Chap2-Elements, Compounds, Chem Equations and Calculations
Transcript of Chap2-Elements, Compounds, Chem Equations and Calculations
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ELEMENTS, COMPOUNDS,CHEMICALEQUATIONS ANDCALCULATIONS
CHAPTER 2
MISS J
CHM 138
Basic Chemistry
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Chapter Outline
2.1 Definitions of atom, ion, molecule, compound 2.2 Symbol, chemical formula and naming of
elements, molecules and compounds.
2.3 Atomic mass, formula mass, molecular mass
Avogadros constant, mole concept, mole
conversions.
2.4 Calculations on compositions, empirical and
molecular formulas. 2.5 Balancing chemical equations.
2.6 Stoichiometric calculations
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Atom:
Neutral particles & the smallest unit of anelement.
Can take part in chemical reaction
Are composed of smaller particles (electrons,
protons & neutrons)Example: Mg, K, Zn, Fe, Al, Ca, Li, S
Molecule:
The collection of two or more atoms bound togetherchemically
Molecules such as H2,N2,O2,Br2are called diatomicmolecules.
Example: CO, NH4, HNO3, CO2
ATOMS & MOLECULES
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THE STRUCTURE OF THE ATOM
electron
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The particles in the nucleus of an atom called NUCLEONS
Sub-atomic particles = protons, neutrons and electrons
Proton number (Z)= the number of protons in the
nucleus of an atom (atomic number)
Nucleon number (A) = the total number of protons &
neutrons in the nucleus (mass number)
part ic le symbol charge
Proton p +1
Neutron n 0
Electron e -1
THE STRUCTURE OF THE ATOM
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Atomic number(Z)
= number of protons in nucleus
Mass number(A)
= number of protons + number of neutrons
= atomic number (Z) + number of neutrons
XAZMass Number
Atomic NumberElement Symbol
Atomic number, Mass number
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Example:
Mass number (A) = 16
Atomic number (Z) = 8 (indicating 8 protons innucleus)
Number of neutrons = 16-8
= 8 Number of electrons = 8 (when the element is
neutral)
O16
8Element Symbol
Mass Number
Atomic Number
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The Modern Periodic Table
Period
Grou
p
AlkaliMe
tal
NobleGas
Halo
gen
AlkaliEarthMe
tal
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A moleculeis an aggregate of two or more atoms in a definite
arrangement held together by chemical forces
H2 H2O NH3 CH4
A diatomic moleculecontains only two atoms
Examples: H2, N2, O2, Br2, HCl, CO
A polyatomic moleculecontains more than two atoms
Examples: O3, H2O, NH3, CH4
diatomic elements
MOLECULES AND IONS
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Anionis an atom, or group of atoms, that has a net positive or
negative charge.
Cationion with a positive charge- If a neutral atom loses one or more electrons it becomes a cation.
anionion with a negative charge
- If a neutral atom gains one or more electrons it becomes an anion.
Na11 protons
11 electrons Na+11 protons
10 electrons
Cl17 protons
17 electrons Cl-
17 protons
18 electrons
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A monatomic ioncontains only one atom
A polyatomic ioncontains more than one atom
Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-
Examples: OH-, CN-, NH4+, NO3
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Common Ions Shown on the Periodic Table
metals tend to form cations
nonmetals tend to form anions
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1) How many protons and electrons are in ?Al2713
3+
2) How many protons and electrons are in ?Se7834
2-
Examples:
No. of protons = 13
Charge = 3+ (loss of 3 electrons)
No. of electrons = 133 = 10
No. of protons = 34
Charge = 2- (accept of 2 electrons)
No. of electrons = 34 + 2 = 36
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A molecular formulashows the exact number of atoms of each element
in the smallest unit of a substance
An empir ical formulashows the simplest whole-number ratio of the
atoms in a substance
H2OH2O
Molecular formula Empirical formula
C6H12O6 CH2OO3 O
N2H4NH2
CHEMICAL FORMULAS
A structural formulashows how atoms are bonded to one another in
a molecule
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Formulas and Models
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ionic compoundsconsist of a combination of cations and an
anions
The formula is usually the same as the empirical formula
The sum of the charges on the cation(s) and anion(s) in
each formula unit must equal zero
Examples: NaCI (consists of equal numbers of Na+and Cl-)
Formula of Ionic Compounds
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The most reactive metals (green) and the most reactive
nonmetals (blue) combine to form ionic compounds.
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Method of Writing Chemical Formula for Ionic
Compounds
1) Aluminium oxide (containing Al3+and O2-)
Al3+ O2-
Charge 3+ 2-
Simplest rationof ion combined 2 3
Sum of charges is 2(+3) + 3(-2) = 0
So, 2 cation Al3+combined with 3 anion O2-to form
aluminium oxide
Formula: Al2O3
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Method of Writing Chemical Formula for Ionic
Compounds
2) Ammonium carbonate (containing NH4
+and CO3
2-)
NH4+ CO3
2-
Charge 1+ 2-
Simplest rationof ion combined 2 1
Sum of charges is 2(+1) + 1(-2) = 0
So, 2 cation NH4+combined with 1 anion CO3
2-to
form ammonium carbonate
Formula: (NH4)2CO3
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1) Elements:
Refer to the periodic table
- Examples:
i) Na = sodium
ii) Si = silicon
CHEMICAL NOMENCLATURE
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2) Ionic Compounds Often a metal (cation) + nonmetal (anion)
Binary compounds (compounds formed from twoelements)
- first element named is the metal cation followed bythe nonmetallic anion.
Anion (nonmetal), add ide to element name Examples:
i) BaCl2= barium chloride
ii)K2O = potassium oxide
iii) Mg(OH)2= Magnesium hydroxide
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Transition metal ionic compounds
- older nomenclature system:
- ending ous cation with fewer positive charges
- ending ic to the cation with more positive charges
- examples: Fe2+ ferrous ion
Fe3+ ferric ion- indicate charge on metal with Roman numerals
i) FeCl2 2 Cl- -2 so Fe is +2 iron(II) chloride
ii) FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride
iii) Cr2S3 3 S-2-6 so Cr is +3 (6/2) chromium(III) sulfide
Examples:
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3) Molecular compounds
- place the name of the first element in theformula first and second element is namedby adding -ide to the root of elementname
- Nonmetals or nonmetals + metalloids
- Common names: H2O, NH3, CH4
- Element furthest to the left in a period and
closest to the bottom of a group on
periodic table is placed first in formula
- If more than one compound can be formedfrom the same elements, use prefixes toindicate number of each kind of atom
- Last element name ends in -ide
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Guidelines in naming
compounds with prefixes
The prefix mono- maybe omitted for the first
element.
For oxides, the ending a in the prefix issometimes omitted.
- for example: N2O4maybe called dinitrogen
teroxide rather than dinitrogen teraoxide.
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HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide
Molecular Compounds
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An acidcan be defined as a substance that yields hydrogen ions
(H+) when dissolved in water.
For example: HCl gas and HCl in water
- Pure substance, hydrogen chloride
- Dissolved in water (H3O+ and Cl), hydrochloric acid
Anions whose names end in -ide form acids with a hydro-
prefix and an -ic ending.
4) Acids and bases
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An oxoacidis an acid that contains hydrogen, oxygen, and
another element.
i) HNO3 nitric acid
ii) H2CO3 carbonic acid
iii) H3PO4 phosphoric acid
iv) HCIO3 chloric acid
v) H2SO4 sulfuric acid
vi) HIO3 iodic acid
vii)HBrO3 bromic acid
Examples:
N i O id d O i
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Naming Oxoacids and Oxoanions
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The rules for naming oxoanions, anions of oxoacids, areas follows:
1. When all the H ions are removed from the -ic acid, theanions name ends with -ate.
2. When all the H ions are removed from the -ous acid,the anions name ends with -ite.
3. The names of anions in which one or more but not all thehydrogen ionshave been removed must indicate thenumber of H ions present.
For example:
H3PO4 Phosphoric acid
H2PO4- dihydrogen phosphate HPO4
2-hydrogen phosphate
PO43- phosphate
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Abasecan be defined as a substance that yields
hydroxide ions (OH-
) when dissolved in water.
Examples:
NaOH sodium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide
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Hydratesare compounds that have a specific number of
water molecules attached to them.Examples:
i) BaCl22H2O barium chloride dihydrate
ii) LiClH2Oiii) MgSO47H2O
iv) Sr(NO3)24H2O
lithium chloride monohydrate
magnesium sulfate heptahydrate
strontium nitrate tetrahydrate
CuSO45H2O CuSO4
5) Hydrates
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By definition:atom 2C weighs 2 amu
On this scale:
1H = 1.008 amu
16O = 16.00 amu
Atomic massis the mass of an atom in atomic mass units (amu)One atomic mass unita mass exactly equal to one-twelfth the
mass of one carbon-12 atom.
ATOMIC MASS
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Average atomic mass (6.941)
AVOGADROS NUMBER AND THE
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The mole (mol)is the amount of a substance that contains
as many elementary entities as there are atoms in exactly
12.00 grams of12
C
1 mol = NA= 6.0221367 x 1023
Avogadros number (NA)
Dozen = 12
Pair = 2
The Mole (mol): A unit to count numbers of particles
AVOGADRO S NUMBER AND THE
MOLAR MASS
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Molar massis the mass of 1 mole of in gramsatoms
1 mole 12C atoms = 6.022 x 1023atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
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NA= Avogadros number = 6.022 x 1023atoms
Mass of
element (m)
No. of
moles (n)
No. of
atoms/molecules (N)
molar mass (g/mol) x NA
NAx molar mass (g/mol)
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How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023atoms K
Example:
No. of moles = 0.551 g
39.10 g/mol
= 0.014 mol
No. of atoms = 0.014 mol x 6.022 x 1023atoms/mol
= 8.43 x 1021atoms K
MOLECULAR MASS
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Molecular mass(or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2= 64.07 amu
1 mole SO2= 64.07 g SO2
SO2
MOLECULAR MASS
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How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023atoms H
= 5.82 x 1024atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O1 mol C3H8O
60 g C3H
8O
x8 mol H atoms
1 mol C3H
8O
x6.022 x 1023H atoms
1 mol H atomsx
Example
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Formula massis the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na 22.99 amu
1Cl + 35.45 amu
NaCl 58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
NaCl
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What is the formula mass of Ca3(PO4)2?
1 formula unit of Ca3(PO4)2
3 Ca 3 x 40.08
2 P 2 x 30.97
8 O + 8 x 16.00
310.18 amu
Example:
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Percent compositionof an element in a compound =
nx molar mass of element
molar mass of compound
x 100%
nis the number of moles of the element in 1 moleof
the compound
C2H6O
%C =2x (12.01 g)
46.07 gx 100% = 52.14%
%H =6x (1.008 g)
46.07 gx 100% = 13.13%
%O =1x (16.00 g)
46.07 gx 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
D t i ti f i i l f l
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Determination of empirical formula
Elements K Mn O
Mass (g) 24.75 34.77 40.51
mol 24.75 g
39.10 g/mol
= 0.6330
34.77 g
54.94 g/mol
= 0.6329
40.51 g
16.00 g/mol
= 2.532
Simplestratio
0.63300.6329
1
0.63290.6329
=1
2.5320.6329
4
Empirical formula = KMnO4
Determine the empirical formula of a compound that has
the following percent composition by mass:
K: 24.75%, Mn: 34.77%, O: 40.51%
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Determination of empirical formula
Elements C H O
Mass (g) 40.92 4.58 54.50
mol 40.92 g
12.01 g/mol
= 3.407
4.58 g
1.008 g/mol
= 4.54
54.50 g
16.00 g/mol
= 3.406Simplest
ratio
3.407
3.406
1 x 3
= 3
4.54
3.406
=1.33 x 3
= 4
3.406
3.406
=1 x 3
= 3
Empirical formula = C3H4O3
Determination of Molecular Formula
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Determination of Molecular Formula
Elements N O
Mass (g) 1.52 3.47
mol 1.52 g
14.01 g/mol
= 0.108
3.47 g
16.00 g/mol
= 0.217
Simplest ratio 0.108
0.108
1
0.217
0.108
2
Empirical formula = NO2
Determination ofempirical formula
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Determination of molecular formula
1) Empirical molar mass
= 14.01 g/mol + 2(16.0g/mol) = 46.01 g/ molmolar mass compound between 90 g/mol-95 g/mol
2) Determine the ratio between the molar mass andempirical formula
Molar mass = 90 g/mol 2
Empirical molar mass 46.01 g/mol
Molecular formula = 2(NO2)= N2O4
Actual molecular molar mass = 2(14.01 g/mol) + 4(16.00)
= 92.02 g/mol
CHEMICAL REACTIONS AND
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3 ways of representing the reaction of H2with O2to form H2O
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equationuses chemical symbols to show what happens
during a chemical reaction
reactants products
CHEMICAL EQUATIONS
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AMOUNTS OF REACTANTS ANDPRODUCTS
Stoichiometry:
- comparison of coefficients in a balanced equation
- The quantitative study of reactants and products
in a chemical reaction
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1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number
of moles of the sought quantity
4. Convert moles of sought quantity into desired units
Example:
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Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2+ 4H2OIf 209 g of methanol are used up in the combustion, what mass of
water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
Example:
1) Moles of CH3OH = 209 g
32 g/mol
= 6.53 mol
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2) From the equation, 2 mol CH3OH is used to give 4 mol H2O, if
we have 6.53 mol CH3OH, how many mole that H2O willproduce?
2 mol CH3OH = 4 mol H2O
6.53 mol CH3OH = ? mol H2O= 4 mol H2O x 6.53 mol CH3OH
2 mol CH3OH
= 13.06 mol H2O
3) Mass of H2O
= mol x molar mass H2O
= 13.06 mol x 18 g/mol
= 235.1 g
2CH3OH + 3O2 2CO2+ 4H2O
LIMITING
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2NO + O2 2NO2
NO is the limiting reagent
O2is the excess reagent
Reactant used up first in the reaction.
Excess reagents: the reactants
present in quantities greater than
necessary to react with the quantity
of the limiting reagent
LIMITING
REAGENT
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3) Divide moles of Al and Fe2O3with their stoichiometric
coefficientsi) Al ii) Fe2O3
= 4.59 mol = 2.295 mol = 3.76 mol = 3.76 mol
2 1
The reagent that show the smallest no. of moleis a limiting
reagent, while another reagent is a excess reagent.
So, Al is a limiting reagent, while Fe2O3is a excess reagent.
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4)From the equation, 2 mol Al is used to give 1 mol Al2O3, if we
have 4.59 mol Al, how many mole that Al2O3will produce?
2 mol Al produce 1 mol Al2O3
4.59 mol Al = 1mol Al2O3 x 4.59 mol Al
2 mol Al
= 2.295 mol Al2O3
5) Mass of Al2O3
= mol x molar mass Al2O3
= 2.295 mol x 102.0 g/mol= 234 g
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Theoretical Yieldis the amount of product that would
result if all the limiting reagent reacted.
Actual Yieldis the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield x 100%
REACTION YIELD
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The End.TQ