Ch03-Mata Kuliah III

8/10/2019 Ch03-Mata Kuliah III

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Ch. 3 The Mole: Relating the Microscopic World of Atoms to

Laboratory Measurements

Mata KuliahKIMIA DASAR

Jurusan Teknik Mesin dan Industri

Universitas Tarumanegara

Pengajar : Gadang Priyotomo,2014 Sumber : Brady & Senese, 5th Ed.


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2

Index

3.1 The mole conveniently links mass to number of atoms ormolecules

3.2 Chemical formulas relate amounts of substances in acompound

3.3 Chemical formulas can be determined from experimental

mass measurements3.4 Chemical equations link amounts of substances in a

reaction3.5 The reactant in shortest supply limits the amount of

product that can form3.6 The predicted amount of product is not always obtained

experimentally


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3.1 The mole conveniently links mass to number of atoms or molecules 3

Particles Have Characteristics Masses

The same mass maynot represent the samenumber of molecules

Suppose one rabbit hasa mass of 250 g. Whatmass in kg would acase of 24 rabbitshave?

1000gkg

rabbitg250rabbits24

6.0 kg


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Counting Atoms By Their Mass

The mass of an atom is called its atomic mass Atomic mass provides a means to count atoms by

measuring the mass of a sample The periodic table gives atomic masses of the

elements in u per atom

to reduce rounding errors, use the most precisevalues possible


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Learning Check

How many atoms of C are there in 3.5 108 u?

What is the mass (in u) of 2.33

1016

atoms ofH?

u 12.0107C atom1u103.5

8

2.35 1016 u

atomic masses: C=12.0107 u; H=1.00794 u

Hatom1

u.007941 atoms102.33

16

2.9 107


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Your Turn!

Given that the atomic mass of Ba is 137.327u, whatis the mass of 23 atoms of Ba?

A. 3.2 103 uB. 3.2 10 -4 u

C. 1.37

102

uD. none of these


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Your Turn!

A new element is discovered that has a mass of3.2 102 u for15 atoms. What is the atomic mass?A. 3.2 102

B. 0.047

C. 21D. not enough informationE. None of these answers


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Relationships

1.66 10 -27 kg = 1 u (from the inside back coverof the book) may also be written as:

6.0223 1023 u = 1 g ( a form you will oftenuse)

We can use this as a conversion factor to convertbetween mass quantities in u, and those in g

grams (g)atomic mass units (u)

u106.0223g1

23

g 1

u106.0223 23


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Relationships

Atomic Mass (AM) u = 1 particle We can use this as a conversion factor to convert

between these quantities.

uAM

particle1

particleuAM

mass (u)particles


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Learning Check

How many u of Na are there in 55.2 kg Na?

How many g Na are there in 3.2 x 10 15 u ofNa?

gu106.0223

kgg1055.2kg 23

3

u106.02231g

u103.2

2315 5.3 10 -9 g

3.32 1028 u


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Your Turn!

Which of the following are not equivalent to asample of 10.5 107 u of Cu?

A. 1.74 10 -16 gB. 1.65 106 atoms

C. 63.54 uD. None of these


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What Is The Formula Mass Of?

Ba3(PO 4)2 :

(NH 4)2CO 3:

atomic masses: Ba: 137.327(7)u; P:30.973761(2)u;O: 15.9994(3)u; H:1.00794u; N:14.00672u; C 12.0107(8)u

96.08603 u/fu

601.9261 u/fu


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Relationships

Formula mass (FM) u = 1 particle We can use this as a conversion factor to convert

between these quantities.

uFM

particle1

particle

uFM

mass (u)particles


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Counting Molecules By Their Masses

The molecular mass allows counting of moleculesby mass

The molecular mass is the sum of atomic massesof the atoms in the compounds formula

Strictly speaking, ionic compounds do not have a molecular mass , we describe an analogousquantity- the formula mass - to cover allpossibilities


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Learning Check:

How many molecules of CO 2 are there in 3.5 108 u?

What is the mass (in u) of 2.33 1016 moleculesof H 2 ?

u44.0095

COmolecule1u103.5

628

4.70

1016

u

atomic masses: C=12.0107 u; H=1.00794 u; O=15.99943 u

2

16

Hmolecule1u.015882

atoms102.33

8.0

106

u


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What Is a Mole?

One mole of any substance contains the samenumber of units, called Avogadros number , N

1 mole formula units = 6.0223 x 10 23 formulaunits

It is a large quantity of particles because theparticles described are so small.


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Why is Molar Mass the Same as Formula Mass?

suppose we start with 12.0107g of C. How manyatoms of C are there?

given that the atomic mass of C is 12.0107 u

12.0107 g C = 6.0223 x 10 23 atoms thus for any substance, the formula mass (in g)

corresponds to the same number of atoms, N

u12.0107atom1

gu106.0223

12.0107gC23


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Molar Mass

One mole contains the same number of particles asthe number of atoms in exactly 12 g of carbon-12

The molar mass of a substance has the samenumeric value as the formula mass

The value is different because the units aredifferent

Thus if the formula mass of Ba 3(PO 4)2 is 610.332 u/fu,the molar mass of Ba 3(PO 4)2 is 610.332 g/mol


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Relationships

MM g = 1 mole Use this as a conversion factor to convert between

these quantities

moleMass (g)

gMM

mole1

mole1gMM


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Learning Check: Converting Between MassAnd MolesGiven that the molar mass of CO 2 is 44.0098 g/mol What mass of CO 2 is found in 1.55 moles?

How many moles of CO 2 are there in 10 g?

molg44.0098

mol1.55

g 44.0098molg10 0.2 mol

68.2 g


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Your Turn!

What is the molar mass of Ca 3(PO 4)2 in g/mol?Ca: 40.078 ; P: 30.973761 ; O:15.9994A. 279.203B. 215.205

C. 310.177D. none of these


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Your Turn!

What mass in g, of Ca 3(PO 4)2 (MM=310.1767)would a 3.2 mole sample have?

A. 1.0 10 -3 gB. 9.9 102 g

C. 6.0

1026

gD. 1.6 10 -21 gE. None of these


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Using Avogadros Number, N

Counting formula units by moles is no differentthan counting eggs by the dozen (12 eggs) or pensby the gross (144 pens)

Since the individual particle is very small, themole is a more practical quantity

It is a group, in which 6.0223 1023 individualscomprise 1 mole

The quantity, N , is Avogadro's number and ismeasured as 6.0223 1023


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Relationships

N particles = 1 mole We can use this as a conversion factor to convert

between these quantities

particlesmole1

N

moleparticles N

Molesparticle


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Learning Check: Mole Conversions

Calculate the formula units of Na 2CO 3 in 1.29moles of Na 2CO 3

How many moles of Na 2CO 3 are there in 1.15 x105 formula units of Na 2CO 3?

molfu106.0223mol1.29 23

fu106.0223

molfu101.15

23

5 1.91 10 -19

mol

7.77 1023fu


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Relationships Between Quantities

moles

mass (u)

# particles

mass (g)

N

N

FM MM


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Your Turn!

Which of the following is not a relationship, but is asample size?

A. molar massB. Avogadros number

C. formula massD. Mass in uE. None of these


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Your Turn!

Given that you have asample of 5.5 g

Na 2CO 3 how manyformula units arepresent?

A. 6.0 1023

B. 5.2 10 -2

C. 3.2 10 -23

D. 3.3 1024

E. None of these

moles

mass (u)

# particles

mass (g)

N

N

FM MM

Na: 22.989770 ; C: 12.011; O:15.9994

3.1 10 22


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3.2 Chemical formulas relate amounts of substances in a compound 29

Using The Chemical Formula

To relate components of a compound to thecompound quantity we look at the chemical

formula In Na 2CO 3 there are 3 relationships:

2 mol Na: 1 mol Na 2CO 31 mol C: 1 mol Na 2CO 33 mol O: 1 mol Na 2CO 3

We can also use these on the atomic scale ,e.g.:1 atom C:1 fu Na 2CO 3


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Learning Check:

Calculate the number of moles of sodium in 2.53moles of sodium carbonate

Calculate the number of atoms of sodium in 2.53moles of sodium carbonate

32CONamol1Namol2mol2.53 5.06 mol Na

Namol1 Naatoms106.0223CONa1mol Namol2mol2.53

23

32

3.05 10 24 atoms Na


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Your Turn!

How many atoms of iron are in a 15.0 g sample ofiron(III) oxide (MM 159.6885 9 g/mol)?

A. 1.13

1023

B. 9.39 10 -2

C. 5.66 1022

D. 1.88 10 -1

E. None of these


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3.3 Chemical formulas can be determined from experimental mass measurements 32

Percent Composition

Percent composition is a list of the mass percent ofeach element in a compound

Na 2CO 3 is43.38% Na11.33% C

45.29% O What is the sum of the percent composition of a

compound?


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Percent Composition: How Is It Calculated?

What is the % C in CO 2? Determine the molar mass of the compound

MM=44.0095 6 g/mol

Multiply the ratio of the mass of the element to themolar mass of the compound by 100

(12.0107/44.00965 6) 100= 27.2911 %C

MM g/mol C:12.0107; O:15.99943


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Learning Check

A sample was analyzed and found to contain 0.1417g nitrogen and 0.4045 g oxygen. What is the

percentage composition of this compound?

100 totalg0.40450.1417gNg0.1417

+

74.06% O25.94% N


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Your Turn!

A 35.5 g sample is analyzed and found to contain23.5% Si. What mass of Si is present in the

sample?A. 6.62 10 -1 gB. 8.88 101 gC. 1.51 102 gD. 8.34 g

E. None of these


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Empirical vs. Molecular Formulas

The empirical formula isthe lowest whole number

ratio of atoms in acompound Note that the molecular

formula is a whole numbermultiple of the empiricalformula.

C 6H 12O 6

glucose

CH 2O

C 1x6 H 2x6 O 1x6


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Strategy

Convert starting quantities to moles Divide all quantities by the smallest number of

moles to get the smallest ratio of moles Convert any non-integers into integers

If any number ends in a common decimal equivalent ofa fraction, multiply by the least common denominatorOtherwise, round the numbers to the nearest integers


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Common Ratios And Their DecimalEquivalents

decimal Fraction

equivalent

multiplier

.25or

.75

or 4

.3333or

.6667

1/3 or 2/3 3

.50 2

For example:

54x25.1 =


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Learning Check:

A 2.012 gsample of a

compoundcontains0.522 g ofnitrogen and

1.490 g ofoxygen.Calculate itsempiricalformula

1.4900.522mass(g)

ON

2 5integerratio

14.00674 15.99943MM

0.0372 68 0.09312 83mol

1 2.50lowestratio


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Determining The Multiplier, n

Ratio of the molecular mass to the mass predicted bythe empirical formula and round to an integer

The actual molecule is larger by this amountIf the empirical formula is A xBy , the molecular formulawill be A n xBn y

massformulaempiricalmassformulamolecularn =

glucose 6 OCH g30.0262OHC g180.1572n

2

6126


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Example:

The empirical formula of hydrazine is NH 2, and itsmolecular mass is 32.0. What is its molecular

formula? A substance is known to be 35.00% N, 5.05% H

and 59.96% O. What is its EF? Determine the

Molecular Formula if the MM of the compound is80.06 g/mol

N2H4n=(32.0/16.02)=2

n=(80.06/80.043)=1

N2H4O3

EF: N 2H4O3

MM: N:14.00674; H:1.00794; O:15.99943


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Your Turn!

Given the composition analysis of lindane (acontroversial pesticide ) what is its empirical

formula?A. C 24H2Cl73B. C 2H2Cl2C. C 142 HCl 126D. CHCl

E. None of these

73.14129%2.07943%24.77928%

ClHC


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Your Turn!

We found that the empirical formula was CHCl.Given that the MM is 290.8316 g/mol, what is

the molecular formula?A. C 6H6Cl6B. C 8H17Cl5C. C 3H5Cl7D. C 5H18Cl7

E. none of these


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Combustion Analysis:

Empirical formulas may also be calculatedindirectly

When a compound made only from carbon,hydrogen, and oxygen burns completely in pureoxygen, only carbon dioxide and water are

produced


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Combustion Analysis:

Empirical formulas may be calculated from theanalysis of combustion information

grams of C can be derived from amount of CO 2grams of H can be derived from amount of H 2Othe mass of oxygen is obtained by difference:

g O = g sample ( g C + g H )


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Learning Check:

The combustion of a 5.217 g sample of a compound ofC, H, and O gave 7.406 g CO 2 and 4.512 g of H 2O.

Calculate the empirical formula of the compound.

22

COg44.00956Cg12.0107

COg.4067

H: 1.00794; C:12.0107; O: 15.99943

OHg18.01531Hg2.01588

OHg512.42

2 .5048 84 g H

2.021 18 g C

5.217g- 2.021 18 g C-.5048 84 g H= 2.690 94 g O


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Learning Check (con.):

Calculate the empirical formula of the compound.

H: 1.00794; C:12.0107; O: 15.99943

2.690940.5048842.02118mass

mol12.0107

C

integerratio

lowratio

15.999431.00794MM

OH

3 11

.500907 .16819.16828

2.97 11

CH 3O


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Your Turn!

Combustion analysis of 3.88 g of a compoundcontaining C, H, and S reveals the following data.

What is the empirical formula of the compound?A. C 6H5SB. C 9H2SC. C 5H5SD. C 3H9S2

E. None of these

1.59 g9.377 g

H2OCO 2


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3.4 Chemical equations link amounts of substances in a reaction 49

What Does The Balanced Equation Mean?

2CO (g) + O 2(g) 2CO 2(g) For every 2 CO reacted, 1 O 2 is also reacted and 2

CO 2 are also reacted


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Using The Balanced Equation:

The balanced equation gives the relationshipbetween amounts of reactants used and amounts of

products likely to be formed The numeric coefficient tells:

how many individual particles are needed in the

reaction on the microscopic levelhow many moles are necessary on the macroscopiclevel

The stoichiometric coefficient


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Stoichiometric Ratios

Consider the reaction N 2 + 3H 2 2NH 3 What is the ratio between N2 and H 2 ?

1 mole N 2: 3 mole H 2 N

2and NH

3?

1mole N 2: 2 mole NH 3 H2 and NH 3?

3 mole H 2 : 2 mole NH 3


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Learning Check:

For the reaction N 2 + 3 H 2 2NH 3, How manymoles of N 2 are used when 2.3 moles of NH 3 are

produced?

If 0.575 mole of CO 2 is produced by thecombustion of propane, C 3H8, how many molesof oxygen are consumed? The balanced equationis C 3H8 + 5 O 2 3 CO 2 + 4 H 2O

3

23

NHmol2Nmol1

NHmol2.3

2

22

COmol3Omol5

COmol0.575 0.958 mol O 2

1.2 mol N 2

L i Ch k


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Learning Check

How many grams of Al 2O3 are produced when 41.5g Al react? 2Al (s) + Fe 2O3(s) Al2O3(s) + 2 Fe (l)

MM (g/mol): Al: 26.9815; Al 2O3:101.9613

32

3232

OmolAl1OgAl101.9613

Almol2OAlmol1

Alg26.9815Almol1

Alg41.5

78.4 g Al 2O3

Y T !


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Your Turn!

Given the reaction:H2SO 4 + 2KOH 2H 2O + K 2SO 4,

How many moles of KOH are required to make 3.0moles of K 2SO 4?

A. 3.0 molesB. 6.0 molesC. 1.5 moles

D. None of these

Y T !


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Your Turn!

Given the reaction:H2SO 4 + 2KOH 2H 2O + K 2SO 4,

How many g of H 2O (18.0153 ) would result fromthe complete reaction of 1.2 g H 2SO 4 (98.08)?

A. 2.4 gB. 1.2 gC. 0.60 g

D. 0.44 gE. none of these

Balancing B Inspection


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Balancing By Inspection

Balance the most complex substance in theequation first

Balance elements, H and O last Use coefficients to adjust quantities, not subscripts Some equations may be balanced using fractions,

but the most common approach allows only forinteger coefficients If polyatomic ions remain intact in a reaction

balance them as a group If you have an even/odd problem dilemma,multiply all previously balanced moieties by 2

Learning Check: Balance The Following:


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Learning Check: Balance The Following:

____Ba(OH) 2(aq) +____ Na 2SO 4(aq) ___BaSO 4(s) + ____NaOH (aq)1 1 2

2___KClO 3(S) ___KCl (s) +___ O 2(g)2 3

1

___H 3PO 4(aq) +___ Ba(OH) 2(aq) ___Ba 3(PO 4)2(s) + ___H 2O (l)3 1 62

Your Turn!


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Your Turn!

Given the following reaction:KCl + Hg 2(NO 3)2 KNO 3 + Hg 2Cl2 , when it is

balanced, what is the coefficient for KCl?A. 1B. 2C. 3D. 4

E. none of these

Limiting Reagent


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3.5 The reactant in shortest supply limits the amount of product that can form 59

Limiting Reagent

Consider the reaction of N 2 with H 2 to form NH 3: N2(g) + 3H 2(g) 2NH 3(g) The stoichiometry suggests that for every mole of

N2 we will need 3 moles of H 2 to form 2 moles ofNH 3.

So what happens if these proportions are not met?The reaction proceeds, to use up one of thereactants (the limiting reagent) and will not use allof the other reactant (it is in excess)

Limiting Reagents


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Limiting Reagents

Note that in this reaction, some of the O 2 is notconsumed. This is because there is not enough

CO to continue consuming the O 2. Thus, CO is the limiting reagent.

Determining The Limiting Reagent (LR)


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Determining The Limiting Reagent (LR)

There are several approaches to this. One methodis to compare the quantities available to the

quantities required. Any substance present in excess of the

requirement cannot be limiting.

Learning Check:


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Learning Check:

Ca(OH) 2(aq) + 2HCl (aq) 2 H 2O (l) + CaCl 2(s)when 1.00 g of each reactant is combined:

What is the theoretical yield of H2O?

The limiting reagent?

TY H2O (mol)

mol

MM (g/mol)18.0152836.4609474.09468mass (g)1.001.00

H2O

0.013496 0.027427

OHmol0274.0 HClmol2

OHmol21

HClmol0.027427

OHmol0269.0Ca(OH)mol1 OHmol21 Ca(OH)mol0.0134

2272

2922

2296

=

=

0.0269 92

0.486 277

Ca(OH) 2 HCl

Ca(OH) 2: 74.09468; HCl: 36.46094; H 2O: 18.01528

Learning Check:


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Learning Check:

How many grams of NO can form when 30.0 g NH 3and 40.0 g O 2 react according to:4 NH 3 + 5 O 2 4 NO + 6 H 2O

TY NO (mol)mol

MM (g/mol)30.006131.998817.03052mass (g)40.030.0

NOO2NH3

1.76 15 1.25 00 1.00 00

30.0

NOmol1.00Omol5NOmol4

1Omol1.25

NOmol1.76NHmol4NOmol4

1NHmol1.76

002

200

153

315

=

=

NH 3: 17.03052; O 2=31.9988; NO: 30.0061

Your Turn!


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Your Turn!

Given 1.0 g each of KCl and Hg 2(NO 3)2, what is theexpected mass of Hg 2Cl2 ?

A. 1.0 gB. 2.0 gC. 0.90 gD. 3.2 gE. none of these

MM (g/mol)472.086525.189974.5513Hg 2Cl2Hg 2(NO 3)2KCl

Actual Yield


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3.6 The predicted amount of product is not always obtained experimentally 65

Actual Yield

Often we do not obtain the quantity expected This may be due to errors, mistakes, side reactions,

contamination or a host of other events Thus we describe the actual yield , the amount

obtained experimentally

Percent Yield


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3.6 The predicted amount of product is not always obtained experimentally 66

The amount of product, predicted by the limitingreagent is termed the theoretical yield

Percent yield relates the actual yield to thetheoretical yield

It is calculated as:

If a cookie recipe predicts a yield of 36 cookiesand yet only 24 are obtained, what is the % yield?

100yieldltheoretica

yieldactual% x

=

1003624

% x

=

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