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Prof.H.A.Mohammed General Chemistry – Chem 101

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Prof.H.A.Mohammed Instruments and Lab Safety

Taibah University

Faculty of Science and Arts

Chemistry Department

For

Chemistry , Physics and Biology Students

General Chemistry Chem. 101


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General Chemistry (101 Chem.)

Chapter: 3 Stoichiometry

3.1 Atomic Masses

3.2 The Mole

3.3 The Molar Mass

3.4 Percent Composition of Compounds

3.5 Determining the Formula of a compound

3.6 Chemical Equations

3.7 Balancing Chemical Equations

3.8 Stoichiometric Calculations

3.9 Calculations involving a limiting Reactant and a percent yield

ULecture 1:

UStoichiometry U : The study of quantities of materials

Consumed and produced in chemical reactions.

3.1 Atomic Masses

3.2 The Mole

3.3 The Molar Mass


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3.4 Percent Composition of Compounds


3.1 Atomic Masses

Atoms are very, very light and very, very small Since atomic dimensions are so small, it would be cumbersome to use units we typically use for length (cm, m) or mass (g). Hence, on the atomic scale we define units appropriate for this scale Mass

unit typically used is an Atomic Mass Unit (amu)

1 amu = 1.66054 x 10–24 g

Particle Charge Mass (g) Mass (amu)

Proton +1 1.6727x10-24 1.0073

Neutron Neutral 1.6750x10-24 1.0087

Electron -1 9.109x10-28 5.486 x 10-4

Length – Angstrom (Å) = 10-10m Typical atomic dimensions are 1 to 5 x10-10 m which corresponds to 1 to 5 Å.

Relative Atomic Mass

A relative scale has been developed to compare the relative masses of atoms. The ATOMIC MASS of an atom is its relative mass on this scale.


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Carbon- 12 (12C) has been set as the standard and assigned a Relative Mass of exactly 12.

Relative atomic masses have no units since they are the ratio of two masses.

Average Relative Atomic Mass

Because the abundance of the isotopes of different elements are essentially constant, we can define an AVERAGE RELATIVE ATOMIC mass Average Relative Atomic Mass = average mass of atoms of an element = (Abundance)A(Mass)A + (Abundance)B(Mass)A + …

Problem

Naturally occurring chlorine has two isotopes ,3517 Cl ,37

17 Cl .The 35-Cl isotope has a relative atomic mass of 34.9688 and an abundance of 75.77% and the 37-Cl isotope has a relative atomic mass of 36.9659 and an abundance of 24.23% .Calculate the average atomic mass of Cl.

Average Atomic Mass of Cl

= (0.7577x34.9659 ) +( 0.2432x36.9659)=35.4527g


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URelative Molecular Mass

The relative molecular mass is the sum of the relative atomic masses of the atoms that make up the molecule.

Example ,the chemical formula for water is H2O

Its relative molecular mass

) =1.00794(2 +15.9994 =18.0153

UAvogadro’s Number

Avogadro’s number: the number of atoms in exactly 12 g of P

12PC.

NRoR = 6.022137 x 10 P

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Sodium (Na) has a relative atomic mass of 22.98977

Hence a sodium atom is (22.98977) times as heavy as P

12PC

If NRoR atoms of P

12PC have a mass of 12g then, the mass of NRoR atoms of

sodium must be

(22.98977) 12g = 22.98977 g

The mass, in grams, of NRoR atoms of ANY element is numerically equal to the relative atomic mass of that element.


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Same applies to molecules.

Since the relative molecular mass of water is 18.0152, the mass of No water molecules is 18.0152g

3.2 The Mole:

UMOLE U :

A mole has been defined as a unit containing 6.022137 x 10 23 ,Avogadro’s number, atoms or molecules,

One mole of any atom or molecule contains the same number of atoms or molecules

The mass ,in grams, of UOne Mole U of atoms or molecules is numerically equal to relative atomic or molecular mass.

Hence 1 mole of Na weighs 22.9898 g, 1 mole of H2O weighs 18.0153 g of g/mol

U 3.3 The Molar Mass

The MASS of one mole of atoms or molecules is called its MOLAR MASS and has UNITS


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Problem

How many moles and atoms of Fe are there

in 8.232 g of Fe?

Moles of Fe = 8.232 g Fe x 1 mole = 0.1474 mol Fe

55.85 g Fe

How many grams of water are there in 0.2000 moles of water?

0.2000 mol HR2RO x 18.015 g HR2RO = 3.603 g HR2RO

1 mol H2O

UMole Relationships in Chemical Formulas

Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound.

Moles of compound Moles of constituents

1 mol NaCl 1 mol Na, 1 mol Cl

1 mol HR2RO 2 mol H, 1 mol O

1 mol CaCOR3 1 mol Ca, 1 mol C, 3 mol O


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1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

3.4 Percent Composition of Compounds :Mass percent of an element:

For iron in iron (III) oxide, (FeR2ROR3R) For iron in iron (III) oxide, (FeR2ROR3R)

UEmpirical Formulas

The simplest, whole-number ratio of atoms in a molecule is called the (empirical formula).

– Can be determined from percent composition or combining masses.

The molecular formula is a multiple of the empirical formula. molecular formula = (empirical formula)RnR [n = integer molecular formula = CR6RHR6R = (CH)R6 empirical formula = CH

Empirical Formulas

Hydrogen Peroxide

Molecular formula = HR2ROR2

Empirical formula = HO

Mass % = mass of element in compound x 100

mass of compound


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Benzene

Molecular formula = C6H6

Empirical formula = CH

Glucose

Molecular formula = C6H12O6

Empirical formula = CH2O


1. Base calculation on 100 grams of compound.

2. Determine moles of each element in 100 grams of compound.

3. Divide each value of moles by the smallest of the values.

4. Multiply each number by an integer to obtain all

whole numbers.


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• Apply the solution map:– Calculate the moles of each element.

C mol 9964C g 12.01

C mol 1C g 0060 .. =×

H mol 444H g 1.01H mol 1H g .484 .=×

O mol 2212O g 16.00

O mol 1O g 5.533 .=×

Solution Map: g C,H,O → mol C,H,O →mol ratio → empirical formula

Apply the solution map:

Write a pseudoformula.

C4.996H4.44O2.221

• Apply the solution map: – Find the mole ratio by dividing by the smallest number of

moles.

122.25

2.2212.221

2.2214.44

2.2214.996

OHC

OHC


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Molecular Formulas

• The molecular formula is a multiple of the empirical formula. • To determine the molecular formula, you need to know the

empirical formula and the molar mass of the compound. •

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8.

1. Determine the empirical formula. – May need to calculate it as previous.

C5H8

1. Determine the molar mass of the empirical formula. 5 C = 60.05, 8 H = 8.064

C5H8 = 68.11 g/mol

3. Divide the given molar mass of the compound by the molar mass of the empirical formula.

– Round to the nearest whole number.

3g/mol 11.68

g/mol 204=


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4. Multiply the empirical formula by the factor above to give the molecular formula.

5. 3(CR5RHR8R)=CR15RHR24

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of CR5RHR3R. What is its Molecular Formula? (C = 12.01, H=1.01)

CR5R = 5( )12.01 g = 60.05 g

HR3RU= 3( 1.01)g = 3.03g

C5H 3 = 63.08 g

4 g/mol 63.08

g/mol 252==n


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Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H,

and the Rest N, Continued

Tro's "Introductory Chemistry", Chapter 6 34

Given: 74.0% C, 8.7% H, {100 – (74.0+8.7)} = 17.3% N ∴in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.

Find: CxHyNzConversion Factors:

1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 gSolution Map:

g C mol C

g H mol H pseudo-formula

empiricalformula

moleratio

wholenumber

ratio

g N mol N


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Apply solution map:

C mol 6.16 g 12.01C mol 1C g 074 =×.

N mol 1.23 g 14.01N mol 1N g 7.31 =×

C6.16H8.6N1.23

NHC NHC 751.231.23

1.238.6

1.236.16 =

H mol 8.6 g 1.01H mol 1H g .78 =×

C5 =5(12.01 g) = 60.05 gN1 =1(14.01 g) = 14.01 g

=7Hg) =01 .1(7g07 .7C5H7N=81.13 g

2 g 81.13

g 162form. emp. mass mol.

nicotine mass mol.==

{C5H7N} x 2 = C10H14N2

Mass spectrometer


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3.6 Chemical Equations

Chemical change involves a reorganization of the atoms in one or more substances.

A representation of a chemical reaction:

C2H5OH + 3O2 2CO2 + 3H2O

reactants products

3.7 Balancing Chemical Equations

CR2RHR5ROH + 3OR2R 2COR2R + 3HR2RO

The equation is balanced.

1 mole of ethanol reacts with 3 moles of oxygen

to produce


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2 moles of carbon dioxide and 3 moles of water

U Some practice problems:

Here are some practice problems. The solutions are in

the section below this one .

1. __NaCl + __BeFR2R --> __NaF + __BeClR2R

2. __FeClR3R + __BeR3R(POR4R)R2R --> __BeClR2R + __FePOR4R


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3. __AgNO3 + __LiOH --> __AgOH + __LiNO3

4. __CH4 + __O2 --> __CO2 + __H2O

5. __Mg + __MnR2ROR3R --> __MgO + __Mn

USolutions for the practice problems:

1 .2 NaCl +1 BeF 2 -- <2 NaF +1 BeCl 2

2 .2 FeCl 3 +1 Be 3) PO 4(2 -- <3 BeCl 2 +2 FePO 4

3 .1 AgNO 3 +1 LiOH -- <1 AgOH +1 LiNO 3

4 .1 CH 4 +2 O 2 -- <1 CO 2 +2 H2O

5 .3 Mg + 1 Mn2O 3 -- <3 MgO +2 Mn

U3.8 Stoichiometric Calculations

1. Balance the equation. 2. Convert mass to moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of

desired substituent.


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5. Convert moles to grams, if necessary.

Limiting Reactant

The limiting reactant is the reactant that is consumed first ,limiting the amounts of products formed.

U3.9 Calculations involving a limiting Reactant

Uand a percent

1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios

to find moles of desired product.

5.Convert from moles to grams

UExample :

Nitrogen gas(NR2R) is prepared by passing ammonia gas(NHR3R) over solid cupper oxide(CuO) at high degrees of temperatures according to the following reaction:

2NHR3(g)R + 3CuOR(s) R®R RNR2(g)R+3CuR(s) R+ 3HR2ROR(g)


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Find the limiting reactant, if 18.1g of ammonia reacted with 90.4 g of cupper oxide. Then calculate the weight of Nitrogen produced

Nitrogen gas(NR2R) is prepared by passing ammonia gas(NH3) over solid cupper oxide(CuO) at high degrees of temperatures according to the following reaction:

2NHR3(g)R + 3CuOR(s) R®R RNR2(g)R+3CuR(s) R+ 3HR2ROR(g)

Find the limiting reactant, if 18.1g of ammonia reacted with 90.4 g of cupper oxide. Then calculate the weight of nitrogen produced

1-Convert masses to moles:

No. of (NH3) moles = 18.1 / (14+3) =18.1/17=1.064 moles

No. of (CuO) moles = 90.4 / (63.55+16)=90.4/79.55= 1.299moles

2-The limiting reactant is the reactant which has least mole ratio (consumed first) according to the balanced equation

Mole ratio of NHR3R=1.064 moles /2 = 0.532

Mole ratio of CuO=1.299moles /3= 0.433


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CuO is the limiting reactant

3-To calculate the weight of NR2R produced: we write the relation between the limiting reactant and the NR2R produced

3 mol CuO 1 mol NR2

1.299 mol X mole NR2

(X mole NR2R)= (1.299 mol CuO) (1 mol NR2R) /(3mol CuO) = 0.433 mol

4-Convert mole to mass by multiplying by the molecular weight of NR2

The weight of NR2R produced = 0.433 x 28 = 12.124 g

UCalculating Atomic Mass

To arrive at the atomic mass presented in the periodic table, you must first multiply the mass of each naturally occurring isotope of an element by the fraction of its abundance, and then add up all the fractions

UExample: U Carbon-12 has a mass of 12.0000 atomic mass units, and it makes up 98.89 percent of naturally occurring carbon. Carbon-13 has a


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mass of 13.0034 atomic mass units, and it makes up 1.11 percent of naturally occurring carbon. Use this information to show that the atomic mass of carbon shown in the periodic table, 12.011 atomic mass units, is correct

Answer :

Actual mass of C = (O.9889 x 12 )+ (o.o111x13.0034) =

11.867 + o.144 = 12.011

UExample :

Gallium is a metallic element found in small lasers used in compact disc players .In a sample of gallium, there is 60.2% of gallium-69 (68.9 amu) atoms and 39.8% of gallium-71 )70.9 amu) atoms .

What is the atomic mass of gallium?

Ga-69 (%/100)

68.9 amu x U 60.2 U = 41.5 amu for P

69PGa

100

Ga-71 (%/100)

70.9 amu x U 39.8 U = 28.2 amu for P

71PGa

100

Atomic mass Ga = 69.7 amu

Copper has two isotopes 63 Cu (62.9 amu) and 65 Cu (64.9 amu). What is the % abundance of each isotope? (Hint: Check periodic table for atomic mass(


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1 (30% 2 (70% 3 (100%

Ans :

2 (70%

62.9X +6490 =64.9 X + 6350

-2.0 X = -140 X = 70%

UChapter :4 Types of Chemical Reactions

Uand Solution Chemistry

4.1 Water, the common solvent.

4.2 The Nature of Aqueous Solutions: Strong and Weak electrolytes.

4.3 The composition of Solutions.

4.4 Types of chemical Reactions.

4.5 Precipitation Reactions.

4.6 Describing Reactions in Solutions

4.7 Stoichiometry of Precipitation Reactions

4.8 Acid-Base Titrations.

4.9 Oxidation-Reduction Reactions.

4.10 Balancing Oxidation-Reduction Reactions.


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U 4.1 Water, the common solvent.

Water is the dissolving medium, or solvent. USome Properties of Water

• Water is “bent” or V-shaped. • The O-H bonds are covalent. • Water is a polar molecule. • Hydration occurs when salts dissolve in water.

Hydration occurs when salts dissolve in water.


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Attraction takes place between H+ (H2O) and OH- of Ethanol( CH3CH2OH)

U 4.2:The Nature of Aqueous Solutions: Strong and Weak electrolytesU.

UElectrolytes :

Strong - conduct current efficiently

NaCl, HNOR3

Weak - conduct only a small current

Vinegar, tap water

Non - no current flows

Pure water, sugar solution


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When NaCl dissolves in water, the Cl - ions and Sodium ions Na + spread randomly

The solution conducts electricity

UAcids


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strong acids - dissociate completely to produce

H+ in solution hydrochloric and sulfuric acid

Weak acids - dissociate to a slight extent to give

H+ in solution acetic and formic acid

Bases

Strong bases - react completely with water

to give OH- ions. sodium hydroxide

Weak bases - react only slightly with

water to give OH- ions.

ammonia


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U 4.3 The composition of Solutions U.

UA Solute

dissolves in water (or other “solvent”)

changes phase (if different from the solvent)

is present in lesser amount (if the same phase as the solvent)


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retains its phase (if different from the solute)

is present in greater amount (if the same phase as the solute)

Molarity:

Molarity (M) = moles of solute per volume

of solution in liters:

Common Terms of Solution Concentration

Standard Solution: a solution of exact known concentration, prepared

by dissolving a definite weight of solute in a volumetric flask

Stock - routinely used solutions prepared in cocentrated form.

Concentrated - relatively large ratio of solute to solvent.

(5.0 M NaCl)

Dilute - relatively small ratio of solute to solvent.

(0.01 M NaCl)

STRENGTH OF REDUCING AND OXIDIZING AGENTS

Zn + Fe 2+ ↔ Zn 2+ + Fe

Which way will the reaction go?


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∆GRrR =-16.3 kcal/mole

Zn is a stronger reducing agent than Fe.

Fe + Cu 2+ ↔ Fe 2+ + Cu


Fe is a stronger reductant than Cu.

UDilution :

Addition of water (solvent) to a concentrated solution to get a less concentrated one

ULaw of Dilution :

MR1R x VR1R = MR2R x VR2R where:


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M1 & V1 = Molarity and volume before dilution

M2 & V2 = Molarity and volume after dilution

U 4.4 Types of chemical Reactions.

Precipitation reactions

AgNOR3R(aq) + NaCl(aq) ® AgCl(s) + NaNOR3R(aq)

Acid-base reactions

NaOH(aq) + HCl(aq) ® NaCl(aq) + HR2RO(l)

Oxidation-reduction reactions

FeR2ROR3R(s) + Al(s) ® Fe(l) + AlR2ROR3R(s)


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U4.5 Precipitation Reactions

UExample:

When Pottasium Chromate K2CrO4 is added to a solution of Barium Nitrate Ba)NO 3(2 , a yellow precipitate of Barium

Chromate BaCrOR4R is formed.

KR2RCrOR4(aq)R + Ba(NOR3R)R2(aq)R → KNOR3(aq)R + BaCrOR4(s)R

2KP

+PR (aq)R + CrOR4RP

2-PR(aq) R+ Ba P

+2PR(aq)R + 2NOR3RP

-PR(aq)R →

KP

+ PR(aq)RP

P+P

PNOR3RP

-PR(aq)R + BaCrOR4(s)R


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When Potassium chromate K2CrO4 is added to a solution of Barium nitrate Ba (NO 3(2 , a yellow precipitate of Barium chromate BaCrOR4R is formed.

UExample :2

AgNOR3(aq)R + KClR(aq)R → KNOR3(aq)R + AgClR(s)R

Ag P

+PR (aq)R + NOR3RP

-PR(aq) R+ KP

+PR(aq)R + Cl P

-PR(aq)R →

KP

+ PR(aq)RP

P+P

PNOR3RP

-PR(aq)R + AgClR(s)R

When Silver nitrate solution AgNOR3R is added to a solution of Potassium chloride KCl ,a white precipitate of Silver chloride is formed AgCl

USimple Rules for Solubility

1. Most nitrate (NOR3RP

-P) salts are soluble.


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2. Most alkali (group 1A) salts and NH4+

are soluble.

3. Most Cl-, Br-, and I- salts are soluble

(NOT Ag+, Pb2+, Hg22+)

4. Most sulfate salts are soluble

(NOT BaSO4, PbSO4, HgSO4, CaSO4)

5. Most OH- salts are only slightly soluble (NaOH, KOH are soluble,

Ba(OH)2, Ca(OH)2

are marginally soluble)

6. Most S2-, CO32-, CrO4

2-, PO43- salts

7. are only slightly soluble.

4.6 Describing Reactions in Solutions:

1.Molecular equation (reactants and products as compounds)

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

2.Complete ionic equation (all strong electrolytes

shown as ions)

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)


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AgCl(s) + Na+(aq) + NO3-(aq)

3. Net ionic equation (show only components that actually react)

Ag+(aq) + Cl-(aq) AgCl(s)

Na+ and NO3- are spectator ions.

4.7 Stoichiometry of Precipitation Reactions

1- A 1.000-g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 1.286 g. Calculate the atomic mass of M.

a) 222.8 g

b) 76.00 g

c) 152.0 g

d) 304.0 g

e) none of these

Ans: c)152.0 g


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4.8 Acid-Base Titrations:

Acid : a proton donor

Base : a proton acceptor

HClR(aq)R + NaOHR(aq)R → HR2ROR(aq)R + NaClR(aq)

HP

+P R(aq)R+ ClP

-PR(aq)R+ Na P

+PR(aq)R+ OHP

-PR(aq)R →HR2ROR(aq)R + Cl P

-PR(aq)R+ Na P

+PR(aq)

HP

+P R(aq)R + OHP

-PR(aq)R → HR2ROR(l)

UAcid-Base Titrations

UVolumetric Analysis

Volumetric analysis is used for determining the concentration of a

certain volume of unknown solution by titration against a solution of

known exact concentration using an indicator to attain the end-point.

1- The solution taken from the burett is called “ Titrant”

2-The solution of unknown concentration is called” Analyte”

3-The end-point or “ equivalent-point ” is the point at which the number

of equivalents of titrant and analyte are equal.


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4-The indicator is a weak organic acid or organic base, its color depends

on acidic or basic nature of the analyte during the titration process

5.Phenolphthaline : is colorless in acidic medium and pink in base

medium while Methyl Orange is orange in a base medium and red in

acidic medium


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Performing Calculations for Acid-Base Reactions

1. List initial species and predict reaction.

2. Write balanced net ionic reaction.

3. Calculate moles of reactants.

4. Determine limiting reactant.

5. Calculate moles of required reactant/product.

6. Convert to grams or volume, as required.

4.9 Oxidation-Reduction Reactions.

- Losing electrons is oxidation, and the substance that loses the electrons is called the reducing agent.

- Gaining electrons is reduction, and the substance that gains the electrons is called the oxidizing agent.

Mg(s) + S(s) MgS(s)

Performing Calculations for Acid-Base Reactions

1. List initial species and predict reaction.


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2. Write balanced net ionic reaction.

3. Calculate moles of reactants.

4. Determine limiting reactant.

5. Calculate moles of required reactant/product.

6. Convert to grams or volume, as required.

• An “oxidation number” is a positive or negative number assigned

to an atom to indicate its degree of oxidation or reduction.

• Generally, a bonded atom’s oxidation number is the charge it

would have if the electrons in the bond were assigned to the atom

of the more electronegative element

CH4(g) + 2O2(q) → CO2(g) + 2H2O(g)

-4 +1 +4 -2 +1 -2

CH4 → CO2 + 8e-

-4 +4

2O2 + 8e- → CO2 + 2H2O

0 4(-2) = -8


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Rules for Assigning Oxidation States

1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic element = charge

3. Oxygen = -2 in covalent compounds (except in peroxides where it = -1)

4. H = +1 in covalent compounds

5. Fluorine = -1 in compounds

6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

General Rules for Assigning Oxidation Numbers

1) The oxidation number of any uncombined element is zero.

2) The oxidation number of a monatomic ion equals its charge.

3) The oxidation number of oxygen in compounds is -2, except in

peroxides, such as H2O2 where it is -1.

4) The oxidation number of hydrogen in compounds is +1, except in

metal hydrides, like NaH, where it is -1.

5) The sum of the oxidation numbers of the atoms in the compound

must equal 0.

6) The oxidation number of oxygen in compounds is -2, except in

peroxides, such as H2O2 where it is -1.


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7) The oxidation number of hydrogen in compounds is +1, except in

metal hydrides, like NaH, where it is -1.


must equal 0.


must equal 0.

6) The sum of the oxidation numbers in the formula of a polyatomic

ion is equal to its ionic charge.

• An increase in oxidation number = oxidation

• A decrease in oxidation number = reduction

Sodium is oxidized – it is the reducing agent

Chlorine is reduced – it is the oxidizing agent

Trends in Oxidation and Reduction

Active metals:

Lose electrons easily Are easily oxidized Are strong reducing agents


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Active nonmetals:

Gain electrons easily Are easily reduced Are strong oxidizing agents

4.10 Balancing Oxidation-Reduction Reactions

Balancing by Half-Reaction Method

1.Write separate reduction, oxidation reactions.

2.For each half-reaction:

Balance elements (except H, O)


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Balance O using H2O

Balance H using H+

Balance charge using electrons

3.If necessary, multiply by integer to equalize electron count.

4.Add half-reactions.

5.Check that elements and charges are balanced.

UHalf-Reaction Method - Balancing in Base

1.Balance as in acid.

2.Add OHP

-P that equals HP

+P ions (both sides!)

3.Form water by combining HP

+P, OH P

-P.

4. Check elements and charges

for balance.

UVARIABLE VALENCE ELEMENTS


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Sulfur: SO42-(+6), SO3

2-(+4), S(0), FeS2(-1), H2S(-2)

Carbon: CO2(+4), C(0), CH4(-4)

Nitrogen: NO3-(+5), NO2

-(+3), NO(+2), N2O(+1), N2(0), NH3(-3)

Iron: Fe2O3(+3), FeO(+2), Fe(0)

Manganese: MnO4-(+7), MnO2(+4), Mn2O3(+3), MnO(+2), Mn(0)

Copper: CuO(+2), Cu2O(+1), Cu(0)

Tin: SnO2(+4), Sn2+(+2), Sn(0)

Uranium: UO22+(+6), UO2(+4), U(0)

Arsenic: H3AsO40(+5), H3AsO3

0(+3), As(0), AsH3(-1)

Chromium: CrO42-(+6), Cr2O3(+3), Cr(0)

Gold: AuCl4-(+3), Au(CN)2

-(+1), Au(0)

BALANCING OVERALL REDOX REACTIONS

Example - balance the redox reaction below:

Fe + Cl2 ↔ Fe3+ + Cl-

Step 1: Assign valences,

Fe0 + Cl20 ↔ Fe3+ + Cl-

Step 2: Determine number of electrons lost or gained by reactants.


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Fe0 + Cl20 ↔ Fe3+ + Cl-

↓ ↑

3e- 2e-

Step 3: Cross multiply.

2Fe + 3Cl20 ↔ 2Fe3+ + 6Cl-

HALF-CELL REACTIONS

The overall reaction:

2Fe + 3Cl20 ↔ 2Fe3+ + 6Cl-

may be written as the sum of two half-cell reactions:

2Fe ↔ 2Fe3+ + 6e- (oxidation)

3Cl20 + 6e- ↔ 6Cl- (reduction)

All overall redox reactions can be expressed as the sum of two half-cell

reactions, one a reduction and one an oxidation.


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Another example - balance the redox reaction:

FeS2 + O2 ↔ Fe(OH)3 + SO42-

Fe+2S20 + O2

0 ↔ Fe+3(OH)3 + S+6O42-

↓ ↑

15e- 4e-

4FeS2 + 15O2 ↔ Fe(OH)3 + SO42-

4FeS2 + 15O2 ↔ 4Fe(OH)3 + 8SO42-

4FeS2 + 15O2 +14H2O ↔ 4Fe(OH)3 + 8SO42- + 16H+

This reaction is the main cause of acid generation in drainage from sulfide ore deposits. Note that we get 4 moles of H+ for every mole of pyrite oxidized!

Final example:

C2H6 + NO3- ↔ HCO3

- + NH4+

C-32H6 + N+5O3

- ↔ 2HC+4O3- + N-3H4

+

↓ ↑


47


14e- 8e-

8C2H6 + 14NO3- ↔ 2HCO3

- + NH4+

8C2H6 + 14NO3- ↔ 16HCO3

- + 14NH4+

8C2H6 + 14NO3- + 12H+ + 6H2O↔ 16HCO3

- + 14NH4+

STRENGTH OF REDUCING AND OXIDIZING AGENTS

Zn + Fe 2+ ↔ Zn 2+ + Fe

Which way will the reaction go?


Zn is a stronger reducing agent than Fe.

Fe + Cu 2+ ↔ Fe 2+ + Cu


Fe is a stronger reductant than Cu.

general chemistry chem. 101 forhamohammed.pbworks.com/f/10134.pdf · prof.h.a.mohammed general...

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