

  12. Examples: C2H6 + O2 -> CO2 + H2O C2H6O + O2 -> CO2 + H2O CaCO3 + H3PO4 -> Ca3(PO4)2 + CO2 + H2O

  13. ANSWER 2C2H6 + 7O2 -> 4CO2 + 6H2O C2H6O + 3O2 -> 2CO2 + 3H2O 3CaH6O3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O

  14. Percent Composition We can describe composition in two ways1. number of atoms 2. % (by mass) of its elements. We can find % mass from formula mass, by comparing each element present in 1 mole of compound to the total mass of 1 mole of compound

  15. Example of % Comp • Calculate the percentage of nitrogen in Ca(NO3)2

  16. Answer % N = # N atoms(m.w N) X 100% m. w Ca(NO3)2 % N = 2(14.02 N amu) X 100% 164.12 Ca(NO3)2amu = 17%

  17. Inter-converting Grams -> moles -> molec -> atoms How many oxygen atoms are present in 4.20 grams NaHCO3?

  18. Answer 4.20 g NaHCO3 (1mole NaHCO3) (6.02e 23molec)3 Oxygen atoms 84 g NaHCO3 1 mol1 molec NaHCO3 = 9.033 x 10 22 atoms of Oxygen in 4.20 grams NaHCO3

  19. Question • Determine the mass in grams of 3.00 x 1020 N2 molecules

  20. Answer 3.00 x 1020 molecN2 (1 mol)(28g N2) 6.02e23molec 1 mol N2 = 0.0140 g

  21. Determining empirical formula from mass percent • Recall: Empirical formula: simplest whole # ratio of atoms in a compound. Example: Vitamin C is composed of 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula?

  22. Answer: CHO • Convert mass % into grams (assume 100g) • Convert grams to moles • Divide each mol by the smallest number of moles present. You may round to nearest whole # 40.92 g C 1mol C= 3.4 moles C /3.4 = 1 C 12 g C 54.40 g O 1mol O = 3.4 moles O / 3.4 = 1 O 16g O 4.48 g H 1mol H = 4.48 moles H/ 3.4 = 1 H 1 g H

  23. Determine Molecular formula from Empirical Formula Recall: Molecular formula: the exact formula of a molec, giving types of atoms and the number of each type. • Using mass % and molar mass, determine mass of each element in 1 mole of compound • Determine number of moles of each element in 1 mole of compound • The integers from the previous step represent the subscripts in the molecular formula

  24. Let’s look back at our work 40.92 g C 1mol C = 3.4 moles C /3.4 = 1 C 12 g C 54.40 g O 1mol O = 3.4 moles O / 3.4 = 1 O 16g O 4.48 g H 1mol H = 4.48 moles H/ 3.4 = 1.3 H 1 g H C3H4O3= molecular formula

  25. Shortcut n = Molecular Weight empirical Formula Weight The molecular weight of butyric acid is 88 amu. If the empirical formula is C2H4O. What is the molecular formula?

  26. C2H4O = 12 + 12+ 1+1+1+1+16 = 44 amu n = 88 amu = 2 44 Molecular formula = (empirical)n (C2H4O)2 = Molecular Formula = C4H8O2

  27. Stoichiometry: mixing exactly enough chemical so that all is used Mass-Mass problems g given  mol given  mol required  g required (grams to moles to moles to grams)

  28. Silicon carbide is made by heating silicon dioxide to high temperatures. SiO2 (s) + 3C (s)  SiC(s) + 2CO (g) How many grams of CO are formed by complete rxn of 5.00 g SiO2? AP EXAM HINT: always make sure your equation is balanced first or mole ratios will be wrong.

  29. Given: 5.00 g SiO2 Find : CO g SiO2 (s) + 3C (s)  SiC(s) + 2CO (g) grams to moles to moles to grams 5.00 g SiO2 1mol SiO22 mol CO28 g CO = 4.6 g CO 60 g SiO2 1 mol SiO2 1 mol CO Mole ratio

  30. How many moles of sulfuric acid would be needed to produce 4.80 moles of molecular iodine (I2) according to the following balanced equation. 10HI + 2KMnO4 + 3H2SO4 5I2 + 2MnSO4 + K2SO4 + 8H2O

  31. 4.80 mol I23 mol H2SO4= 2.88 mol H2SO4 5 mol I2

  32. Limiting reagent This will be at least 1 AP question! The number of products that can form is limited by the amount of reactant present. The limiting reactant is the one that gives the least amount of product. Reactants  Products

  33. When a mixture of silver and sulfur is heated, silver sulfide is formed: 16 Ag (s) + S8 (s)  8 Ag2S (s) What mass of Ag2S is produced from a mixture of 2.0 g of Ag and 2.0 g of S?

  34. 2.0 g Ag 1mol Ag 8 mol Ag2S 247.9 g Ag2S 107.9 g Ag 16 mol Ag 1 mol Ag2S = 2.3 g Ag2S 2.0 g S8 1mol S8 8 mol Ag2S 247.9 g Ag2S 256.8 S8 1 mol S8 1 mol Ag2S = 15 g Ag2S

  35. Theoretical / Percent yield • The amount of product that is calculated based on the limiting reactant. % Yield = Actual yield X 100% theoretical yield \

  36. HW • 7, 19, 21, 31,33 • 35, 56,57, 58, 62,63, 67, 69. 71,74 81,82,83, 95, 98, 100a-b,105

  37. Types of reactions

  38. Chemical Reactivity Combination/Synthesis Reaction: 2 or more substances react to form one new product A + B  C + 

  39. solid magnesium and oxygen gas react to produce solid magnesium oxide 2Mg (s) + O2(g)  2MgO (s) Metal nonmetal ionic compound Diatomic 2+ 2-

  40. Decomposition Rxn • One substance undergoes a reaction to produce two or more substances. • Typically occurs when things are heated. AX  A + X  +

  41. Solid calcium carbonate reacts to produce solid calcium oxide and carbon dioxide gas CaCO3 (s)  CaO (s) + CO2 (g) 2+ (2-) 2+ 2- 4+ 2(2-)

  42. Single displacement • One element replaces a similar element in a compound A + BX  AX + B BX + Y  BY + X +  +

  43. Solid copper is dissolved in aqueous silver nitrate to produce solid silver and aqueous copper nitrate. Cu(s) + AgNO3 (aq) Ag(s) + Cu(NO3)2 (aq) Write the sentence for this reaction: Fe (s) + Cu(NO3)2 (aq)Fe(NO3)2 (aq)+ Cu (s)

  44. Double Replacement Rxn/ Metathesis • The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX • One of the compounds formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

  45. Double Replacement Rxn/ Metathesis AX + BY  AY + BX +  +

  46. Write the sentence for these double replacement reactions KOH (aq) + H2SO4 (aq) K2SO4 (aq) + H2O (l) FeS (aq) + HCl (aq) FeCl2 (aq) + H2S (aq)

  47. Combustion Reaction A substance combines with oxygen, releasing a large amount of energy in the form of light and heat. C3H8 (g)+5O2 (g)  3CO2 (g) + H2O (g) Usually CO2 (carbon dioxide) / CO (carbon monoxide) and water are produced.

  48. Reactive elements combine with oxygen P4(s) + 5O2(g)  P4O10 (s) (This is also a synthesis reaction) • The burning of natural gas, wood, gasoline C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)