General Chemistry

M. R. Naimi-Jamal

Faculty of Chemistry

Iran University of Science & Technology

فصلچهارم:

IIاستوکیومتری

شیمیائی معادالت و

Contents

4-1 Chemical Reactions and Chemical Equations

4-2 Chemical Equations and Stoichiometry

4-3 Chemical Reactions in Solution

4-4 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 AlBr3

Chemical Reaction

Nitrogen monoxide + oxygen → nitrogen dioxide

Step 1: Write the reaction using chemical symbols.

NO + O2 → NO2

Step 2: Balance the chemical equation.

2 1 2

Molecular Representation

Balancing Equations

• Never introduce extraneous atoms to balance.

NO + O2 → NO2 + O

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

NO + O2 → NO3

Nitrogen monoxide + oxygen → nitrogen dioxide

Example:

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)

P4 (s) + 6Cl2 (g) 4PCl3 (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

P4 + 6 Cl2 4 PCl3

Initial amount (mol)

1.00 mol (124 g) 6.00 mol (425 g) 0 mol (0 g)

Change in amount (mol)

- 1.00 mol -6.00 mol +4.00 mol

After complete reaction (mol)

0 mol (0 g) 0 mol (0 g) 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 4-3Relating the Numbers of Moles of Reactant and Product.

How many moles of H2O are produced by burning 2.72 mol H2 in an excess of O2?

H2 + O2 → H2O

Write the Chemical Equation:

Balance the Chemical Equation:

2 2

Use the stoichiometric factor or mole ratio in an equation:

nH2O = 2.72 mol H2 × = 2.72 mol H2O2 mol H2O2 mol H2

Example 4-6Additional 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/cm3. A 0.691 cm3 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 H2 obtained?

Al + HCl → AlCl3 + H2

Write the Chemical Equation:

Example 4-6

Balance the Chemical Equation:

2 6 2 3

2 Al + 6 HCl → 2 AlCl3 + 3 H2

Example 4-6

Plan the strategy:

cm3 alloy → g alloy → g Al → mole Al → mol H2 → g H2

We need 5 conversion factors!

× ×

Write the Equation

mH2 = 0.691 cm3 alloy × × ×2.85 g alloy

1 cm3

97.3 g Al100 g alloy

1 mol Al26.98 g Al

3 mol H2

2 mol Al2.016 g H2

1 mol H2

= 0.207 g H2

and Calculate:

Combustion Reactions

• Combustion reaction – burning of a substance. Substance combines with oxygen to form carbon dioxide and water

• C8H18 (l) + O2(g) CO2(g) + H2O (l)

Balancing Combustion Reactions

1. Write correct formulas for the reactants and products

2. Balance the carbon atoms

3. Balance the hydrogen atoms

4. Balance the oxygen atoms

5. 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, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion.

152 6 7C6H14O4 + O2 → CO2 + H2O 6

2. Balance H.

2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O

4. Multiply by two

Example 4-2

3. Balance O.

and check all elements.

Chemical Equation:

1. Balance C.

6 7

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

واکنش بازده

آن بازده با معموال واکنش یک مطلوبیت میزانشود می .گزارش

مقدار عملی

100 x درصد بازده واکنش = مقدار نظری

واکنش بازده :مثال

خوردگی برابر در مقاوم و وزن سبک قوی، فلزی تیتانیمسازی دوچرخه و هواپیما صنایع برای آن از که است . با کلرید تیتانیم واکنش در تیتانیم شود می استفاده

بین مذاب .C 1150تا C 950منیزیم آید می دست به

TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l)

صنعتی فرآیند یک با 3.54x107 gدر کلرید تیتانیم نمک ازاضافی . مقدار است شده واکنش وارد منیزیم

است؟( گرم چند تیتانیم تولید الفعمل( در اگر درصد 7.91x106 gب باشد، شده تولید تیتانیم

. کنید حساب را بازده

:الف

65

5

4

1096.89.471087.1

1087.1.

11

Ti

Ti

g

molno

TimolTiClmol

3.88%1001096.8

1091.7%

5

5

R

: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

(NH4)2PtCl4(s) + 2 NH3(aq) 2 NH4Cl(aq) + Pt(NH3)2Cl2(s)

$100/g $0.01/g

All of the expensive reagent is used up, leaving the cheap unreacted ammonia

عامل محدود کننده

آن مWاده ای اسWت کWه بWه عامWل محWدود کننWده•تمWامی مصWرف شWده و درپایWان واکنش چWیزی از آن بWاقی نمانWد. و آن محWدود کننWده تولیWد یWا

مصرف بقیه مواد واکنش است.در حWل مسWئله ابتWدا بایWد عامWل محWدود کننWده •

آن بWه نسWبت مWواد بقیWه تWا شWود مشWخص سنجیده شوند.

عامل محدود کننده کدام است؟

4N2 6H2

4NH3

2N2

عامل محدود کننده

بین • واکنش و 73در اسWید کلریWدریک گWرم می 80 ایجWاد نمWک مقWدار چWه سWود گWرم

شود؟

گWرم سWولفوریک اسWید و 98در واکنش بین • گرم سود چقدر نمک ایجاد می شود؟ 160

و داده واکنش استات سرب و اسید سولفوریک محلولمی محلول اسید استیک و جامد سولفات سرب

اگر. باهم 15.0دهند اولیه مواد از هریک از گرمسولفات سرب گرم چند کنید حساب شوند، مخلوط . انجام از پس کنید حساب همچنین شود می تولید

ماند؟ می باقی اضافی ماده گرم چند واکنش

مثال

H2SO4 + Pb(CH3COO)2 PbSO4 + 2 CH3COOH

حل:

PbSO g 14.0

reactant limiting COO)Pb(CH

PbSO g 0.41PbSO mol 1

g 303.3

COO)Pb(CH mol 1

PbSO mol 1

g 25.33

COO)Pb(CH mol 1COO)Pb(CH g 15.0

PbSO g 46.4PbSO mol 1

g 303.3

SOH mol 1

PbSO mol 1

g 1.98

SOH mol 1SOH g 15.0

4

23

4423

4

2323

4442

4

4242

excess. is SOH of g 10.5 g 4.52 - g 15.0

used is SOH of g 4.52

SOH g 52.4SOH mol 1

g 98.1

COO)Pb(CH mol 1

SOH mol 1

g 25.33

COO)Pb(CH mol 1COO)Pb(CH g 15.0

42

42

424223

42

2323

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.

CH3CO2H(aq) + NaOH(aq) NaCH3CO2(aq) + H2O(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.

C7H5NO3S + X convert to SO42- Na2SO4

+ other

Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2 NaCl(aq)

1 mol Na2SO4 1 mol S 1 mol SO42- 1 mol BaSO4

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

•. شود می انجام محلول در واکنشها از بسیاریگونه • این برای استوکیومتری محاسبات

مبنای بر های حجمواکنشها محلول کار های بهو .غلظترفته است ها محلول این

، غلظت• محلول شده یک حل ماده در مقدارمعینی شده حاللمقدار حل ماده مقدار یا ،

. است محلول از معینی مقدار در موجود•. دارد وجود غلظت بیان برای روش چند

Molarity

Molarity (M) = Volume of solution (L) # Mole of solute       

If 0.444 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is:

curea = 1.000 L

0.444 mol urea = 0.444 M CO(NH2)2

M موالریته -

توجه داشته باشید که تعریف موالریته بر مبنای •یک لیتر محلول است ، و نه بر مبنای یک لیتر

حالل .

محلول • در 1.0شامل M 1.0یک شده حل ماده L 1مولاست . محلول

محلول • در 1.5شامل M 1.5یک شده حل ماده L 1مول. است محلول

محلول • در 3.0شامل M 3.0یک شده حل ماده L 1مول. است محلول

( M–موالریته ادامه)

محلول • یک :M 3.0برای

•1000 mL شامل ،3.0. است شده حل ماده مول

•500 mL شامل ،1.5. است شده حل ماده مول

•2000 mL شامل ،6.0. است شده حل ماده مول

غلظت • نمونه .است M 3.0درهرسه

M موالریته -

Preparation of a Solution

Weigh the solid sample.

Dissolve it in a volumetric flask partially filled with solvent.

Carefully fill to the mark.

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 K2CrO4 (MW=194.02) solution in water. What mass of K2CrO4 should we use?

Plan strategy:

Example 4-6

Volume → moles → mass

We need 2 conversion factors!Write equation and calculate:

mK2CrO4 = 0.2500 L × × = 12.1 g

0.250 mol 1.00 L

194.02 g1.00 mol

Solution Dilution

Mi × Vi = ni

Mi × ViMf × Vf

= nf = Mf × Vf

Mi × ViMf = Vf

= Mi

Vi

Vf

M = n

V

Preparing a solution by dilution.

A particular analytical chemistry procedure requires 0.0100 M K2CrO4. What volume of 0.250 M K2CrO4 should we use to prepare 0.250 L of 0.0100 M K2CrO4?

Calculate:

VK2CrO4 = 0.2500 L × × = 0.0100 L

0.0100 mol 1.00 L

1.000 L0.250 mol

Example 4-10

Plan strategy: Mf = Mi

Vi

Vf

Vi = Vf

Mf

Mi

Solution formation by Dilution

Oxidation States

Metals tend to lose electrons.

Na Na+ + e-

Non-metals tend to gain electrons.

Cl + e- Cl-

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

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. Neutral species is 0.ii. 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

6. In compounds, the OS of hydrogen (H) is usually +1

7. In compounds, the OS of oxygen (O) is usually –2.

8. 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.

Assigning Oxidation States.

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 an element. P OS = 0

b) Al2O3: O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3.

c) MnO4-: net OS = -1, O4 is –8. Mn OS = +7.

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

Example:

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

NaCl = sodium chloride

name is unchanged

+ “ide” endingelectrically neutral

MgI2 = magnesium iodide

Al2O3 = aluminum oxide

Na2S = sodium sulfide

Binary Compounds of Two Non-metals

Molecular compoundsusually write the positive OS element first.HCl hydrogen chloride

mono 1 penta 5

di 2 hexa 6

tri 3 hepta 7

tetra 4 octa 8

Some pairs form more than one compound

Binary Acids

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

HCl hydrogen chloride hydrochloric acid

HF hydrogen fluoride hydrofluoric acid

Acids produce H+ when dissolved in water.

They are compounds that ionize in water.

Polyatomic Ions

Polyatomic ions are very common.

The following table gives a list of some of them. Here are a few:

ammonium ion NH4+ acetate ion C2H3O2

-

carbonate ion CO32- hydrogen carbonate HCO3

-

hypochlorite ClO- phosphate PO43-

chlorite ClO2- hydrogen phosphate HPO4

2-

chlorate ClO3- sulfate SO4

2-

perchlorate ClO4- hydrogensulfate HSO4

-

Electrolytes

• Some solutes can dissociate into ions.

• Electric charge can be carried.

Types of Electrolytes

• Weak electrolyte partially dissociates.– Fair conductor of electricity.

• Non-electrolyte does not dissociate. – Poor conductor of electricity.

• Strong electrolyte dissociates completely.– Good electrical conduction.

Representation of Electrolytes using Chemical Equations

MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)

A strong electrolyte:

A weak electrolyte:

CH3CO2H(aq) ← CH3CO2-(aq) + H+(aq)→

CH3OH(aq)

A non-electrolyte:

MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)

[Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M [Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M

Notation for Concentration

In 0.0050 M MgCl2:

Stoichiometry is important.

Example:

Calculating Ion concentrations in a Solution of a Strong Electolyte.

What are the aluminum and sulfate ion concentrations in 0.0165 M Al2(SO4)3?.

Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)

Balanced Chemical Equation:

[Al] = × =1 L

2 mol Al3+

1 mol Al2(SO4)3

0.0165 mol Al2(SO4)3 0.0330 M Al3+

Example:

0.0495 M SO42-[SO4

2-] = × =1 mol Al2(SO4)3

Sulfate Concentration:

1 L

3 mol SO42-0.0165 mol Al2(SO4)3

Aluminum Concentration:

Precipitation Reactions

• Soluble ions can combine to form an insoluble compound.

• Precipitation occurs.

Ag+(aq) + Cl-(aq) → AgCl(s)

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →

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

Spectator ionsAg+(aq) + NO3

-(aq) + Na+(aq) + I-(aq) →

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

Net Ionic Equation

AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)

Overall Precipitation Reaction:

Complete ionic equation:

Ag+(aq) + I-(aq) → AgI(s)

Net ionic equation:

Solubility Rules

Compounds that are soluble:

Li+, Na+, K+, Rb+, Cs+ NH4+

NO3- ClO4

- CH3CO2-

– Alkali metal ion and ammonium ion salts

– Nitrates, perchlorates and acetates

Solubility Rules

– Chlorides, bromides and iodides Cl-, Br-, I-

• Except those of Pb2+, Ag+, and Hg22+.

– Sulfates SO42-

• Except those of Sr2+, Ba2+, Pb2+ and Hg22+.

• Ca(SO4) is slightly soluble.

•Compounds that are mostly soluble:

Solubility Rules

–Hydroxides and sulfides HO-, S2-

Except alkali metal and ammonium salts

Sulfides of alkaline earths are soluble

Hydroxides of Sr2+ and Ca2+ are slightly soluble.

–Carbonates and phosphates CO32-, PO4

3-

Except alkali metal and ammonium salts

•Compounds that are insoluble:

Chapter 4 Questions

1, 4, 7, 10, 17, 20, 24, 29, 31, 35, 38,

51