General Chemistry

M. R. Naimi-Jamal

Faculty of Chemistry Iran University of Science & Technology

II یرتمویکوتسا یئایمیش تلاداعم و

: مراهچ لصف

Contents

4-1 4-2 4-3 4-4 Chemical Reactions and Chemical Equations Chemical Equations and Stoichiometry Chemical Reactions in Solution Determining the Limiting reagent

4-1 Chemical Reactions and Chemical Equations

As reactants are converted to products, we observe: – Color change – Precipitate formation – Gas evolution – Heat absorption or evolution Chemical evidence may be necessary.

Formation of AlBr 3

Chemical Reaction

Nitrogen monoxide + oxygen → nitrogen dioxide Step 1: Write the reaction using chemical symbols.

Step 2: Balance the chemical equation.

2 NO + O 2 → NO 2

Molecular Representation

Balancing Equations

Example: Nitrogen monoxide + oxygen → nitrogen dioxide • Never introduce extraneous atoms to balance.

NO + O

2

→ NO

2 + O • Never change a formula for the purpose of balancing an equation.

NO + O

2

→ NO

3

Balancing Equation Strategy

• Balance elements that occur in only one compound on each side first.

• Balance free elements last.

• Balance unchanged polyatomics as groups.

• Fractional coefficients are acceptable and can be cleared at the end by multiplication.

Chemical Equations

• The physical states: – Solid (s) – liquid (l) – gas (g) – and aqueous (aq) P 4 (s) + 6Cl 2 (g) 4PCl 3 (l)

4-2 Chemical Equations and Stoichiometry

• Stoichiometry includes all the

quantitative

relationships involving: – atomic and formula masses – chemical formulas.

– The coefficients in front of the compounds in a balanced equation are called stoichiometric coefficients •

Mole ratio

is a central conversion factor.

Mass Relationships

Initial amount (mol) Change in amount (mol) After complete reaction (mol) P 4 + 1.00 mol (124 g) - 1.00 mol 0 mol (0 g) 6 Cl 2 6.00 mol (425 g) -6.00 mol 0 mol (0 g) 4 PCl 3 0 mol (0 g) +4.00 mol 4.00 mol (549 g) From this we can calculate mass of one compound required to complete the reaction if the mass of the other compound is given

Example 4-3

Relating the Numbers of Moles of Reactant and Product.

How many moles of H 2 O are produced by burning 2.72 mol H 2 in an excess of O 2 ?

Write the Chemical Equation: Balance the Chemical Equation: 2

H 2 + O 2 → H 2 O

Use the stoichiometric factor or mole ratio in an equation: n H 2 O = 2.72 mol H 2 × 2 mol H 2 O 2 mol H 2 = 2.72 mol H 2 O

Example 4-6

Additional Conversion Factors in a Stoichiometric Calculation: Volume, Density, and Percent Composition.

An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm 3 . A 0.691 cm 3 piece of the alloy reacts with an excess of HCl(aq). If we assume that

all

the Al but

none

of the Cu reacts with HCl(aq), what is the mass of H 2 obtained?

Example 4-6

Write the Chemical Equation: Balance the Chemical Equation: 2 2

3 + H 2

Example 4-6

2 Al + 6 HCl → 2 AlCl 3 + 3 H 2

Plan the strategy:

cm 3 alloy → g alloy → g Al → mole Al → mol H 2 → g H 2

We need 5 conversion factors !

Write the Equation and Calculate: m H2 = 0.691 cm 3 alloy × 2.85 g alloy 1 cm 3 × 97.3 g Al 100 g alloy × 1 mol Al 26.98 g Al × 3 mol H 2 2 mol Al × 2.016 g H 2 1 mol H 2 = 0.207 g H 2

Combustion Reactions

• Combustion reaction – burning of a substance.

Substance combines with oxygen to form carbon dioxide and water • C 8 H 18 (l) + O 2 (g) CO 2 (g) + H 2 O (l)

Balancing Combustion Reactions

1.

2.

3.

4.

5.

Write

correct

formulas for the reactants and products Balance the carbon atoms Balance the hydrogen atoms Balance the oxygen atoms Verify that the number of atoms of each element is balanced

Example 4-2

Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound.

Liquid triethylene glycol, C 6 H 14 O 4 , is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion.

Example 4-2

Chemical Equation:

C 6 H 14 O 4 15 2 2 6 7 2 O

1. Balance C.

2. Balance H.

3. Balance O.

4. Multiply by two

2 C 6 H 14 O 4 + 15 O 2 → 12 CO 2 + 14 H 2 O

and check all elements.

Yields

• Theoretical yield – the maximum amount of product that can be formed from a chemical reaction • Actual yield – the amount of product that is formed in the laboratory

شنکاو هدزاب

.

دوش یم شرازگ نآ هدزاب اب لاومعم شنکاو کی تیبولطم نازیم شنکاو هدزاب دصرد = یلمع رادقم x 100 یرظن رادقم

شنکاو هدزاب

: لاثم زا ه ک تسا یگدروخ ربارب رد مواقم و نزو کبس ،یوق یزلف میناتیت رد می هب ناتیت .

دوش یم هدافتسا یزاس هخرچود و امیپاوه عیانص یارب نآ 1150  C ات 950  C نیب باذم میزینم اب دیرلک میناتیت شنکاو .

دیآ یم تسد TiCl 4 (g) + 2 Mg(l)  Ti(s) + 2 MgCl 2 (l) رادقم اب دیرلک میناتیت کمن زا 3.54x10

7 g یتعنص دنیآرف کی رد .

تسا هدش شنکاو دراو میزینم یفاضا ؟تسا مرگ دنچ میناتیت دیلوت ) فلا ار هد زاب دصرد ،دشاب هدش دیلوت میناتیت 7.91x10

6 g لمع رد رگا .

) ب دینک باسح

1

m ol TiCl

4  1

m ol Ti



no

.

m ol Ti

 1 .

87  10 5 

g Ti

 1 .

87  10 5  47 .

9  8 .

96  10 6 %

R

 7 .

91  10 5 8 .

96  10 5  100  % 88 .

3 : فلا : ب

Theoretical, Actual and Percent Yield

• When actual yield = % 100, then the reaction is said to be quantitative .

• Side reactions reduce the percent yield.

• By-products are formed by side reactions.

Limiting Reagents

• The goal of chemical reactions is to produce the highest amount of product possible – So, one reagent will probably be in excess – This allows for the complete reaction of one reagent, even though some others remain unreacted (NH 4 ) 2 PtCl 4 (s) + 2 NH 3 (aq) 2 NH 4 Cl(aq) + Pt(NH 3 ) 2 Cl 2 (s) $100/g $0.01/g All of the expensive reagent is used up, leaving the cheap unreacted ammonia

هدننک دودحم لماع

نا یاپرد و هدش فرصم یمامت هب هک تسآ یآ هدام ن آ هدننک دودحم لماع دآوم هیق ب فرصم ای دیلوت هدننک دودحم ن آ و .

دنامن یقاب ن آ زآ یزیچ شنکآو .

تسآ شنکآو تبسن دآوم هیقب ات دوش صخشم هدننک دودحم لماع دیاب آدتبآ هلئسم لح رد .

دنوش هدیجنس ن آ هب • •

4NH 3 4N 2 6H 2 2N 2

؟تسا مادک هدننک دودحم لماع

هدننک دودحم لماع

داجیآ کمن رآدقم هچ دوس مرگ 80 و دیسآ کیردیرلک مرگ 73 نیب شنکآو ؟دوش رد یم • یم داجیآ کمن ردقچ دوس مرگ 160 و دیسآ کیروفلوس مرگ 98 نیب شنکآو رد ؟دوش •

لاثم

برس زا مرگ و هداد 15.0

مرگ ماجن ا دنچ زا دینک سپ شنکاو رگا .

تاتسا دنهد یم باسح دینک ،دنوش باسح طولخم نینچمه .

مهاب دوش یم هیلوا دیلوت داوم زا تافلوس کیره برس ؟دنام لولحم یم برس یقاب و دیسا دیسا کیروفلوس کیتسا و دماج یفاضا هدام مرگ دنچ لولحم تافلوس شنکاو

H 2 SO 4 + Pb(CH 3 COO) 2



PbSO 4 + 2 CH 3 COOH

15.0

g H 2 SO 4  1 mol H 2 SO 4 98 .

1 g  1 mol P bSO 4 1 mol H 2 SO 4  303.3

g 1 mol P bSO 4  46.4

g P bSO 15.0

g P b(CH 3 COO) 2  1 mol P b(CH 3 COO) 2 3 25.3

g 4  1 mol P bSO 4 1 mol P b(CH 3 COO) 2  303.3

g 1 mol P bSO 4  1 4 .

0 g P bSO P b(CH 3 COO) 2 limit ing react ant  14.0

g P bSO 4 4 : لح

15.0

g P b(CH 3 COO) 2  1 mol P b(CH 3 COO) 2 3 25.3

g  1 mol H 2 SO 4 1 mol P b(CH 3 COO) 2  98.1

g 1 mol H 2 SO 4  4 .

52 g H 2 SO 4 4.52

g of H 2 SO 4 is used 15.0

g 4.52

g  10.5

g of H 2 SO 4 is excess.

Limiting Reagent

Chemical Equations & Chemical Analysis • • • Analytical chemists try to identify substances in a mixture, and try to measure the quantities of the components.

Mostly it is done with instrumental methods It is essential to use chemical reactions and stoichiometry

Quantitative Analysis of a Mixture • Usually depends on one of the two following ideas.

Idea 1: – A substance, present in unknown amount, can be allowed to react with a known quantity of another substance. If the stoichiometric ratio for their reaction is known, the unknown can be determined.

CH 3 CO 2 H(aq) + NaOH(aq) NaCH 3 CO 2 (aq) + H 2 O(l) Know the amount of NaOH so we can determine the amount of acetic acid

Quantitative Analysis of a Mixture Idea 2 – A material of unknown composition can be converted to one or more substances of known composition. Those substances can be identified, their amounts determined, and these amounts related to the amount of the original, unknown substance.

C 7 H 5 NO 3 S +

X

convert to SO 4 2 Na 2 SO 4 + other Na 2 SO 4 (aq) + BaCl 2 (aq) BaSO 4 (s) + 2 NaCl(aq) 1 mol Na 2 SO 4 1 mol S 1 mol SO 4 2 1 mol BaSO 4

Chemical Reactions in Solution

• Close contact between atoms, ions and molecules necessary for a reaction to occur.

• Solvent – We will usually use

aqueous

(aq) solution.

• Solute – A material dissolved by the solvent.

Chemical Reactions in Solution

.

دوش یم ماجنا لولحم رد اهشنکاو زا یرایسب نیا رب اهشنکاو هنوگ نیا یارب یرتمویکوتسا تابساحم تظلغ و هتفر راک هب یاه لولحم یاه مجح یانبم .

تسا اه لولحم رادقم رد هدش لح هدام رادقم ، لولحم کی تظلغ رادقم رد دوجوم هدش لح هدام رادقم ای ، .

للاح ینیعم تسا لولحم زا ینیعم .

دراد دوجو تظلغ نایب یارب شور دنچ • • • •

Molarity

Molarity (

M

) = # Mole of solute Volume of solution (L) If 0.444 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is:

c

urea = 0.444 mol urea 1.000 L = 0.444 M CO(NH 2 ) 2

M

هتیرلاوم

و ، تسآ لول حم رتیل کی یانبم رب هتیرلاوم فیرعت هک دیشاب هتشآد هجوت .

للاح رتیل کی یانبم رب هن • لولحم 1 L رد هدش لح هدام لوم 1.0

لماش 1.0 M لولحم کی .

تسا لولحم 1 L رد هدش لح هدام لوم 1.5

لماش 1.5 M لولحم کی .

تسا لولحم 1 L رد هدش لح هدام لوم 3.0 لماش 3.0 M لولحم کی .

تسا • • •

M

هتیرلاوم

همادا

( M –

هتیرلاوم

: 3.0 M لولحم کی یارب .

تسا هدش لح هدام لوم 3.0

لماش ، 1000 mL .

تسا هدش لح هدام لوم 1.5

لماش ، 500 mL .

تسا هدش لح هدام لوم 6.0

لماش ، 2000 mL • • • • .

تسا 3.0 M تظلغ هنومن هسرهرد

•

Preparation of a Solution

Weigh the solid sample.

Dissolve it in a volumetric flask partially filled with solvent.

Carefully fill to the mark.

Example 4-6

Calculating the mass of solute in a solution of known molarity.

We want to prepare exactly 0.2500 L (250 mL) of an 0.250 M K 2 CrO 4 (MW=194.02) solution in water. What mass of K 2 CrO 4 should we use?

Plan strategy:

Volume → moles → mass

We need 2 conversion factors !

Write equation and calculate: m K2CrO4 = 0.2500 L × 0.250 mol × 1.00 L 194.02 g 1.00 mol = 12.1 g

Solution Dilution

M

i × V i

M

f × V f

M

= n V

M

i × V i = n i = n f =

M

f × V f

M

f =

M

i × V i V f

= M

i V i V f

Example 4-10

Preparing a solution by dilution.

A particular analytical chemistry procedure requires 0.0100 M K 2 CrO 4 . What volume of 0.250 M K 2 CrO 4 prepare 0.250 L of 0.0100 M K 2 CrO 4 ?

should we use to

Plan strategy: M

f =

M

i V i V f V i = V f

M

f

M

i

Calculate:

V K2CrO4 = 0.2500 L × 0.0100 mol × 1.00 L 1.000 L 0.250 mol = 0.0100 L

Solution formation by Dilution

Oxidation States

Metals tend to lose electrons.

Na Na + +

e

Non-metals tend to gain electrons.

Cl + We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element.

e

Cl -

Rules for Oxidation States

1.

The oxidation state (OS) of an individual atom in a free element is 0.

2.

The total of the OS in all atoms in: i.

ii.

Neutral species is 0.

Ionic species is equal to the charge on the ion.

3.

In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively.

4.

In compounds the OS of fluorine (F) is

always

–1

Rules for Oxidation States

5.

In compounds, the OS of hydrogen (H) is

usually

+1 6.

In compounds, the OS of oxygen (O) is

usually

–2.

7.

In binary (two-element) compounds with metals: i.

Halogens have OS of –1, ii.

Group 6A have OS of –2 and iii. Group 5A have OS of –3.

Example:

Assigning Oxidation States.

What is the oxidation state of the underlined element in each of the following? a) P 4 ; b) Al 2 O 3 ; c) MnO 4 ; d) NaH a) P 4 is an element. P OS = 0 b) Al 2 O 3 : O is –2 . O 3 is –6. Since (+6)/2=(+3), Al OS = +3 .

c) MnO 4 : net OS = -1, O 4 is –8. Mn OS = +7 .

d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and H OS = -1 .

Naming Compounds

Trivial names are used for common compounds.

A systematic method of naming compounds is known as a system of

nomenclature .

Inorganic compounds Organic compounds

Inorganic Nomenclature

Binary Compounds of Metals and Nonmetals: first the name of the metal, then of the non-metal Na Cl electrically neutral = sodium chloride name is unchanged + “ide” ending Mg I 2 Al 2 O 3 Na 2 S = = = magnesium iodide aluminum oxide sodium sulfide

Binary Compounds of Two Non-metals

Molecular compounds

usually

write the positive OS element first.

H Cl hydrogen chloride Some pairs form more than one compound mono 1 di 2 tri tetra 3 4 penta hexa hepta octa 5 6 7 8

Binary Acids

Acids produce H + when dissolved in water.

They are compounds that ionize in water.

Emphasize the fact that a molecule is an acid by altering the name.

HCl HF hydrogen chloride hydrogen fluoride hydrochloric acid hydrofluoric acid

Polyatomic Ions

Polyatomic ions are

very

common.

The following table gives a list of some of them. Here are a few: ammonium ion carbonate ion hypochlorite chlorite chlorate perchlorate NH 4 + CO 3 2 ClO ClO 2 ClO 3 ClO 4 acetate ion hydrogen carbonate phosphate hydrogen phosphate sulfate hydrogensulfate C 2 H 3 O 2 HCO 3 PO 4 3 HPO 4 2 SO 4 2 HSO 4 -

Electrolytes

• • Some solutes can

dissociate

into ions.

Electric charge can be carried.

Types of Electrolytes

•

Strong electrolyte

dissociates completely.

– Good electrical conduction.

•

Weak electrolyte

partially dissociates.

– Fair conductor of electricity.

•

Non-electrolyte

does not dissociate. – Poor conductor of electricity.

Representation of Electrolytes using Chemical Equations

A strong electrolyte:

MgCl 2 (s) → Mg 2+ (aq) + 2 Cl (aq)

A weak electrolyte:

CH 3 CO 2 H(aq) CH 3 CO 2 (aq) + H + (aq)

A non-electrolyte:

CH 3 OH(aq)

Notation for Concentration

MgCl 2 (s) → Mg 2+ Cl (aq) In 0.0050 M MgCl 2 : Stoichiometry is important.

[Mg 2+ ] = 0.0050 M [Cl [MgCl 2 ] = 0 M

Example:

Calculating Ion concentrations in a Solution of a Strong Electolyte.

What are the aluminum and sulfate ion concentrations in 0.0165 M Al 2 (SO 4 ) 3 ?

.

Balanced Chemical Equation:

Al 2 (SO 4 ) 3 (s) → 2 Al 3+ (aq) + 3 SO 4 2 (aq)

Example:

Aluminum Concentration:

0.0165 mol Al 1 L 2 (SO 4 ) 3 2 mol Al 3+ 1 mol Al 2 (SO 4 ) 3 = 0.0330 M Al 3+

Sulfate Concentration:

[SO 4 2 0.0165 mol Al ] = × 1 L 2 (SO 4 ) 3 3 mol SO 4 2 1 mol Al 2 (SO 4 ) 3 = 0.0495 M SO 4 2-

Precipitation Reactions

• • Soluble ions can combine to form an

insoluble

compound.

Precipitation occurs.

Ag + (aq) + Cl (aq) → AgCl(s)

Net Ionic Equation

Overall Precipitation Reaction:

AgNO 3 (aq) +NaI (aq) → AgI(s) + NaNO 3 (aq)

Complete ionic equation:

Spectator ions Ag + (aq) + NO 3 (aq) + Na + (aq) + I (aq) → AgI(s) + Na + (aq) + NO 3 (aq)

Net ionic equation:

Ag + (aq) + I (aq) → AgI(s)

Solubility Rules

Compounds that are

soluble

: – Alkali metal ion and ammonium ion salts Li + , Na + , K + , Rb + , Cs + NH 4 + – Nitrates , perchlorates and acetates NO 3 ClO 4 CH 3 CO 2 -

Solubility Rules

• Compounds that are

mostly soluble

: – Chlorides, bromides and iodides Cl , Br , I • Except those of Pb 2+ , Ag + , and Hg 2 2+ .

– Sulfates SO 4 2 • Except those of Sr 2+ , Ba 2+ , Pb 2+ and Hg 2 2+ .

• Ca(SO 4 ) is slightly soluble.

Solubility Rules

• Compounds that are

insoluble

: –Hydroxides and sulfides Except alkali metal and ammonium salts Sulfides of alkaline earths are soluble Hydroxides of Sr 2+ and Ca 2 + are slightly soluble.

HO , S 2 –Carbonates and phosphates Except alkali metal and ammonium salts CO 3 2 , PO 4 3-