Chap2-Elements, Compounds, Chem Equations and Calculations

8/13/2019 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

Chap2-Elements, Compounds, Chem Equations and Calculations

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