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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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3.1 The mole conveniently links mass to number of atoms or molecules 4
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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3.1 The mole conveniently links mass to number of atoms or molecules 5
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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3.1 The mole conveniently links mass to number of atoms or molecules 6
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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3.1 The mole conveniently links mass to number of atoms or molecules 7
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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3.1 The mole conveniently links mass to number of atoms or molecules 8
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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3.1 The mole conveniently links mass to number of atoms or molecules 9
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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3.1 The mole conveniently links mass to number of atoms or molecules 10
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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3.1 The mole conveniently links mass to number of atoms or molecules 11
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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3.1 The mole conveniently links mass to number of atoms or molecules 12
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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3.1 The mole conveniently links mass to number of atoms or molecules 13
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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3.1 The mole conveniently links mass to number of atoms or molecules 14
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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3.1 The mole conveniently links mass to number of atoms or molecules 15
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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3.1 The mole conveniently links mass to number of atoms or molecules 16
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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3.1 The mole conveniently links mass to number of atoms or molecules 17
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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3.1 The mole conveniently links mass to number of atoms or molecules 18
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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3.1 The mole conveniently links mass to number of atoms or molecules 19
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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3.1 The mole conveniently links mass to number of atoms or molecules 20
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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3.1 The mole conveniently links mass to number of atoms or molecules 21
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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3.1 The mole conveniently links mass to number of atoms or molecules 22
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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3.1 The mole conveniently links mass to number of atoms or molecules 23
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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3.1 The mole conveniently links mass to number of atoms or molecules 24
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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3.1 The mole conveniently links mass to number of atoms or molecules 25
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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3.1 The mole conveniently links mass to number of atoms or molecules 26
Relationships Between Quantities
moles
mass (u)
# particles
mass (g)
N
N
FM MM
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3.1 The mole conveniently links mass to number of atoms or molecules 27
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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3.1 The mole conveniently links mass to number of atoms or molecules 28
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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3.2 Chemical formulas relate amounts of substances in a compound 30
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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3.2 Chemical formulas relate amounts of substances in a compound 31
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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3.3 Chemical formulas can be determined from experimental mass measurements 33
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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3.3 Chemical formulas can be determined from experimental mass measurements 34
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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3.3 Chemical formulas can be determined from experimental mass measurements 35
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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3.3 Chemical formulas can be determined from experimental mass measurements 36
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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3.3 Chemical formulas can be determined from experimental mass measurements 37
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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3.3 Chemical formulas can be determined from experimental mass measurements 38
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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3.3 Chemical formulas can be determined from experimental mass measurements 39
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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3.3 Chemical formulas can be determined from experimental mass measurements 40
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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3.3 Chemical formulas can be determined from experimental mass measurements 41
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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3.3 Chemical formulas can be determined from experimental mass measurements 42
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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3.3 Chemical formulas can be determined from experimental mass measurements 43
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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3.3 Chemical formulas can be determined from experimental mass measurements 44
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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3.3 Chemical formulas can be determined from experimental mass measurements 45
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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3.3 Chemical formulas can be determined from experimental mass measurements 46
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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3.3 Chemical formulas can be determined from experimental mass measurements 47
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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3.3 Chemical formulas can be determined from experimental mass measurements 48
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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3.4 Chemical equations link amounts of substances in a reaction 50
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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3.4 Chemical equations link amounts of substances in a reaction 51
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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3.4 Chemical equations link amounts of substances in a reaction 52
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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3.4 Chemical equations link amounts of substances in a reaction 53
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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3.4 Chemical equations link amounts of substances in a reaction 54
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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3.4 Chemical equations link amounts of substances in a reaction 55
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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3.4 Chemical equations link amounts of substances in a reaction 56
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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3.4 Chemical equations link amounts of substances in a reaction 57
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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3.5 The reactant in shortest supply limits the amount of product that can form 60
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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3.5 The reactant in shortest supply limits the amount of product that can form 61
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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3.5 The reactant in shortest supply limits the amount of product that can form 62
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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3.5 The reactant in shortest supply limits the amount of product that can form 63
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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65/66
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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66/66
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
=