Presentation is loading. Please wait.

Presentation is loading. Please wait.

The Mole and Stoichiometry H Chemistry I Unit 8. Objectives #1-4 The Mole and its Use in Calculations I.Fundamentals 1 mole of an element or compound.

Similar presentations


Presentation on theme: "The Mole and Stoichiometry H Chemistry I Unit 8. Objectives #1-4 The Mole and its Use in Calculations I.Fundamentals 1 mole of an element or compound."— Presentation transcript:

1 The Mole and Stoichiometry H Chemistry I 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 particles and has a mass = to its molar mass “The Triad of the Mole” 1 mole = 6.02 X particles = molar mass *henceforth the number 6.02 X 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 II.Introduction to Mole Problems (Follow the procedures outlined in Unit 1 for dimensional analysis problems) 1.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 atoms *Use conversion factor solve problem:.500 moles Fe X 6.02 X atoms/1 mole

6 Check answer for units, sig. figs,, and reasonableness 3.01 X atoms Fe 2.Calculate the number of atoms in.450 moles of zinc..450 moles Zn X 6.02 X atoms Zn / 1 mole = 2.71 X atoms Zn

7 3.Calculate the number of moles in 2.09 X atoms of sulfur X atoms S X 1mole S / 6.02 X atoms S = 34.7 moles S 4.Calculate the number of moles in 3.06 X atoms of chlorine X atoms Cl X 1 mole Cl / 6.02 X atoms Cl =.0508 moles Cl

8 5.Calculate the number of atoms in 35.7 g of silicon g Si X 6.02 X atoms Si / 28.1 g = 7.65 X atoms Si

9 II.Molar Mass 1.Calculate the molar mass of H 3 PO 4 : 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 2.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 3.Calculate the molar mass of BaCl 2. 2H 2 O: 1 Ba at g = g 2 Cl at 35.5 g = 71.0 g 2 H 2 O at 18.0 g = 36.0 g Total = g

12 III.Mole-Mass Problems 1.How many grams are in 7.20 moles of dinitrogen trioxide? *determine known: 7.20 moles N 2 O 3 *determine unknown: grams N 2 O 3

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 N 2 O 3 X 76.0 g N 2 O 3 / 1 mole = g

14 *check answer for units, sig. figs., and reasonableness: 547 g N 2 O 3

15 2.What is the mass in grams of 4.52 g of barium chloride? 4.52 g BaCl 2 X g / 1 mole = 941 g 3. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate g (NH 4 ) 3 PO 4 X 1 mole (NH 4 ) 3 PO 4 / g X 3 NH 4 +1 / 1 mole X 6.02 X ions / 1 mole NH 4 +1 = 4.24 X NH 4 +1 ions

16 4.Calculate the mass of carbon in 7.88 X molecules of C 8 H X molecules C 8 H 18 X 1 mole C 8 H 18 / 6.02 X molecules C 8 H 18 X 8 moles C / 1 mole C 8 H 18 X 12.0 g C / 1 mole C = 1.26 X 10 5 g C

17 5.Calculate the number of molecules present in 2.50 moles of water moles H 2 O X 6.02 X molecules / 1 mole = 1.51 X molecules H 2 O

18 6.Calculate the mass in grams of 4.50 X molecules of C 6 H X molecules C 6 H 10 X 82.0 g / 6.02 X molecules = 6130 g C 6 H 10

19 7.Calculate the mass of 1 molecule of propane. 1 molecule C 3 H 8 X 44.0 g / 6.02 X molecules = 7.31 X g C 3 H 8

20 Objective #5 Characteristics of Solutions Characteristic Componentssolute and solvent Particle Size.01 – 1 nm (atoms, ions, and molecules) Tyndall Effect Resultnegative Ability to be separated by filtration no Homogeneous or heterogeneoushomogeneous Examples in different phasessolid – brass liquid – saltwater gas - air

21 Objective #5 Characteristics of Solutions II.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)H 2 O (polar) HCl (polar) H 2 O (polar) I 2 (nonpolar)CCl 4 (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: 1.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 g NaOH X 1 mole / 40.0 g =.288 moles M =.288 moles / 1.50 L =.192 M

27 2.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 K 2 Cr 2 O 7

28 .200 moles K 2 Cr 2 O 7 X g / 1 mole = 59 g K 2 Cr 2 O 7 3.What is the molarity of each ion in the following solutions (assuming all are strong electrolytes).15 M calcium chloride: CaCl 2 1 Ca to 2 Cl.15 M Ca +2 2(.15 M) =.30 M Cl -1

29 .22 M calcium perchlorate Ca(ClO 4 ) 2 1 Ca to 2 ClO 4.22 M Ca +2 2 (.22 M) =.44 M ClO 4 -1

30 4.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 M 1 V 1 = M 2 V 2

31 #1 = concentrated #2 = diluted M 2 = M 1 V 1 / V 2 = (3.50 L) (.200 M) / 5.00 L =.140 M

32 5.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? M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 / M 1 = (.400 M) (2.00 L) / (2.00 M) =.400 L = 400. ml How would one prepare solution?

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

34 *examples: 1.Calculate the percentage composition phosphate (K 3 PO 4 ) *determine molar mass of compound: g 2.Determine total mass of each component: 3 K at 39.1 g each = g 1 P at 31.0 g each = 31.0 g 4 O at 16.0 g each = 64.0 g

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

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

37 2.Calculate the percentage composition of sodium carbonate (Na 2 CO 3 ) molar mass = g % Na = (46.0 g / g) X 100 = 43% % C = (12.0 g / g) X 100 = 11% % O = (48.0 g / g ) X 100 = 45%

38 Objective #8 Determining Empirical Formulas *empirical vs. molecular formulas: H 2 O 2 --› HO H 2 O --› H 2 O *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: K2SK2S

42 2.A compound of iron and oxygen when analyzed showed 70.0% iron and 30.0% oxygen by mass. Determine the empirical formula 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 = / 1.25 = Fe X 2 = 2 Fe 1.5 O X 2 = 3 O Fe 2 O 3

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

44 Objective #9 Determining Molecular Formulas *examples: 1.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 CH 3 *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: C2H6C2H6

47 2.If the empirical formula for a compound is C 2 HCl, determine the molecular formula if the gram molecular mass is g. emp. mass 60.5 g g / 60.5 g = 2 C 4 H 2 Cl 2

48 3.A compound contains 58.5% carbon, 9.8% hydrogen, 31.4% oxygen and the gram molecular mass is 102 g. Determine the molecular formula 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 = C 5 H 10 O 2 emp. mass = 102 g molecular formula = C 5 H 10 O 2

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 + 3O 2 --› 2Al 2 O 3 *the following information can be determined from this reaction: 1.Number of particles 2.Number of moles 3. Mass

51 1.Particles: 4Al + 3O 2 --› 2Al 2 O 3 atom molecule formula unit 4 atoms Al 3 molecules O 2 2 formula units Al 2 O 3

52 2.Moles 4Al + 3O 2 --› 2Al 2 O 3 4 moles Al 3 moles O 2 2 moles Al 2 O 3

53 3.Mass: 4Al + 3O 2 --› 2Al 2 O g Al 96 g O g Al 2 O 3

54 4Al + 3O 2 --› 2Al 2 O 3 *playing with the mole ratios: 4 moles Al --› ? Al 2 O 3 4 moles Al --› 2 moles Al 2 O 3 8 moles Al --› ? Al 2 O 3 8 moles Al --› 1 mole Al 2 O 3 2 moles Al --› ? Al 2 O 3 2 moles Al --› 1 mole Al 2 O 3

55 4Al + 3O 2 --› 2Al 2 O moles Al --› ? moles Al X 2 moles Al 2 O 3 / 4 mole Al = mole Al

56 Objective #12 Stoichiometric Calculations *examples: 1.If 20.0 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced? *write balanced chemical equation if not provided: Mg + 2HCl --› MgCl 2 + H 2

57 *determine known: 20.0 g *determine unknown: g MgCl 2 *convert given to moles: 20.0 g Mg X 1 mole Mg / 24.3 g *select and utilize appropriate mole-mole to convert to moles of unknown: X 1 mole MgCl 2 / 1 mole Mg

58 *if unknown is to be measured in moles skip to step 7. If unknown is to be measured in grams, multiply by unknown’s molar mass. If unknown is to be measured in particles, multiply by Avogadro’s number: X 95.3 g MgCl 2 / 1 mole MgCl 2

59 *perform math: 78.4 g MgCl 2 *check answer for units, sig figs, and reasonableness

60 2.How many moles of water can be formed from 7.5 moles of ethyne gas (C 2 H 2 ) reacting with excess oxygen gas? 2C 2 H 2 + 5O 2 --› 4CO 2 + 2H 2 O 7.5 moles C 2 H 2 X 2 moles H 2 O / 2 moles C 2 H 2 = 7.5 moles H 2 O

61 3.If g of Rb reacts with excess S 8 how many formula units of Rb 2 S can be formed? 16 Rb + S 8 --› 8Rb 2 S g Rb X 1 mole Rb / 85.5 g Rb X 8 mole Rb 2 S / 16 mole Rb X 6.02 X f. units Rb 2 S / 1 mole Rb 2 S = X f. units Rb 2 S

62 Objective #13 Limiting Reagents *General Terms: excess reagent reactant that is leftover limiting reagent reactants that is used up *Examples: 1.How many grams of carbon dioxide can be obtained by the action of 50.0 g of sulfuric acid on g of calcium carbonate? (follow general steps for stoichiometry, just do it twice)

63 H 2 SO 4 + CaCO 3 --› CaSO 4 + CO 2 +H 2 O 50.0 g H 2 SO 4 X 1 mole H 2 SO 4 / 98.1 g X 1 mole CO 2 / 1 mole H 2 SO 4 X 44.0 g CO 2 / 1 mole CO 2 = 22.4 g CO g CaCO 3 X 1 mole CaCO 3 / g X 1 mole CO 2 / 1 mole CaCO 3 X 44.0 g CO 2 / 1 mole CO 2 = g CO 2

64 (examine two answers produced; the smaller of the two is the right answer of product produced) *the reactant quantity that provides the smaller correct answer is called the limiting reagent *the reactant quantity that provides the larger incorrect answer is called the excess reagent

65 2.How many grams of hydrogen gas can be produced by the reaction of 100. g of phosphoric acid on 25.0 g of aluminum? 2H 3 PO 4 + 2Al --› 2AlPO 4 + 3H g H 3 PO 4 X 1 mole H 3 PO 4 / 98.0 g X 3 mole H 2 / 2 mole H 3 PO 4 X 2.0 g H 2 / 1 mole H 2 = 3.06 g H 2

66 25.0 g Al X 1 mole Al / 27.0 g Al X 3 mole H 2 / 1 mole H 2 X 2.0 g H 2 / 1 mole= 2.78 g H g H 2 (smaller value)

67 Determine how much of the excess reagent is left over. *determine limiting reagent: Al *use stochiometry to convert from mass of limiting reagent to find mass of excess reagent actually used: 25.0 g Al X 1 mole Al / 27.0 g Al X 2 mole H 3 PO 4 / 2 mole Al X 98.0 g H 3 PO 4 / 1mole = 90.7 g H 3 PO 4 used

68 *Subtract amount of excess reagent used from original amount of excess reactant given in problem to determine mass of excess reactant left over: 100. g – 90.7 g = 9 g left over

69 3.How many grams of sodium chloride can be produced from then reaction of 15.0 g of chlorine gas and 15.0 g of sodium bromide in the above reaction? Determine how much of the excess reagent is left over? Cl 2 + 2NaBr --› Br 2 + 2NaCl

70 15.0 g Cl 2 X 1 mole Cl 2 / 71.0 g Cl 2 X 2 mole NaCl / 1 mole Cl 2 X 58.5 g NaCl / 1 mole NaCl = 24.7 g NaCl g NaBr X 1 mole NaBr / g NaBr X 2 mole NaCl / 2 mole NaBr X 58.5 g NaCl / 1 mole NaCl = 8.53 g NaCl

71 15.0 g NaBr X 1 mole NaBr / g NaBr X 1 mole Cl 2 / 2 mole NaBr X 71.0 g Cl 2 / 1 mole Cl 2 = 5.17 g Cl 2 (used) g – 5.17 g = 9.8 g Cl 2 leftover

72 Objective #15 Percentage Yield *General Formula: Percent Yield = (actual yield / theoretical yield) X 100% *examples: 1.When 9.00 g of aluminum react with an excess of phosphoric acid, 30.0 g of aluminum phosphate are produced. What is the percentage yield of this reaction?

73 *write balanced chemical equation: 2Al + 2 H 3 PO 4 --› 3H 2 + 2AlPO 4 *determine actual yield of a product in reaction given in the problem: 30.0 g AlPO 4 *use given reactant quantity and general stochiometry procedures to determine mass of appropriate product:

74 9.00 g Al X 1 mole Al / 27.0 g Al X 2 mole AlPO 4 / 2 mole Al X g / 1 mole = 40.7 g AlPO 4 *divide actual yield from problem by the theoretical yield calculated to find percentage yield: %Y = (30.0 g / 40.7 g) X 100 = 74%

75 2.Calculate the percentage yield if 6.00 g of aluminum are produced from the decomposition of 25.0 g of aluminum oxide. 2Al 2 O 3 --› 4Al + 3O g Al 2 O 3 X 1 mole Al 2 O 3 / g X 4 mole Al / 2 mole Al 2 O 3 X 27.0 g Al / 1 mole Al = 13.2 g Al %Y= (6.00 g / 13.2 g) X 100% = 45%


Download ppt "The Mole and Stoichiometry H Chemistry I Unit 8. Objectives #1-4 The Mole and its Use in Calculations I.Fundamentals 1 mole of an element or compound."

Similar presentations


Ads by Google