Presentation on theme: "Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole"— Presentation transcript:
1Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole 10.3 Moles of Compounds10.4 Empirical and Molecular Formulas10.5 The Formula for a Hydrate
2Section 10.1 Measuring Matter Chemists use the mole to count atoms, molecules, ions, and formula units.Describe how the mole is definedExplain 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 numberConvert between moles and number of representative particles and vice-versa using the factor-label process.
3Section 10.1 Measuring Matter Key ConceptsThe 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.
4Words That Refer to # of Items NumberPair2Dozen12Gross144Ream500
5The Mole and Avogadro’s Number Abbreviation is molSI base unit for amount of substanceDefined as number of representative particles (carbon atoms) in exactly 12 g of pure carbon-12Mole of anything contains X 1023 representative particles6.022 X 1023 = Avogadro’s number
6The Mole Examples of representative particles: Atoms Ions Electrons MoleculesFormula Units
7Representative Particle Formula UnitMoleculeAtom18 mL H2O63.5 g Cu58.4 g NaCl
8Converting Moles to # of Particles and # of Particles to Moles rep. part. = representative particlesMoles to # of rep. part.# mol 6.02x1023 rep. part./mol= # of rep. part.# rep. part. to moles# rep. part. X 1 mol/6.02x1023 rep. part.= # mol
9Converting Moles to # of Particles and # of Particles to Moles 4.00 Moles Water # Molecules of Water4.00 mol H2O 6.02x1023 molecules H2O /mol H2O= ???= 2.41x1024 molecules H2O1.2x1024 Atoms Iron Moles Iron1.2x1024 atoms Fe X 1 mol Fe/ 6.02x1023 atoms Fe= ???= 2.0 mol Fe
10Practice Problems 1- 4, page 323 Problems 5 (a-b), 6 (a-b), p 324 (Put up next slide as reference)
11Converting Moles to # of Particles and # of Particles to Moles rep. part. = representative particlesMoles to # of rep. part.# mol 6.02x1023 rep. part./mol= # of rep. part.# rep. part. to moles# rep. part. X 1 mol/6.02x1023 rep. part.= # mol
13Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole 10.3 Moles of Compounds10.4 Empirical and Molecular Formulas10.5 The Formula for a Hydrate
14Section 10.2 Mass and the Mole A mole always contains the same number of particles; however, moles of different substances have different masses.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.
15Section 10.2 Mass and the Mole Key ConceptsThe 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.
16Mass vs Number of Objects mass 1 dozen limes mass 1 dozen eggs
17Mass of a Mole of AtomsBy definition (see page 119), an atomic mass unit (amu) is exactly 1/12 the mass of a carbon-12 atomAtomic mass carbon-12 = 12 amuBy definition, mol = number of carbon atoms in 12 g of carbon-12Mass of 1 mol carbon-12 = 12 gMass (g) of 1 mol element numerically same as atomic mass in amu
18Molar Mass Mass in grams of one mole of any pure substance Numerically equal to mass of substance in amuHas units of g/mol
19Molar Mass Hydrogen (element) Manganese Atomic mass = 1.01 amu Molar mass H = 1.01 g/mol6.02x1023 H atoms has mass of 1.01 gManganeseAtomic mass = amuMolar mass Mn = g/mol6.02x1023 Mn atoms has mass of g
20Mole to Mass Conversion 3.00 Moles Mn Mass Mn3.00 mol Mn g Mn/mol Mn= ???= 165 g MnMoles Cr Mass Cr4.50x10-2 mol Cr x g Cr/mol Cr= ???= 2.34 g Cr
21Mass to Mole Conversion 525 g Ca Moles Ca525 g Ca 1 mol Ca/40.08 g Ca= ???= 13.1 mol Ca
23Mass to # Atoms Conversion Need two conversion factorsMass Gold (Au) # of atoms of Gold25.0 g Au 1 mol Au/ g Au 6.02x1023 atoms Au/mol Au= ???= 7.65x1022 atoms AuCould do in two steps if prefer:Calculate # moles Au = molCalculate # atoms from # mol
24# Atoms to Mass Conversion Need two conversion factors# Atoms of He Mass of He5.50x1022 atoms He 1mol He/6.02x1023 atoms He 4.00 g He/ mol He= ???= g HeCould do in two steps if prefer:Calculate # moles He = molCalculate mass from # moles
25Conversion Pathways mass moles represent- ative particles molar mass Avaga- dro’s number
27Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole 10.3 Moles of Compounds10.4 Empirical and Molecular Formulas10.5 The Formula for a Hydrate
28Section 10.3 Moles of Compounds 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.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.
29Section 10.3 Moles of Compounds Key ConceptsSubscripts 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.
30Mole and Chemical Formulas Chemical Formula indicates types and number of each atom contained in compoundFreon – CCl2F2
31Mole and Chemical Formulas Freon – CCl2F2In every molecule of freon, there are:1 carbon atom2 chlorine atoms2 fluorine atomsIn one mole of freon, are 6.02x1023 times as many atoms as aboveIn one dozen molecules, there are 12 times as many atoms as above
32Mole and Chemical Formulas If have 1 mole Freon molecules (CCl2F2) have:1 mole of C atoms2 moles of Cl atoms2 moles of F atoms
35Molar Mass of Compounds Mass of one mole of compoundSum of masses of every particle that makes up the compoundMolar Mass = massimassi = mass of particles contained in one mole of the compound(mass is conserved)i
36Molar Mass of Compounds massi = mass of particles contained in one mole of the compoundIf particles are atoms, thenmassi = molar mass of atom number of moles of atom contained in one mole of the compound
37Molar Mass of Compounds Example: K2CrO4 – 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 cpd2 mol K/mol KC39.10 g K/mol K =78.20 g K/mol KC1 mol Cr /mol KC52.00 g Cr/mol Cr=52.00 g Cr/mol KC4 mol O /mol KC16.00 g O/mol O = g O /mol KCMolar mass K2CrO4 = g/mol KC
39Mole to Mass Conversion Allyl sulfide = (C3H5)2SMass of 2.50 moles of allyl sulfide?Step 1 – calculate molar mass1 mol S g S/mol S = g S6 mol C g C/mol C = g C10 mol H g H/mol H = g HMolar mass = g/mol (C3H5)2S
40Mole to Mass Conversion Allyl sulfide = (C3H5)2SMass of 2.50 moles of allyl sulfide?Step 2 – convert moles to massMolar mass = g/mol (C3H5)2S2.50 mol (C3H5)2S g (C3H5)2S/ mol (C3H5)2S= 286 g (C3H5)2SNote: 3 SF in answer
45Mass to # Particles Conversion Problem AlCl3 Sample 35.6 g# Al+3 ions? # Cl- ions?Mass of 1 formula unit of AlCl3?Step 1 – calculate molar mass1 mol Al g Al/mol Al = g Al3 mol Cl g Cl/mol Cl = g CMolar mass AlCl3 = g/mol AlCl3
46Mass to # Particles Conversion Problem AlCl3 Sample 35.6 g# Al+3 ions? # Cl- ions?Mass of 1 formula unit of AlCl3?Step 2 – Convert mass to moles35.6 g AlCl3 1mol AlCl3/ g AlCl3 =mol AlCl3
47Mass to # Particles Conversion Problem AlCl3 Sample 35.6 g# Al+3 ions? # Cl- ions?Mass of 1 formula unit of AlCl3?Step 3 – Convert moles to # particles0.267 mol AlCl3 x 6.02x1023 formula units AlCl3/mol AlCl3= 1.61x1023 formula units AlCl3
48Mass to # Particles Conversion Problem AlCl3 Sample 35.6 g# Al+3 ions? # Cl- ions?Mass of 1 formula unit of AlCl3?Step 4 – Convert # form. units to # ions1.61x1023 formula unit AlCl3 x 1 Al+3/formula unit AlCl3= 1.61x1023 Al+3 ions1.61x1023 formula unit AlCl3 x 3 Cl-/formula unit AlCl3= 4.83x1023 Cl- ions
49Mass to # Particles Conversion Problem AlCl3 Sample 35.6 g# Al+3 ions? # Cl- ions?Mass of 1 formula unit of AlCl3?Step 5 – Convert molar mass to g per formula unitg AlCl3/mol x 1 mol/6.0221x1023 formula unit= x10-22 g AlCl3/ formula unitPage 339 of book incorrect – need 5 SF
50Conversions For Compounds (Cpd) moles atoms or ions in cpdchemical formularepresent- ative particlesmassmolescpdmolar massAvgad.#
52Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole 10.3 Moles of Compounds10.4 Empirical and Molecular Formulas + Laws of Definite Proportions, Multiple Proportions from section 3.410.5 The Formula for a Hydrate
53Section 3.4 Elements and Compounds A compound is a combination of two or more elements.Explain how all compounds obey the laws of definite and multiple proportions.Demonstrate using calculations that these laws apply to specific compounds.
54Section 3.4 Elements and Compounds Key ConceptsThe 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.
55Section 10.4 Empirical & Molecular Formulas A molecular formula of a compound is a whole-number multiple of its empirical formula.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.
56Section 10.4 Empirical & Molecular Formulas Key ConceptsThe 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.
57Percent CompositionUnknown compound’s composition may be determined by elemental analysisResults reported as percent by mass% by mass =100 mass of element/mass of compoundPercent by mass for all elements in the compound called percent composition
58Laws of Definite and Multiple Proportions These laws were developed as result of careful experimental measurements prior to the full development of the mole concept
59Law 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 presentH2O = formula for water no matter how much water you have – proportions always same, mole ratio same, mass ratios same
60Law of Definite (or Constant) Proportion (or Composition) Both sources of calcium carbonate (CaCO3) have same % composition
61Law of Definite Proportions Focus here is % composition from some chemical analysis% by mass = 100 mass element mass compoundMasses are given rather than derived from a formula
62Law of Definite Proportions % by mass = 100 mass element mass compoundExamine analyses of sucrose in table 3.4, page 88Analysis of g sample same as analysis of g sample have the same compositionBook says same compound (sort of)
64Law of Multiple Proportions If elements form more than one compound, those compounds will have compositions that are small, whole-number multiples of each otherFocus: ratio of mass ratios = integerWater vs Hydrogen PeroxideH2O (O:H 16:2) vs H2O2 (O:H 32:2)Ratio of Mass ratios O:H (H2O2) / O:H (H2O) = 2:1
68% Composition from Formula Elemental analysis to determine percent compositionDone entirely by experimentDon’t need formulaEasier to do if know elements presentIf already know chemical formula, can compute percent composition
69% Composition from Formula Step 1 – Assume have 1 mol of compoundStep 2 – Look up the molar mass of element jStep 3 – Calculate the mass of element j present using formulaStep 4 – Do steps 2 & 3 until all elements done. masses = molar massStep 5 – Use mass from step 3 and molar mass to get % for element j
70% Composition from Formula NaHCO3Step 1 – assume have 1 mol NaHCO3Steps 2 & 3 – get masses of each element1 mol Na g Na/mol Na = g Na1 mol H g H/mol H = g H1 mol C g C/mol C = g C3 mol O g O/mol O = g OStep 4 – sum masses to get molar massMolar mass = g/mol NaHCO3
71% Composition from Formula Step 5 – use molar mass and individual masses to calculate % composition% Na = 100 g Na/84.01 g NaHCO3= %% H = 100 g H/ g NaHCO3= %% C = 100 g C/ g NaHCO3= %% O = 100 g O/ g NaHCO3= %
72% Composition from Formula Cross – check: % should sum to 10027.37% % % % = %Note: only can keep 2 places after decimal point even though 1.200% has 3 places after decimal
74Empirical FormulaSmallest whole number ratio of elements in a compoundIf ionic, same as formula unitIf non-ionic, may or may not be molecular formulaNaCl – ionic, formula unitHO - non ionic (H2O2, H3O3, H4O4, etc)CHO – non ionic (C2H2O2, C3H3O3, etc)
75Empirical Formula Can determine from: Percent composition information Assume 100 g of compoundActual masses of elements for given mass of the compound
76Empirical FormulaStep 1 – If given % composition, determine g of each element in 100 gStep 2 – Calculate moles of each element from the mass of the elementStep 3 – Calculate mole ratio of elementsStep 4 – Convert mole ratio to whole numbersSince compounds have whole numbers of atoms, mole ratio of compound must involve whole numbers
77Empirical Formula Given % Comp Methyl AcetateC 48.64% H 8.16% O 43.20%Steps 1&2 – Assume 100 g and get # moles48.64 g C 1 mol C/12.01 g C = mol C8.16 g H 1 mol H/1.008 g H = 8.10 mol H43.20 g O 1 mol O/16.00 g O = mol OStep 3 – Compute mole ratio4.05 mol C : 8.10 mol H : mol O
78Empirical Formula Given % Comp Methyl Acetate4.05 mol C : 8.10 mol H : mol OStep 4 – Convert mole ratio to whole #Divide all values by smallest4.050 mol C mol O = C/O ratio8.10 mol H mol O = H/O ratio2.700 mol O mol O = O/O ratioFind multiplier to make whole number ratios2 x 1.5 = x 3.0 = x 1.0 = 2
79Empirical Formula Given % Comp Methyl Acetate2 x 1.5 = x 3.0 = x 1.0 = 2Empirical formula C3H6O2
80PracticePractice includes both percent composition given and masses given type problemsProblems , page 346Problems 161, 165, 167, pages 360-1
81Molecular formula = n x (empirical formula) CH – empirical formulaC2H2 – molecular formula for acetyleneC6H6 – molecular formula for benzeneMolecular formula = n x (empirical formula)Need to know molar mass of actual compound to determine molecular formula from molar mass of empirical formula
82Molecular Formula CH – empirical formula Molar mass: 13.02 g/mol 1. Molar mass of compound = g/molRatio molar masses = 2.000Molecular formula = C2H22. Molar mass of compound = g/molRatio molar masses = 6.000Molecular formula = C6H6
83Empirical and Molecular Formulas Process Map – Fig 10.15, page 347 Express % by mass in gFind # moles each elementExamine mole ratioWrite empirical formulaFind integer n relating emp. & molecular formulasMultiply subscripts by nWrite molecular formula
84Practice Problems 62-66, page 350 Problem 164, page 361
85Chapter 10 – The Mole 10.1 Measuring Matter 10.2 Mass and the Mole 10.3 Moles of Compounds10.4 Empirical and Molecular Formulas10.5 The Formula for a Hydrate
86Section 10.5 Formulas of Hydrates Hydrates are solid ionic compounds in which water molecules are trapped.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.
87Section 10.5 Formulas of Hydrates Key ConceptsThe 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.
88Formulas for HydratesHydrate – compound that has a specific number of water molecules bound to the atomsNumber of water molecules associated with each formula unit is written following a dotCaCl22H2O Calcium chloride dihydrate
90Hydrates Single compound can form a number of different hydrates CaCl2 – mono-, di-, and hexa- hydratesCaCl2H2O, CaCl22H2O, CaCl26H2O“Water of hydration” can be driven off by heating – produces anhydrous formMass of water associated with formula unit must be included in molar mass calculations
91Color Change in Hydrated Compounds Heating: drives off water and produces anhydrous formHydratedAnhydrousCopper (II) sulfate
92Determining Hydrate Formula BaCl2•xH2O - what is value of x ?Initial mass = 5.00 gFinal mass after heating = 4.26 gMass water = = 0.74 g H2OMol H2O = 0.74 g 1 mol/18.02 g H2O= mol H2OMol BaCl2 = 4.26 g BaCl2 1 mol/ g BaCl2 = mol BaCl2
93Determining Hydrate Formula BaCl2•xH2O - what is value of x ?0.041 mol H2O mol BaCl2x = moles H2O /moles BaCl2x = mol H2O / mol BaCl2x = 2.0 mol H2O / 1.00 mol BaCl2x= 2/1= 2BaCl2•2H2O