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Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate.

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Presentation on theme: "Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate."— Presentation transcript:

1 Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate

2 Section 10.1 Measuring Matter Describe how the mole is defined Explain how a mole is used to indirectly count the number of particles of matter. Relate the mole to a common everyday counting unit. Explain the relationship between the mole and Avogadro’s number Convert between moles and number of representative particles and vice-versa using the factor-label process. Chemists use the mole to count atoms, molecules, ions, and formula units.

3 Key Concepts The mole is a unit used to count particles of matter indirectly. One mole of a pure substance contains Avogadro’s number of particles. Representative particles include atoms, ions, molecules, formula units, electrons, and other similar particles. One mole of carbon-12 atoms has a mass of exactly 12 g. Conversion factors written from Avogadro’s relationship can be used to convert between moles and number of representative particles. Section 10.1 Measuring Matter

4 Words That Refer to # of Items WordNumber Pair2 Dozen12 Gross144 Ream500

5 The Mole and Avogadro’s Number Abbreviation is mol SI base unit for amount of substance Defined as number of representative particles (carbon atoms) in exactly 12 g of pure carbon-12 Mole of anything contains X representative particles X = Avogadro’s number

6 The Mole Examples of representative particles: AtomsIonsElectronsMolecules Formula Units

7 18 mL H 2 O 63.5 g Cu 58.4 g NaCl Representative Particle Molecule Atom Formula Unit

8 Converting Moles to # of Particles and # of Particles to Moles rep. part. = representative particles Moles to # of rep. part. # mol  6.02x10 23 rep. part./mol = # of rep. part. # rep. part. to moles # rep. part. X 1 mol/6.02x10 23 rep. part. = # mol

9 = ??? Converting Moles to # of Particles and # of Particles to Moles 4.00 Moles Water  # Molecules of Water 4.00 mol H 2 O  6.02x10 23 molecules H 2 O /mol H 2 O = 2.41x10 24 molecules H 2 O 1.2x10 24 Atoms Iron  Moles Iron 1.2x10 24 atoms Fe X 1 mol Fe/ 6.02x10 23 atoms Fe = 2.0 mol Fe = ???

10 Practice Problems 1- 4, page 323 Problems 5 (a-b), 6 (a-b), p 324 (Put up next slide as reference)

11 Converting Moles to # of Particles and # of Particles to Moles rep. part. = representative particles Moles to # of rep. part. # mol  6.02x10 23 rep. part./mol = # of rep. part. # rep. part. to moles # rep. part. X 1 mol/6.02x10 23 rep. part. = # mol

12 Practice Section Assessment, page 324 Problems Chapter Assessment, page 358 Problems 90(a-d), 91(a-d), 92(a-d), 93(a-d), 94(a-d), 95(a-c),

13 Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate

14 Section 10.2 Mass and the Mole Determine the molar mass of an element given its atomic mass in amu. Convert between number of moles and the mass of an element and vice-versa using using the factor-label process. Convert between number of moles and number of atoms of an element and vice-versa using the factor- label process. A mole always contains the same number of particles; however, moles of different substances have different masses.

15 Key Concepts The mass in grams of 1 mol of any pure substance is called its molar mass. The molar mass of an element is numerically equal to its atomic mass. The molar mass of any substance is the mass in grams of Avogadro’s number of representative particles of the substance. Molar mass is used to convert from moles to mass. The inverse of molar mass is used to convert from mass to moles. Section 10.2 Mass and the Mole

16 Mass vs Number of Objects mass 1 dozen limes  mass 1 dozen eggs

17 Mass of a Mole of Atoms By definition (see page 119), an atomic mass unit (amu) is exactly 1/12 the mass of a carbon-12 atom Atomic mass carbon-12 = 12 amu By definition, mol = number of carbon atoms in 12 g of carbon-12 Mass of 1 mol carbon-12 = 12 g Mass (g) of 1 mol element numerically same as atomic mass in amu

18 Molar Mass Mass in grams of one mole of any pure substance Numerically equal to mass of substance in amu Has units of g/mol

19 Molar Mass Hydrogen (element) Atomic mass = 1.01 amu Molar mass H = 1.01 g/mol 6.02x10 23 H atoms has mass of 1.01 g Manganese Atomic mass = amu Molar mass Mn = g/mol 6.02x10 23 Mn atoms has mass of g

20 = ??? Mole to Mass Conversion 3.00 Moles Mn  Mass Mn 3.00 mol Mn  g Mn/mol Mn = 165 g Mn Moles Cr  Mass Cr 4.50x10 -2 mol Cr x g Cr/mol Cr = 2.34 g Cr = ???

21 Mass to Mole Conversion 525 g Ca  Moles Ca 525 g Ca  1 mol Ca/40.08 g Ca = 13.1 mol Ca = ???

22 Practice Problems 15 (a-b), 16 (a-b) page 328 Problems 17 (a-b), 18 (a-b) page 329

23 Mass to # Atoms Conversion Need two conversion factors Mass Gold (Au)  # of atoms of Gold 25.0 g Au  1 mol Au/ g Au  6.02x10 23 atoms Au/mol Au = 7.65x10 22 atoms Au Could do in two steps if prefer: Calculate # moles Au = mol Calculate # atoms from # mol = ???

24 # Atoms to Mass Conversion Need two conversion factors # Atoms of He  Mass of He 5.50x10 22 atoms He  1mol He/6.02x10 23 atoms He  4.00 g He/ mol He = g He Could do in two steps if prefer: Calculate # moles He = mol Calculate mass from # moles = ???

25 Conversion Pathways massmolesrepresent- ative particles molar mass Avaga- dro’s number

26 Practice Problems 19 (a-c), 21 (a-b) page 331 Problems 20 (a-e) page 331

27 Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate

28 Section 10.3 Moles of Compounds Recognize the mole relationships shown by a chemical formula. Calculate the molar mass of a compound. Convert between the number of moles and mass of a compound and vice-versa using the factor-label process. Apply conversion factors to determine the number of atoms or ions in a known mass of a compound. The molar mass of a compound can be calculated from its chemical formula and can be used to convert from mass to moles of that compound.

29 Key Concepts Subscripts in a chemical formula indicate how many moles of each element are present in 1 mol of the compound. The molar mass of a compound is calculated from the molar masses of all of the elements in the compound. Conversion factors based on a compound’s molar mass are used to convert between moles and mass of a compound. Section 10.3 Moles of Compounds

30 Mole and Chemical Formulas Chemical Formula indicates types and number of each atom contained in compound Freon – CCl 2 F 2

31 Mole and Chemical Formulas Freon – CCl 2 F 2 In every molecule of freon, there are: 1 carbon atom 2 chlorine atoms In one dozen molecules, there are 12 times as many atoms as above 2 fluorine atoms In one mole of freon, are 6.02x10 23 times as many atoms as above

32 Mole and Chemical Formulas If have 1 mole Freon molecules (CCl 2 F 2 ) have: 1 mole of C atoms 2 moles of Cl atoms 2 moles of F atoms

33 Moles Compound to Moles Atoms Aluminum oxide (alumina), Al 2 O moles alumina  moles Al mol alumina  2 mol Al +3 /1 mol alumina = 2.50 mol Al +3 = ???

34 Practice Problems page 335

35 Molar Mass of Compounds Mass of one mole of compound Sum of masses of every particle that makes up the compound Molar Mass =  mass i mass i = mass of particles contained in one mole of the compound (mass is conserved) i

36 Molar Mass of Compounds mass i = mass of particles contained in one mole of the compound If particles are atoms, then mass i = molar mass of atom  number of moles of atom contained in one mole of the compound

37 Molar Mass of Compounds Example: K 2 CrO 4 – abbreviate as “KC” Revised from p. 335 text (credit: Allison Mazur) elm = element cpd = compound # mol elm/mol cpd  molar mass elm = mass elm/mol cpd 2 mol K/mol KC  g K/mol K =78.20 g K/mol KC 1 mol Cr /mol KC  g Cr/mol Cr=52.00 g Cr/mol KC 4 mol O /mol KC  g O/mol O = g O /mol KC Molar mass K 2 CrO 4 = g/mol KC

38 Practice Problems (all are a –c) page 335

39 Mole to Mass Conversion Allyl sulfide = (C 3 H 5 ) 2 S Mass of 2.50 moles of allyl sulfide? Step 1 – calculate molar mass 1 mol S  g S/mol S = g S 10 mol H  g H/mol H = g H 6 mol C  g C/mol C = g C Molar mass = g/mol (C 3 H 5 ) 2 S

40 Mole to Mass Conversion Allyl sulfide = (C 3 H 5 ) 2 S Mass of 2.50 moles of allyl sulfide? Step 2 – convert moles to mass = 286 g (C 3 H 5 ) 2 S Note: 3 SF in answer Molar mass = g/mol (C 3 H 5 ) 2 S 2.50 mol (C 3 H 5 ) 2 S  g (C 3 H 5 ) 2 S / mol (C 3 H 5 ) 2 S

41 Practice Problems page 336

42 Mass to Mole Conversion Ca(OH) 2 Moles Ca(OH) 2 in 325 g? Step 1 – calculate molar mass 1 mol Ca  g Ca/mol Ca= g Ca 2 mol O  g O/mol O = g O 2 mol H  g H/mol H = g H Molar mass = g/mol Ca(OH) 2

43 Mass to Mole Conversion Ca(OH) 2 Moles Ca(OH) 2 in 325 g? Step 2 – convert to moles Molar mass = g/mol Ca(OH) g Ca(OH) 2  1 mol Ca(OH) 2 /74.10 g Ca(OH) 2 = 4.39 mol Ca(OH) 2 Note: 3 SF in answer

44 Practice Problems 40(a-c) – 41(a-b) page 337

45 Mass to # Particles Conversion Problem 10.9 AlCl 3 Sample 35.6 g # Al +3 ions? # Cl - ions? Mass of 1 formula unit of AlCl 3 ? Step 1 – calculate molar mass 1 mol Al  g Al/mol Al = g Al Molar mass AlCl 3 = g/mol AlCl 3 3 mol Cl  g Cl/mol Cl = g C

46 Mass to # Particles Conversion Problem 10.9 AlCl 3 Sample 35.6 g # Al +3 ions? # Cl - ions? Mass of 1 formula unit of AlCl 3 ? Step 2 – Convert mass to moles 35.6 g AlCl 3  1mol AlCl 3 / g AlCl 3 = mol AlCl 3

47 Mass to # Particles Conversion Problem 10.9 AlCl 3 Sample 35.6 g # Al +3 ions? # Cl - ions? Mass of 1 formula unit of AlCl 3 ? Step 3 – Convert moles to # particles mol AlCl 3 x 6.02x10 23 formula units AlCl 3 /mol AlCl 3 = 1.61x10 23 formula units AlCl 3

48 Mass to # Particles Conversion Problem 10.9 AlCl 3 Sample 35.6 g # Al +3 ions? # Cl - ions? Mass of 1 formula unit of AlCl 3 ? Step 4 – Convert # form. units to # ions 1.61x10 23 formula unit AlCl 3 x 1 Al +3 /formula unit AlCl 3 = 1.61x10 23 Al +3 ions 1.61x10 23 formula unit AlCl 3 x 3 Cl - /formula unit AlCl 3 = 4.83x10 23 Cl - ions

49 Mass to # Particles Conversion Problem 10.9 AlCl 3 Sample 35.6 g # Al +3 ions? # Cl - ions? Mass of 1 formula unit of AlCl 3 ? Step 5 – Convert molar mass to g per formula unit = x g AlCl 3 / formula unit g AlCl 3 /mol x 1 mol/6.0221x10 23 formula unit Page 339 of book incorrect – need 5 SF

50 Conversions For Compounds (Cpd) mass moles cpd represent- ative particles molar mass Avgad.# moles atoms or ions in cpd chemical formula

51 Practice Problems 42 (a-c), 43 (a-c), 44 (a-c), 45, 46 (a-d) page 339 Problems 128 – 152 pages 359 – 60

52 Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas + Laws of Definite Proportions, Multiple Proportions from section The Formula for a Hydrate

53 Explain how all compounds obey the laws of definite and multiple proportions. Demonstrate using calculations that these laws apply to specific compounds. A compound is a combination of two or more elements. Section 3.4 Elements and Compounds

54 Key Concepts The law of definite proportions states that a compound is always composed of the same elements in the same proportions. The law of multiple proportions states that if elements form more than one compound, those compounds will have compositions that are whole-number multiples of each other.

55 Section 10.4 Empirical & Molecular Formulas Explain what is meant by the percent composition of a compound and calculate it from the formula for the compound. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data using calculations. A molecular formula of a compound is a whole-number multiple of its empirical formula.

56 Key Concepts The percent by mass of an element in a compound gives the percentage of the compound’s total mass due to that element. The subscripts in an empirical formula give the smallest whole-number ratio of moles of elements in the compound. The molecular formula gives the actual number of atoms of each element in a molecule or formula unit of a substance. The molecular formula is a whole-number multiple of the empirical formula. Section 10.4 Empirical & Molecular Formulas

57 Percent Composition Unknown compound’s composition may be determined by elemental analysis Results reported as percent by mass % by mass = 100  mass of element/mass of compound Percent by mass for all elements in the compound called percent composition

58 Laws of Definite and Multiple Proportions These laws were developed as result of careful experimental measurements prior to the full development of the mole concept

59 Law of Definite Proportions Regardless of amount, compound composed of same elements in same proportion by mass  statement that compound’s formula doesn’t change with amount of compound present H 2 O = formula for water no matter how much water you have – proportions always same, mole ratio same, mass ratios same

60 Law of Definite (or Constant) Proportion (or Composition) Both sources of calcium carbonate (CaCO 3 ) have same % composition

61 Law of Definite Proportions Focus here is % composition from some chemical analysis % by mass = 100  mass element mass compound Masses are given rather than derived from a formula

62 Law of Definite Proportions % by mass = 100  mass element mass compound Examine analyses of sucrose in table 3.4, page 88 Analysis of g sample same as analysis of g sample  have the same composition Book says  same compound (sort of)

63 Practice Law of Definite Proportions Problems , page 88 Problems 72, 74-78, pages Problem 3, page 977

64 Law of Multiple Proportions If elements form more than one compound, those compounds will have compositions that are small, whole- number multiples of each other Focus: ratio of mass ratios = integer Water vs Hydrogen Peroxide H 2 O (O:H 16:2) vs H 2 O 2 (O:H 32:2) Ratio of Mass ratios O:H (H 2 O 2 ) / O:H (H 2 O) = 2:1

65 Atomic Basis of the Law of Multiple Proportions

66 Law of Multiple Proportions Compound 1 Compound 2 Copper Chloride Compounds Cpd% Cu% ClRatio Cu : Cl Ratio of cpd #1 ratio to #2 ratio : : 1

67 Practice Law of Multiple Proportions Problems 73, 79, 80 pages 95-6 Problem 4, page 977

68 % Composition from Formula Elemental analysis to determine percent composition Done entirely by experiment Don’t need formula Easier to do if know elements present If already know chemical formula, can compute percent composition

69 % Composition from Formula Step 1 – Assume have 1 mol of compound Step 2 – Look up the molar mass of element j Step 3 – Calculate the mass of element j present using formula Step 4 – Do steps 2 & 3 until all elements done.  masses = molar mass Step 5 – Use mass from step 3 and molar mass to get % for element j

70 % Composition from Formula NaHCO 3 Step 1 – assume have 1 mol NaHCO 3 Steps 2 & 3 – get masses of each element 1 mol Na  g Na/mol Na= g Na 3 mol O  g O/mol O = g O 1 mol H  g H/mol H = g H Molar mass = g/mol NaHCO 3 1 mol C  g C/mol C = g C Step 4 – sum masses to get molar mass

71 % Composition from Formula Step 5 – use molar mass and individual masses to calculate % composition % Na = 100  g Na/84.01 g NaHCO 3 % O = 100  g O/ g NaHCO 3 % H = 100  g H/ g NaHCO 3 % C = 100  g C/ g NaHCO 3 = % = % = % = %

72 % Composition from Formula Cross – check: % should sum to % % % % = % Note: only can keep 2 places after decimal point even though 1.200% has 3 places after decimal

73 Practice Problems , page 344 Problems 162, 168 – 171, 174 pages

74 Empirical Formula Smallest whole number ratio of elements in a compound If ionic, same as formula unit If non-ionic, may or may not be molecular formula NaCl – ionic, formula unit HO - non ionic (H 2 O 2, H 3 O 3, H 4 O 4, etc) CHO – non ionic (C 2 H 2 O 2, C 3 H 3 O 3, etc)

75 Empirical Formula Can determine from: Percent composition information  Assume 100 g of compound Actual masses of elements for given mass of the compound

76 Empirical Formula Step 1 – If given % composition, determine g of each element in 100 g Step 2 – Calculate moles of each element from the mass of the element Step 3 – Calculate mole ratio of elements Step 4 – Convert mole ratio to whole numbers Since compounds have whole numbers of atoms, mole ratio of compound must involve whole numbers

77 Empirical Formula Given % Comp Methyl Acetate C 48.64% H 8.16% O 43.20% Steps 1&2 – Assume 100 g and get # moles g C  1 mol C/12.01 g C = mol C 8.16 g H  1 mol H/1.008 g H = 8.10 mol H g O  1 mol O/16.00 g O = mol O Step 3 – Compute mole ratio 4.05 mol C : 8.10 mol H : mol O

78 Empirical Formula Given % Comp Methyl Acetate 4.05 mol C : 8.10 mol H : mol O Step 4 – Convert mole ratio to whole # Divide all values by smallest mol C  mol O = C/O ratio 8.10 mol H  mol O = 3.00 H/O ratio mol O  mol O = O/O ratio Find multiplier to make whole number ratios 2 x 1.5 = 3 2 x 3.0 = 6 2 x 1.0 = 2

79 Empirical Formula Given % Comp Methyl Acetate 2 x 1.5 = 3 2 x 3.0 = 6 2 x 1.0 = 2 Empirical formula C 3 H 6 O 2

80 Practice Practice includes both percent composition given and masses given type problems Problems , page 346 Problems 161, 165, 167, pages 360-1

81 Molecular Formula CH – empirical formula C 2 H 2 – molecular formula for acetylene C 6 H 6 – molecular formula for benzene Molecular formula = n x (empirical formula) Need to know molar mass of actual compound to determine molecular formula from molar mass of empirical formula

82 CH – empirical formula Molar mass: g/mol 1. Molar mass of compound = g/mol  Ratio molar masses = Molecular Formula  Molecular formula = C 2 H 2 2. Molar mass of compound = g/mol  Ratio molar masses =  Molecular formula = C 6 H 6

83 Empirical and Molecular Formulas Process Map – Fig 10.15, page 347 Express % by mass in g Find # moles each element Examine mole ratio Write empirical formula Find integer n relating emp. & molecular formulas Multiply subscripts by n Write molecular formula

84 Practice Problems 62-66, page 350 Problem 164, page 361 Problems 29, 30 page 982

85 Chapter 10 – The Mole 10.1Measuring Matter 10.2 Mass and the Mole 10.3Moles of Compounds 10.4Empirical and Molecular Formulas 10.5The Formula for a Hydrate

86 Section 10.5 Formulas of Hydrates Explain what a hydrate is and name a hydrate based on its composition. Explain what the term anhydrous means. Determine the formula of a hydrate from laboratory data. Explain what a desiccant is and how it is used. Hydrates are solid ionic compounds in which water molecules are trapped.

87 Key Concepts The formula of a hydrate consists of the formula of the ionic compound and the number of water molecules associated with one formula unit. The name of a hydrate consists of the compound name and the word hydrate with a prefix indicating the number of water molecules in 1 mol of the compound. Anhydrous compounds are formed when hydrates are heated. Section 10.5 Formulas of Hydrates

88 Formulas for Hydrates Hydrate – compound that has a specific number of water molecules bound to the atoms Number of water molecules associated with each formula unit is written following a dot CaCl 2  2H 2 O Calcium chloride dihydrate

89 Formulas for Hydrates – page 351

90 Hydrates Single compound can form a number of different hydrates CaCl 2 – mono-, di-, and hexa- hydrates CaCl 2  H 2 O, CaCl 2  2H 2 O, CaCl 2  6H 2 O “Water of hydration” can be driven off by heating – produces anhydrous form Mass of water associated with formula unit must be included in molar mass calculations

91 Color Change in Hydrated Compounds Heating: drives off water and produces anhydrous form Copper (II) sulfate Hydrated Anhydrous

92 Determining Hydrate Formula BaCl 2xH 2 O - what is value of x ? Initial mass = 5.00 g Final mass after heating = 4.26 g Mass water = = 0.74 g H 2 O Mol H 2 O = 0.74 g  1 mol/18.02 g H 2 O = mol H 2 O Mol BaCl 2 = 4.26 g BaCl 2  1 mol/ g BaCl 2 = mol BaCl 2

93 Determining Hydrate Formula BaCl 2xH 2 O - what is value of x ? mol H 2 O mol BaCl 2 x = moles H 2 O /moles BaCl 2 x = mol H 2 O / mol BaCl 2 x = 2.0 mol H 2 O / 1.00 mol BaCl 2 x= 2/1= 2 BaCl 22H 2 O

94 Practice Problems , page 353 Problems , page 361 Problems , page 982

95 Hydrates Anhydrous solids used as drying agents (desiccants)

96 End of Chapter Problems 63-64, page 340 Problems , page 349 Problems 33-34, page 877


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