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The Mole and Stoichiometry

The Mole and Stoichiometry. 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 and Stoichiometry

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  1. The Mole and Stoichiometry 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 1023 atoms 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 g of barium chloride? 4.52 g 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 X 6.02 X 1023 ions / 1 mole NH4+1 = 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 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. 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

  27. 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

  28. .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

  29. .22 M calcium perchlorate Ca(ClO4)2 1 Ca to 2 ClO4 .22 M Ca+2 2 (.22 M) = .44 M ClO4-1

  30. 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

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

  32. 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?

  33. Objective #7 Percentage Composition • Formula % element in compound = mass of element in sample of compound / mass of sample of compound

  34. *examples: • Calculate the percentage composition 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

  35. 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

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

  37. 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%

  38. Objective #8 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.

  39. *If given percentages, assume 100 gram sample and convert percents to grams: 70.9% K --› 70.9 g K 29.1% S --› 29.1 g S

  40. *convert masses to moles: 70.9 g K X 1 mole K / 39.1 g = 1.81 moles K 29.1 g S X 1 mole S / 32.1 g =.907 moles S *divide all mole values by the smallest mole value in the set: .907 / .907 = 1 K 1.81 / .907 =1.995 S

  41. *round resulting values to the nearest whole number or multiply by factor: 1 K 2 S *use final values as subscripts and write formula with the elements in order of increasing electronegativity; if compound is organic, list carbon first, then hydrogen, and the remainder by electronegativity: K2S

  42. A compound of iron and oxygen when analyzed showed 70.0% iron and 30.0% oxygen by mass. Determine the empirical formula. 70. 0 g Fe X 1 mole/55.8 g = 1.25 moles Fe 30.0 g O X 1 mole/16.0 g = 1.88 moles O 1.25 / 1.25 = 1 1.88 / 1.25 = 1.5 1 Fe X 2 = 2 Fe 1.5 O X 2 = 3 O Fe2O3

  43. Determine the empirical formula for a compound that contains 10.88 g of calcium and 19.08 g of chlorine. 10.88 g Ca X 1 mole / 40.1 g = .271 mole 19.08 g Cl X 1 mole / 35.5 g = .537 mole .271 / .271 = 1 Ca .537 / .271 = 2 Cl CaCl2

  44. Objective #9 Determining Molecular Formulas *examples: • Analysis of a compound showed it to consist of 80.0% carbon and 20.0% hydrogen by mass. The gram molecular mass is 30.0 g. Determine the molecular formula. *determine empirical formula if needed:

  45. 80.0 g C X 1 mole / 12.0 g = 6.67 moles C 20.0 g H X 1 mole / 1.0 g = 20.0 moles H 6.67 / 6.67 = 1 C 20.0 / 6.67 = 3 H CH3 *determine empirical formula mass (efm): 15 g

  46. *if efm matches gram molecular mass (gmm), then empirical formula is the same as molecular 15 g ≠ 30 g *if masses don’t match, divide gmm by efm to determine factor: 30 g / 15 g = 2 *multiply subscripts of empirical formula by factor to obtain molecular formula: C2H6

  47. If the empirical formula for a compound is C2HCl, determine the molecular formula if the gram molecular mass is 181.5 g. emp. mass 60.5 g 181.5 g / 60.5 g = 2 C4H2Cl2

  48. A compound contains 58.5% carbon, 9.8% hydrogen, 31.4% oxygen and the gram molecular mass is 102 g. Determine the molecular formula. 58.5 g C X 1 mole / 12.0 g = 4.88 mole C 9.8 g H X 1 mole / 1.0 g = 9.8 mole H 31.4 g O X 1 mole / 16.0 g = 1.96 mole O 4.88 / 1.96 = 2.5 C 9.8 / 1.96 = 5 H 1.96 / 1.96 = 1 O

  49. 2.5 C X 2 = 5 C 5 H X 2 = 10 H 1 O X 2 = 2 O emp. formula = C5H10O2 emp. mass = 102 g molecular formula = C5H10O2

  50. Objectives #10-11 Introduction to Stochiometry *Stoichiometry is a method of calculating amounts in a chemical reaction I. Interpreting Chemical Equations Quantitatively *example: 4Al + 3O2 --› 2Al2O3 *the following information can be determined from this reaction: • Number of particles • Number of moles 3. Mass

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