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Chemistry 100(01) Fall 2000

Chemistry 100(01) Fall 2000. Dr. Upali Siriwardane CTH 311 Phone 257-4941 Office Hours: M, Tu, W, Th, F 9:00-11:00 a.m. Test 1 : Chapters 1, 2: September 27 Test 2: Chapters 3, 4: October 23 Test 3: Chapters 4, 5: November 8 Make-up, Comprehensive, November 13.

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Chemistry 100(01) Fall 2000

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  1. Chemistry 100(01) Fall 2000 Dr. Upali Siriwardane CTH 311 Phone 257-4941 Office Hours: M, Tu, W, Th, F 9:00-11:00 a.m. Test 1 : Chapters 1, 2: September 27 Test 2: Chapters 3, 4: October 23 Test 3: Chapters 4, 5: November 8 Make-up, Comprehensive, November 13

  2. Chemistry 100(05) Fall 2000 Dr. Upali Siriwardane CTH 311 Phone 257-4941 Office Hours: M, Tu, W, Th, F 9:00-11:00 a.m. Test 1 : Chapters 1, 2: September 28 Test 2: Chapters 3, 4: October 19 Test 3: Chapters 4, 5: November 9 Make-up, Comprehensive, November 14

  3. Chapter 3. Stoichimetry • Calculations with chemical formula and reactions are called Stoichiometry

  4. KEY CONCEPTS • Atomic, molecular and formula mass • Mole, Avogadro's number and Molar mass • Mass percent of elements • Mass % from analytical data • Mass % from formula • Empirical formula from Mass % • Chemical formula from composition • Chemical formula from mass % • Chemical Equations • Stoichiometric coefficients • Balancing chemical equations Stoichimetry • Stoichiometric coefficients and Limiting reactant • Yields of chemical reactions • Actual yield and Thoretical yield • Solutions • concentration and dilution of solutions • Solution stoichiometry

  5. Question • You were given a chance to pick 100 kg of gold(I)periodate - AuIO4 or gold(I)nitrate- AuNO3 Which one you would pick? • Rationalize your choice.

  6. Gold • 1 oz = $ 400 • 1 oz = 28.35 g • 1 kg = 35.27 oz • 1 kg Au = $ 14,109

  7. Molecular mass vs. formula mass • Formula mass • Add the masses of all the atoms in formula • - for molecular and ionic compounds. • Molecular mass • Calculated the same as formula mass • - only valid for molecules. • Both have units of either u or grams/mole.

  8. Examples of M.W. H3PO4: H = 1.01 g/mol, P = 30.97 g/mol, O = 16.00 g/mol 1 amu is equal to 1 g/mol m.w. H3PO4 = 3 x 1.01+1 x 30.97+ 4 x 16.00 = 98.00 g/mol

  9. Examples of F.W. • K2CO3: • K = 39.10 g/mol; • C = 12.01 g/mol; • O = 16.00 g/mol • f.w. K2CO3 = • 2 x 39.10 + 1 x 12.01 + 3 x 16.00 • = 138.2 g/mol

  10. Mole Concept • We use masses weighed in grams in chemical reaction. Need a conversion factor to convert grams to atoms and molecules. • atomic weight or molecular weight taken in grams contains • 6.022 x 1023 • atoms, molecules or particles. • The number 6.022 x 1023 is called Mole or Avagadro’s Number

  11. Can you guess this? • How long is 1 mole of seconds? • 1.9096 x 1019 years • 2 x 1010 billion years • How deep is the layer of marbles, if 1mole of marbles are spread over the surface of earth? • ~50 miles high

  12. The Mole and Avogadro's number • Number of atoms in 12.000 grams of 12C • 1 mol = 6.022 x 1023 atoms • 1 g = 6.022 x 1023 u • 1 u = 1 g/mol • Mole is the converssion factor between M.W and grams • 1 mol = grams / molecular weight • mole = g/m.w. Or g/f.w.

  13. Conversion factors in Stoichiometry? • 6.022 x 1023 atoms = gram atomic weight • 6.022 x 1023 molecules = gram molecular weight • 6.022 x 1023 atoms C = 12.01 grams of carbon (C) • 6.022 x 1023 molecules H2O = 18.02 g of H2O • 6.022 x 1023 = 1 mol • 1 g = 6.022 x 1023 amu (u) • 1 amu = 1 g/mol

  14. Atomic Masses • Atomic mass in grams ? • Mass of a copper atom in grams • 63.55 g Cu 1 mole • 1 mole 6.022 x 1023 • = 1.055 x 10-22 g • Atomic mass in the periodic table divided by Avagadro’s Number

  15. An atom weighs 7.47 x 10-23 g. What is the name of the element this atom belongs to?

  16. An atom weighs 7.47 x 10-23 g. What is the name of the element this atom belongs to? • Conversion factor: 1g = 6.022 x 1023 u • 7.47 x 10-23 g x 6.022 x 1023 u • 1g • = 44.98 u or g/mole • The element is Sc.

  17. 1 mol 132.14 g moles = 20.0 g x = 0.151 mol atoms unit units mol Example - (NH4)2SO4 • How many atoms are in 20.0 grams of ammonium sulfate? • Formula weight = 132.14 grams/mol • Atoms in formula = 15 atoms / unit atoms = 0.151 mol x 15 x 6.02 x1023 atoms = 1.36 x1024

  18. How do you convert grams to mole and vice versa? • Grams ---> mole • grams /molecular (formula) weight • Moles ----> grams • moles x molecular (formula) weight

  19. Calculations • Calculate how many moles are in 100 g of sulfur-S8 • Calculate the grams of sugar in 3 moles of -glucose-C6H12O6

  20. % Element Composition in Compounds • n x Atomic weight • % mass = --------------------- x 100 • molecular weight • n = subscript of the element in the formula

  21. Calculation • Al2(SO4)3 • calculate % Al, % S and % O

  22. Examples • a) Al2(SO4)3: f.w. = 342.14 g/mol, • Atomic weight of oxygen = 16.00 g/mol, n = 12 • 12 x 16.00 • % O mass = ------------- x 100 = 56.12% oxygen. 342.14 • c) CuSO4: f.w. = 159.61 g/mol • Atomic weight of oxygen = 16.00 g/mol n = 4 • 4 x 16.00 • % O mass =-------------- = 40.01% oxygen. • 159.61

  23. Calculate % composition from analysis • A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur. • Calculate the percent composition of elements.

  24. Examples Glucose has a molecular formula of C6H12O6 (M.W. 180.16 g/mol). a) How many grams of C, H and O are available in 1 mole of glucose? b) Calculate mass percents of elements C, H and O in glucose.

  25. How many grams of C, H and O are available in 1 mole of glucose? • 1 mol C6H12O6= 6 mol C = 6 x 12.01 =72.06g C • 1 mol C6H12O6=12 mol H= 12 x 1.01 • = 12.12g H • 1 mol C6H12O6=6 mol O =6 x 16.00 • = 96.00g O

  26. Mass percent of element. C6H12O6 = 180.16 g/mol • 6 x 12 • %C = --------- x 100 = 40.00% C 180.16 • 12 x 1.01 • % H = ------------- x 100 = 6.73% H • 180.16 • 6 x 16.00 • %O = ------------ x 100 = 53.29% O 180.16 ----------- • 100.02%

  27. What is Empirical Formula • Simple whole number ratio of each atom expressed in the subscript of the formula. • Molecular Formula = C6H12O6 of glucose • Empirical Formula = CH2O • Emiprical formula is calculated from % composition

  28. How do you get Empirical Formula from % composition and vice versa?

  29. % composition to Empirical formula • Divide percent composition by atomic weight. • Divide answer by lowest number to get simple ratio of moles or atoms. • Multiply by a factor to get whole number ratio. • Write a formula with whole number ratio as subscripts to get empirical formula.

  30. Examples • A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur. • a) Calculate the percent composition of elements. • b) The empirical formula of the compound

  31. Calculation • mass of element • % Element = ------------------- x 100 • mass of sample • 7.04 • % P = --------- x 100 = 56.3% P 12.5 • 5.46 • % S = ------- x 100 = 43.7% S • 12.5 100.0

  32. Calculation • % of P and S: • 56.3 43.7 • Moles of P and S: • 56.3g 43.7g • ------- = 1.82 mol P ------ = 1.36 mol S • 30.97 32.06 • Atom ratio: • 1.82 atoms P 1.36 atoms S • 1.33 atoms P 1.00 atoms S

  33. Calculation..continued • 1.82 atoms P 1.36 atoms S • Divide by lowest number: • 1.33 atoms P 1.00 atoms S • simple atom ratio: • 1.33 atom P 1 atom S • to get a simple whole number ratio. Multiply both numbers by 2, 3, and 4 etc. until 1.33 becomes close to a whole number.

  34. Calculation..continued • To get simple whole number ratio: • P S • 1.33 atom 1 atom • 2 x 1.33 = 2.66 2 x 1 • 2.66 2 • 3 x 1.33= 3.99 3 x 1 • 4.00 3.0 • Therefore, Formula should contain 4 P atoms and 3 S atoms. • Therefore, the empirical Formula of the compound is: P4S3

  35. More Example • Glucose contains the elements C, H, and O in 39.99% C, 6.71% H and 53.28% O, respectively, by mass. • a) Calculate the empirical formula of glucose. • b) Molecular weight determination in solution for glucose has shown a molecular weight close to 180 g/mol. What is the molecular formula of glucose?

  36. Calculation • Empirical formula • from % elemental composition. • % composition: 39.99% C, 6.71% H, 53.28% O • ii) mole ratio: • C H O • 39.99 6.71 53.28 • ------- = 3.33; -------- = 6.64; -------- = 3.33 • 12.01 1.01 16.00 • 3.33 mol C: 6.64 mol H: 3.33 mol O

  37. Calculation..continued • Atom ratio • 3.33 mol C: 6.64 mol H: 3.33 mol O • Simple atom: • 3.33 6.64 3.33 • ------- ------- ------- • 3.33 3.33 3.33 • 1 1.99 1 • Simple whole • number 1 2 1 • number ratio of atom • The empirical formula of the compound is • CH2O

  38. Calculation..continued • Molecular Formula = n x empirical Formula • Molecular weight 180 • n = -------------------------------------- = ------ = 6 • Empirical Formula Weight 30 • Molecular Formula = (CH2O)n = (CH2O)6 • Molecular Formula = C6H12O6 of glucose

  39. Question • A molecular compound contains 92.3% carbon and 7.7% hydrogen by weight. What is its empirical formula?

  40. Question • A molecular compound has empirical formula CH. If 0.050 mol of the compound weighs 3.90 g, what is its molecular formula?

  41. Chemical Equation • P4O10 (s) + 6H2O (l) = 4 H3PO4(l) • reactants enter into a reaction. • products are formed by the reaction. • Parantheses represent physical state • stoichiometric coefficients are numbers in front of chemical formula formula • gives the amounts (moles) of each substance used and each substance produced. • Equation Must be balanced!

  42. Chemical Reaction • Could be described in words • Chemical equation: • Reactants? • Products? • reaction conditions? • =, ---> , <==> or ? • stoichiometric coefficients? • Number in front of substances representing moles, atoms, molecules

  43. Steps in Stoichiometric Calculations • Check whether chemical equation is balanced • get the moles from grams of materials • find the limiting reactant • calculate moles of products from the limiting reactant • convert moles of the products to grams • find the actual yield of the reaction • calculate % yield of the reaction

  44. Question • How much hydrogen gas is produced when 1 kg of sodium reacts with water?

  45. Examples • Calculate the following using the chemical equation given below: • 4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g) • a) moles of NO(g) from 2 moles of NH3(g) and excess O2(g). • b) moles of H2O(g) from 3 moles of O2(g) and excess NH3(g).

  46. Examples • How many moles of H2O will be produced by 0.80 mole of O2 with excess H2 according to the equation? • 2H2(g) + O2(g) = 2 H2O(l)

  47. Question • 2Al(s) + 6HCl(aq)--> 2AlCl3(aq) +3H2(g) According to the equation above, how many grams of aluminum are needed to react with 0.582 mol of hydrochloric acid?

  48. 5.23 g Al

  49. What is the limiting reagent? • Limiting reagent is the reactant, which is used up first. • To find the limiting reactant you have to compare the amounts of reactants in moles.

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