3.3 Atomic and Molecular Weights

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Demo: large bag of Styrofoam peanuts, bottle of sand, 1 L beaker, 18 mL of water

How can we count very large numbers of How can we count very large numbers of particles?particles?

Demo: weigh pennies of various agesDemo: weigh pennies of various ages

3.3 Atomic and Molecular Weights

3.3 Atomic and Molecular Weights

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Masses of CoinsMasses of Coins

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Masses of CoinsMasses of Coins

If we have a large number of particles of two close but different masses, how do we describe the mass of these particles?

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How do we accommodate the masses of How do we accommodate the masses of isotopes of an element?isotopes of an element?

Atomic and Molecular Weights


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Why do we use a C-12 standard for the Why do we use a C-12 standard for the mass of atoms of elements? mass of atoms of elements?

How can we determine these masses?How can we determine these masses?

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Mass Spectrometry of Cl Atoms

Mass Spectrometry of Cl Atoms

How do we get How do we get average atomic average atomic weights?weights?

When would the When would the average atomic average atomic weight not be a weight not be a useful number?useful number?

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atomic weight: average mass of 1 atom of an atomic weight: average mass of 1 atom of an element, expressed in amuelement, expressed in amu

formula weight: sum of the atomic weights formula weight: sum of the atomic weights of each atom in a chemical formulaof each atom in a chemical formula

What is the formula weight of CaClWhat is the formula weight of CaCl22??

molecular weight: same as formula weight molecular weight: same as formula weight when the chemical formula is a molecular when the chemical formula is a molecular formulaformula

What is the molecular weight of HWhat is the molecular weight of H22COCO33??

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Percentage Composition from Percentage Composition from FormulasFormulas

Percentage Composition from Percentage Composition from FormulasFormulas

Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

Atoms of Element AW% Element 100

FW of Compound

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3.4 The Mole3.4 The Mole

Demo: 1 mole Demo: 1 mole of different of different substancessubstances

What do these What do these substances all substances all have in have in common?common?

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Avogadro’s number: NAvogadro’s number: NAA = 6.022 x 10 = 6.022 x 1023 23 particles/moleparticles/mole

Hg

CuSO4.5H2O

NaCl

Sn

Fe2O3

S8

Fe

CuH2O

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Describe the difference between molar Describe the difference between molar mass, molecular weight, and atomic weight.mass, molecular weight, and atomic weight.

How do we get these quantities?How do we get these quantities?Calculations: Calculations:

mass mass moles moles number of particles number of particlesHow do we carry out these conversions?How do we carry out these conversions?mass mass moles: moles: Use molar massUse molar mass

moles moles number of particles: number of particles: Use NUse NAA

Molar Mass and MolesMolar Mass and Moles

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Write on theWrite on the

blackboard and blackboard and

calculate thecalculate the

number of CaCOnumber of CaCO33

units in the writing. units in the writing. How do we determine the How do we determine the

mass of the writing? mass of the writing? How much Ca, C, and O is How much Ca, C, and O is

in the writing?in the writing?

Mole CalculationsMole Calculations

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Weigh the chalk before and after

writing

Weigh the chalk before and after

writing

Mass of writing = 5.473 g - 5.448 g = 0.025 gMass of writing = 5.473 g - 5.448 g = 0.025 g

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How much Ca, C, O in the writing?

How much Ca, C, O in the writing?

0.025 g x 1 mol/100.08 g = 0.00025 mol CaCO0.025 g x 1 mol/100.08 g = 0.00025 mol CaCO33

0.00025 mol CaCO0.00025 mol CaCO33 contains 0.00025 mol Ca, contains 0.00025 mol Ca,

0.00025 mol C, 0.00075 mol O0.00025 mol C, 0.00075 mol O0.00025 mol Ca x 40.08 g/mol = 0.0100 g Ca0.00025 mol Ca x 40.08 g/mol = 0.0100 g Ca0.00025 mol C x 12.01 g/mol = 0.00300 g C0.00025 mol C x 12.01 g/mol = 0.00300 g C0.00075 mol O x 16.00 g/mol = 0.0120 g O0.00075 mol O x 16.00 g/mol = 0.0120 g O sum = 0.0100 + 0.00300 + 0.0120 = 0.0250 gsum = 0.0100 + 0.00300 + 0.0120 = 0.0250 g

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3.5 Empirical Formulas from Analyses

3.5 Empirical Formulas from Analyses

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Calculate the empirical formulas:Calculate the empirical formulas: 50% O, 50% S50% O, 50% S 60% O, 40% S60% O, 40% S

Assume 100 g of the material:Assume 100 g of the material: 50 g O x 1 mol/16.0 g = 3.125 mol O50 g O x 1 mol/16.0 g = 3.125 mol O 50 g S x 1mol/32.066 g = 1.559 mol S50 g S x 1mol/32.066 g = 1.559 mol S mol O/mol S = 3.125/1.559 = 2.004 or 2mol O/mol S = 3.125/1.559 = 2.004 or 2 SOSO22

60 g O, 40 g S 60 g O, 40 g S 3.75 mol O, 1.25 mol S 3.75 mol O, 1.25 mol S SOSO33

Empirical FormulasEmpirical Formulas

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Molecular formula from empirical formula

Molecular formula from empirical formula

How to calculate a molecular formula if a How to calculate a molecular formula if a molar mass is known?molar mass is known?

Empirical formula = CHEmpirical formula = CH22O, MM = 90 g/molO, MM = 90 g/molWhat is the molecular formula?What is the molecular formula?

Empirical formula mass = 12 + 2 + 16 = 30Empirical formula mass = 12 + 2 + 16 = 30The molar mass is 3 x as great, so the The molar mass is 3 x as great, so the

molecular formula is three empirical molecular formula is three empirical formulas: (CHformulas: (CH22O)O)33 or C or C33HH66OO33

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Group Quiz FormatGroup Quiz Format

Place TA name and section letters (BA, or Place TA name and section letters (BA, or BB, or … , or BH) in upper left corner of BB, or … , or BH) in upper left corner of paper.paper.

Place name of group members Place name of group members participatingparticipating in the quiz in the upper right corner. Please in the quiz in the upper right corner. Please use correct spelling and write or print use correct spelling and write or print legibly.legibly.

Show work for partial credit.Show work for partial credit.

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Group Quiz 4Group Quiz 4

A compound contains 71.0% potassium A compound contains 71.0% potassium and 29.0 % oxygen. The molar mass of and 29.0 % oxygen. The molar mass of the compound is about 110 g/mol.the compound is about 110 g/mol.What is the empirical formula of the What is the empirical formula of the

compound?compound?What is the molecular formula of the What is the molecular formula of the

compound?compound?

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Answers to group quizAnswers to group quiz

Assume a sample of 100 g, which then contains Assume a sample of 100 g, which then contains 71.0 g K and 29.0 g O.71.0 g K and 29.0 g O.

71.0 g K x 1 mol/39.1 g = 1.82 mol K71.0 g K x 1 mol/39.1 g = 1.82 mol K29.0 g O x 1 mol/16.0 g = 1.81 mol O29.0 g O x 1 mol/16.0 g = 1.81 mol O ratio = 1.82 mol K/1.81 mol O = 1.00ratio = 1.82 mol K/1.81 mol O = 1.00 empirical formula = KOempirical formula = KO formula weight = 39.1 + 16.0 = 55.1formula weight = 39.1 + 16.0 = 55.1molar mass/formula weight = 110/55.1 = 2molar mass/formula weight = 110/55.1 = 2 formula = (KO)formula = (KO)22 = K = K22OO22

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3.6 Quantitative Information 3.6 Quantitative Information from Balanced Equationsfrom Balanced Equations

3.6 Quantitative Information 3.6 Quantitative Information from Balanced Equationsfrom Balanced Equations

Mass-Mole ConversionsMass-Mole Conversions

10 g CaCO10 g CaCO33 How many moles? How many moles?

MM = 100 g/molMM = 100 g/mol10 g x 1 mol/100 g = 0.10 mol10 g x 1 mol/100 g = 0.10 mol

How many moles in 20 g?How many moles in 20 g?How many moles in 25 g of NaOH?How many moles in 25 g of NaOH?

MM = 40 g/molMM = 40 g/mol

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CalculationsCalculations

Mole-Mole ConversionsMole-Mole Conversions

CaClCaCl22 + Na + Na22COCO33 CaCO CaCO33 + 2NaCl + 2NaCl

0.105 mol xs ? ?0.105 mol xs ? ?How many moles of each product?How many moles of each product?

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Mass - Mass ConversionsMass - Mass Conversions

Mass-Mass ConversionsMass-Mass Conversions

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Mass Conversions in a Single ReactionMass Conversions in a Single ReactionCaClCaCl22 + Na + Na22COCO33 CaCO CaCO33 + 2NaCl + 2NaCl

5.45 g ? ? 5.45 g ? ?111 g/mol 100.1 g/mol 58.4 g/mol111 g/mol 100.1 g/mol 58.4 g/mol

Mass Conversions in Sequences of ReactionsMass Conversions in Sequences of ReactionsFollow the same sequence of conversions, Follow the same sequence of conversions, using the amounts of products from the first using the amounts of products from the first reaction as the amount of reactant in the reaction as the amount of reactant in the second reaction.second reaction.

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3.7 Limiting Reactants3.7 Limiting Reactants

Limiting Reactant Limiting Reactant Demo: Mg or Zn Demo: Mg or Zn in HClin HCl

Analogy: making Analogy: making cheese sandwichescheese sandwiches

05m05vd105m05vd1

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Limiting ReactantLimiting Reactant

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2H2H22 + O + O22 2H 2H22OO

What is the limiting reactant?What is the limiting reactant?

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2H2H22 + O + O22 2H 2H22OO


Limit_React.exe.lnk

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2H2H22 + O + O22 2H 2H22OO


Limit_React.exe.lnk

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Limiting Reactant ProblemsLimiting Reactant Problems

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CaClCaCl22 + Na + Na22COCO33 CaCO CaCO33 + 2NaCl + 2NaCl 0.105 mol 0.085 mol ? ?0.105 mol 0.085 mol ? ?How many moles of each product?How many moles of each product?

CaClCaCl22 + Na + Na22COCO33 CaCO CaCO33 + 2NaCl + 2NaCl

5.45 g 4.55 g ? ?5.45 g 4.55 g ? ?111 g/mol 106 g/mol 100.1 g/mol 58.4 g/mol111 g/mol 106 g/mol 100.1 g/mol 58.4 g/mol

How much CaCOHow much CaCO33 (in g) is formed? (in g) is formed?

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3.3 Atomic and Molecular Weights