CHEMISTRYChemistry studies properties and transformation of the matter.

Matter is formed by atoms. More atoms can form molecules.

Elements are formed of the same atoms. They can be metals or non metals.

Molecules can be formed of same or different atoms.

Compounds are formed by molecules constituted by different atoms.

Periodical Table collects all the elements.

ATOM

It is formed by a central nucleus (positive charge) and some surrounding electrons (charge -1).

Nucleus has positive charge and is formed essentially of proton (charge 1+) and neutrons (no charge).

Protons give the atomic number of the element, while Neutrons contribute to the atomic mass.

Atomic mass is the sum of protons (mass 1) + neutrons (mass 1). Electrons have no mass.

Periodic Table reports elements and their atomic (Z = atomic number) and mass number (W).

Protons and Electrons are responsible of a lot of properties of elements, ions and molecules.

Elements of VIII group – Noble gasesHe = Helium (Z =2), Ne = Neon (Z = 10), Ar = Argon (Z = 18), Kr = Kripton (Z = 36), Xe = Xenon (Z = 54), Rn = Radon (Z = 86).The name noble gas indicates their high stability. Reactions involving noble gas are really very unusual.

Each electron has a specific energy and occupies a particular shellaround the nucleus. External shells containing 8 electrons are particularly stable, like noble gases.Compounds are formed by elements tending to reach the noble gas corresponding. For example Chlorine having 7 electrons in the last shell tends to catch 1 electron more to assume the Ar configuration.Sodium (Z=11) having 1 alone electron in the last shell, has difficulty to hold it and tends to release it to assume to Ne configuration.Ions Cl- and Na+ are respectively formed. They attract each otherforming the compound NaCl = sodium chloride.

HOW DIFFERENT ATOMS FORM MOLECULES:

THE CHEMICAL BOND

• Elements belonging to the same group of periodical table have similar properties.

• Ionic or Electrostatic bond: Na+ + Cl- =NaCl (sodium gives an electron to chlorine, ions of opposite charge attract each other).

• Homopolar or Covalent bond: Cl + Cl =Cl2 (Two chlorine atoms put together a couple of electrons

• Coordination bond: Cu(NH3)42+. Each :NH3 put in

common two electrons bonding Cu2+

• Others less important (Hydrogen, Van der Walls...)

NOMENCLATURE 1

•Basic Oxides = Metals + Oxygen; i.e. CaO = calcium oxide

•Bases or Hydroxides = Oxides + H2O; i.e. CaO+ H2O = Ca(OH)2 = calcium hydroxide

•Acid Oxides = Non Metals + Oxygen; i.e. CO2 = carbon dioxide

•Acids = acids oxides + H2O; CO2 + H2O= H2CO3 = carbonic acid (acids containing oxygen)

•Acids = Non metal + hydrogen = i.e. HCl = Hydrochloric acid

NOMENCLATURE 2

•Acids containing oxygen = the name changes depending on the kind of oxides:

SO2 (sulphur dioxide) forms solphorous acid, whileSO3 (sulphur trioxide) forms solphoric acid.Cl2O + H2O = 2 HClO (ipochlorous acid)Cl2O3 + H2O = 2 HClO2 (chlorous acid)Cl2O5 + H2O = 2 HClO3 (chloric acid)Cl2O7 + H2O = 2 HClO4 (perchloric acid)

NOMENCLATURE 3

Acid + Bases = Salt

HCl + NaOH = NaCl (sodium chloride) + H2O

HClO+ NaOH = NaClO (sodium ipochlorite) + H2OHClO2+ NaOH = NaClO2 (sodium chlorite) + H2OHClO3+ NaOH = NaClO3 (sodium chlorate) + H2OHClO4+ NaOH = NaClO4 (sodium perchlorate) + H2O

N2O3+H2O=2HNO2 (nitrous acid)+2 NaOH=2 NaNO2 (sodium nitrite)N2O5+H2O=2HNO3 (nitric acid)+2 NaOH= 2 NaNO3 (sodium nitrate)

NOMENCLATURE 4

Acid + Bases = Salt

H2S (sulphydric acid) + 2 NaOH = Na2S (sodium sulphide) + 2H2O

H2SO3 (Sulphurous acid)+ 2 NaOH = Na2SO3 (sodium solphite) + 2H2O

H2SO4 (Sulphuric acid)+ 2NaOH = Na2SO4 (sodium solphate)+ 2 H2O

What is a formula?

•The formula gives the identity to aCompound.•It tell us which elements are present in the compound•It gives the quantitative composition of the compound•It allows to calculate the molecular mass•From its formula it is possible to know the properties of the compound (acid, bases, salt, oxidant, reducing,…)Example: Na2SO4

It is a salt, formed by sodium, sulphur and oxygen with a ratio 2:1:4, its mm is 23x2+32+4x16=142.It is not an oxidant reagent, but it can be reduced to Na2SO3 or toS or to Na2S.

What is a reaction?

The transformation of a few compounds to others. Reactions evolve to a lower energy status

In any reaction, the material balance must be verified

Frequent types of reactions are:1) Acid – base reactions. 2) Precipitation reactions.3) Complex formation reactions. 4) Redox reactions.

Property of compoundsMolecular compoundsionic compounds

Molecular compounds remain no dissociatedIonic compounds dissociate in ions

Ions are pieces of compounds which have an electrical chargeThey can be positive (cations) and negative (anions), both are able to carry electrical current.Acids and bases dissociate in Protons (H+) or hydroxyl ions (OH-) and anions and cations respectively.i.e.: HCl (hydrochloric acid) dissociates in H+ and Cl-, while

NaOH (sodium hydroxide) dissociates in OH- and Na+. Salts dissociate in cations and anions (i.e. NaCl = Na+ + Cl-)

Some easy rules

Oxidation numbers have a sign too, it can be + or -.The highest oxidation number corresponds to the periodical table group.The lowest one can be obtained by subtracting 8 from the highest (– 8).i.e. The highest oxidation number of chlorine is +7, the lowest is –1. Metals have generally only zero or the oxidation number corresponding to the periodical table group. Compounds have no charge. Ions have a charge.

Each element can have one or more oxidation numbers.

Some examples of easy reactions

Reaction between aluminium sulphate and barium nitrate.

Formulae: aluminium belongs to the third group and has +3;Sulphate is the ion of sulphuric acid then is SO4

= (charge –2),The formula of aluminium sulphate is Al2(SO4)3.

Barium belongs to the second group then can be Ba2+, while nitrate being the ion of nitric acid is NO3

-. Formula is Ba(NO3)2

Reaction is:Al2(SO4)3 + 3 Ba(NO3)2 = 3 BaSO4 + 2 Al (NO3)3

This means that (2x27+3x96) g of aluminium sulphate reacts with 3(137,3+ 2x62) g of Ba(NO)2 to give what????By the way, compare the number with the atomic mass.

Some easy calculations 1

In any reaction, the material balance must be verified

By putting in a becker 10 g of zinc and by adding HCl, hydrogenDevelops. Write the reaction and calculate how much zinc chloride Will be formed.Write the reaction: zinc belongs to the second group, its oxidation numbers can be 0 or +2. Chlorine is in the VII group, in the lowest oxidation state is –1.Zn + 2 HCl = ZnCl2 + H2 (gas)

Calculations: 1atomic mass of zinc (65,37) reacts with 2 molecularMass of HCl (35,45=70,90) to give 1 molecular mass of ZnCl2 (134,27) and 1 molecular mass of H2 (2).

Some easy calculations 2 (follows from 1)But zinc put in the reaction was 10 g (not 65,3) then to know how much ZnCl2 is formed, it is necessary to divide 10 / mw of Zinc (65,37) to have the number of moles (= 0,153). The same number of moles of ZnCl2 (see reaction coefficients). The mass of ZnCl2 will be = number of moles (0,153) x mm of the salt (134,27) = 20,5 gIn the above reaction H2 (gas) was formed. Some rules for gas exist.They connect volume, pressure and temperature of a gas through a general equation : PV = nRT, where P is pressure, V volume and TThe absolute temperature (°C + 273,14°K). n = the number of molesOf the gas. R is a constant. When Volume and temperature are expressed in litres and atmospheres, respectively R = 0,0821. Avogadro law: equal volumes of different gases at the same pressureAnd temperature contain the same number of molecules. A mole =6 x 1023 molecules at 1 atm and 273°C occupies 22,4 litres.

Some calculations 3 (follows from 2)1) The compound ferric alum has the formula: K2SO4Fe2(SO4)324H2O. Calculate the percentage of iron present.By knowing the formula, the mm (mole mass) can be obtained. Atomic mass are yield from the tables. K = 39,1; S = 32; O = 16; Fe = 55,85; H = 1.MM of the compound is = 1005,7. In it 2 Fe atoms (2 x 55,85=111.7) are present. By dividing 111,7/1005,7 and multiplying by 100, the percentage of Fe in the alum is obtained.2) Calculate how many moles of CO2 can be obtained by heating 150 g of CaCO3 (calcium carbonate).By heating CaCO3, the following reaction takes place:CaCO3 = CaO + CO2. This means that 1 mole of CO2 (i.e. 44 g) is produced by 1 mole of CaCO3 (100 g). By dividing 150 g by 100 (1,5), the number of moles of CaCO3 and then of CO2 is obtained.To obtain the weight of CO2, 1,5 is multiplied by 44 (mm of CO2).

Some calculations 4 (follows from 3)Reaction with formation of a slight soluble compound. Somecompounds are not too much soluble in water and their solubilityis regulated as discussed in next slides.Slight soluble substances are BaSO4 (it is used for x-ray inspection),Fe2O3 (rust), Pb3O4 (minium), CaCO3 (marmor), CaC2O4 (calcium oxalate, is an organic substance, responsible to the formation ofrenal stones), CaSO4 (gypsum or chalk), and others. Problem: An operator has to prepare 50 g of Barium sulphate. Calculate what amount of BaCl2 and Na2SO4 must mixer to obtainthe required quantity of BaSO4.The reaction is: BaCl2 + Na2SO4 = BaSO4 + 2 NaCl.One mole of BaSO4(329,34) is obtained from 1 of BaCl2 and Na2SO4

50 /329,34 = 0,152. This number by 208,24 and by 142, respectively,Gives the weight of BaCl2 and Na2SO4 to use.

Some definitionsSystem is a part of matter subject of study. It can be homogeneous(if in all part has the same properties) or heterogeneous (when its parts have different properties.A phase is a homogeneous part of a system. When the compositionof a phases vary, but homogeneity remains, the phase is called Solution.

A solution is a homogeneous mixture of two or more substances.The main component is the solvent, the minor components arecalled solute.Concentration can be defined as the quantity of a solute present in a fixed quantity of solvent. There are different possibility :% weight; % volume; Molarity; Normality; Molality.

Total or free (at equilibrium) concentrations will be distinguished.

Total Concentration 1% weight = g of solute per 100 g of solvent.% volume = g of solute per 100 cc of solvent.Molarity (M or mol dm-3)= number mol of solute per litre of solution.Normality (N) = number equivalents (to define) of solute per litre(dm3) of solution.Molality (m) = number mol of solute per kg of solvent.

Easy calculations1) Calculate the molarity of a solution of NaCl 5% in volume.The problem indicates that 5 g of NaCl are solved in 100 cm3 of asolvent (Water). It is the same to say 50 g per dm3. It is sufficient todivide 50 per mm of NaCl (mm =23+35,45=58,45) to have 0.855 M

2) How many grams of Na2SO4 (mm=142) are in 10 cc of its solution 0.25 M? Answer: 0.01x0.25x142 = 0.355.

Total Concentration 2Molarity corresponds to the presence of 1 mm(molecular mass)in 1.00 dm3 of solution. The mm is exactly defined.

Normality corresponds to the presence of 1 equivalent in 1.00 dm3

of solution. The equivalent weight is not a constant, but it dependson the reaction taking place, then it can assume different values.However, some general rules exist: (weq = equivalent weight)• For an atom, weq = ma/oxidation number. I.e. weqAl= Al/3=26.98 /3, corresponding to the reaction 2Al+6HCl=2AlCl3+3H2.Weq is the weight corresponding in the reaction to 1 H.• For an acid, weq = mm of acid / number of H• For a basis, weq = mm of basis / number of OH• For H3PO4, weq can be mm/1; mm/2 or mm/3 depending on the protons involved in the reaction.Weq calculation is different for an oxidant or a reducing (see bellow)

Total Concentration 3 -REDOX

Oxidant is a compound which takes electrons and passes in its reduced form: Ox + n1 e = red.Reducing is a compound which releases electrons and turns in its oxidated form: Red = Ox + n2 e.

As free electrons cannot be present in a condensed state of matter,an oxidation cannot take place if, contemporary, a reduction does notoccur, by involving the same number of electrons:The reaction can be: Ox1 +Red2 = Red1 + Ox2, if n1 = n2

Weq of an oxidant or of a reducing is calculated as follows: Weq = Mm oxidant (or reducing) / electrons number involved in the reaction.

For example: IO3- can be reduced to I2 or to I-, according to the

half -reactions: 2 IO3- +10 e= I2; weq = 2 mm/10 electrons or

: IO3- + 5 e = I-; weq = mm/5 electrons

Redox reaction balance rules1) All elements not combined have oxidation number zero2) Oxygen has oxidation number –2, (except peroxydes, i.e. H2O2)3) Hydrogen has oxidation number +1, (except hydrures, i.e. LiH)4) A compound has always charge zero.5) An ion has always the charge corresponding to its properties6) It is possible to have fractional oxidation numbers7) In a redox reaction the released electron number must be equal tothose taken by the oxidant.

To balance a redox reaction1.Find the oxidant and the reducing in the proposed reaction

2.Write separately half reaction for both

3.Balance the ion where is the oxidant.

4.Calculate the redox number variation and add the corresponding electrons number on the part of the ox species

5.Charge balance: if the reaction occurs in acid field balance with H+, in alkaline field balance with OH-

6.Mass balance: write the corresponding H2O on the other side of H+, or OH-. Now the half reaction of oxidant is balanced.

7.Treat in the same way the half reaction of the reducing.

8.Find the minimum common multiple between the acquired and released electrons and balance definitively both half reaction.

EXAMPLES 1Reaction between Cu2+ + Zn = Cu + Zn2+

Copper is the oxidant because reduces its oxidation number from2+ to zero, while zinc is the reducing by passing from Zn to Zn2+

Half reactions are:Cu2+ +2e = Cu; Zn = Zn2+ + 2e. No balance of charge and mass isnecessary because they are already balanced. The number of exchanged electrons is 2 in both cases then the minimum commonmultiple is 2 and the complete reaction is: Cu2+ + Zn = Cu + Zn2+

Reaction between Ag++ Cu = Ag + Cu2+(not balanced)Half reaction of the oxidant: Ag++ e = Ag Half reaction of the reducing: Cu = Cu2+ + 2eThe half reactions are balanced for charge and mass, but the electronnumber is not the same, then half reaction of Ag must be multipliedby two, to have: 2Ag++ Cu = 2Ag + Cu2+

EXAMPLES 2 – acid fieldReaction between: MnO4

- + Fe2+ + H+ = Mn2+ + Fe3+

Oxidant is permanganate ion and reducing is iron (II).H. reaction ox: MnO4

- = Mn2+, Manganese passes from +7 to +2, by Acquiring 5e, and then: MnO4

- + 5 e = Mn2+.Right there is charge –6, left is +2 and H+ is used to balance charges,so that it is necessary to add 8 H+ right and balance with H2O left:MnO4

- + 5 e + 8 H+ = Mn2+ + 4 H2O (half reaction balanced)H. reaction red: Fe3+ + e = Fe2+ (half reaction balanced).Minimum common multiple between 1 and 5 is 5, then at end:MnO4

- +5 Fe2+ + 8 H+ = Mn2+ + 5 Fe3+ + 4 H2O

EXAMPLES 3 – acid field1) Reaction between: dichromate ion and iron (II) in acid field:Cr2O7

= + Fe2+ + H+ = Cr2+ + Fe3+

S. reaction ox:Cr2O7= = Cr3+, to balance chrome, it is Cr2O7

= = 2Cr3+,Chrome assumes 3 e x 2Cr = 6 e to have: Cr2O7

= + 6 e= 2 Cr3+, there are 8- right and 6+ left and H+ is used to balance charges, it is necessary to add 14 H+ right (to balance charges) and 7 H2O leftTo balance mass, i.e. Cr2O7

= + 6 e + 14 H+ = 2 Cr3+ + 7 H2O.S. reaction red: For iron is easy: Fe2+ = Fe3+ +e. We need to multiplyby six the iron half reaction, to have definitely:Cr2O7

= + 6 Fe2+ + 14 H+ = 2 Cr2+ + 6 Fe3+ + 7 H2O2) Reaction between thiosulphate and iodine: S2O3= + I2 = S4O6

= + I-

Ox. Number of S right is +2, but left is + 2.5, but there are 4 S, so that, 2 S2O3

= = S4O6= + 2 e and I2 + 2 e = 2 I-.

Complete reaction is: 2 S2O3= + I2 = S4O6

= + 2 I-

EXAMPLES 4 – acid fieldDissolution of copper (II) sulphur by means of nitric acid.S in CuS has oxidation number –2 and can be oxydated by HNO3, Nitrogen has ox number +5 and it can be reduced to NO (+2).Oxidation half reaction: CuS = Cu2+ + S + 2 e (balanced for charge and mass)Reduction half reaction: NO3

- + 3 e + 4 H+ = NO + 2 H2O (balan.)Minimum common multiple between 3 and 2 is 6, so that complete reaction is: 3 CuS + 2 NO3

- + 8 H+ = 2 NO + 4 H2O + 3 Cu2+ + 3 SThe reaction can be written also in molecular form, as follows:3 CuS + 8 HNO3 = 2 NO + 4 H2O + 3 Cu(NO3)2 + 3 S.

EXAMPLES 4 – alkaline fieldThe mechanism is the same that in acid. Only the charge is balanced By OH-, instead than H+

Example 1

Arsenic (As) oxidation to arseniate (AsO43-) by means of sodium

hypochloride (ClO-) in alkaline solution.As + 8 OH- = AsO4

3- + 5 e + 4 H2OClO- + 2 e + H2O = Cl- + 2 OH- (Between 5 and 2 is 10)Complete ionic reaction:2 As + 5 ClO- + 6 OH- = 2 AsO4

3- + 5 Cl- + 3 H2OComplete molecular reaction:2 As + 5 NaClO + 6 NaOH = 2 Na3AsO4 + 5 NaCl + 3 H2O

EXAMPLES 5 – alkaline fieldExample 2

Reduction of bismuth hydroxide [Bi(OH)3] by means of sodium stannite Na2SnO2 in alkaline solution.

Bismuth can have oxidation numbers: +5, +3, 0; Tin can have oxidation number +4, +2, 0. The reagents are Bi(+3) and Sn(+2). If bismuth is reduced it will beBi0, while tin will be oxidated, becoming Sn(+4), that means stannate SnO3

=.The half – reactions:Bi(OH)3 + 3 e = Bi0 + 3 OH-

SnO2= + 2 OH- = SnO3

= +2 e + H2OReaction: 2Bi(OH)3 + 3 SnO2

= = 2 Bi0 + 3 SnO3= + 3H2O

Molecular reaction: 2Bi(OH)3 + 3 Na2SnO2 = 2 Bi0 + 3 Na2SnO3 + 3H2O

Chemical equilibrium – Reactions 1

In the previously studied slides, the mass of the reaction productswere calculated, by assuming the complete transformation of thereagents in the corresponding products.This is not always true. Most of the reactions are of equilibrium, i.e.The transformation of the reagents into the products is not complete,But part of the reagents remain together with part of products.

Put in a bottle CO and H2O in appreciable concentration. At start,they react quickly. The speed decreases by increasing the reactionproducts concentration. The following reaction takes place:CO + H2O CO2 + H2. The symbol indicates a double arrow, i.e. the reaction can give from right to left or vice versa.

The equilibrium reactions occur frequently, overall if the reactiontakes place in a single phases (homogeneous), for example for gases

Chemical equilibrium – Reactions 2

Chemical reactions are equilibrium reaction, except those involvingFormation of a gas or a precipitate (a slight soluble compound). The separation of phases makes complete a reaction.Equilibrium reaction means that after the start, the reaction productReact among them to have again the reagents. In the above exampleAt start CO and H2O produce CO2 + H2, but when a appreciableConcentration of CO2 + H2 is formed, they react to form again CO + H2O. A point exists when the speed of the direct reaction willBe equal to the reverse one. This is the equilibrium point. This meansthat no static equilibrium, but a dynamic equilibrium exists.Really, chemical reactions occur on the basis of energetic content.If reagents have more energy than the products the reaction takesPlace, till both energy contents will be equal. This means that a mathematic relation regulates the trend of the reaction.

The mass action lawAt equilibrium and at constant temperature, the ratio among the concentrations of the reaction products and the reagents, each of them at exponent equal to the stoichiometric coefficient, is aconstant, generally indicated by k

The equilibrium reaction: a A + b B c C + d Dis regulated by the constant: k = (Cc Dd) / (Aa Bb)

The value of k is constant and it is independent of the reagents orproducts concentration. It is only depending on the temperature.

The mass action law – calculation 1

Example 1:2.94 moles of I2 and 8.1 of H2 form at equilibrium at a constanttemperature 5.64 moles of HI. Calculate the equilibrium constant.The reaction is: I2 + H2 2 HI and the constant k=cHI

2/(cH2 cI2)

It is possible to draw up the following table:

Table H2 I2 HIInitial moles 8.1 2.94 ---Formed moles --- ---- 5.64Used moles 2.82(5.64/2) 2.82 ----At equilibrium 5.28 0.12 5.64

K = (5.64)2 / (5.28 x 0.12) = 50.2

The mass action law – calculation 2

Example 2: At a constant temperature, the constant of the reaction:I2 + H2 2 HI is k = 0.02. Calculate the equilibrium concentrationFor all the present species, starting by 0.48 moles of both I2 + H2.

Prepare a table similar to that above formulated. Table H2 I2 HIInitial moles 0.48 0.48 ---Formed moles --- ---- 2 xUsed moles x x ----At equilibrium 0.48-x 0.48-x 2 x

Write the constant k = 0.02= (2x)2/[(0.48-x) (0.48-x)]. It can be calculated x = 0.032 moles

Concentrations: cHI=0.064; cH2= 0.448 = cI2

Total and free concentration 1It is unusual that a chemical species alone is present in a solution.

As previously described, different types of bonds can take placeAmong atoms or ions, according to the properties of them.

Elements put right in the periodical table tend to catch an electronTo reach a stable configuration. They become anions, i.e. Cl-. The presence of the negative charge or the presence of two electrons(like :NH3) give them the property to hook a cation to form a new Species, called a complex ion. It can be charged or neutral.

The formation of a complex follows the equilibrium rules. For instance, a complex can be formed between Fe3+ and Cl-, as follows:Fe3+ + Cl- FeCl2+ with a k = c FeCl2+ / (cFe3+ cCl-).The same occurs between Cu2+ + NH3 Cu (NH3)2+.

Total and free concentration 2In the example of the above slide, it is possible to write:CFe = cFe + cFeCl and in the similar way: CCu = cCu + cCu(NH3)

Charges are omitted. C = total concentration and c = the free one orEquilibrium concentration. Sometimes [Cu2+] is used to indicate theFree concentration of the ion copper (II).

Pay attention in all the equilibrium constants and / or in all the thermodynamic expression, the free concentration must be used.

Example: Calculate the free concentration of silver and cyanide in a solution 0.1 M of KAg(CN)2, knowing k = 1 x 1021.As KAg(CN)2 completely dissociates in K+ and Ag(CN)2

-, the equilibrium : Ag+ + 2 CN- Ag(CN)2

-, is to study.It has k = [Ag(CN)2

-] [Ag+]-1 [CN)-]-2 = 1 x 1021 (follows)

Total and free concentration 3 (follows)

The table relative to the species can be built:

Table Ag(CN)2- Ag+ CN-

Initial moles 0.1 --- ---Formed moles --- x 2xUsed moles x --- ----At equilibrium 0.1 -x x 2x

k = [Ag(CN)2-] [Ag+]-1 [CN-]-2 = (0.1-x)/(x) (2x)2= 1 x 1021,

By resolving the above expression, it is obtained [Ag+]= 2.9x10-8M[CN-]= 5.85x10-8 and [Ag(CN)2

-] remain 0.1M, because 2.9x10-8 isnegligible with respect to 0.1M.