Chapter 2Chapter 2

BASIC CHEMISTRY

CHM 160

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:Example:

Mass number (A) = 16Mass number (A) = 16 Atomic number (Z) = 8 (indicating 8 protons in Atomic number (Z) = 8 (indicating 8 protons in

nucleus)nucleus) Number of neutrons = 16-8Number of neutrons = 16-8 = 8= 8 Number of electrons = 8 (when the element is Number of electrons = 8 (when the element is

neutral)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:1) Elements: Refer to the periodic tableRefer to the periodic table

- Examples: - Examples:

i) Na = sodiumi) Na = sodium

ii) Si = siliconii) Si = silicon

CHEMICAL NOMENCLATURE

2) Ionic Compounds2) Ionic Compounds Often a metal (cation) + nonmetal (anion)Often a metal (cation) + nonmetal (anion) Binary compounds (compounds formed from two Binary compounds (compounds formed from two

elements)elements)

- first element named is the metal cation followed by - first element named is the metal cation followed by the nonmetallic anion.the nonmetallic anion.

Anion (nonmetal), add “Anion (nonmetal), add “ideide” to element name” to element name Examples:Examples:

i) BaCli) BaCl22 = barium chlor = barium chlorideide

ii)Kii)K22O = potassium oxO = potassium oxideide

iii) Mg(OH)iii) Mg(OH)22 = Magnesium hydrox = Magnesium hydroxideide

Transition metal ionic compoundsTransition metal ionic compounds

- older nomenclature system:- older nomenclature system:

- ending “ous” cation with fewer positive charges- ending “ous” cation with fewer positive charges

- ending “ic” to the cation with more positive charges- ending “ic” to the cation with more positive charges

- examples: Fe- examples: Fe2+ 2+ ferrous ion ferrous ion

FeFe3+ 3+ ferric ion ferric ion

- indicate charge on metal with Roman numerals- 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:

3) Molecular compounds3) Molecular compounds

- place the name of the first element in the - place the name of the first element in the formula first and second element is named formula first and second element is named by adding “-ide” to the root of element nameby adding “-ide” to the root of element name

- Nonmetals or nonmetals + metalloids- Nonmetals or nonmetals + metalloids

- Common names: H- Common names: H22O, NHO, NH33, CH, CH44

- Element furthest to the left in a period and - Element furthest to the left in a period and closest to the bottom of a group on closest to the bottom of a group on periodic table is placed first in formulaperiodic table is placed first in formula

- If more than one compound can be formed - If more than one compound can be formed from the same elements, use prefixes to from the same elements, use prefixes to indicate number of each kind of atomindicate number of each kind of atom

- Last element name ends in “-- Last element name ends in “-ideide””

Guidelines in naming compounds Guidelines in naming compounds with prefixeswith prefixes

The prefix ‘mono-’ maybe omitted for the first The prefix ‘mono-’ maybe omitted for the first element.element.

For oxides, the ending ‘a’ in the prefix is For oxides, the ending ‘a’ in the prefix is sometimes omitted.sometimes omitted.

- for example: N- for example: N22OO44 maybe called dinitrogen maybe called dinitrogen

teroxide rather than dinitrogen teraoxide.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 The rules for naming oxoanions, anions of oxoacids, oxoanions, anions of oxoacids, are as are as follows:follows:

1. When all the H ions are removed from the 1. When all the H ions are removed from the “-ic” acid“-ic” acid, the , the anion’s name ends with anion’s name ends with “-ate”.“-ate”.

2. When all the H ions are removed from the 2. When all the H ions are removed from the “-ous” acid“-ous” acid, the , the anion’s name ends with anion’s name ends with “-ite.”“-ite.”

3. The names of anions in which 3. The names of anions in which one or more but not allone or more but not all the the hydrogen ionshydrogen ions have been removed must indicate the number have been removed must indicate the number of H ions present. of H ions present.

For example:For example: HH33POPO4 4 Phosphoric acidPhosphoric acid HH22POPO44

-- dihydrogen phosphatedihydrogen phosphate HPOHPO44 2-2- hydrogen phosphatehydrogen phosphate POPO44

3-3- phosphatephosphate

• 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 The average atomic massaverage atomic mass is the weighted average of is the weighted average of all of the naturally occurring isotopes of the all of the naturally occurring isotopes of the element. 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 formulaDetermination of empirical formula

Elements Elements KK MnMn OO

Mass (g)Mass (g) 24.7524.75 34.7734.77 40.5140.51

molmol 24.7524.75 g g

39.10 39.10 g/molg/mol

= 0.6330= 0.6330

34.7734.77 g g

54.94 54.94 g/molg/mol

= 0.6329= 0.6329

40.5140.51 g g

16.00 16.00 g/molg/mol

= 2.532= 2.532

Simplest Simplest ratioratio

0.63300.6330

0.63290.6329

≈≈11

0.63290.6329

0.63290.6329

=1=1

2.5322.532

0.63290.6329

≈≈44Empirical formula = KMnOEmpirical formula = KMnO44

Determine the empirical formula of a compound that has the following percent composition by mass:

* Assume we have 100 g of the compound, then each percentage can be converted directly to grams.

K: 24.75%, Mn: 34.77%, O: 40.51%

Determination of empirical Determination of empirical formulaformula

Elements Elements CC HH OO

Mass (g)Mass (g) 40.9240.92 4.584.58 54.5054.50

molmol 40.9240.92 g g

12.01 12.01 g/molg/mol

= 3.407= 3.407

4.584.58 g g

1.008 1.008 g/molg/mol

= 4.54= 4.54

54.5054.50 g g

16.00 16.00 g/molg/mol

= 3.406= 3.406

Simplest Simplest ratioratio

3.4073.407

3.4063.406

≈≈1 x 31 x 3

= 3= 3

4.544.54

3.4063.406

=1.33 x 3=1.33 x 3

= 4= 4

3.4063.406

3.4063.406

==1 x 31 x 3

= 3= 3

Empirical formula = CEmpirical formula = C33HH44OO33

Determination of Molecular FormulaDetermination of Molecular Formula

Elements Elements NN OO

Mass (g)Mass (g) 1.521.52 3.473.47

molmol 1.521.52 g g

14.01 g/mol14.01 g/mol

= 0.108= 0.108

3.473.47 g g

16.00 g/mol16.00 g/mol

= 0.217= 0.217

Simplest Simplest ratioratio

0.1080.108

0.1080.108

≈≈11

0.2170.217

0.1080.108

≈≈22

Empirical formula = NOEmpirical formula = NO22

Determination of empirical formula

Determination of molecular formulaDetermination of molecular formula

1) Empirical molar mass 1) Empirical molar mass = 14.01 g/mol + 2(16.0g/mol) = 46.01 g/ mol= 14.01 g/mol + 2(16.0g/mol) = 46.01 g/ mol molar mass compound between 90 g/mol-95 g/molmolar mass compound between 90 g/mol-95 g/mol

2) Determine the ratio between the molar mass and 2) Determine the ratio between the molar mass and empirical formulaempirical formula

Molar mass = 90 g/mol Molar mass = 90 g/mol ≈ 2≈ 2Empirical molar mass 46.01 g/molEmpirical molar mass 46.01 g/mol

Molecular formula = 2(NOMolecular formula = 2(NO22))

= = NN22OO44

Actual molecular molar mass = 2(14.01 g/mol) + 4(16.00)Actual molecular molar mass = 2(14.01 g/mol) + 4(16.00) = 92.02 g/mol= 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 EquationsHow 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 CH2) From the equation, 2 mol CH33OH is used to give 4 mol HOH is used to give 4 mol H22O, if we O, if we

have 6.53 mol CHhave 6.53 mol CH33OH, how many mole that HOH, how many mole that H22O will produce?O will produce?

2 mol CH2 mol CH33OH = 4 mol HOH = 4 mol H22OO

6.53 mol CH6.53 mol CH33OH = ? mol HOH = ? mol H22OO

= 4 mol H= 4 mol H22O x 6.53 mol CHO x 6.53 mol CH33OHOH

2 mol CH2 mol CH33OHOH

= 13.06 mol H= 13.06 mol H22OO

3) Mass of H3) Mass of H22OO

= mol x molar mass H= mol x molar mass H22OO

= 13.06 mol x 18 g/mol= 13.06 mol x 18 g/mol

= = 235.1 g235.1 g

2CH3OH + 3O2 2CO2 + 4H2O

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 LIMITING REAGENTREAGENT

LIMITING LIMITING REAGENTREAGENT

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)4) From the equation, 2 mol Al is used to give 1 mol AlFrom the equation, 2 mol Al is used to give 1 mol Al22OO33 , if we , if we

have 4.59 mol Al, how many mole that Alhave 4.59 mol Al, how many mole that Al22OO33 will produce? will produce?

2 mol Al produce 1 mol Al2 mol Al produce 1 mol Al22OO33

4.59 mol Al = 4.59 mol Al = 1mol Al1mol Al22OO33 x 4.59 mol Al x 4.59 mol Al

2 mol Al2 mol Al

= 2.295 mol Al= 2.295 mol Al22OO33

5) Mass of Al5) Mass of Al22OO33

= mol x molar mass Al= mol x molar mass Al22OO33

= 2.295 mol x 102.0 g/mol= 2.295 mol x 102.0 g/mol

= = 234 g234 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