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TOPIC 5-CHEMICAL COMPOUNDS

CONTENTS

•Types of Chemical Compounds and Their Formulas

•The Mole Concept and Chemical Compounds

•Composition of Chemical Compounds

•Oxidation States

•Naming Chemical Compounds

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CHEMICAL COMPOUNDS• They are composed of two or more

different elements

• A molecule of a compound is a group of bonded atoms that actually exists and can be identified as a distinct entity

• Chemical formula indicates

• The elements present and

• The relative numbers of atoms of each element in the compound

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Molecular CompoundsA molecular compound is a compound

• Comprised of discrete molecules• The forces that hold atoms together in molecules are

known covalent bonds

–A formula unit is the smallest collection of atoms on which a formula can be based.

–Water H2O –Carbon dioxide CO2

–Sodium chloride NaCl–Iron (II) sulfide FeS

Subscript numbers indicate the relative numbers of atoms

Types of Chemical Compounds and Their Formulas

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Empirical Formula,Molecular FormulaAnd Structural Formula

The empirical formula is the simplest formula of a compound.It tells us the types of atoms present and their relative numbers Exp: The empirical formula of glucose is CH2O.

The molecular formula is based on an actual molecule of the compound. In some cases the empirical and molecular formulas are identical such as Formaldehyde(CH2O). In other cases the molecular formula is a «multiple» of the empirical formula such as acetic acid(C2H4O2) and glucose(C6H12O6)

A structural formula show the order in which atoms are bonded together in a molecule and by what types of bonds

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Empirical formula CH2O

Molecular formula C2H4O2

Structural formula

Molecular Model(Ball and stick)

Molecular Model(Space filling)

CH3COOH

CH3CO2H

Linear formula

Oleic acid (Omega-9)

C18H36

Acetic acid

CH3CO2H

C

H

H

H

C

O

O H

O

O

H

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Ionic Compounds An important feature of the metallic elements isthe tendency

of their atoms to lose one or more electrons when they combine with nonmetal atoms. In these combinations the nonmetal atoms display a tendency to gain one or more electrons.

The ionic compounds are the compounds comprised of positive and negative ions joined together by electrostatic forces of attraction.

As a result of the electron transfer, the metal atom becomes a positive ion and is a cation .

And the nonmetal atom becomes a negative ion and is called an anion.

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In some cases electrons are transferred from one atom to another. Positive and negative ions are formed and attract each other through electrostatic forces called Ionic bonds

Na [Ne] 3s1 + Cl [Ne] 3s2 3p5

Na [Ne]+ + Cl [Ar]-

Cl-Na+

“Cation” - Metal atoms lose electrons and build up the “ + ” charged ions.

“Anion” – Nonmetal atoms gain electrons and build up the “ - ” charged ions

Sodium combines with chlorine.As a result of this combination occurs an ionic compound Sodiumchloride

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Na:Cl = 1:1 formula unit = NaCl

Mg+2

Na+ + Cl - Na Cl

NaCl is a compound with a neutral electrical charge

Cl-

Cl-

Mg+2 + 2Cl - MgCl2

MgCl2 is a compound with a

neutral electrical charge

formula unit

Formula unit is the smallest collection of the atoms on which the formula is based

Formula Unit

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The Mole Concept and Chemical Compounds

• Formula Mass: The mass of a compound in a formula unit as in amu(atomic mass unit).

• Molecular mass: The mass of a molecule relative to the amu. (Formula mass= Molecular mass)

• 1 mole of compound is an amount of compound containing 6,02214 x1023 formula unit or molecules

• Molar mass: The mass of one mol of the compound- one mol of molecules of a molecular compound and one mol of formula units of an ionic compound

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Formula mass: is the total mass of the atoms, building up a compound, in amu.

Calculate the formula mass of these compounds: NaCl, H2O, ve H3PO4

NaCl

1 Na 22,9898 amu= 22,9898 amu1 Cl 35,4527 amu= 35,4527 amu

1 NaCl= 58,4425 amu

2 H 1,0079 amu= 2,0158 amu1 O 15,9994 amu= 15,9994 amu

1 H2O= 18,0152 amu

H3PO4

3 H 1,0079 amu= 3,0237 amu1 P 30,9738 amu= 30,9738 amu4 O 15,9994 amu= 63,9976 amu

1 H3PO4= 97,9951 amu

H2O

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Matter Formula mass Molecular mass

H2

Fe

H3PO4

SO3

NaCl

2,0158 amu 2,0158 g

55,8470 amu 55,8470 g

97,9952 amu 97,9952 g

80,0642 amu 80,0642 g

58,4425 amu 58,4425 g

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Halotan C2HBrClF3

M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF

= (2 x 12,01) + 1,01 + 79,90 + 35,45 + (3 x 18,99)

= 197,38 g/mol

Composition of Chemical Compounds

Mole rate nC/nhalotan

Mass rate MC/Mhalotan

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• 2 mol N x 14,0067 g/mol = 28,0134 g N• 4 mol H x 1,0079 g/mol = 4,0316 g H• 3 mol O x 15,9994 g/mol = 47,9982 g O

80,0432 g/mol

%N = x 100% = % 35,0028,0134 g N2

80,0432 g

%H = x 100% = % 5,04 4,0316 g H2

80,0432 g

%O = x 100% = % 59,9647,9982 g O2

80,0432 g% 100,00

The molar mass of NH4NO3 and the percent composition of the atoms

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Molar mass of sulfuric acid = 2(1,008g) + 1(32,07g) + 4(16,00g) = 98,09 g/mol

%H = x 100% = % 2,06 H2(1,008g H2) 98,09g

%S = x 100% = % 32,69 S1(32,07g S) 98,09g

%O = x 100% = % 65,25 O4(16,00g O) 98,09 g

% 100,00

The percent composition of Sulfuric acid

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1. Assume that you have a sample of 100 g 2. Convert the masses of the elements in the 100 g

sample to amounts into moles3. Write a tentative formula based on the number of

moles just determined.4. Attempt to convert the subscripts in the tentative

formula to small whole numbers. This require dividing each of the subscripts by the smallest one

5. If the subscript at this point differ only very slightly from whole numbers, round them off to whole numbers.

6. Multiply all subscripts by a small whole number chosen to make all subscripts integral.

Solution in 6 steps:

Determination of the empirical and molecular formula

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Step 1 : Determine the mass of each element in a 100 g sample

C: 62,58 g H: 9,63 g O: 27,79 g

Dibutyl succinate is an insect repellant used against household ants and roaches. Its composition is 62,58 % carbon, 9,63% hydrogen and 27,9% oxygen. Its experimentally determined molecular mass is 230 amu. What are the empirical and molecular formulas of dibutyl succinate

Example

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Step 2: Convert each of these masses to an amount in moles.

OmolOg

OmolOgn

HmolHg

HmolHgn

CmolCg

CmolCgn

O

H

C

737,1999,15

179,27

55,9008,1

163,9

210,5011,12

158,62

Step 3 : Write a tentative formula based on these numbers of moles

Step 4: Divide each of the subscripts of the tentative formula by the smallest.

C5,21H9,55O1,74

C2,99H5,49O

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Step 5: Multiply all subscripts by a small number chosen to make all subscripts integral(here by 2).

C5,98H10,98O2

Empirical formula: C6H11O2

Step 6: Determination of the molecular formula.

If the empirical formula is 115 amu(12,011 x 6 +11x1,008+2x15,99),Molecular formula is 230 amu just twice as the empirical formula:

Molecular formula: C12H22O4

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• Assume that the sample is 100 g !

• C = 56,8 g C/(12,01 g C/ mol C) = 4,73 mol C

• H = 6,50 g H/( 1,008 g H / mol H) = 6,45 mol H

• O = 28,4 g O/(16,00 g O/ mol O) = 1,78 mol O

• N = 8,28 g N/(14,01 g N/ mol N) = 0,591 mol N

• Divide by 0,591 =

• C = 8,00 mol C = 8,0 mol C

• H = 10,9 mol H = 11,0 mol H

• O = 3,01 mol O = 3,0 mol O C8H11O3N

• N = 1,00 mol N = 1,0 mol N

It is found that adrenalin an important compound for our body has a mass percent composition as 56,8% C, 6,5% H, 28,4% O an 8,28% N. Calculate and write down the empirical formula of adrenalin

Example

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Combustion Analysis

CnHm + (n+ ) O2 = n CO(g) + H2O(g)m 2

m 2

Oxygen flow

The sample containing C, H and O

MagnesiumperchlorateH2O absorber

Sodiumhydroxide

CO2 absorber

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The combustion of a 6,49 mg sample of ascorbic acid containing C, H and O yields 9,74 mg CO2 and 2,64 mg H2O . Determine the empirical formula of ascorbic acid.

C: 9,74 x10-3g CO2 x(12,01 g C/44,01 g CO2)

= 2,65 x 10-3 g C

H: 2,64 x10-3g H2O x (2,016 g H2/18,02 gH2O)

= 2,92 x 10-4 g H

Mass of Oxygen = 6,49 mg – 2,65 mg – 0,30 mg

= 3,54 mg O

• C = 2,65 x 10-3 g C / ( 12,01 g C / mol C ) =

= 2,21 x 10-4 mol C• H = 0,295 x 10-3 g H / ( 1,008 g H / mol H ) =

= 2,92 x 10-4 mol H• O = 3,54 x 10-3 g O / ( 16,00 g O / mol O ) =

= 2,21 x 10-4 mol O• Each result is divided by 2,21 x 10-4 :

• C = 1,00 multiplied by 3 = 3,00 = 3,0• H = 1,32 = 3,96 = 4,0• O = 1,00 = 3,00 = 3,0

C3H4O3

Example

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Metals tend to lose electron

Na Na+ + e-

Nonmetals tend to gain electron

Cl + e- Cl-

Oxidation state designates the number of electrons that an atom loses, gains or otherwise uses in joining with other atoms in compounds.

Oxidation States

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1. The oxidation state(O.S.) of an atom in the free(uncombined) element is 0 .

2. The total of the oxidation states of all the atoms in a molecule or formula unit is 0. For an ion this total is equal to the charge of the ion.

3. In their compounds the alkali metals have O.S. +1 and the alkaline earth metals +2.4. In its compounds, hydrogen has generally O.S. +1(but sometimes -1); fluorine, -1.

5. In its compounds oxygen has O.S. -2.

6. In their binary(two-element) compounds with metals;

i. Halogenes (Group 7A elements ) have O.S. = -1,ii. Group 6A elements have O.S. = -2 andiii. Group 5A elements have O.S. = -3 .

Rules for assigning oxidation state

Whenever two rules appear to contradict one another(which they often will) , follow the rule that appears higher in the list.

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Determination of O.S. :

What is the oxidation state of the underlined element in each of the following?

a) P4; b) Al2O3; c) MnO4-; d) NaH

a) P4 is the formula of a molecule of elemental phosphorus. For an atom of a free element the O.S. = 0

b) Al2O3: The total of the oxidation states of all the atoms in this formula unit is 0(Rule 2)

The O.S. for oxygen = -2 . The total for three O atoms is -6 .The total for two Al atoms is -6. The O.S. of Al = +3

c) MnO4-: The total of the oxidation states of all the atoms in the ion

is -1, the total for the four O atoms is -8. The O.S. of Mn = +7.

d) NaH: Rule 3 states Na should have O.S +1. Rule 4indicates that H should also have O.S. +1. If both atoms had O.S. +1, the total for the formula unit would be +2.This violates Rule 2. Rule 2 and rule 3 take precedence over rule 4. Na has O.S. +1; the total for the formula unit is 0and the O.S. of H is -1.

Examples

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Naming Inorganic Compounds

Organic Compounds: The compounds formed by carbon and hydrogen or carbon and hydrogen together with oxygen,nitrogen and a few other elements.

Inorganic Compounds: Compounds that do not fit the description above

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Naming Inorganic CompoundsBinary compounds are those formed between two elements (If one of the elements is a metal and the other a nonmetal, the binary compound is usually comprised of ions, that is a binary ionic compound)

NaCl = sodium chloride

The name of metal is unchanged

Ionic compounds must be electrically neutral

MgI2 = Magnesium iodide

Al2O3 = Aluminium oxide

Na2S = Sodium sulfide

The name of the nonmetal modified to end in «ide»

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Some Simple IonsCations Anions

Charge Symbol Name Charge Symbol Name

1+ H+ Hydrogen 1- H - Hydride Li+ Lithium F - Floride Na+ Sodium Cl - Chloride K+ Potassium Br - Bromide Cs+ Cesium I - Iodide Ag+ Silver2+ Mg2+ Magnesium 2- O 2 - Oxide Ca2+ Calcium S2 - Sulfide Sr2+ Strontium Ba2+ Barium Zn2+ Zink Cd2+ Cadmium3+ Al3+ Aluminium 3- N 3 - Nitride

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Sodium and Oxygen

Zink and Chlorine

Calcium and Fluorine

Strontium and Nitrogen

Hydrogen and Iodine

Scandium and Sulfur

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a) Sodium and Oxygen Na2O Sodium oxide

b) Zink and Chlorine ZnCl2 Zink chloride c) Calcium and Fluorine CaF2 Calcium fluoride

d) Strontium and Nitrogen Sr3N2 Strontium nitride

e) Hydrogen and Iodine HI Hydrogen iodide

f) Scandium and Sulphur Sc2S3 Scandium sulfide

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Some Metals with more than one O.S.

Element Symbol Nomenclature

Chromium Cr+2 Chromium (II)Cr+3 Chromium (III)

Cobalt Co+2 Cobalt (II)Co+3 Cobalt (III)

Copper Cu+1 Copper (I)Cu+2 Copper (II)

Iron Fe+2 Iron (II)Fe+3 Iron (III)

Lead Pb+2 Lead (II)Pb+4 Lead (IV)

Manganese Mn+2 Manganese (II)Mn+3 Manganese (III)

Mercury Hg2+2 Mercury (I)

Hg+2 Mercury (II)Tin Sn+2 Tin (II)

Sn+4 Tin (IV)

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Binary Compounds of Two Nonmetals

Molecular CompoundWe first write the element with the (+) O.S.

HCl Hydrogen chloride

mono 1 penta 5

di 2 hexa 6

tri 3 hepta 7

tetra 4 octa 8

Some pairs of nonmetals form more than a single binary molecular compound. We use the prefixes written below:

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Exercise-2• BCl3

• CCl4• CO

• CO2

• NO

• NO2

• N2O

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Exercise-3• Dinitrogen trioxide

• Dinitrogen tetroxide

• Dinitrogen pentoxide

• Phosphorus trichloride

• Phosphorus pentachloride

• Sulphur hexafluoride

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Formula Name

BCl3 Boron trichloride

CCl4 Carbon tetrachloride

CO Carbon monoxide

CO2 Carbon dioxide

NO Nirogen monoxide

NO2 Nitrogen dioxide

N2O Dinitrogen monoxide

N2O3 Dinitrogen trioxide

N2O4 Dinitrogen tetroxide

N2O5 Dinitrogen pentoxide

PCl3 Phosphorus trichloride

PCl5 Phosphorus pentachloride

SF6 Sulphur hexafluoride

Nomenclature of Binary Compounds

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NitriteNitrateOxalatePermanganatePhosfateHydrogen phosphateDihydrogen phosphateSulfiteHydrogen sulfiteor bisulfiteSulfate

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Hydrogen sulfateor bisulfateThiosulfateAcetateCarbonateHydrogen carbonateor bicarbonateHypochlorite ChloriteChloratePerchlorateChromate DichromateCyanideHydroxideAmmonium

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Some Common Polyatomic IonsName Formula Typical Compound

Cation

Ammonium NH4+ NH4Cl

Anions

Acetate C2H3O2- NaC2H3O2

Carbonate CO32- Na2CO3

Hydrogen carbonateor bicarbonate

HCO3- NaHCO3

Hypochlorite ClO- ClO

Chlorite ClO2- NaClO2

Chlorate ClO3- NaClO3

Perchlorate ClO4- NaClO4

Chromate CrO42- Na2CrO4

Dichromate Cr2O72- Na2Cr2O7

Cyanide CN- NaCN

Hydroxide OH- NaOH

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Some Common Polyatomic IonsName Formula Typical Compound

Nitrite NO2- NaNO2

Nitrate NO3- NaNO3

Oxalate C2O42- Na2C2O4

Permanganate MnO4- NaMnO4

Phosfate PO43- Na3PO4

Hydrogen phosphate

HPO42- Na2HPO4

Dihydrogen phosphate

H2PO4- NaH2PO4

Sulfite SO32- Na2SO3

Hydrogen sulfiteor bisulfite

HSO3- NaHSO3

Sulfate SO42- Na2SO4

Hydrogen sulfateor bisulfate

HSO4- NaHSO4

Thiosulfate S2O32- Na2S2O3

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Nomenclature of some Oxoacids and their salts

Oxidation StateFormula of acid

Name of acidFormula of salt

Name of salt

Cl: +1 HClOHypochlorous acid

NaHClO Sodium hypochlorite

Cl: +3 HClO2 Chlorous acid NaClO2 Sodium chlorite

Cl: +5 HClO3 Chloric acid NaClO3 Sodium chlorate

Cl: +7 HClO4 Perchloric acid NaClO4 Sodium perchlorate

N: +3 HNO2 Nitrous acid NaNO2 Sodium nitrite

N: +5 HNO3 Nitric acid NaNO3 Sodium nitrate

S: +4 H2SO3 Sulfurous acid Na2SO3 Sodium sulfite

S: +6 H2SO4 Sulfuric acid Na2SO4 Sodium sulfate

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Binary acids

Definition: Certain compounds of hydrogen with other nonmetal atoms

Examples: HF= hydrofluoric acid

HCl= hydrochloric acid

HBr= hydrobromic acid

HI = hydroiodic acid

H2S = hydrosulfuric acid

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Some Compounds of Greater Complexity

Hydrate:Each formula unit of the compound has associated with it a certain number of water molecules. •The water molecules are incorporated in the solid structure of the compound

•Example: CoCl2 . 6 H2O (Cobalt(II) chloride hexahydrate)

•Formula mass= 129.8 u+(6x18.02 u)=237.9 u•When the water is totally removed from the hydrates,the resulting compound is said to be anhydrous(without water). Anhydrous compounds can be used as water absorbers,as in the use of anhydrous magnesium perchlorate in combustion analysis

•Anhydrous CoCl2 is blue, the hexahydrate is pink