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The Mole, Stoichiometry , and Solution Chemistry

The Mole, Stoichiometry , and Solution Chemistry. H Chemistry I Unit 8. Objectives #1-4 The Mole and its Use in Calculations. Fundamentals 1 mole of an element or compound contains 6.02 X 10 23 particles and has a mass = to its molar mass “The Triad of the Mole”

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The Mole, Stoichiometry , and Solution Chemistry

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  1. The Mole, Stoichiometry, and Solution Chemistry H Chemistry I Unit 8

  2. Objectives #1-4 The Mole and its Use in Calculations • Fundamentals 1 mole of an element or compound contains 6.02 X 1023 particles and has a mass = to its molar mass “The Triad of the Mole” 1 mole = 6.02 X 1023 particles = molar mass *henceforth the number 6.02 X 1023 will be known as the Avogadro’s number

  3. *examples of mole quantities: 1 mole of iron = 55.8 grams 1 mole of copper = 63.5 grams 1 mole of water = 18.0 grams

  4. 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. *Determine known: .500 atoms Fe

  5. *Determine unknown: atoms Fe *Use Triad of the Mole to determine conversion factor: 1 mole = 6.02 X 1023 atoms *Use conversion factor solve problem: .500 moles Fe X 6.02 X 1023 atoms/1 mole

  6. Check answer for units, sig. figs,, and reasonableness 3.01 X 1023 atoms Fe • 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

  7. Calculate the number of moles in 2.09 X 1025atoms of sulfur. 2.09 X 1025 atoms S X 1mole 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

  8. 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

  9. Molar Mass • Calculate the molar mass of H3PO4: 3 H at 1.00 g = 3.00 g 1 P at 31.0 g = 31.0 g 4 O at 16.0 g = 64.0 g Total = 98.0 g

  10. Calculate the molar mass of Al(OH)3: 1 Al at 27.0 g = 27.0 g 3 O at 16.0 g = 48.0 g 3 H at 1.0 g = 3.0 g Total = 78.0 g

  11. Calculate the molar mass of BaCl2.2H2O: 1 Ba at 137.3 g = 137.3 g 2 Cl at 35.5 g = 71.0 g 2 H2O at 18.0 g = 36.0 g Total = 244.2 g

  12. Mole-Mass Problems • How many grams are in 7.20 moles of dinitrogen trioxide? *determine known: 7.20 moles N2O3 *determine unknown: grams N2O3

  13. *determine molar mass of compound if grams are involved: 76.0 g *use “Triad of the Mole” to determine conversion factor: 1 mole = 76.0 g *use conversion factor to solve problem: 7.20 moles N2O3 X 76.0 g N2O3 / 1 mole = 547.2 g

  14. *check answer for units, sig. figs., and reasonableness: 547 g N2O3

  15. What is the mass in grams of 4.52 moles of barium chloride? 4.52 moles BaCl2 X 208.2 g / 1 mole = 941 g 3. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate. 3.50 g (NH4)3PO4 X 1 mole (NH4)3PO4 / 149. 0 g X 3 NH4+1 / 1 mole (NH4)3PO4 X 6.02 X 1023 ions / 1 mole NH4+1 ions = 4.24 X 1022 NH4+1 ions

  16. 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

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

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

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

  20. Objective #5 Characteristics of Solutions

  21. 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, raising temperature, and powdering

  22. 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 C6H12O6 - C6H12O6 nonelectrolytes vs. electrolytes *”like dissolves like”: materials that have similar bonds will dissolve each other NaCl (ionic) H2O (polar) HCl (polar) H2O (polar) I2 (nonpolar) CCl4 (nonpolar)

  23. Objective #5 Characteristics of Solutions *miscible vs. immiscible liquids: miscible liquids dissolve in each other and form one phase immiscible liquids don’t dissolve in each other and form two phases *unsaturated vs. saturated vs. supersaturated solutions: Unsaturated (less solute dissolved than possible) Saturated (limit of solute dissolving) Supersaturated (beyond limit of solute dissolving)

  24. 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 (interpreting solubility graphs)

  25. Interpreting Solubility Graphs

  26. Example Problems: • Which chemical is the most soluble at 20oC? _________ • Which chemical is the least soluble at 40oC? ________ • What is the solubility of KCl at 25oC? ________ • At 90oC you dissolved 10 g of KCl in 100 g of water. Is this solution saturated or unsaturated? _________ 5. Amass of 80 g of KNO3 is dissolved in 100 g of water at 50oC. The solution is heated to 70oC. How many more grams of potassium nitrate must be added to make the solution saturated. _________

  27. Objective #6 Molarity *formula: Molarity = moles of solute / liter of solution *examples: • 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 / 40.0 g = .288 moles M = .288 moles / 1.50 L = .192 M

  28. A chemist requires 1.00 L of .200 M potassium dichromate solution. How many grams of solid potassium dichromate will be required? Moles = M X L = (.200 M) (1.00 L) = .200 moles K2Cr2O7 .200 moles K2Cr2O7 X 294.2 g / 1 mole = 58.8 g K2Cr2O7

  29. .200 moles K2Cr2O7 X 294.2 g / 1 mole = 59 g K2Cr2O7 • What is the molarity of each ion in the following solutions (assuming all are strong electrolytes) .15 M calcium chloride: CaCl2 1 Ca to 2 Cl .15 M Ca+2 2(.15 M) = .30 M Cl-1

  30. .22 M calcium perchlorate Ca(ClO4)2 1 Ca to 2 ClO4 .22 M Ca+2 2 (.22 M) = .44 M ClO4-1 .18 M sodium chloride .18 M Na and .18 M Cl

  31. 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? moles before dilution = moles after dilution recall moles = MV M1V1 = M2V2

  32. M1V1 = M2V2 #1 = concentrated #2 = diluted M2 = M1V1 / V2 = (3.50 L) (.200 M) / 5.00 L = .140 M

  33. 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 solution?

  34. Objs. #7-10 Acids, Bases, and pH/pOH • 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: HCl, H2SO4, HNO3, HClO4, NaOH, KOH, Ca(OH)2

  35. *weak acids and bases don’t completely 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

  36. Self-Ionization of Water *water has the ability to act as both a Bronsted-Lowery acid and base 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

  37. *these values for the H3O+1 and OH-1 concentrations are valid for room temperature conditions; K does vary with temperature however III. pH and pOH *pH and pOH are convenient methods for expressing the relative H3O+1or OH-1of a solution *pH is defined as the –log{H3O+1} *pOH is defined as the –log{OH-1} *at 25oC pH +pOH = 14

  38. *pH and pOH scales compared:

  39. *examples: Calculate the pH for each of the following solutions: • 1.0 M H+1 • 1.0 X 10-9 M H+1 3. 2.5 X 10-4M H+1

  40. Calculate the hydronium ion concentration of a solution having a pH of 4.5. 5. Calculate the hydronium ion concentration of a solution having a pH of 7.52

  41. Calculate the H+1 and OH-1 for a 1.0 X 10-3 M NaOH solution. Calculate the pH and pOH.

  42. 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

  43. *example: 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

  44. *Examples: Which of the following mixtures would form buffers? • HCl and NaCl • HNO2 and NaNO2 3. HNO2 and NaCl

  45. Objective #11 Percentage Composition • Formula % element in compound = mass of element in sample of compound / mass of sample of compound

  46. *examples: • Calculate the percentage composition potassium phosphate (K3PO4) *determine molar mass of compound: 212.4 g • Determine total mass of each component: 3 K at 39.1 g each = 117.3 g 1 P at 31.0 g each = 31.0 g 4 O at 16.0 g each = 64.0 g

  47. Divide total mass of each component by molar mass of compound: % K = 117.3 g / 212.4 g = .55 % P = 31.0 g / 212.4 g = .15 % O = 64.0 g / 212.4 g = .30

  48. Multiply each result by 100 % and round appropriately. .55 X 100 = 55 % K .15 X 100 = 15 % P .30 X 100 = 30 % O

  49. Calculate the percentage composition of sodium carbonate (Na2CO3) molar mass = 106.0 g % Na = (46.0 g / 106.0 g) X 100 = 43% % C = (12.0 g / 106.0 g) X 100 = 11% % O = (48.0 g / 106.0 g ) X 100 = 45%

  50. Objective #12 Determining Empirical Formulas *empirical vs. molecular formulas: H2O2 --› HO H2O --› H2O *examples: 1. Determine the empirical formula for a compound that when analyzed contained 70.9% potassium and 29.1% sulfur by mass.

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