Download Compounds

Principles of Chemistry:
A Molecular Approach,
1st Ed.
Nivaldo Tro
Chapter 3
Molecules, Compounds, and
Chemical Equations
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Chapt 3
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Compounds
„
„
The properties of the compound are totally different
from the constituent elements.
Compounds are composed of atoms held together
by chemical bonds.
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1
Chemical Bond and bond types
„
„
„
Bonds are forces of attraction between atoms.
The bonding attraction comes from attractions
between protons and electrons
Two general types of bonding: ionic and covalent
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Ionic bond
„
„
„
„
Ionic bonds result when electrons have been transferred
between atoms, resulting in oppositely charged ions that attract
each other e.g.Na+Cl„ between metal atoms and nonmetal atoms (ionic
compounds NaCl)
Anion: An atom(a nonmetal) that picks up an extra electron
becomes a negatively charged ion e.g. ClCation: An atom(metal) that loses electrons becomes a
positively charged ion e.g. Na+.
An ionic compound is a compound composed of cations and
anions attracted to one another by electrostatic forces(ionic
bonds)e.g. NaCl.
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Ionic bond
2
Covalent bond
•Covalent bonds result when two atoms share some
of their electrons.
9generally found when nonmetal atoms bond
together (Molecular compounds)
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Representing Compounds
with Chemical Formula
„
„
„
Compounds are generally represented with a
chemical formula.
All formulas and models convey a limited amount of
information—none are perfect representations.
All chemical formulas indicates the elements present in the
compound.
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Types of chemical Formulas
Ionic compound s
1.
Empirical Formula- Gives the relative number of
atoms of each element in a compound
„
„
„
„
Does not describe how many atoms, the order of
attachment, or the shape.
The formulas for ionic compounds are empirical
The empirical formula for the ionic compound fluorspar is
CaCl2.(one Ca2+for every 2 Cl- ions)
The empirical formula for the molecular compound oxalic
acid is CHO2.(actual formula is C2H2O4)
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3
Types of Formulas
2. Molecular Formula-Gives the actual number of
atoms of each element in a molecule of a compound
„
„
„
„
Molecular formula is always whole-number
multiple of empirical formula
For some compound molecular formula is
empirical formula e.g H2O
does not describe the order of attachment, or the
shape
The molecular formula is C2H2O4, (E.F =CH2O)
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Types of chemical Formulas
„
„
Structural Formula-shows how the atoms in a
molecule are connected or bonded to each other.
Uses lines to represent the covalent bonds
„
Single covalent bond
„
a single line = 2 shared electrons,
„
Double covalent bond
„
double line = 4 shared electrons,
„
Triple covalent bond
„
triple line ≡ 6 shared electrons,
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Representing Compounds—
Molecular Models
„
„
„
„
Amore accurate and complete way to specify a
compound is with molecular models
Ball-and-Stick Models
Oxalic acid
Space-Filling Models use interconnected spheres
to show the electron clouds of atoms connecting
together
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4
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An Atomic –level view of elements
and compounds
„
Atomic elements-exist in nature with single atom
as their basic unit
„
Molecular elements- exist as molecules (diatomic
or polyatomic
„
Molecular compounds-usually composed of two or
more covalently bonded non-metals
„
e.g. water is composed of H2O molecules, propane
gas is composed of C3H8 molecules
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Molecular View of
Elements and Compounds
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5
Molecular Elements
„
„
Certain elements occur as two-atom molecules.
„ Rule of 7’s
Other elements occur as polyatomic molecules.
7A
„ P4, S8, Se8
H2
N2
7
O2 F2
Cl2
Br2
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I2
Molecular and Ionic compounds
„
„
Ionic compounds- compounds whose particles are
cations(metal) and anions (nonmetal) bound together
by ionic bonds
Basic unit of ionic compound is the formula unit,
„
„
the smallest, electrically neutral collection of of
ions
Polyatomic ions are composed of a group of
covalently bonded atoms with an overall charge e.g.
ClO- (hypochlorite ion)
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Sample Problem
Classify each of the following as either an atomic
element, molecular element, molecular compound, or
ionic compound.
aluminum, Al
aluminum chloride, AlCl3
chlorine, Cl2
acetone, C3H6O
carbon monoxide, CO
cobalt, Co
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6
Writing formula for Ionic compound
1.
2.
3.
4.
5.
¾
¾
Write the symbol for the metal cation and its charge.
Write the symbol for the nonmetal anion and its charge.
Charge (without sign) becomes subscript for other ion.
Reduce subscripts to smallest whole number ratio.
Check that the sum of the charges of the cation cancels the
sum of the anions.
Compound must have no total charge; therefore, we must
balance the numbers of cations and anions in a compound to
get 0 charge.
If Na+ is combined with S2−, you will need 2 Na+ ions for every
S2− ion to balance the charges; therefore, the formula must be
Na2S.
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Sample Problem
„
„
Write the formula of a compound made from
aluminum ions and oxide ions.
Solution:
„ Write the symbol for the metal cation
„ Write the symbol for the nonmetal anion
„ Charge (without sign) becomes subscript for other
ion.
„ Reduce subscripts to smallest whole number ratio
„ Check that the total charge of the cations cancels
the total charge of the anions.
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Practice—What are the formulas for
compounds made from the following
ions?
„
„
„
potassium ion with a nitride ion
+
3–
„ K with N
calcium ion with a bromide ion
2+ with Br–
„ Ca
aluminum ion with a sulfide ion
3+ with S2–
„ Al
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7
Naming Ionic compounds
„
„
„
Write systematic name by simply naming the ions.
„ if cation is:
„ metal with only one type of ion = metal name
„ metal with variable charge = metal name(charge in
roman numeral)
„ polyatomic ion = name of polyatomic ion
„ if anion is:
„ nonmetal = stem of nonmetal name + -ide
„ polyatomic ion = name of polyatomic ion
E.g. CuCl2 = copper (II)chloride
CaO calcium oxide
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Metal cations
Naming Monatomic Nonmetal Anion
„
Determine the charge from position on the Periodic
Table.
„ Most of the main group metals form cations with
the charge equal to their group number
„ The charge on a monatomic anion for a nonmetal
equals the group number minus 8
„
Most transition elements form more than one ion,
each with a different charge
„
To name anion, change ending on the element name
to -ide.
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8
Naming Binary Ionic Compounds for
Metals with Invariant Charge
1. name metal cation first, name nonmetal anion second
by changing the ending with –ide
For example
NaF -Sodium Fluoride
LiCl Lithium Chloride
MgO Magnesium Oxide
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Common cations of the Transitions
metals
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Practice problem
„
Name the following compounds.
KCl
„
MgBr2
„
Al2S3
„
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9
Naming Binary Ionic compounds
containing metals that form cations with
variable charges
CrBr3
Chromium(III) bromide
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Determining the Charge on a Cation with
Variable Charge Au2S3
1.
2.
3.
4.
Determine the charge on the anion.
Au2S3—The anion is S. Since it is in Group 6A, its charge is
2−.
Determine the total negative charge.
Since there are 3 S in the formula, the total negative
charge is −6.
Determine the total positive charge.
Since the total negative charge is −6, the total positive
charge is +6.
Divide by the number of cations.
Since there are 2 Au in the formula and the total positive
charge is +6, each Au has a 3+ charge.
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Find the charge on the cation
1. TiCl4
2. CrO3
3. Fe3N2
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10
Naming binary Ionic compounds
1. TiCl4
titanium(IV) chloride
2. Fe2S3
iron(III) sulfide
3. PbBr2
lead(II) bromide
Example—Writing Formula for Binary Ionic Compounds
Containing Variable Charge Metal
manganese(IV) sulfide Mn = (1) × (4+) = +4
S = (2) × (2–) = −4
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Naming Polyatomic ions
„
„
A polyatomic ion is an ion consisting of two or
more atoms chemically bonded together and carrying
a net electric charge.
Name any ionic compound by naming cation first and
then anion.
−
SO 4
NO 3 nitrate
NO 2 nitrite
See table 3.5 in the textbook
Chapt 3
sulfate
2−
−
„
2−
SO 3 sulfite
A. Ghumman
Some Common Polyatomic Ions
Name
Formula
Name
Formula
acetate
C2H3O2–
hypochlorite
ClO–
carbonate
CO32–
chlorite
ClO2–
chlorate
ClO3–
hydrogen carbonate
(aka bicarbonate)
HCO3–
perchlorate
ClO4–
hydroxide
OH–
sulfate
SO42–
nitrate
NO3–
sulfite
SO32–
nitrite
NO2–
chromate
CrO42–
hydrogen sulfate
(aka bisulfate)
HSO4–
dichromate
Cr2O72–
ammonium
NH4+
hydrogen sulfite
(aka bisulfite)
HSO3–
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11
Naming Ionic compounds containing
polyatomic Ions
„
„
„
Ionic compounds containing polyatomic ions are named in the
same way except the name of polyatiomic ion is used when ever
it occurs.
For example
NaNO2
sodium nitrite
•
Na2SO4
•
Na2SO3
•
AgCN
•
Ca(OCl)2
•
Cd(OH)2
•
KClO4
sodium sulfate
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Oxyanions
„
„
„
Oxyanions- polyatomic ions containing oxygen and
another element.
They are named systematically according to the
number of oxygen atoms in the ion
If there are only two ions in the series, the one
„ with more oxygen atoms is given the ending –ate.
„ With fewer oxygen atoms is given the ending –
ite. e.g.
„
NO2NO3-
Nitrite ion
Nitrate ion
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OXyanions
„
„
„
More than two ions in the series are named using
following prefixes
hypo means less than and
per means more than
ClOHypochlorit
ClO2Chlorite
ClO3Chlorate
ClO4Perchlorate
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12
Writing Formula for Ionic Compounds
Containing Polyatomic Ion
Write the formula for Iron(III) phosphate
1.
Write the symbol for the cation and its charge.
2.
Write the symbol for the anion and its charge.
3.
Charge (without sign) becomes subscript for other
ion.
4.
Reduce subscripts to smallest whole number ratio.
5.
Check that the total charge of the cations cancels
the total charge of the anions.
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PracticeProblem
What are the formulas for compounds made from the
following ions?
• aluminum ion with a sulfate ion
• chromium(II) with hydrogen carbonate
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Hydrates
„
A hydrate is a compound that contains water
molecules weakly bound in its crystals.
„
„
„
„
Hydrates are named from the anhydrous
(dry) compound, followed by the word
“hydrate” with a prefix to indicate the
number of water molecules per formula unit
of the compound.
in formula, attached waters follow ·
„ CoCl2·6H2O
CoCl2·6H2O = cobalt(II) chloride hexahydrate
CaSO4·½H2O = calcium sulfate hemihydrate
Chapt 3
Prefix
No. of
Waters
hemi
½
mono
1
di
2
tri
3
tetra
4
penta
5
hexa
6
hepta
7
octa
8
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13
Hydrates
Hydrate
Anhydrous
CoCl2·6H2O
CoCl2
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Molecular compounds: Formulas and
Names
„
„
1.
2.
3.
Binary compounds composed of two nonmetals are usually
molecular and are named using a prefix system.
Common names- water (H2O) and ammonia(NH3)
Write the full name of the first element in the formula.
Write the name of the second element in the formula with an ide suffix.
Use a prefix in front of each name to indicate the number of
atoms.
a)
Never use the prefix mono- on the first element.
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Subscript—Prefixes
„
„
„
„
„
„
„
„
„
„
1 = mono„ not used on first nonmetal
2 = di3 = tri4 = tetra5 = penta6 = hexa7 = hepta8 = octa
9 = nona10 = deca-
•Drop last “a” if name begins with a vowel.
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14
Example—Naming Binary Molecular
„
Here are some examples of prefix names for binary
molecular compounds:
„ SF
sulfur tetrafluoride
4
„ ClO
chlorine dioxide
2
„ SF
sulfur hexafluoride
6
„ Cl O
dichlorine heptoxide
2 7
„ BF3
--„ NO2
--„ PCl5
--„ I2F7
--Chapt 3
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Acids
„
„
„
„
Acids are molecular compounds that form H+ when
dissolved in water.
„ To indicate the compound is dissolved in water,
(aq) is written after the formula.
„ not named as acid if not dissolved in water
sour taste
dissolve many metals
„ like Zn, Fe, Mg; but not Au, Ag, Pt
formula generally starts with H
„ e.g., HCl, H2SO4
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Acids
„
„
Binary acids consist of a H+ ion and any single
anion. For example, HCl is hydrochloric acid.
An oxoacid is an acid containing hydrogen,
oxygen, and another element. An example is
HNO3, nitric acid.
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15
Naming Binary Acids
„
„
„
„
Write a hydro prefix.
Base name of nonmetal name.
Change ending on nonmetal name to -ic.
Write the word acid at the end of the name.
Acids: HCl(aq) hydrochloric acid
„
Examples
„ Naming compounds from its formula
„ Writing the formula from name of the compounds
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Naming Oxyacids
„
1.
2.
3.
Naming Oxyacids H2SO4(aq)
Identify the anion.
SO42− = sulfate
If the anion has -ate suffix, change it to -ic. If the
anion has -ite suffix, change it to -ous.
SO42− = sulfate ⇒ sulfuric
Write the name of the anion followed by the word
acid.
sulfuric acid
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Oxoacids and their anions
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16
3.7-Formula mass and mole concept
for compounds
„
Atomic mass of an element is the average mass of
„
Formula mass-the average mass of a molecule (or
„
an atom of the element
formula unit) of a compound
„ also known as molecular mass or molecular weight
sum of the masses of the atoms in a single molecule
or formula unit
mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
Chapt 3
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Calculating Formula mass
Calculate formula mass of calcium nitrate, Ca(NO3)2.
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Molar Mass of a compound
„
Molar Mass- is the mass in grams of one mole of
„
Molar Mass of a compound- the mass in grams of
„
atoms of an element = its atomic mass
one mole of its molecules or formula units
„ Numerically equal to its formula mass in grams
„ Examples e.g. H2O
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
since 1 mole of H2O contains 2 moles of H and 1 mole of O
Molar Mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
so the Molar Mass of H2O is 18.02 g/mole
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17
Molar mass Calculation
„
Calculate the molar mass of sucrose,
C12H22O11.
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Mole and mass conversion:
converting grams to molecules
„
nA = mA(g)/MmA (g mol-1)
a) Find the number of CO2 molecules in 10.8 g of dry
ice.
b) What is the mass of a drop of water containing 3.55 × 1022 H2O
molecules?
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How many moles are in 50.0 g of PbO2?
(Pb = 207.2, O = 16.00)
Given: 50.0 g PbO2
Find: moles PbO2
Conceptual
Plan:
g PbO2
mol PbO2
Relationships: 1 mol PbO2 = 239.2 g
Solution:
Check: Since the given amount is less than 239.2 g,
the moles being <1 makes sense.
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18
Calculating Mass from moles
„
Conversely, suppose we have 3.25 moles of
glucose, C6H12O6 (molecular wt. = 180.0 g/mol).
What is its mass?
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Practice Problem
„
What is the mass of 4.78 × 1024 NO2 molecules?
molecules
mol NO2
Chapt 3
g NO2
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3.8-Percent Composition of
compounds
„
The percent composition of a compound is the
mass percentage of each element in the compound.
„
We define the mass percentage of “A” as the
parts of “A” per hundred parts of the total, by
mass. That is,
Mass % of A= mass of A in 1 mole of compound x 100%
mass of 1 mol of the compound
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19
Mass Percentages from Formulas
„
Let’s calculate the percent composition of butane, C4H10.
„ First, we need the molecular mass of C4H10.
4 carbons @ 12.0 amu/atom = 48.0 amu
10 hydrogens @ 1.00 amu/atom = 10.0 amu
1 molecule of C4 H10 = 58.0 amu
Now, we can calculate the percentage
amu C
% C = 5848.0
.0 amu total × 100% = 82.8%C
amu H
% H = 5810.0
.0 amu total × 100% = 17.2%H
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Example 3.13- Find the mass percent
of Cl in C2Cl4F2 a FCF refrigerant
Given:
Find:
C2Cl4F2
% Cl by mass
Conceptual
Plan:
Relationships:
Solution:
Check:
Since the percentage is less than 100 and Cl is much
heavier than the other atoms, the number makes sense.
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Mass Percent as a conversion Factor
„
The mass percent tells you the mass of a constituent
element in 100 g of the compound.
The fact that CCl2F2 is 58.64% Cl by mass means that
100 g of CCl2F2 contains 58.64 g Cl.
This can be used as a conversion factor.
„ 100 g CCl F : 58.64 g Cl
2 2
„
„
Chapt 3
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20
Practice problem
„
Benzaldehyde is 79.2% carbon. What mass of
benzaldehyde contains 19.8 g of C?
gC
g benzaldehyde
Chapt 3
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Conversion factor from chemical
formula
Chemical formula gives us relationship between the
moles(amounts) of substances not between the
masses (in grams) of them
Mass of compound
mass element
Chapt 3
moles
moles element
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Sample problems
a)What mass of hydrogen is contained in 1.00 gallon of water?
(dH2O = 1.00g ml-1 and I gallon =3.785L)
Given: 1.00 gal H2O, dH2O = 1.00 g/mL
Plan:
gal H2O
g H2 O
L H2 O
mol H2O
Find: m= ?g H
mL H2O
g H2 O
moL H
gH
b)Determine the mass of oxygen in 7.2 –g sample of aluminum
sulfate,Al2(SO4)3.
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21
Empirical Formula
„
„
Empirical Formula-simplest, whole-number ratio
of the atoms of elements in a compound
can be determined from elemental analysis
„
masses of elements formed when decompose or
react
a. combustion analysis
b. percent composition
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Finding an Empirical Formula from
% composition
1) Convert the percentages to grams.
a) assume you start with 100 g of the compound
2) Convert grams to moles.
a) use molar mass of each element as conversion factor
3) Write a tentative formula using moles as subscripts.
4) Divide all subscripts (number of moles) by smallest subscript.
a) If result is within 0.1 of whole number, round to whole
number.
5) Multiply all mole ratios by a number to make all whole
numbers.
If ratio is
a)
0.5, multiply all by 2
b)
0.33 or 0.67, multiply all by 3
c)
0.25 or 0.75, multiply all by 4
b) skip if already whole numbers
Chapt 3
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Example 3.15
Laboratory analysis of aspirin determined the
following mass percent composition. Find the
empirical formula
„ C = 60.00%, H = 4.48%, O = 35.53%
Solution: Given: C = 60.00%, H = 4.48% ,O =
35.53%
Find: empirical formula, CxHyOz
Assume 100 g of sample.
„
In 100 g sample of aspirin there are 60.00 g C, 4.48 g
H, and 35.53 g O
Chapt 3
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22
„
Write a conceptual Plan
.
gg
whole
mole number
empirical
ratio empirical
pseudo- ratio
pseudoformula
formula
mol
mol
formula
formula
Convert g of C, H and O to moles using molar mass as
conversion factor
„
Chapt 3
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Example Cont’d
„
„
„
write a tentative formula C
4.996H4.44O2.220
find the mole ratio by dividing by the smallest
number of moles
multiply subscripts by factor to give whole number
{C2.25H2O1} x 4
C9H8O4
Chapt 3
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Practice example
„
Determine the empirical formula of stannous fluoride,
which contains 75.7% Sn (118.70g/mol) and the rest
fluorine (19.00g/mol).
Sn
ggSn
molSn
Sn
mol
ggFF
molFF
mol
mole
pseudo- ratio
pseudoformula
formula
Chapt 3
empirical
empirical
formula
formula
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23
Molecular Formula
„
„
„
Molecular formula (M.F)-gives the actual number
of each type of atoms in a molecule
„ Molecular formula may be same as the empirical
formula
„ May be a multiple of the empirical formula
„ M.F = (E.F) x n
„ Multiple (n ) = M.F. mass/ E.F mass
Can derive % composition from a chemical formula
To determine the molecular formula, you need to
know the empirical formula and the molar mass of
the compound.
Chapt 3
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Molecular Formula
„
„
„
For example, suppose the empirical formula of a
compound is CH2O and its molecular weight is
60.0 g/mol.
The molar mass of the empirical formula (the
empirical formula mass) is only 30.0 g/mol.
n = 60.0/30.0=2
This would imply that the molecular formula is
actually the empirical formula doubled, or
C2H4O2
Chapt 3
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Practice—Benzopyrene has a molar mass of
252 g/mol and an empirical formula of C5H3.
What is its molecular formula?
1. Determine E.F mass
C5 = 5(12.01 g) = 60.05 g
3.03 g
H3 = 3(1.01 g) =
C5 H3
=
63.08 g
2. Molar mass is given
so calculate multiple ‘n’
3. Mulitply E.F with n to obtain M.F
Molecular Formula = {C5H3} × 4 = C20H12
Chapt 3
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24
Empirical formula from combustion
analysis
Chapt 3
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Example 3.19
Combustion of a 0.8233-g sample of a compound
containing only carbon, hydrogen, and oxygen
produced 2.445 g CO2 and 0.6003 g H2O. Determine
the empirical formula of the compound.
1. Write down the given quantity and its units.
Given: compound = 0.8233 g
CO2 = 2.445 g
H2O = 0.6003 g
Find:
empirical formula, CxHyOz
„
Chapt 3
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Write a conceptual plan:
g
CO2, H2O
mol
C, H, O
mol
CO2, H2O
mol
C, H
pseudo
formula
Chapt 3
g
C, H
mol
ratio
g
O
mol
O
empirical
formula
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25
Calculate the grams of C and H in CO2 and H2O
Calculate the grams of O
„ = mass of sample-(mass of O+mass H)
Calculate the # of moles of C,H and O using their
molar masses as conversion factor
Write a tentative formula
Find the mole ratio by dividing by the smallest
number of moles
Multiply subscripts by factor to give whole number, if
necessary
Write the empirical formula
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Chapt 3
A. Ghumman
Sample Problem
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Combustion of 0.844 g of caproic acid(containing
only C, H and O) produced 0.784 g of H2O and 1.92 g
of CO2. Calculate the empirical formula for caproic
acid.
If the molar mass of caproic acid is 116.2 g/mol,
what is the molecular formula of caproic acid?
Chapt 3
A. Ghumman
Chemical Reactions
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Reactions involve rearrangement and exchange of
atoms to produce new molecules.
Chemical reactions are represented by chemical
equations
Reactants →Products
For example combustion of methane gas CH4
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Chemical Equations
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shorthand way of describing a reaction
provides information about the reaction
„ formulas of reactants and products
„ states of reactants and products
„ relative numbers of reactant and product
molecules that are required
To show the reaction obeys the Law of Conservation
of Mass, the equation must be balanced.
Chapt 3
A. Ghumman
Writing chemical equations
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Methane gas burns to produce carbon dioxide gas
and gaseous water.
„ Whenever something burns, it combines with
O2(g).
CH4(g) + O2(g) → CO2(g) + H2O(g)
We adjust the numbers of molecules so there are
equal numbers of atoms of each element on both
sides of the arrow.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Chapt 3
A. Ghumman
Balanced Chemical Equations
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This equation is balanced, meaning that there are
equal numbers of atoms of each element on the
reactant and product sides.
„ To obtain the number of atoms of an element,
multiply the subscript by the coefficient.
1←C→1
4←H→4
4←O→2+2
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Chapt 3
A. Ghumman
27
Symbols Used in Equations
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symbols used to indicate state after chemical
„ (g) = gas; (l) = liquid; (s) = solid
„ (aq) = aqueous = dissolved in water
energy symbols used above the arrow for
decomposition reactions
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Δ = heat and hν = light
Chapt 3
A. Ghumman
Example 3.21 Write a balanced equation for the
combustion of butane, C4H10.
Write a skeletal
equation:
Balance atoms in more
complex substances
first:
Balance free elements
by adjusting coefficient
in front of free element:
C4H10(l) + O2(g) → CO2(g) + H2O(g)
4⇐C⇒1×4
C4H10(l) + O2(g) → 4 CO2(g) + H2O(g)
10 ⇐ H ⇒ 2 × 5
C4H10(l) + O2(g) → 4 CO2(g) + 5 H2O(g)
13/2 × 2 ⇐ O ⇒ 13
C4H10(l) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(g)
If fractional coefficients, {C4H10(l) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(g)} × 2
multiply thru by
2 C4H10(l) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
denominator:
Check:
8 ⇐ C ⇒ 8; 20 ⇐ H ⇒ 20; 26 ⇐ O ⇒ 26
Chapt 3
A. Ghumman
Practice Example
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Acetic acid reacts with the metal aluminum to make
aqueous aluminum acetate and gaseous
hydrogen.Write a balanced equation for the reaction
„ Acids are always aqueous.
„ Metals are solid except for mercury.
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Practice
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Balance the following equations.
O2 +
P4
+
As2S3 +
PCl3 →
POCl3
N2O →
P4O6 + N2
O2 → As2O3 + SO2
Ca3(PO4)2
+ H3PO4
Chapt 3
→
Ca(H2PO4)2
A. Ghumman
Key Skills
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Writing empirical and Molecular formulas
Classification of substances as atom, molecules, Ionic or
molecular compounds
Naming ionic compounds, compounds containing polyatomic
ion, Molecular compounds, Acids.
Calculating Formula mass, using formula mass to count
molecules by weighing
Calculating mass percent composition
Using chemical formula as a conversion(Mole concept)
Obtaining empirical formula from (a) experimental data (b) from
combustion analysis
Obtaining molecular formula from empirical formula and molar
mass
Writing and Balancing chemical equations
Chapt 3
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