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The Mole, Stoichiometry, and Solution Chemistry

The Mole, Stoichiometry, and Solution Chemistry. Chemistry I Honors/ACP C101-121 Unit 8. Objectives #1-4: The Mole and its Use in Calculations. I . Fundamentals 1 mole of an element or compound contains

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The Mole, Stoichiometry, and Solution Chemistry

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  1. The Mole, Stoichiometry, and Solution Chemistry Chemistry I Honors/ACP C101-121 Unit 8

  2. Objectives #1-4: The Mole and its Use in Calculations I. Fundamentals 1 mole of an element or compound contains 6.02 X 1023particles and has a mass = to its molar mass,which is calculated based on chemical formula & PT values “The Triad of the Mole” 1 mole Molar mass of elem/cmpd 6.02 x 1023 particles 1 mole = 6.02 X 1023 particles* = molar mass

  3. Henceforth, the number 6.02 X 1023 will be known as Avogadro’snumber

  4. Examples of gram quantities that are the EQUIVALENT of one mole: 1 mole of iron = 55.8 grams 1 mole of copper =63.5 grams 1 mole of water = 18.0 grams

  5. Introduction to Mole Problems (Follow the procedures outlined in Unit 1 for dimensional analysis problems) • Calculate the number of atoms in .500 moles of iron. Step 1: Determine known value. Look for the unit!! .500 moles Fe Step 2: Determine unknown: _____atoms Fe

  6. Step 3: Use Triad of the Mole to determine conversion factor: 1 mole = 6.02 X 1023 atoms Step 4: Use conversion factor to solve problem: .500 moles Fe x (6.02 x 1023 atoms) 1 mole Step 5: Check answer for units, sig. figs,, and reasonableness 3.01 X 1023 atoms Fe Reminders: *Remember to set problem up so that the units cancel!! **Use ( ) when plugging in 6.02 x 1023

  7. Calculate the number of atoms in .450 moles of zinc. .450 moles Zn x 6.02 X 1023 atoms Zn 1 mole = 2.71 X 1023 atoms Zn

  8. Calculate the number of moles in 2.09 x 1025 atoms of sulfur. 2.09 x 1025 atoms S x 1 mole S 6.02 X 1023 atoms S = 34.7 moles S • Calculate the number of moles in 3.06 x 1022 atoms of chlorine. 3.06 x 1022 atoms Cl x 1 mole Cl 6.02 X 1023 atoms Cl = .0508 moles Cl

  9. Calculate the number of atoms in 35.7 g of silicon. 35.7 g Si x 6.02 X 1023 atoms Si 28.1 g = 7.65 X 1023 atoms Si Make any additional notes you need about these types of problems at this time! 

  10. Molar Mass Recall that the molar mass is the mass of the chemical that is equivalent to 6.02 x 1023 particles…it is based on the average atomic mass units for each chemical as listed on the PT, but accounts for the ratio of atoms within the fomula…. 6. Calculate the molar mass of H3PO4: 3 moles H @ 1.00 g/mole = 3.00 g 1 moles P @ 31.0 g/mole = 31.0 g 4 moles O @ 16.0 g/mole = 64.0 g Total = 98.0 g

  11. 7. Calculate the molar mass of Al(OH)3: 1 mole Al @ 27.0 g/mol = 27.0 g 3 moles O @ 16.0 g/mol = 48.0 g 3 moles H @ 1.0 g/mol = 3.0 g Total = 78.0 g

  12. 8. Calculate the molar mass of BaCl2 2H2O: 1 mole Ba @ 137.3 g/mol = 137.3 g 2 moles Cl @ 35.5 g/mol = 71.0 g 2 moles H2O @ 18.0 g/mol = 36.0 g Total = 244.2 g Note that the total mass of the water is ADDED to the mass of the barium chloride!!!

  13. Mole-Mass Problems • Use molar mass to calculate the number of grams in amounts that are more or less than 1 mole… 9.How many grams are in 7.20 moles of dinitrogen trioxide? Step 1: Determine known value. 7.20 moles N2O3 Step 2: Determine unknown value. _____ grams N2O3

  14. Step 3: Determine molar mass of compound if grams are involved: 76.0 g Step 4: Use “Triad of the Mole” to determine conversion factor: 1 mole = 76.0 g Step 5: Use conversion factor to solve problem: 7.20 moles N2O3 x 76.0 g N2O3 1 mole = 547.2 g N2O3 Step 6: Check answer for units, sig. figs., and reasonableness:

  15. Molar mass 10. What is the mass in grams of 4.52 moles of barium chloride? 4.52 moles BaCl2 x 208.3 g = 942 g 1 mole 11. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate. Think of the pattern first!! grams  moles  ions 3.50 g (NH4)3PO4 x 1 mole (NH4)3PO4 149. 0 g x 3 NH4+1 1 mole x 6.02 X 1023 ions 1 mole NH4+1 = 4.24 X 1022 NH4+1 ions Internal Molar Ratio from formula

  16. 12. Calculate the mass of carbon in 7.88 X 1026 molecules of C8H18. 7.88 x 1026 molecules C8H18 x 1 mole C8H18 6.02 X 1023 molecules C8H18 x 8 moles C 1 mole C8H18 x 12.0 g C 1 mole C = 1.26 X 105 g C Internal Molar Ratio from formula

  17. 13. Calculate the number of molecules present in 2.50 moles of water. 2.50 moles H2O x 6.02 X 1023 molecules 1 mole H2 O = 1.51 X 1024 molecules H2O

  18. 14. Calculate the mass in grams of 4.50 X 1025 molecules of C6H10. 4.50 x 1025 molecules C6H10 x 82.0 g C6H10 6.02 X 1023 molecules = 6130 g C6H10

  19. 15. Calculate the mass of 1 molecule of propane, C3H8 . 1 molecule C3H8X 44.0 g 6.02 X 1023 molecules = 7.31 X 10-23 g C3H8

  20. Unit 8 Objectives 5 & 6

  21. Objective #5 Characteristics of Solutions

  22. Objective #5 Characteristics of Solutions • The Solution Process • Dissociation and Hydration of solutes: Solute is split apart by solvent (dissociation) Solute particles are surrounded by solvent particles (hydration or solvation) • The rate of solution formation can be increased by: Stirring—increases the contact b/t solute & solvent Raising temperature—increases KE of the pariticles Powdering solute--increasing the surface area of the solute for greater rate of contact w solvent

  23. Objective #5: Characteristics of Solutions • Behavior of ionic, polar, and nonpolar solutes in water: Water is polar and will dissolve many ionic and polar solutes NaCl → Na+1 + Cl-1 HCl → H+1 + Cl-1 Nonelectrolytes vs. Electrolytes • “Like dissolves like”: materials that have similar bonds will dissolve each other NaCl (ionic) dissolves in H2O (polar) HCl (polar) dissolves in H2O (polar) I2 (nonpolar) dissolves in CCl4 (nonpolar) Dissociation will release ions

  24. Objective #5 Characteristics of Solutions • Miscible vs. immiscible liquids: Miscible liquids dissolve in each other and form one phase Ex: Alcohol in H2O Immiscible liquids don’t dissolve in each other and form two phases Ex: Oil & water • Unsaturated vs. saturated vs. supersaturated solutions: Unsaturated-less solute dissolved than possible Saturated -limit of solute dissolving Supersaturated –beyond the usual limit of solute dissolving

  25. Objective #5 Characteristics of Solutions • Effect of pressure on solubility: Gases will be more soluble in a liquid if the atmospheric pressure is increased • Effect of temperature on solubility: For solids in liquids – increasing temp. usually increases solubility For gases in liquids – increasing temp. decreases solubility

  26. Interpreting Solubility Graphs See lecture notes for practice questions…

  27. Objective #6: Molarity A. Formula: Molarity = moles of solute 1 liter of solution B. Examples: 17. Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. 11.5 g NaOH x 1 mole = .288 moles 40.0 g M = .288 moles = .192 M 1.5 L H2O

  28. 18. A chemist requires 1.00 L of .200 M potassium dichromate solution. How many grams of solid potassium dichromate will be required? Molarity = Moles .200 M = Moles 1 Liter 1 Liter Moles = (M) (L) = (.200 M) (1.00 L) = .200 moles K2Cr2O7 .200 moles K2Cr2O7 x 294.2 g = 59 g K2Cr2O7 1 mole

  29. 19. What is the molarity of each ion in the following solutions (assuming all are strong electrolytes) Assume a 1:1 ratio of ions:molarity, and adjust the ratio based on the chemical formula… a.) .15 M calcium chloride: CaCl2 mole ratio = 1 Ca to 2 Cl Ca +2 = .15 M Cl-1 = 2 (.15 M) = .30 M

  30. 19. What is the molarity of each ion in the following solutions (assuming all are strong electrolytes) b.) .22 M calcium perchlorate Ca(ClO4)2 mole ratio = 1 Ca+2 to 2 ClO4-1 Ca+2: .22 M ClO4-1 : 2 (.22 M) = .44 M

  31. 20. What is the molarity of an HCl solution made by diluting 3.50 L of a .200 M solution to a volume of 5.00 L? Does the number of moles change?? No…only the molarity, since volume increases Therefore… moles before dilution = moles after dilution Molarity = Moles 1 Liter … So, moles = (M)(V) M1V1 = M2V2

  32. 20. What is the molarity of an HCl solution made by diluting 3.50 L of a .200 M solution to a volume of 5.00 L? M1V1 = M2V2 #1 = concentrated solution #2 = diluted solution M2 = M1V1 V2 = (3.50 L) (.200 M) 5.00 L = .140 M

  33. 21. An experiment calls for 2.00 L of a .400 M HCl, which must be prepared from 2.00 M HCl. What volume in milliliters of the concentrated acid is needed to be diluted to form the .400 M solution? M1V1 = M2V2 V1 = M2V2 M1 = (.400 M) (2.00 L) (2.00 M) = .400 L = 400. ml How would one prepare the solution? …Add 1600 ml of water to bring volume to 2L

  34. Unit 8 Objectives 7-10

  35. Objs. #7-10 Acids, Bases, and pH/pOH I. Strong vs. Weak Acids and Bases • Strong acids and bases completely ionize in water HCl(aq) H+1(aq) + Cl-1(aq) NaOH(aq)  Na+1(aq) + OH-1(aq) • Common examples: Acids: HCl, H2SO4, HNO3, HClO4, Bases: NaOH, KOH, Ca(OH)2 Memorize these!

  36. Weak acids and bases don’t ionize completely in water HCN(aq) + H2O(aq) CN-1(aq) + H3O+1(aq) NH3(aq) + H2O(aq) NH4(aq)+1 + OH-1(aq) • Common examples: HC2H3O2, NH3 • Assume all acids/bases are weak unless they appear on the Strong List on slide 35.

  37. Self-Ionization of Water • Water has the ability to act as both a Bronsted-Lowery acid and base by spontaneously releasing a H+ which is then attracted to the negative oxygen in another water molecule H2O(l) + H2O(l) H3O(aq)+1 + OH(aq)-1 Kw = [H3O+1] [OH-1] where, at 25oC, the Kw = 1.00 X 10-14 so 1.00 X 10-14 = [H3O+1] [OH-1] 1.00 X 10-14 = x2 1.00 X 10-7 = x, SO….Kw = [1.00 X 10-7 ][1.00 X 10-7 ]

  38. These values for the [H3O+1] and [OH-1] concentrations are valid for room temperature conditions; K does vary with temperature, however • Kw can be used to find [H3O+1] and [OH-1] , if one of the concentrations is known… • [H3O+1] = Kw and [OH-1] = Kw [OH-1] [H3O+1]

  39. pH & pOH • pH and pOH are convenient methods for expressing the relative [H3O+1] or [OH-1] of a solution • pH is defined as the –log [H3O+1] • pOH is defined as the –log[OH-1] • at 25oC, pH + pOH = 14

  40. pH and pOH Scales Compared Inverse relationship If pH is known, pOH can be inferred…

  41. pH & pOH Calculations Example 21: Calculate the pH for each of the following solutions: a) 1.0 M H+1 = -log (1.0) = 0 b) 1.0 x 10-9 M H+1 = -log (1.0 x 10-9) = 9 c) 2.5 x 10-4 M H+1 = -log (2.5 x 10-4) = 3.6 pOH… 14 5 10.4

  42. To find the concentration from a pH value… Process: Press the blue 2nd function button, press log key to activate the antilogfunction, enter the negative version of pH, close parentheses, hit equal…answer is displayed… 22. Calculate the hydronium ion concentration of a solution having a pH of 4.5. 2nd, log Antilog (-4.5) = 3.16 X 10-5 M 23. Calculate the hydronium ion concentration of a solution having a pH of 7.5 2nd, log Antilog (-7.52) = 3.0 X 10-8 M

  43. [H3O+], [OH-],pH & pOH Calculations

  44. Calculate the [H+1] and [OH-1] for a 1.0 X 10-3 M NaOH solution. Calculate the pH and pOH. NaOH is a strong base so the [OH-1] is 1.0 x 10-3 Step 1:pOH = -log [OH-1]  -log [1.0 X 10-3] = 3 Step 2: pH + pOH = 14 pH = 14-3 = 11 Step 3: [H+1] = Antilog [-11] = 1 X 10-11 M

  45. Buffers • Buffers are solutions that resist changes in pH when small amounts of acid or base are added • Buffer solutions can be produced from a combination of a weak acid and a salt containing a common ion or a combination of a weak base and a salt containing a common ion

  46. Example 25: Using a Buffer Adding acid to a mixture of sodium acetate and acetic acid: CH3COO-1 + H+1 CH3COOH + H2O The hydrogen ion of the additional acid will be “captured” by the acetate ion Adding base to a mixture of sodium acetate and acetic acid: CH3COOH + OH-1  CH3COO-1 + H2O The hydroxide ion of the base will be “neutralized” by the acid

  47. Practice: Which of the following mixtures would form buffers? • HCl and NaCl No—HCl is a strong acid • HNO2 and NaNO2 Yes, HNO2 is a weak acid and NaNO2 contains a common ion • HNO2 and NaCl No, HNO2 is a weak acid but NaCl does not contain a common ion

  48. Unit 8 Objectives 11-13

  49. Objective 11 • Percent Composition Answers the question “how much of each chemical is in the compound?”

  50. Objective #11: Percentage Composition A. Formula % element in compound = mass of element in sample of compound mass of sample of compound

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