elements compounds chemical equations calculations
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chemistryTRANSCRIPT
Chapter 2
BASIC CHEMISTRY
CHM 138
1. Elements are composed of extremely small particles called atoms.
2. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements.
3. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction.
4. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction.
8 X2Y16 X 8 Y+
Law of Conservation of Mass
- Matter can be neither created nor destroyed
Element- A substance that cannot be separated into simpler substances by chemical means.
Atom-The basic unit of an element that can enter into chemical combination
Proton- The positively charged particles in the nucleus
Neutron- Electrically neutral particles having a mass slightly greater than that of protons
Electron- Negatively charged particles
THE STRUCTURE OF THE ATOM
THE STRUCTURE OF THE ATOM
electron
6
mass p ≈ mass n ≈ 1840 x mass e-
Atomic number (Z)
= number of protons in nucleus
Mass number (A)
= number of protons + number of neutrons
= atomic number (Z) + number of neutrons
XAZ
Mass Number
Atomic NumberElement Symbol
Atomic number, Mass number and Isotopes
Example:
Mass number (A) = 16Atomic number (Z) = 8 (indicating 8 protons in
nucleus)Number of neutrons = 16-8 = 8Number of electrons = 8 (when the element is
neutral)
O16
8Element Symbol
Mass Number
Atomic Number
Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei or atoms that have the same atomic number but different mass number.
Examples:
H11
H (D)21
H (T)31
U23592
U23892
1) Hydrogen
2) Uranium
The Modern Periodic Table
Period
Group
Alkali M
etal
Noble G
as
Halogen
Alkali E
arth Metal
• A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical forces
H2 H2O NH3 CH4
• A diatomic molecule contains only two atoms
• Examples: H2, N2, O2, Br2, HCl, CO
• A polyatomic molecule contains more than two atoms
• Examples: O3, H2O, NH3, CH4
diatomic elements
MOLECULES AND IONS
• An ion is an atom, or group of atoms, that has a net positive or negative charge.
• Cation – ion with a positive charge - If a neutral atom loses one or more electrons it becomes a cation.
• anion – ion with a negative charge - If a neutral atom gains one or more electrons it becomes an anion.
Na 11 protons11 electrons Na+ 11 protons
10 electrons
Cl 17 protons17 electrons Cl-
17 protons18 electrons
• A monatomic ion contains only one atom
• A polyatomic ion contains more than one atom
• Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-
• Examples: OH-, CN-, NH4+, NO3
-
Common Ions Shown on the Periodic Table
• metals tend to form cations
• nonmetals tend to form anions
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 = 13 – 3 = 10
No. of protons = 34
Charge = 2- (accept of 2 electrons)
No. of electrons = 34 + 2 = 36
A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance
Allotrope: one of two or more distinct forms of an element.
- Example: two allotropic forms of the element carbon which
are diamond and graphite.
An empirical formula shows the simplest whole-number ratio of the atoms in a substance
H2OH2OMolecular formula Empirical formula
C6H12O6 CH2O
O3 O
N2H4NH2
CHEMICAL FORMULAS
A structural formula shows how atoms are bonded to one another in a molecule
Formulas and Models
• ionic compounds consist 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
The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds.
Method of Writing Chemical Formula for Ionic Compounds
1) Aluminium oxide (containing Al3+ and O2-)
Al3+ O2-
Charge 3+ 2-
Simplest ration of 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
Method of Writing Chemical Formula for Ionic Compounds
2) Ammonium carbonate (containing NH4+ and CO3
2-)
NH4+ CO3
2-
Charge 1+ 2-
Simplest ration of 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
1) Elements:Refer to the periodic table
- Examples:
i) Na = sodium
ii) Si = silicon
CHEMICAL NOMENCLATURE
2) Ionic CompoundsOften a metal (cation) + nonmetal (anion)Binary compounds (compounds formed from two
elements)
- first element named is the metal cation followed by the nonmetallic anion.
Anion (nonmetal), add “ide” to element nameExamples:
i) BaCl2 = barium chloride
ii)K2O = potassium oxide
iii) Mg(OH)2 = Magnesium hydroxide
27
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:
31
3) Molecular compounds
- place the name of the first element in the formula first and second element is named by adding “-ide” to the root of element name
- 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 formed from the same elements, use prefixes to indicate number of each kind of atom
- Last element name ends in “-ide”
Guidelines in naming compounds with prefixes
The prefix ‘mono-’ maybe omitted for the first element.
For oxides, the ending ‘a’ in the prefix is sometimes omitted.
- for example: N2O4 maybe called dinitrogen teroxide rather than dinitrogen teraoxide.
HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide
Molecular Compounds
36
• An acid can 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
• An oxoacid is 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:
39
Naming Oxoacids and Oxoanions
The rules for naming oxoanions, anions of oxoacids, are as follows:
1. When all the H ions are removed from the “-ic” acid, the anion’s name ends with “-ate”.
2. When all the H ions are removed from the “-ous” acid, the anion’s name ends with “-ite.”
3. The names of anions in which one or more but not all the hydrogen ions have been removed must indicate the number of H ions present.
For example:H3PO4 Phosphoric acidH2PO4
- dihydrogen phosphateHPO4 2- hydrogen phosphatePO4
3- phosphate
• A base can 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
• Hydrates are compounds that have a specific number of water molecules attached to them.
• Examples:
i) BaCl2•2H2O barium chloride dihydrate
ii) LiCl•H2O
iii) MgSO4•7H2O
iv) Sr(NO3)2 •4H2O
lithium chloride monohydrate
magnesium sulfate heptahydrate
strontium nitrate tetrahydrate
CuSO4•5H2O CuSO4
5) Hydrates
By definition: 1 atom 12C “weighs” 12 amu
On this scale:
1H = 1.008 amu
16O = 16.00 amu
• Atomic mass is the mass of an atom in atomic mass units (amu)
• One atomic mass unit – a mass exactly equal to one-twelfth the mass of one carbon-12 atom.
ATOMIC MASS
The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.
Average atomic mass of natural carbon
= (0.9890)(12.00000 amu)+(0.0110)(13.00335 amu)
= 12.01 amu
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
(7.42 x 6.015) + (92.58 x 7.016)100
= 6.941 amu
Average atomic mass of lithium:
Example:
Average atomic mass (6.941)
• The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
Dozen = 12
Pair = 2
• The Mole (mol): A unit to count numbers of particles
AVOGADRO’S NUMBER AND THE MOLAR MASS
Molar mass is the mass of 1 mole of in gramsatoms
1 mole 12C atoms = 6.022 x 1023 atoms = 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)
NA = Avogadro’s number = 6.022 x 1023 atoms
Mass of element (m)
No. of moles (n)
No. of atoms/molecules (N)
÷ molar mass (g/mol) x NA
÷ NAx molar mass (g/mol)
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms 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 1023 atoms/mol
= 8.43 x 1021 atoms K
• Molecular mass (or molecular weight) is the sum ofthe 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
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 1023 atoms H
= 5.82 x 1024 atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O1 mol C3H8O
60 g C3H8Ox
8 mol H atoms
1 mol C3H8Ox
6.022 x 1023 H atoms
1 mol H atomsx
Example
• Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.
1Na 22.99 amu
1Cl + 35.45 amuNaCl 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
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:
Percent composition of an element in a compound =
n x molar mass of elementmolar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
C2H6O
%C =2 x (12.01 g)
46.07 gx 100% = 52.14%
%H =6 x (1.008 g)
46.07 gx 100% = 13.13%
%O =1 x (16.00 g)
46.07 gx 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
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
Simplest ratio
0.6330
0.6329
≈1
0.6329
0.6329
=1
2.532
0.6329
≈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%
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
= 3.406
Simplest 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
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 of empirical formula
Determination of molecular formula1) Empirical molar mass = 14.01 g/mol + 2(16.0g/mol) = 46.01 g/ mol molar mass compound between 90 g/mol-95 g/mol
2) Determine the ratio between the molar mass and empirical formula
Molar mass = 90 g/mol ≈ 2Empirical 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
3 ways of representing the reaction of H2 with O2 to form H2O
• A process in which one or more substances is changed into one or more new substances is a chemical reaction
• A chemical equation uses chemical symbols to show what happens during a chemical reaction
reactants products
CHEMICAL REACTIONS AND CHEMICAL EQUATIONS
How to “Read” Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6 NOT C4H12
3. Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbonon left
1 carbonon right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogenon left
2 hydrogenon right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
Balancing Chemical Equations
4. Balance those elements that appear in two or more reactants or products.
2 oxygenon left
4 oxygen(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen(3x1)
multiply O2 by 72
= 7 oxygenon right
C2H6 + O2 2CO2 + 3H2O72 remove fraction
multiply both sides by 22C2H6 + 7O2 4CO2 + 6H2O
Balancing Chemical Equations
5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C12 H14 O
4 C12 H14 O
4 C (2 x 2) 4 C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)
Balancing Chemical Equations
AMOUNTS OF REACTANTS AND PRODUCTS
• Stoichiometry:
- comparison of coefficients in a balanced equation
- The quantitative study of reactants and products in a chemical reaction
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
Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 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 massCH3OH
coefficientschemical equation
molar massH2O
Example:
1) Moles of CH3OH = 209 g
32 g/mol
= 6.53 mol
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 will produce?
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
• How many grams and molecules that CO2 will produced?
2NO + O2 2NO2
NO is the limiting reagent
O2 is 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
LIMITING REAGENT
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
1) Mole of Al
= 124 g
27.0 g/mol
= 4.59 mol
2) Mole of Fe2O3
= 601 g
160 g/mol
= 3.76 mol
Determination of limiting reagent and excess reagent
3) Divide moles of Al and Fe2O3 with their stoichiometric coefficients
i) 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 mole is a limiting reagent, while another reagent is a excess reagent.
• So, Al is a limiting reagent, while Fe2O3 is a excess reagent.
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 Al2O3 will 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
• Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.
• Actual Yield is the amount of product actually obtained from a reaction.
% Yield = Actual Yield
Theoretical Yieldx 100%
REACTION YIELD