Transcript COMPOUNDS AND MOLECULES

1
Molecules and Compounds
Read Chapter 3.
Study all examples and complete all
exercises.
Complete all bold numbered
problems.
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Chapter 3 Outline
• Molecular Formula
• Molar Mass
• Empirical and Molecular
Formula
• Nomenclature
3
Compounds & Molecules
Buckyball, C60
NaCl, salt
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Compounds & Molecules
• COMPOUNDS are a combination of 2 or
more elements in definite ratios by mass.
• The character of each element is lost when
forming a compound.
• MOLECULES are the smallest unit of a
compound that retains the characteristics of
the compound.
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MOLECULAR FORMULAS
• Formula for glycine is C2H5NO2
• In one molecule there are
– 2 C atoms
– 5 H atoms
– 1 N atom
– 2 O atoms
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WRITING FORMULAS
• Formula
HOCH2CH2OH
to show atom ordering
• or in the form of a structural
formula
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Molecular Modeling
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Molecular Modeling
H H O
H N C C O H
H
Ball & stick
Drawing of glycine
Space-filling
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Resources for
Molecular Modeling
• Oxford Molecular/CAChe Scientific
software on Saunders General
Chemistry CD-ROM
• Rasmol and Chime on the Internet
• See http://www.saundercollege.com
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ELEMENTS THAT EXIST
AS MOLECULES
Allotrope of C
Buckyball, C60
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IONS AND IONIC COMPOUNDS
• IONS are atoms or groups of atoms
with a positive or negative charge.
• Taking away an electron from an atom
gives a CATION
with a positive
charge.
• Adding an electron to an atom gives an
ANION with a negative charge.
Formation of
Cations & Anions
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Formation of
Cations & Anions
Mg --> Mg2+ + 2 e-
F + e- --> F-
A cation forms
when an atom
loses one or
more electrons.
An anion forms
when an atom
gains one or
more electrons
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PREDICTING ION CHARGES
MONATOMIC IONS
In general
• metals (Mg) lose electrons ---> cations
• nonmetals (F) gain electrons ---> anions
Li1+
F1-
LiF
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Figure 3.7
Also
Sn4+ Pb4+
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METALS
M ---> n e- + Mn+
where n = periodic group
Na+
Mg2+
Al3+
Transition metals --> M2+ or M3+
are most common
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NONMETALS
NONMETAL + n e- ------> Xnwhere n = 8 - Group number
C4N3O2F-
carbide
nitride
oxide
fluoride
Bromine
POLYATOMIC IONS
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Groups of atoms with a charge.
(See back of periodic chart)
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Some Common
Polyatomic Ions
HNO3
nitric acid
NO3nitrate ion
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Some Common
Polyatomic Ions
NH4+
ammonium ion
One of the few common
polyatomic cations
Some Common
Polyatomic Ions
CO32carbonate ion
HCO3bicarbonate ion
hydrogen
carbonate
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Some Common
Polyatomic Ions
PO43phosphate ion
CH3CO2acetate ion
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Some Common
Polyatomic Ions
SO42Sulfate ion
SO32Sulfite ion
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Some Common
Polyatomic Ions
NO3Nitrate ion
NO2Nitrite ion
COMPOUNDS FORMED FROM
IONS
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CATION + ANION
COMPOUND
Na+ + Cl-
A neutral compound requires equal
number of + and - charges.
NaCl
IONIC COMPOUNDS
ammonium chloride, NH4Cl
NH4+
Cl-
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Some Ionic Compounds
Ca2+ + 2 FCaF2
Calcium fluoride
Mg2+ + 2 NO3-
Mg(NO3)2
Magnesium nitrate
Calcium Fluoride
3 Fe2+ + 2 PO43Iron(II) phosphate
Fe3(PO4)2
Sample Questions
• Predict the charges for the ions formed
from:
Se
P
Ga
Sr
• Give the formula for each ion in Al2(SO4)3
• Give the formula for the ionic compound
that forms between
Na and S
Ga and O
Ba and N
Answers
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Properties of Ionic Compounds
Forming NaCl from Na and Cl2
• A metal atom can
transfer an
electron to a
nonmetal.
• The resulting
cation and anion
are attracted to
each other by
electrostatic
forces.
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Electrostatic Forces
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The oppositely charged ions in ionic
compounds are attracted to one another by
ELECTROSTATIC FORCES.
These forces are governed by
COULOMB’S LAW.
Electrostatic Forces
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COULOMB’S LAW
Force of attraction =
(charge on+)(charge on-)
(distance between ions)2
As ion charge increases, the attractive
force _______________.
As the distance between ions increases,
the attractive force ________________.
This idea is important and will come
up many times in future
discussions!
Importance of Coulomb’s Law
NaCl, Na+ and Cl-,
m.p. 801 oC
MgO, Mg2+ and O2m.p. 2800 oC
AlN, Al3+ and N3- m.p. 2900 oC
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Names of Compounds
• Rules for nomenclature are
found in section 3.5
• STUDY them carefully!!
• We will be studying
nomenclature in the
laboratory in Experiment MA.
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Counting Atoms
• Mg burns in air (O2) to
produce white
magnesium oxide,
MgO.
• How can we figure out
how much oxide is
produced from a
given mass of Mg?
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Counting Atoms
• Chemistry is a quantitative science
— we need a “counting unit.”
• The MOLE
• 1 mole is the amount of substance
that contains as many particles
(atoms, molecules) as there are in
12.0 g of 12C.
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Particles in a Mole
Avogadro’s Number
6.02 x 1023
Amedeo Avogadro
1776-1856
There is Avogadro’s number of
particles in a mole of any substance.
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Mole in Chemistry is NOT:
•An informer / spy
•Dark spot on Cindy Crawford’s upper lip
•Rodent that burrows in the ground
•A tunneling machine
•Wave break
•Spicy Mexican sauce
A mole is a convenient
measuring tool.
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Pair
Dozen
Baker’s Dozen
Ream
Baseballs
2 baseballs
12 baseballs
13 baseballs
500
baseballs
Pineapples
2
Pineapples
12
Pineapples
13 Pineapples
500
Pineapples
Calculators
2
Calculators
12
Calculators
13 Calculators
500
Calculators
Planets
2 Planets
12 Planets
13 Planets
500 Planets
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Mole = mol =
23
6.022  10 “particles”
Mole
Baseballs
6.022  1023 baseballs
Pineapples
6.022  1023 Pineapples
Calculators
6.022  1023 Calculators
Planets
6.022  1023 Planets
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Just as
1 doz eggs = 12 eggs
or
1 mol of eggs = 6.022  1023 eggs
A mole is a “Number”
1 mol of H = 6.022  1023 atoms of H
1 mol of O = 6.022  1023 atoms of O
1 mol of Al = 6.022  1023 atoms of Al
1 mol of Cr = 6.022  1023 atoms of Cr
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In chemistry, the mol is
more then just a number
1 mol  amu = 1.00 g
1
12
24
1amu  mass C  1.661 10 g
12
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1 mol  amu = 1.00 g
Proof:
1 mol = 6.022  1023
amu = 1.661  10-24 g
1 mol amu = (6.022  1023)  (1.661  10-24g)
= 1.00 g
Molar Mass
1 mol of 12C
= 12.00 g of C
= 6.02 x 1023 atoms
of C
12.00 g of 12C is its
MOLAR MASS
Taking into account all
of the isotopes of C,
the molar mass of C is
12.011 g/mol
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Molar Mass
1 mol of 12C = 12.00 g of C
= 6.02 x 1023 atoms of C
12.00 g of 12C is its MOLAR MASS
Taking into account all of the isotopes of
C, the molar mass of C is 12.011 g/mol
Find molar
mass from
periodic
table
13
Al
26.9815
atomic number
symbol
atomic weight
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PROBLEM: How many moles
are represented by 0.200 g of
Mg?
0.200 g
1 mole
24.3 g
= 0.00823 mole
How many atoms in this piece of Mg?
0.00823 mole
6.02 x 1023 atom
1 mole
= 4.95 x 1021 atom
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MOLECULAR WEIGHT
AND MOLAR MASS
Molecular weight is the sum
of the atomic weights of all atoms
in the molecule.
Molar mass = molecular weight
in grams
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What is the molar mass of
ethanol, C2H6O?
1 mol contains
2 mol C (12.0 g C/1 mol) = 24.0 g C
6 mol H (1.0 g H/1 mol) = 6.0 g H
1 mol O (16.0 g O/1 mol) = 16.0 g O
TOTAL = molar mass = 46.0 g/mol
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Tylenol
• Formula = C8H9NO2
• Molar mass = 151.0 g/mol
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How many moles of alcohol are
there in a “standard” can of beer if
there are 21.3 g of C2H6O?
21.3 g
1 mole
46.0 g
=
0.463 mole
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How many molecules of alcohol
are there in a “standard” can of
beer if there are 21.3 g of C2H6O?
0.463 mole
6.02 x 1023 molecule
1 mole
= 2.79 x 1023 molecule
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How many atoms of C are there
in a “standard” can of beer if
there are 21.3 g of C2H6O?
2.79 x 1023 molecule
2 atom C
1 molecule
= 5.58 x 1023 atom C
Sample problems
Empirical and Molecular
Formulas
A pure compound always consists of the
same elements combined in the same
proportions by weight.
Therefore, we can express molecular
composition as PERCENT BY
WEIGHT
Ethanol, C2H6O
52.2% C, 13% H,
34.8% O
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Percent Composition
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Consider some of the family of
nitrogen-oxygen compounds:
NO2, nitrogen dioxide and closely
related, NO, nitrogen monoxide
(or nitric oxide)
Structure of NO2
Chemistry of NO, nitrogen
monoxide (nitric oxide)
Percent Composition
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Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt . % N =
14.0 g N
46.0 g NO 2
• 100% = 30.4 %
(16 .0 g O per mole )
2
x 100 % = 69 .6%
Wt. % O =
46 .0 g
Percent Composition
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What are the weight percentages of N
and O in NO?
14.0 g N
Wt. % N =
30.0 g NO
Wt. % O =
16.0 g O
30.0 g NO
• 100% = 46.7 %
• 100%= 53.3%
Percent Composition
Samples Problems
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1. Calculate the percent composition of H2O.
2.0 g H
Wt. % H =
• 100% = 11 %
18.0 g H2O
gO
16.0
Wt. %O =
18.0 g H2O
•100%=88.9%
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Percent Composition
Samples Problems
2. Calculate the percent O in NaOH.
16.0 g O
Wt. %O 
•100%  40.0%
40.0 g NaOH
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Percent Composition
Samples Problems
3. Calculate the percent O and the
percent water in CuSO4.5H2O.
gO
144.0
Wt. % O =
• 100% = 57.69 %
249.6 g CuSO4 • 5H2O
90.0 g H2O
• 100% = 36.1 %
Wt. % H2O =
249.6 g CuSO4.5H2O
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Determining
Formulas
In chemical analysis we determine
the % by weight of each element in a
given amount of pure compound and
derive the EMPIRICAL or
SIMPLEST formula.
A compound of B and H is
81.10% B. What is its empirical
formula?
B
81.10 g
H
1 mole
18.90 g
10.8 g
1 mole
1.0 g
7.51 mole B
19 mole H
7.51 mole B
7.51 mole B
1.00 mole B
2.00 mole B
2.5 mole H
5.0 mole H
B2H5
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A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
B2H6
B2H6 is one example of this class of compounds.
A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula?
We need to do an EXPERIMENT to find the
MOLAR MASS.
Here experiment gives 53.3 g/mol.
Compare with the mass of B2H5 , 26.66 g/unit
Find the ratio of these masses.
2 units of B2 H 5
53.3 g/mol
=
26.66g/unit of B2 H5
1 mol
Molecular formula = B4H10
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Determine the formula of a
compound of Sn and I using
the following data.
• Reaction of Sn and I2 is done using
excess Sn.
• Mass of Sn in the beginning = 1.056 g
• Mass of iodine (I2) used
= 1.947 g
• Mass of Sn remaining
= 0.601 g
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Tin and Iodine
Compound
Find the mass of Sn that combined
with 1.947 g I2.
Mass of Sn initially =
1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used =
0.455 g
Tin and Iodine
Compound
Sn
0.455 g
1 mole
I
1.947 g
118.7 g
1 mole
126.9 g
0.0383 mole Sn
0.1534 mole I
0.0383 mole Sn
0.0383 mole Sn
1.00 mole Sn
4.01 mole I
SnI4
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More Problems
2. Calculate the formula for the iron sulfide
that forms when 53.73g Fe react with
46.27 g of sulfur.
Fe
S
53.73 g 1 mole
46.27 g 1 mole
55.8 g
32.1 g
0.963
1.44
0.963
0.963
1.00
1.50
2.00
3.00
Fe2S3
More Problems
Calculate the empirical formula a compound
containing 90.7% Pb and 9.33% O.
Pb
1 mole
90.7 g
O
9.33 g
207.2 g
1 mole
16.0 g
0.438
0.583
0.438
0.438
1.00
1.33
3.00
3.99
Pb3O4
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More Problems
Calculate the empirical formula for a
compound containing 36.5% Na, 25.4% S
and 38.1% O.
Na
S
O
36.5 g 1 mole
25.4 g 1 mole
38.1 g 1 mole
23.0 g
32.1 g
16.0 g
1.59
0.791
2.38
0.791
0.791
0.791
2.01
1.00
3.01
Na2SO3
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More Problems
Calculate the empirical and molecular
formulas for nicotine, 74.0% C, 8.7% H and
17.3% N, with a molar mass of 160 g/mole.
C
H
N
74.0 g 1 mole
12.0 g
8.7 g 1 mole
17.3 g 1 mole
1.0 g
14.0 g
6.17
8.7
1.24
1.24
1.24
1.24
4.98
7.0
1.00
Empirical formula
C5H7N
More Problems
7. Calculate the empirical and molecular
formulas for nicotine, 74.0% C, 8.7% H and
17.3% N, with a molar mass of 160 g/mole.
Empirical formula
160
C5H7N
=2
81
Molecular formula C10H14N2
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Practice Problems
Names/Formulas
FeO
Pb(C2H3O2)2
magnesium bromide
sodium chromate
calcium phosphate
ammonium carbonate
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Practice Problems
(NH4)2S
As2O3
SO2
silicon disulfide
As2S5
dinitrogen monoxide
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Practice Problems
As2S3
dinitrogen pentoxide
silicon tetrabromide
diphosphorus pentoxide
HBrO3
H3PO4(aq)
H2CO3
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Practice Problems
Calculations
1. 26 g H2 is how many moles H2?
2. 4.25 x 1021 molecules NH3 is how
many grams NH3?
3. 1.5 x 102 formula units KClO3 is how
many moles KClO3
4. 0.0042 mole Fe2O3 is how many
formula units Fe2O3?
5. 2.15 moles MgSO4.7H2O is how many g
MgSO4.7H2O?
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Practice Problems
6. 25 molecules HBr is how many mole
HBr?
7. .00002 g Sn is how many atoms Sn?
8. 7.25 mole H2S is how many g H2S?
9. 5.2 g Sr(OH)2is how many formula
units Sr(OH)2?
10. How many moles of CCl4 will contain
2.4 g of chlorine?
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Practice Problems
11. 19 g of HNO3 contains how
many
a) molecules of HNO3? b) grams of
O?
12. Calculate the percent
composition of NaCl.
13. Calculate the empirical and
molecular formulas for a
compound containing 43.7g P and
56.3g O, with a molar mass of 140
g/mole.
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Practice Problems Answers
Iron(II) oxide, ferrous oxide
lead(II) acetate, plumbous acetate
MgBr2 Na2CrO4 Ca3(PO4)2
(NH4)2CO3 ammonium sulfide
diarsenic trioxide sulfur dioxide
SiS2 As2S5 N2O
diarsenic trisulfide N2O5 SiBr4
P2O5 bromic acid phosphoric acid
carbonic acid
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Practice Problems Answers
1. 13 mole
2. 0.120 g
3. 2.5 x 10-22 mole
4. 2.5 x 1021 atom
5. 530. g
6. 4.2 x 10-23 mole
7. 1 x 1017 atom
8. 247 g
9. 2.6 x 1022 formula units
10. 0.017 mole
11. a) 1.8 x 1023 molecule
b) 14 g
12. 39.3%, 60.7%
13. C5H7N
14. P2O5
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Sample Questions
• Predict the charges for the ions formed
from:
Se
P
-2
-3
Ga
Sr
+3
+2
Al3+
SO42-
• Give the formula for each ion in Al2(SO4)3
• Give the formula for the ionic compound
that forms between
Na and S
Ga and O
Ba and N
Na2S
Ga2O3
Ba3N2
81
Sample Problems
1. 2.5 mole S = ? atom S
2.5 mole
6.02 x 1023 atom
1 mole
= 1.5 x 1024 atom S
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Sample Problems
2. 2.1 mole Zn = ? g Zn
2.1 mole
65.4 g
1 mole
=
140 g Zn
83
Sample Problems
3. 1.42 g Mg = ? atom Mg
1.42 g
1 mole
6.02 x 1023 atom
24.3 g
1 mole
=
3.55 x 1022 atom
84
Sample Problems
4. 125.2 g O2 = ? mole O2
125.2 g
1 mole
32.0 g
=
3.91 mole O2
85
Sample Problems
1. 1.5 mole H2O = ? g H2O
1.5 mole
18.0 g
1 mole
= 27 g H20
86
Sample Problems
2. 1.50 mole CCl4 = ? molecules CCl4
1.5 mole
6.02 x 1023 molecule
1 mole
= 9.03 x 1023 molecule
87
Sample Problems
3. 1.25 mole CaCl2 = ? formula units CaCl2
1.25 mole
6.02 x 1023 formula unit
1 mole
= 7.52 x 1023 formula unit
88
Sample Problems
4. 2.5 g K = ? g KOH
2.5 g K 1 mole K
39.1 g K
1 mole KOH 56.1 g KOH
1 mole K
1 mole KOH
= 3.6 g KOH
89
Sample Problems
5. 34.5 g CaCO3 = ? g O
34.5gCaCO3 1 moleCaCO3 3 moleO
16.0 gO
100.1gCaCO3 1 moleCaCO3 1 moleO
=
16.5
g O