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Chemical substances: Formulas and names

The chemical formula of a substance is a notation that uses atom

symbols with numerical subscripts to express the relative proportions of

atoms of the different elements in the substance.

e.g. Sodium chloride = NaCl; no subscript is written for the symbol, it is

assumed to be 1.

e.g. Calcium chloride = CaCl2

e.g. Aluminum oxide = Al2O3

(1) Molecular substances A molecule is a definite group of atoms that are chemically bonded

together, i.e. tightly bonded by attractive forces.

A molecular formula gives the exact number of different atoms of an element in a molecule (e.g. H2O2, H2O, NH3, CO2 and C2H6O).

A structural formula is a chemical formula that shows how the atoms are bonded to one another in a molecule (e.g. HOH). e.g. Molecular formula of ethanol is C2H6O, while its structural

formula is CH3CH2OH

An empirical formula is the formula of a substance written with the smallest integer (whole number) subscripts. The molecular

formula tells you the precise number of atoms of different elements

in the molecule. The empirical formula tells you the ratio of

numbers of atoms in the compound.

e.g. The empirical formula of hydrogen peroxide (H2O2) is HO.

Molecular models are aids in visualizing the shapes and sizes of molecules.

1- Ball-and-stick type shows bonds and bond angles clearly.

2- Space-filling type gives a more realistic feeling of the space occupied by the atoms.

Polymers represent an important class of molecular substances.

They are very large molecules that are made of a number of smaller

molecules repeatedly linked together covalently. A monomer is a

compound that is used to make a polymer and from which the polymer's

unit arises.

1) Synthetic polymers Dacron (polyester) Nylon (polyamide) Plastic (PVC) Teflon (monomer is CF2CF2).

2) Natural polymers wool and silk (amino acids).

(2) Ionic substances An ion is an electrically charged particle obtained from an atom or

chemically bonded group of atoms by adding or removing

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electrons. The ion may be a cation or an anion, e.g. Na+, Mg

2+, Cl

-,

SO42-

,

Therefore, an ionic compound is a compound composed of cations

and anions, present in the form of a crystal, and whose formula is written

by giving the smallest possible integer number of different ions in the

substance, i.e. formula unit, e.g. NaCl & Fe2 (SO4)3.

(3) Organic compounds It is a class of molecular substances that contain C combined with

other elements, such as H, O and N, e.g. proteins, amino acids,

DNA, enzymes, table sugar and antibiotics.

The simplest organic compounds are hydrocarbons which are compounds that contain only H and C, e.g., C2H2 and benzene.

A functional group is a reactive portion of an organic molecule that undergoes predictable reactions, e.g. OH, -CO-, -COOH, -NH2, -CONH-, -CHO-, -C-O-C, etc.

(4) Chemical nomenclature

A) Ionic compounds i- Monoatomic ions a monoatomic ion is an ion formed from a

single atom.

Cations : Monoatomic cations are named after the element, but for transition elements which have more than one cation, we use

the Stock system of nomenclature, e.g. Fe2+

= iron (II), Fe3+

=

iron (III), Cu+ = copper (I) or the suffix system, e.g. ferrous,

ferric and cuprous.

Anions: stem name of the element + ide, e.g. bromide.

ii- Polyatomic ions a polyatomic ion is an ion consisting of 2 or more atoms chemically bonded together and carrying a net electric

charge, e.g. Hg22+

, NH4+ and CN

-.

Oxoanions / oxyanions: are anions consisting of oxygen, with another element called characteristic or central element.

Those with the greater number of O atoms have the suffix -ate,

while those with the lesser number of O atoms have the suffix -

ite, e.g. SO42-

& SO32-

. For more than 2 oxoanions of a given

characteristic element, e.g. Cl

ClO-

hypochlorite ion ClO3- chlorate

ClO2- chlorite ClO4

- perchlorate

Acid anions : are oxoanions attached to one (mono-) or more (di-) hydrogen atom(s), e.g. HPO4

2- monohydrogen phosphate,

HCO3- hydrogen carbonate (bicarbonate), HSO4

- (hydrogen

sulfate (bisulfate).

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Thiosulfate (S2O32-

): one O atom in the root ion name (SO42-

)

has been replaced by a S atom.

- Name the following: Mg3N2 and Cr2 (SO4)3 - Write formulas for the following compounds: iron (II) phosphate and titanium

(IV) oxide.

B) Binary molecular compounds A binary compound is a compound composed of only 2 elements.

Those which are composed of 2 nonmetals or metalloids are

usually molecular and are named using a prefix system.

The order of elements in the formula of a binary molecular compound is established by convention, i.e. the nonmetal or

metalloid occurring first in the following sequence is written first

in the formula of the compound.

Element : B Si C Sb As P N H Te Se S I Br Cl O F

Group : IIIA IVA VA VIA VIIA

This order lists the elements from the bottom of the group upward,

then you place H between groups VA and VIA and move so that O is just

before F. This order places the nonmetals and metalloids approximately

in order of increasing nonmetallic character, e.g. NF3 and not F3N.

Subscripts are denoted by Greek prefixes: mono, di, tri, tetra, penta, hexa,

hepta, octa, nona, deca.

Examples:

N2O3 dinitrogen trioxide

SF4 sulfur tetrafluoride

ClO2 chlorine dioxide

Cl2O7 dichlorine heptoxide

H2S dihydrogen sulfide (hydrogen sulfide is old & common)

NO nitrogen monoxide (nitric oxide)

1- Name the following compounds: N2O4, P4O6, PCl3, SiC and Cl2O6. 2- Give the formulas of the following compounds: boron trifluoride,

carbon disulfide and sulfur trioxide.

C) Acids and corresponding anions An acid is a molecular compound that yields H ions, H+, and an

anion for each acid molecule when the acid dissolves in H2O.

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An oxyacid or oxoacid is an acid comtaining H, O and another element called the central atom. The names of the oxoacids are

related to the names of the corresponding oxoanions.

Examples:

Acid

Anion suffix Name

HNO3 nitrate nitric acid

HNO2 nitrite nitrous acid

HClO hypochlorite hypochlorous acid

HClO2 chlorite chlorous acid

HClO3 chlorate chloric acid

HClO4 perchlorate perchloric acid

H2SO4 sulfate Sulfuric acid

H3PO4 phosphate Phosphoric acid

Some binary compounds of H and nonmetals yield acid solutions when dissolved in H2O.

e.g.

H2O HCl (g) HCl (aq)

Hydrogen chloride hydrochloric acid

Q Selenium has an oxoacid, H2SeO4, called selenic acid. What is the

formula and name of the corresponding anion ?

A Selenate ion & SeO42-

D) Hydrates A hydrate is a compound that contains water molecules weakly

bound in its crystals. These substances are often obtained by evaporating

an aqueous solution of the compound.

evaporation

aq sol. of copper (II) CuSO4 . 5H2O CuSO4 + 5H2O sulfate (CuSO4) copper (II) sulfate anhydrous copper (II)

pentahydrate (blue) sulfate (white crystals)

Therefore, we can say that hydrates are named from the anhydrous

compound, followed by the word hydrate with a prefix to indicate the no.

of H2O molecules per formula unit if the compound.

e.g. Epsom salt = MgSO4 . 7 H2O: magnesium sulfate heptahydrate

Washing soda has the formula Na2CO3. 10H2O. what is the chemical

name?

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E) Writing and balancing chemical equations A chemical equation is the symbolic representation of a

chemical reaction in terms of chemical formulas.

Ex. 1 : State/phase (4 types) coefficient

2 Na(s) + Cl2 (g) 2 NaCl (s) yield

reactant product

A reactant is a starting substance in a chemical reaction. A product is a

substance that results from the reaction. Labels to indicate the phase are:

(g) = gas, (l) = liquid, (s) = solid, (aq) = aqueous (water) solution. Ex. 2 : decomposition of NaNO3(s) 2 NaNO2 (s) + O2 (g) Pt 2 H2O2 (aq) 2 H2O (l) + O2 (g)

In order to balance an equation, you select coefficients that will make the number of atoms of each element equal on both sides of

the equation. It is preferable to write the coefficients so that they are

the smallest whole numbers possible. The following method is

called balancing by inspection which is essentially a trial-and-error

method.

Ex. Combustion of propane

third first second

C3H8 + 5 O2 3 CO2 + 4 H2O

Rule: Balance first the atoms for elements that occur in only one substance on each side of the equation

1) 4 H3PO3 3 H3 PO4 + PH3 The key lies in O which occurs in only a substance on each side of

the equation. Use the no. of atoms on the left side of the arrow (3) as

the coefficient of the substance containing that element on the right

side, and vice versa.

2) Ca + 2H2O Ca (OH)2 + H2 3) Fe2 (SO4)3 + 6NH3 + 6H2O 2 Fe(OH)3 + 3(NH4)2 SO4

Balance the following equations:

1) Ca3(PO4)2 + H3PO4 Ca(H2PO4)2

2) As2S3 + O2 As2O3 + SO2

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Chapter III

Chemical Calculation and Stoichiometry

(1) Mass and moles of substance Molecular weight (MW) is the sum of the atomic weights of all

the atoms in a molecule of the substance. It is, therefore, the

average mass of a molecule of that substance expressed in amu.

e.g.

MW of H2O = 18.0 amu (2 x 1.0 amu H + 16.0 amu O)

Formula weight (FW) is the sum of the atomic weights of all the atoms in a formula unit of the compound, whether molecular or

not.

e.g.

Formula weight of NaCl, FW = 58.44 amu (22.99 amu Na + 35.45 amu Cl). It is ionic and so strictly speaking the expression

molecular weight has no meaning. On the other hand, the

molecular weight and the formula weight calculated from the

molecular formula of a substance are identical.

Q , Calculate the FW of chloroform, CHCl3, and iron (III) sulfate,

Fe2(SO4)3 to 3 significant figures using a table of atomic weights.

A ,

CHCl3 Fe2(SO4)3

1 x AW C = 12.0 amu 2 x AW Fe = 2 x 55.8 = 111.6 amu

1 x AW H = 1.0 amu 3 x AW S = 3 x 32.1 = 96.3 amu

3 x AW Cl = 106.4 amu 12 x AWO = 12 x 16.00 = 192.0 amu

______________ _____________

119.4 amu 399.9 amu

Rounding to 3 sf 119 4.00 x 102 amu

Mole (mol) is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly

12 g of carbon-12. For example, 1 mole of C2H5OH contains the

same number of C2H5OH molecules as there are C atoms in 12 g of

C-12. The number of atoms in a 12 g sample of C-12 is called

Avogadro's number (NA) being equal to 6.02 x 1023

. Therefore, a

mole of a substance contains Avogadro's number (6.02 x 1023

) of

molecules or formula units.

e.g. 1 :

Na2CO3 2 Na+ + CO3

2-

1 mol 2 mol 1 mol

1 x 6.02 x 1023

2 x 6.02 x 1023

ions 1 x 6.02 x 1023

ions

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e.g. 2 :

1 mol of oxygen atoms contains 6.02 x 1023

oxygen atoms

1 mol of oxygen molecules contains 6.02 x 1023

oxygen molecules,

i.e. 2 x 6.02 x 1023

oxygen atoms

Molar mass is the mass of 1 mole of the substance. C-12 has a molar

mass of exactly 12 g/mol. For all substances, the molar mass in g/mol is

numerically equal to the FW in amu. For example, C2H5OH has a MW =

46.1 amu and a molar mass = 46.1 g/mol

Q , What is the mass in g of a Cl atom and of a HCl molecule ?

A ,

1 mol of Cl atom 35.5 g/mol 6.02 x 1023 atoms ? 1 atom

35.5 g

rn = ________________

= 5.90 x 10-23

g

6.02 x 1023

36.5 g

For 1 molecule of HCl = ___________________

= 6.06 x 10-23

g

6.02 x 1023

mass

No. of moles = ____________________

molar mass

Q , How many grams of ZnI2 are there in 0.0654 mol ?

A ,

mass

No. of mol = ____________________

molar mass

mass

0.0654 = ___________

mass = 20.9 g ZnI2 319

Q , How many molecules are there in a 3.46 g sample of HCl ?

A , 1 mole of HCl molecules = 6.02 x 1023

molecules = 36.5 g

n = 3.46 g

3.46

n = 6.02 x 1023

x _________

= 5.71 x 1022

HCl molecules

36.5

(2) Determining chemical formulas Percentage composition is the mass percentage of each element in

the compound. Therefore, the mass percentage of A is the parts of A per

100 parts of the total, by mass.

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mass of A in the whole

Mass % A = ___________________________________

x 100

mass of the whole

e.g.

Calculate the mass percentages of the elements in HCHO to 3

significant figures. Then calculate how many grams of C are there in

83.5 g of HCHO.

1 mol CH2O 1 mol C, 2 mol H, 1 mol O = 30.0 g Therefore,

12.0 g

% C = ______________

x 100 = 40.0%

30.0 g

2 x 1.01 g

% H = _________________

x 100 = 6.73%

30.0 g

16.0 g

% O = ______________

x 100 = 53.3%

30.0 g

The mass of C in 83.5 g HCHO is = 83.5 x 0.400 = 33.4 g

Elemental analysis % of C, H and O You burn a sample of the compound of known mass and get CO2

and H2O. Every mole of C in the compound ends up as a mole of CO2

and every mole of H ends up as 1/2 mole of H2O. You calculate the mass

% of C and H, and find the mass % of O by subtracting the mass % of C

and H from 100.

e.g.

Acetic acid contains only C, H and O. A 4.24-mg of the acid is

completely burned. It gives 6.21 mg of CO2 and 2.54 mg of H2O. What

is the mass percentage of each element in acetic acid ?

A ,

1 mol CO2 1 mol C 1 mol H2O 2 mol H

44.0 g 12 g 18.0 g 2 x 1.01 g

6.21 mg X 2.54 mg X

6.21 x 12 2.54 x (2 x 1.01)

x = _____________

= 1.69 mg C x = ______________________

= 0.285 mg H

44.0 18.0

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1.69 0.285

% mass C = ________

x 100 = 39.9 % % mass H = _______

x 100 = 6.72 %

4.24 4.24

% mass O = 100- (39.9 + 6.72) = 53.4% Therefore, the % composition of acetic acid is 39.9% C, 6.7% H

and 53.4% O.

The % composition of a compound leads directly to its empirical formula (or simplest formula) which is the formula of a substance

written with the smallest integer (whole number) subscripts. It

merely tells you the ratio of numbers of atoms in the compound.

You can find the empirical formula from the composition of the

compound by converting masses of the elements to moles.

Compounds with different molecular formulas can have the same empirical formula, and such substances will have the same %

composition. For example, acetylene C2H2, and benzene C6H6

have the same empirical formulas and, therefore, they have the

same % composition : 92.3% C and 7.7% H, by mass. To obtain

the molecular formulas from the latter data, molecular weights

should be known.

Q , In the previous example of acetic acid, determine the empirical

formula and the molecular formula (MW 60.0 amu).

A ,

C H O

% composition 39.9 6.7 53.4

AW 12 1 16

Mol 3.33 6.6 3.34

3.33 3.33 3.33

1 : 2 : 1

Then,

The empirical formula of acetic acid is CH2O.

But the molecular formula of a compound is a multiple of its

empirical formula. Therefore, the MW is some multiple of the empirical

formula weight.

Molecular weight = n x empirical formula weight

molecular weight

n = _________________________________

empirical formula weight

where "n" is the number of empirical formula units in the molecule.

n = 60/30 = 2 , the molecular formula is (CH2O)2, or C2H4O2

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e.g. 1 : A compound of N and O is analyzed, and a sample weighing

1.587 g is found to contain 0.483 g N and 1.104 g O. What is the

empirical formula of the compound ?

A ,

N O

Mass % 0.483 g 1.104

AW 14 16

Mol 0.0345 0.06900

0.0345 0.0345

1 : 2 NO2

e.g. : An analysis of sodium dichromate gives the following mass% :

17.5% Na, 39.7% Cr and 42.3% O. What is the empirical formula of the

compound (note that it is ionic and hence has no molecular formula).

A ,

Na Cr O

Mass % 17.5 39.7 42.8

AW 23.0 52.0 16.0

Mol 0.761 0.763 2.68

0.761 0.761 0.761

1 : 1 : 3.52

you have to multiply all by 2

2 : 2 7 Na2Cr2O7

(3) Stoichiometry : Quantitative relations in chemical reactions Stoichiometry is the calculation of the quantities of reactants and

products involved in a chemical reaction.

You may interpret a chemical equation either in terms of number of

molecules (or ions or formula units) or in terms of number of moles,

depending on your needs.

For example, in the Haber process for producing NH3 we summarize

these interpretations as follows :

N2 + 3H2 2NH3 1 molecule 3 molecules 2 molecules (molecular interpr.)

1 mol 3 mol 2 mol (molar interpr.)

28.0 g 3 x 2.02 g 2 x 17.0 g (mass intepr.)

Rule: The number of moles involved in a reaction is proportional to the coefficients in the balanced chemical equation.

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e.g. 1: Suppose that 4.8 mol H2 reacts with N2 to produce NH3. How

many mol of NH3 can you produce ?

A ,

H2 : NH3

3 : 2

4.8 3.2 mol

e.g. 2: How much H2 (in kg) is needed to yield 907 kg of NH3 by the

Haber process ?

A ,

9.07 x 105 g

mol of NH3 = __________________

= 5.34 x 104 mol NH3

17.0

then,

H2 : NH3

3 : 2

8.01 x 104 mol 5.34 x 10

4

and,

mass (g)

8.01 x 104 =

______________

2.02

mass = 1.62 x 105 g H2, or 162 kg H2

e.g. 3 : Hematite, Fe2O3, is an important ore of Fe. The free metal is

obtained by reacting hematite with CO in a blast furnace. How

many grams of Fe can be produced from 1 kg Fe2O3 ?

A ,

Fe2O3(s) + 3 CO(g) 2Fe(s) + 3 CO2 (g) given 1.00 x 10

3 g

160 g mass

mol 12.5 = __________

55.8

mass = 698 g Fe

e.g. 4 : Cl2 may be prepared in the laboratory by heating HCl with MnO2.

How many grams of HCl react with 5.00 g of MnO2 ?

A ,

4 HCl (aq.) + MnO2(s) 2 H2O (l) + MnCl2 (aq) + Cl2 (g) given

5.00 g = 0.0575374 mol

86.9

mass (g) = 0.23 mol

36.5

mass = 8.40g HCl

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Limiting reactant; theoretical and % yields The limiting reactant (or limiting reagent) is the reactant that is

entirely consumed when a reaction goes to completion. A reactant

that is not completely consumed is referred to as an excess

reactant. Once one of the reactants is used up, the reaction stops,

i.e. the moles of product are always determined by the starting

moles of limiting reactant.

e.g. 1 : In the burning of H2 in O2, suppose you put 1 mol H2 and 1 mol

O2 into a reaction vessel. How many moles of H2O will be

produced?

A ,

2H2(g) + O2(g) 2H2O (g) given 1 mol 1 mol

This would give : 1 mol 2 mol

H2O H2O

excess reagent

least it is the limiting reagent Therefore, by the time 1 mol H2O is produced, all of the H2 is used

up and the reaction stops

e.g. 2: Zn metal reacts with HCl to give H2. If 0.30 mol Zn is added to

HCl containing 0.52 mol HCl, how many mol of H2 are produced ?

A,

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) given 0.30 mol 0.52 mol

0.30 mol H2 0.26 mol H2

excess reactant limiting reactant 0.26 mol H2

e.g. 3: In the production of acetic acid, O2 gas is bubbled into CH3CHO

containing Mn (Ac)2 catalyst under P at 60oC. If 20.0 g CH3CHO and

10.0 g O2 were put into a reaction vessel, how many grams of HAc can be

produced, and of the excess reactant remain after the reaction is

complete?

A,

2 CH3CHO (l) + O2 (g) 2 CH3COOH (l) given mass 20.0 g 10.0 g

MW 44.1 32

mol 0.454 mol 0.3125 mol

0.454 mol HAc 0.625 mol HAc

limiting reactant excess reactant 0.454 mol HAc

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mass (g)

0.454 mol HAc = ______________

60.1

mass = 27.3 g CH3COOH The amount of the excess reactant O2 left over based on the

conversion of the moles of HAc to grams of O2 (the quantity of O2

needed to produce this amount of HAc).

O2 : HAc

1 2

0.227 mol 0.454 mol

mass (g)

0.227 = _______________ 32

The mass of O2 consumed is = 7.26 g The mass of O2 remaining is = 10.0 7.26 = 2.7 g O2

The theoretical yield of product is the maximum amount of product that can be obtained by a reaction from given amounts of

reactants. It is the amount that you calculate from the

stoichiometry based on the limiting reactant. In practice, the actual

yield of a product may be much less for several reasons :

1- Some product may be lost during the process of separating it from the final reaction mixture.

2- There may be other competing reactions that occur simultaneously with the reactant on which the theoretical yield is based.

3- Many reactions appear to stop before they reach completion, i.e. they give mixtures of reactants and products.

It is important to know the actual yield from a reaction in order to

make economic decision about a preparation method. The reactants for a

given method may be too costly per kg, but if the actual yield is very low,

the final cost can be very high.

The percentage yield of product is the actual yield (experimentally

determined) expressed as a percentage of the theoretical yield (calculated)

actual yield

% yield = ________________________

x 100

theoretical yield

For the e.g. of HAc

23.8

% yield = _________

x 100 = 87.2%

27.3

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Chapter IV

Chemical reactions (I) Ions in aqueous solution

Ionic theory of solutions by Arrhenius states that certain substances produce freely moving ions when they dissolve in H2O, and these

ions conduct an electric current in an aqueous solution.

1- Pure water consists of molecules each of which is electrically neutral. Since each molecule carries no net electric charge, it

carries no overall electric charge when it moves. Thus, pure water

is a nonconductor of electricity.

2- If you dissolve NaCl in water, the Na+ and Cl- held strongly in the crystal lattice go into solution as freely moving ions. Suppose you

dip electric wires that are connected to the poles of a battery into a

solution of NaCl, the ions in solution begin to move, and these

moving charges form the electric current in solution (note that in a

wire, it is moving electrons that constitute the electric current).

Therefore, an aqueous solution of ions is electrically conducting.

Electrolytes and nonelectrolytes An electrolyte is a substance that dissolves in water to give an

electrically conducting solution. A nonelectrolyte is a substance

that dissolves in water to give a nonconducting or very poorly

conducting solution.

Electrolytes may be ionic, e.g. NaCl, and/or molecular as some

molecular substances dissolve in water to form ions, e.g. HCl gas that

dissolves to give HCl (aq), which in turn produces H+ & Cl

- in aqueous

solution (the solution of H+ & Cl

- ions called hydrochloric acid).

Nonelectrolytes are molecular substances, e.g. sucrose and methanol, the

solution process occurs because molecules of the substance mix with

molecules of H2O. Molecules are electrically neutral and cannot carry

out electric current so the solution is electrically non-conducting.

When electrolytes dissolve in water they produce ions, but to

varying extent A strong electrolyte is an electrolyte that exists in solution almost

entirely as ions. Most ionic solids that dissolve in H2O do so by going

into the solution almost completely as ions, so they are strong

electrolytes, e.g. NaCl whose dissolution in water is as follows :

H2O

NaCl (s) Na+ (aq) + Cl- (aq)

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A weak electrolyte is an electrolyte that dissolves in H2O to give a

relatively small percentage of ions. They are generally molecular

substances, e.g. NH3. Pure ammonia is a gas that readily dissolves in

water and goes into solution as ammonia molecules NH3 (aq). The latter

react with water to form NH4+ & OH

- ions.

NH3 (aq) + H2O (l) NH4+ (aq) + OH

- (aq)

Both NH4+ & OH

- ions react with each other to give back NH3 &

H2O molecules.

NH4+ (aq) + OH

- (aq) NH3 (aq) + H2O (l)

Both reactions, the original (forward) and its reverse, occur

constantly and simultaneously.

NH3 (aq) + H2O (l) NH4+ (aq) + OH

- (aq)

From these reactions, just a small % of NH3 molecules ( 3%) have reacted at any given moment to form ions. Thus, NH3 is a weak

electrolyte.

Most soluble molecular substances are either nonelectrolytes or

weak electrolytes except HCl (g) that dissolves in H2O to produce H+ &

Cl- ions.

HCl (aq) H+ (aq) + Cl- (aq)

Since HCl dissolves to give almost entirely ions, HCl (or

hydrochloric acid) is a strong electrolyte.

Solubility rules a- All Li+, Na+, K+ & NH4

+ salts are soluble.

b- All acetates and nitrates are soluble. c- All chlorides, bromides and iodides are soluble

Except Ag+, Hg22+

and Pb2+

salts together with HgBr2 & HgI2.

d- All sulfates are soluble

Except Ca2+, Sr2+, Ba2+& Pb2+ together with Ag2SO4 & Hg2SO4. e- All carbonates, phosphates and sulfides are insoluble

Except group IA & NH4+

f- All hydroxides are insoluble

Except group IA & Ca(OH)2, Sr(OH)2 & Ba(OH)2

Summary: Compounds that dissolve in water are soluble; those that

dissolve little or not at all are insoluble. Soluble substances are either

electrolytes or nonelectrolytes. Electrolytes can be strong or weak.

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Almost all soluble ionic substances are strong electrolytes. Soluble

molecular substances usually are nonelectrolytes or weak electrolytes.

NH3 is a molecular substance that is a weak electrolyte. Few molecular

substances such as HCl dissolve almost completely as ions and are

therefore strong electrolytes.

Molecular and ionic equations (i) Molecular equation is a chemical equation in which the reactants and

products are written as if they were molecular substances, even though

they may actually exist in solutions as ions. It closely describes what

you actually do in the laboratory or in an industrial process.

Ca(OH)2 (aq) + Na2CO3 (aq) CaCO3 (s) + 2NaOH (aq)

(ii) Complete ionic equation is a chemical equation in which strong

electrolytes (such as soluble ionic compounds) are written as separate

ions in the solution. The purpose of such an equation is to represent

each substance by its predominant form in the reaction mixture.

If the substance is a soluble ionic compound, it dissolves as individual ions (so it is a strong electrolyte) and you represent the

compound as separate ions.

If the substance is a weak electrolyte, it is present in solution primarily as molecules, so you represent it by its molecular

formula.

If the substance is an insoluble ionic compound, you represent it by the formula of the compound, not by the formulas of the separate

ions in solution.

Then,

Ca2+

(aq) + 2OH- (aq) + 2Na

+ (aq) + CO3

2- (aq) CaCO3(s) + 2Na

+ (aq) + 2OH

- (aq)

(iii) Net ionic equation is an ionic equation from which spectator ions

have been cancelled. A spectator ion is an ion in an ionic equation that

does not take part in the reaction. In the above example, Na+ and CO3

2-

ions appear on both sides of the equation. This means that nothing

happens to these ions as the reaction occurs, so they are cancelled and the

resulting equation is :

Ca2+

(aq) + CO32-

(aq) CaCO3 (s)

If you react Ca (NO3)2 and K2CO3 the net ionic equation will be the

same as in the case of reacting Ca(OH)2 and Na2CO3. Thus, the value of

the net ionic equation is its generality. For example seawater contains

Ca2+

& CO32-

ions from various sources. Whatever the sources of these

-17-

ions, you expect them to react to form a CaCO3 ppt. In seawater, this ppt

results in sediments of CaCO3, which eventually form limestone.

e.g.

Write a net ionic equation for each of the following molecular

equations

(1) 2 HClO4 (aq) + Ca(OH)2 (aq) Ca (ClO4)2 (aq) + 2 H2O (l)

strong electrolyte soluble ionic compound

(2) CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O(l)

molecular substance and weak electrolyte.

A,

You will need to convert the molecular equation to the complete

ionic equation, and then cancel spectator ions to obtain the net ionic

equation. For each ionic compound in the reaction use the solubility rules

to determine if the compound will be soluble (in the solution as ions) or

insoluble (present as an undissolved solid). An ionic compound should

have (aq) after its formula if it is soluble or (s) if it is insoluble.

(1) 2H+(aq) + 2ClO4-(aq) + Ca

2+(aq) + 2OH

-(aq) Ca2+(aq) + 2ClO4

-(aq) + 2H2O (l)

strong electrolyte soluble ionic compound non-or very

strong electrolyte weak electrolyte

H+ (aq) + OH- (aq) H2O (l)

(2) CH3COOH (aq) + Na+ (aq) + OH

- (aq) Na+ (aq) + CH3COO

- (aq) + H2O (l)

weak electrolyte soluble ionic compound strong electrolytes

CH3COOH (aq) + OH- (aq) CH3COO

- (aq) + H2O (l)

(II) Types of chemical reactions 1- Precipitation reactions you mix solutions of 2 ionic substances to

form a solid ionic substance (a precipitate).

2- Acid-base reactions involve the transfer of a proton (H+) between reactants.

3- Oxidation-reduction reactions involve the transfer of electrons between reactants.

A- Precipitation reactions A precipitation reaction occurs in aqueous solution because one

product is insoluble. A precipitate is an insoluble solid compound

formed during a chemical reaction in solution.

You can predict whether a precipitation reaction will occur by writing it as a molecular equation and then it will have the form of

an exchange reaction. Reference is then made to solubility rules

followed by writing both complete and net ionic equations for the

reaction.

-18-

An exchange (or metathesis) reaction is a reaction between compounds that, when written as a molecular equation, appears to

involve the exchange of parts between the 2 reactants. In a

precipitation reaction, the anions exchange between the 2 cations.

e.g.

Reaction between MgCl2 and AgNO3

a) Write a balanced molecular equation

MgCl2 + 2AgNO3 2 AgCl + Mg (NO3)2 b) Refer to solubility rules and append the appropriate phase labels

MgCl2 (aq) + 2 AgNO3 (aq) 2 AgCl (s) + Mg (NO3)2 (aq) c) Write complete ionic equation Mg

2+(aq) + 2Cl

-(aq) + 2Ag

+(aq) + 2NO3

-(aq) 2 AgCl(s) + Mg2+(aq) + 2NO3

-(aq)

d) Write net ionic equation

Ag+ (aq) + Cl

- (aq) AgCl (s)

N.B. :

If the reactants MgCl2 & AgNO3 are added in correct amounts and

the white AgCl solid is filtered, the solution that passes through (the

filtrate) will contain Mg(NO3)2, which you could obtain by evaporation of

the H2O.

Q ,

For each of the following, decide whether a precipitation occurs. If

it does, write the balanced molecular equation, and then the net ionic

equation. If no reaction occurs, write the compounds followed by an

arrow and then NR (no reaction).

a) aq. solutions of sodium chloride and iron (II) nitrate are mixed.

b) aq. solutions of aluminum sulfate and sodium hydroxide are mixed.

A ,

a) NaCl (aq) + Fe (NO3)2 (aq) NR

b) Al3+

(aq) + 3 OH- (aq) Al (OH)3 (s)

B- Acid-base reactions Acids have a sour taste while bases have a bitter taste and a soapy

feel. They cause color changes in certain dyes called acid-base

indicators which are dyes used to distinguish between acidic and

basic solutions by means of the color changes they undergo in

these solutions. Such dyes are common in natural materials, e.g.

indicator acid base

litmus red blue

phenolphthalein colorless pink

bromothymol blue yellow blue

Acids and bases are defined according to different theories.

-19-

1) Arrhenius theory

An acid is a substance that produces H+ when it dissolves in H2O

while a base is a substance that produces hydroxide ions, OH-.

e.g.

H2O

HNO3 (aq) H+ (aq) + NO3

- (aq)

H2O

NaOH (s) Na+ (aq) + OH- (aq)

NH3 (aq) + H2O (l) NH4+ (aq) + OH

- (aq)

The limitation of this theory is that it tends to single out the OH-

ion as the source of base character, when other ions or molecules can play

a similar role (e.g. Na2CO3).

2) Bronsted-Lowry theory

They consider acid-base reactions as proton-transfer reactions. An

acid is the species (molecule or ion) that donates a H+ to another species

in a proton-transfer reaction. A base is the species (molecule or ion) that

accepts a proton in a proton-transfer reaction.

e.g. 1 :

In the reaction of NH3 with H2O, the latter is the acid and the

former is the base because of H+

donation and acceptance, respectively.

NH3 (aq) + H2O (l) NH4+ (aq) + OH

- (aq)

base acid

e.g. 2 :

In the dissolution of nitric acid in water

HNO3 (aq) + H2O (l) NO3- (aq) + H3O

+ (aq)

acid base hydronium ion

Bear in mind that the H+ (aq) (hydrogen ion) and the H3O

+ (aq)

(hydronium ion) represent precisely the same physical ion.

Acids and bases strength Acids and bases are classified as strong or weak, depending on

whether they are strong or weak electrolytes.

A strong acid is an acid that ionizes completely in water, it is a

strong electrolyte, e.g. HCl (aq) and HNO3 (aq). A weak acid is an acid

that only partly ionizes in water; it is a weak electrolyte, e.g HCN (aq)

-20-

and HF (aq). These molecules react with H2O to produce a small % of

ions in solution leaving the majority of the acid molecules unreacted.

* HCl (aq) + H2O (l) H3O+ (aq) + Cl

- (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3

- (aq)

* HCN (aq) H2O (l) H3O+ (aq) + CN

- (aq)

HF (aq) + H2O (l) H3O+ (aq) + F

- (aq)

A strong base is a base that is present in aqueous solution entirely

as ions, one of which is OH-, it is a strong electrolyte, e.g. NaOH or

generally the hydroxides of groups IA & IIA elements except Be (OH)2.

A weak base is a base that is only partly ionized in H2O; it is a weak

electrolyte, e.g. ammonia.

H2O

NaOH (s) Na+ (aq) + OH- (aq) NH3 (aq) + H2O (l) NH4

+ (aq) + OH

- (aq)

When you write an ionic equation, you represent strong acids and

bases by the ions they form and weak acids and bases by the formulas of

the compounds.

Q Identify each of the following compounds as a strong or weak acid or base

LiOH - CH3COOH HBr HNO2 A, s.b. w.a. s.a. w.a.

Neutralization reactions A neutralization reaction is a reaction of an acid and a base that

results in an ionic compound (salt) and possibly water. Note that

most ionic compounds other than hydroxides and oxides are salts.

The salt formed consists of cations obtained from the base and

anions obtained from the acid.

We write molecular, complete ionic and net ionic equations

* 2 HCl (aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2H2O (l) s.a. s.b.

2 H+(aq) + 2Cl-(aq) + Ca2+(aq) + 2OH-(aq) Ca2+(aq) + 2Cl-(aq) + 2H2O (l)

and

H+

(aq) + OH- (aq) H2O (l)

* HCN (aq) + KOH (aq) KCN (aq) + H2O (l) their

HCN (aq) + OH- (aq) CN- (aq) + H2O (l)

w.a.

-21-

Although H2O is one of the products in most neutralization reactions, the reaction of an acid with the base NH3 provides a prominent

exception.

* H2SO4 (aq) + 2 NH3 (aq) (NH4)2 SO4 (aq) acid base salt

then

H+ (aq) + NH3 (aq) NH4

+ (aq)

Q ,

Write the molecular equation and then the net ionic equation for

the neutralization of nitrous acid, HNO2, by sodium hydroxide,

both in aqueous solution.

A ,

* molecular equation :

HNO2 (aq) + NaOH (aq) NaNO2 (aq) + H2O (l) w.a. s.b.

* complete ionic equation :

HNO2(aq) + Na+(aq) + OH

-(aq) Na+(aq) + NO2

-(aq) + H2O(l)

* net ionic equation :

HNO2 (aq) + OH- (aq) NO2

- (aq) + H2O (l)

Acids may be monoprotic (only one acidic H atom per acid molecule) e.g. HCl & HNO3, or polyprotic (yielding 2 or more acidic

hydrogens per molecule) e.g. H3PO4 is a triprotic acid.

H3PO4 + NaOH NaH2PO4 + H2O

H3PO4 + 2NaOH Na2HPO4 + 2H2O

H3PO4 + 3NaOH Na3PO4 + 3H2O Salts such as NaH2PO4 and Na2HPO4 that have acidic hydrogen

atoms and can undergo neutralization with bases are called acid

salts.

Acid base reactions with gas formation Certain salts, notably CO3

2-, SO3

2- and S

2- react with acids to form a

gaseous product. The resulting reaction is considered an exchange, or

metathesis, reaction.

Na2CO3 + 2HCl 2NaCl + H2O + CO2

Na2SO3 + 2HCl 2NaCl + H2O + SO2

Na2S + H2SO4 Na2SO4 + H2S

Q , Write the molecular and the net ionic equations for the reaction of

ZnS & HCl

A , ZnS (s) + 2H+ (aq) Zn2+ (aq) + H2S (g)

-22-

C- Oxidation-reduction reactions These are reactions involving a transfer of electrons from 1 species

to another. If you dip an iron nail into a blue solution of CuSO4 it

will become coated with a reddish-brown tinge of metallic copper.

Fe (s) + CuSO4 (aq) FeSO4 (aq) + Cu(s)

Fe (s) + Cu2+ (aq) Fe2+ (aq) + Cu(s) Iron was oxidized (lost electrons) and copper ions were reduced

(gained electrons).

The oxidation number (or oxidation state) of an atom in a substance is the actual charge of the atom if it exists as a

monoatomic ion, or a hypothetical charge assigned to the atom in

the substance by simple rules. An oxidation-reduction reaction (or

redox reaction) is one in which one or more atoms change

oxidation number, implying that there has been a transfer of

electrons, i.e. it is a reaction in which electrons are transferred

between species or in which atoms change oxidation number.

e.g. The combustion of Ca metal in O2 gas

2 Ca (s) + O2 (g) 2 CaO (s)

Oxidation-number rules In molecular substances, we use these rules to give the approximate

charges on the atoms. For example, in SO2 oxygen atoms tend to attract

electrons, pulling them from other atoms (S in the case of SO2). As a

result, an oxygen atom in O2 takes a ve charge relative to the S atom. The magnitude of the charge on an oxygen atom in a molecule is not a

full -2 charge as in the O2-

ion, but it is given an oxidation number of -2.

For compounds and ions, the sum of oxidation numbers of the

atoms is zero. This rule follows from the interpretation of oxidation

numbers as "hypothetical" charges on the atoms.

1- Because any compound is electrically neutral, the sum of the

charges on its atoms must be zero.

2- The sum of the oxidation number (hypothetical charges) of the

atoms in a polyatomic ion equals the charge on the ion.

3- Oxygen is -2 in most of its compounds except H2O2 where the

oxidation number of O = -1

4- Hydrogen is +1 in most of its compounds except in compounds

with metals where the oxidation number of H = -1, e.g. NaH, CaH2 e.g.

SO2 S + (-2 x 2) = 0 S = +4

HClO4 +1 + Cl (-2 x 4) = 0 Cl = +7

ClO3- Cl + (3 x -2) = -1 Cl = +5

-23-

Describing oxidation-reduction reactions In the reaction of Fe metal with CuSO4 solution

Fe (s) + Cu2+

(aq) Fe2+ (aq) + Cu (s) 0 +2 +2 0

reducing oxidizing

agent agent

This equation/reaction is the resultant of 2 half-reactions. A half-

reaction is one of 2 parts of an oxidation-reduction reaction, one part of

which involves a loss of electrons (or increase in oxidation number) and

the other a gain of electrons (or decrease in oxidation number).

Fe (s) Fe2+ (aq) + 2 electron (electrons lost by Fe)

Cu2+

(aq) + 2 electron Cu (s) (electrons gained by Cu2+) hence,

Oxidation is the half-reaction in which there's a loss of electrons

by a species (or an increase of oxidation number of an atom). Reduction

is the half-reaction in which there's a gain of electrons by a species (or a

decrease in the oxidation number of an atom).

An oxidizing agent is a species that oxidizes another species; it is

itself reduced, while a reducing agent is a species that reduces another

species; it is itself oxidized.

Examples of oxidation-reduction reactions a) Combination reaction it is a reaction in which 2 substances

combine to form a third substance. Note that not all combination

reactions are oxidation-reduction reactions.

e.g.

2 Na (s) + Cl2 (g) 2 NaCl (s) redox

2 Sb (s) + 3Cl2 2 SbCl3 redox

CaO (s) + SO2 (g) CaSO3 (s) not redox

b) Decomposition reaction it is a reaction in which a single compound reacts to give 2 or more substances. Also, not all

decomposition reactions are oxidation-reduction reactions. Often

these reactions occur when the temperature is raised.

e.g. 2HgO (s) 2 Hg (l) + O2 (g) redox

2KClO3 (s) 2KCl (s) + 3O2 (g) redox

MnO2

-24-

CaCO3 (s) CaO (s) + CO2 (g) not redox

c) Displacement reaction (or single-replacement reaction) it is a reaction in which an element reacts with a compound displacing an

element from it.

Cu (s) + 2 AgNO3 (aq) Cu (NO3)2 (aq) + 2Ag (s) Strip greenish-blue

i.e. Cu displaces Ag in AgNO3

Cu (s) + 2Ag+ (aq) Cu2+ (aq) + 2Ag (s)

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)

Zn (s) + 2 H+ (aq) Zn2+ (aq) + H2 (g)

These reactions obey to the activity series of the elements, being a

list of the elements in decreasing order of their ease of losing

electrons during reactions in aqueous solution. The metals listed at the

top are the strongest reducing agents (they lose electrons easily); those at

the bottom, the weakest. Moreover, a free element reacts with the

monoatomic ion of another element if the free element is above the other

element in the activity series.

Li

React vigorously with water and

acids to give H2

K

Na

Ca

Mg

React with acids to give H2

Al

Zn

Fe

Co

Ni

Pb

H2

Cu Do not react with acids to give H2

Hg

Ag

Au

d) Combustion reaction it is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame.

The products include one or more oxides, i.e. oxygen changes

oxidation number from 0 to -2.

-25-

e.g.

1- Organic compounds burn in O2 or air to yield CO2 & H2O

2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10H2O (l) 2- Many metals burn in air, too, such as Fe wool

4Fe (s) + 3O2 (g) 2 Fe2O3 (s) The rusting of iron wool is a similar reaction, although slower.

Balancing oxidation-reduction equations a) Half-reaction method (ion-electron method)

It consists of first separating the equation into 2 half-reactions, one

for oxidation and the other for reduction. You balance each half-reaction,

then combine them to obtain a balanced oxidation-reduction reaction.

e.g.

Zn (s) + Ag+ (aq) Zn2+ (aq) + Ag (s ) not balanced

0 +1 +2 0

Balance the charge in each equation by adding electrons to the

more +ve side to create balanced half-reactions. This is followed by

multiplying each half-reaction by a factor (integer) so that when we add

them together, the electrons cancel.

Thus,

1 x (Zn Zn2+ + 2 electron) oxidation half-reaction

2 x (Ag+ + electron Ag) reduction half-reaction

_____________________________________________

Zn + 2Ag+ Zn2+ + 2 Ag

b) Electron-change method

Zn + 2 Ag+ Zn2+ + 2 Ag

0 +1 +2 0

Q , Apply the half-reaction method to balance this equation

Mg (s) + N2 (g) Mg3 N2 (s) A, Here, a molecular compound, N2, is undergoing reduction. When a

species undergoing reduction or oxidation is a molecule, write the

formula of the molecule in the half-reaction (do not split it up).

Thus,

3 x (Mg Mg2+ + 2 electron) oxidation half-reaction

1 x (N2 + 6 electron 2N3-

) reduction half-reaction ________________________________________________

3 Mg + N2 (g) 3Mg2+

+ 2N3-

or 3 Mg (s) + N2 (g) Mg3N2 (s)

-26-

(III) Working with solutions When we dissolve a substance in a liquid, we call the substance the

solute and the liquid the solvent. Consider NH3 solution. NH3 gas

dissolves readily in H2O, and aqueous NH3 solutions are often used

in the lab. In such solutions, NH3 gas is the solute and H2O is the

solvent.

The general term concentration refers to the quantity of solute in a standard quantity of solution. Qualitatively, we say that a solution

is dilute when the solute concentration is low and concentrated

when the solute concentration is high.

a) Molarity (M) is the moles of solute dissolved in 1 L (1 dm3) of

solution.

moles of solute

M = ____________________________

liters of solution

The advantage of molarity as a concentration unit is that the

amount of solute is related to the volume of solution. Rather than having

to weigh out a specified mass of substance, you can instead measure out a

definite volume of solution of the substance, which is usually easier.

e.g. 1 :

A sample of NaNO3 weighing 0.38 g is placed in a 50.0-mL

volumetric flask. The flask is then filled with H2O to the mark on the

neck, dissolving the solid. What is the molarity of the resulting solution ?

A ,

mol of solute 0.38/85

Molarity = _______________________

= ______________

= 0.089 M

vol. of solution 0.05

N.B.: Although very dilute solutions are possible, there is a limit as to

how concentrated solutions can be. Therefore, any answer that

leads to solution concentrations that are in excess of 20 M should

be suspect.

e.g. 2 :

An experiment calls for the addition of 0.184 g of NaOH in

aqueous solution. How many mL of 0.150 M NaOH should be added ?

A ,

mol of solute

Molarity = _______________________

vol of solution

0.150 = (0.184/40) / V V = 3.07 x 10-2 L = 30.7 mL

-27-

Diluting solutions Commercially available aqueous NH3 (28.0% NH3) is 14.8 M NH3.

In order to prepare a solution that is 1.00 M NH3, you need to dilute the

concentrated solution with a definite quantity of H2O. Note that the

number of moles of solute in the container does not change when

performing the dilution, only the concentration changes.

Moles of solute = molarity x volume in liters = M V

The relationship is: Initial Mi Vi = Mf Vf Final

Note: You can use any volume units but both Vi and Vf must be in the

same unit.

e.g.

You are given a solution of 14.8 M NH3. How many milliliters of

this solution do you require to give 100.0 mL of 1.00 M NH3 when

diluted?

A ,

14.8 x VmL = 1.00 x 100.0

V = 6.75 mL

b) Normality (N) is the number of equivalents of solute dissolved in 1 L

of solution.

equivalents of solute

N = ____________________________

liters of solution

weight

equivalent = __________________________

equivalent weight

The equivalent weight for

M. wt

1- acids and bases = ______________________________________ no. of replaceable H or OH

M. wt

2- redox = _____________________________________ no. of transferred electrons

e.g. 1 : Calculate the N of a solution containing 2.45 g of H2SO4 in 2.00 L

of solution.

-28-

A ,

wt. 2.45

N = ______________

= _________________

= 0.025

Eq wt V 98/2 x 2

e.g. 2 : How many grams of H2SO4 are contained in 3.00 L of 0.500 N

solution ?

A ,

wt.

0.5 = ________________

wt = 73.5 g 98/2 x 3

c) Molality (m) it is the number of moles of solute in 1 kg of solvent.

moles of solute

m = ________________________

kg of solvent

Volume changes with changes in T oC whereas weight is

independent of temperature; hence the molarity of a solution changes

with T oC while the molality does not. In dilute solutions, molar and

molal solutions are very nearly identical, but in more concentrated

solutions, wide differences may be expected.

e.g.

What is the molality of a solution in which 49 g of H2SO4 (M wt =

98) is dissolved in 250 g of H2O ?

A ,

49/98

m = ______________

= 2.0

0.25

d) Mole fraction

moles of solute

mole fraction of solute = ______________________________________________

moles of solute + moles of solvent

and

moles of solvent

mole fraction of solvent = ______________________________________________

moles of solute + moles of solvent

The sum of both should be 1

e.g. 1 : What are the mole fractions of solute and solvent in a solution

prepared by dissolving 98 g H2SO4 in 162 g H2O ?

A ,

-29-

mole fraction of H2SO4 = 1/10 = 0.1 and H2O = 0.9

e.g. 2 : What is the mole fraction of H2SO4 in a 7.0-molar solution of

H2SO4 which has a density of 1.39 g/mL ?

A ,

1 L of this solution weighs 1390 g

7 moles of H2SO4 weigh 7 x 98 g 686 g

weight of H2O in 1 L of solution 704 g moles of H2O = 704/18 = 39 moles

mole fraction of H2SO4 = 7/39 + 7 = 0.15

4) Quantitative analysis Analytical chemistry deals with the determination of composition

of materials, i.e. the analysis of materials such as air, H2O, food, hair,

body fluids, pharmaceutical preparations, and so forth. The analysis of

materials is divided into :

1- Qualitative analysis involves the identification of substances or species present in a material, e.g. you might determine that a

sample of H2O contains Pb2+

ion.

2- Quantitative analysis involves the determination of the amount of a substance or species present in a material, e.g. you

might determine that the amount of Pb2+

ion in the sample of H2O

is 0.067 mg/L.

1) Gravimetric analysis

It is a type of quantitative analysis in which the amount of a species

in a material is determined by converting the species to a product that can

be isolated completely and weighed. Precipitation reactions are frequently

used in gravimetric analysis. You determine the amount of an ionic

species by precipitating it from solution. The formed precipitate is

filtered from the solution, dried and weighed. The advantages of a

gravimetric analysis are its simplicity (at least in theory) and its accuracy.

The chief disadvantage is that it requires time-consuming work.

e.g.

Pb2+

in a sample of drinking H2O is determined gravimetrically in

the form of white crystalline PbSO4 by the addition of Na2SO4 solution.

The ppt is then dried and weighed.

SO42-

+ Pb2+

PbSO4 e.g.: A 1.000-L sample of polluted H2O was analyzed for Pb

2+ ion, by

adding an excess of Na2SO4 to it. The mass of PbSO4 that precipitated

-30-

was 229.8 mg. What is the mass of Pb in a liter of the H2O ? Give the

answer as mg of Pb per liter of solution.

A ,

1st method Obtain the mass % of Pb in PbSO4 by dividing the molar

mass of Pb by that of PbSO4, then multiply by the weight of PbSO4

207.2 g/mol

% Pb = _______________________

x 100 = 68.32 %

303.3 g/mol

amount of Pb in sample = 229.8 mg x 0.6832 = 157.0 mg Pb

The H2O sample contains 157.0 mg Pb/L 2

nd method

Pb2+

+ SO42-

PbSO4 weight ? 229.8 mg

MW 207.2 303.3

mol 0.758 0.758

wt (mg)

0.758 = ___________

207.2

2) Volumetric analysis

It is a method of analysis based on titration. Titration is a procedure for determining the amount of substance A by adding a

carefully measured volume of a solution with known concentration

of B until the reaction of A and B is just complete. An indicator

should be present to detect the end point (point at which the

reaction is complete). The indicator is a substance that undergoes a

color change when a reaction approaches completion.

e.g.

Consider the reaction of H2SO4 with NaOH. How many mL of

0.250 M NaOH titrant must be added to react completely with 35.0 mL of

0.175 M H2SO4 ?

A ,

H2SO4 (aq) + 2 NaOH (aq) Na2SO4 (aq) + 2H2O (l) 35.0 mL 49.0 mL

0.175 M 0.250 M

m mol = 6.125 m mol = 12.25

Whenever you perform a titration calculation using molar concentration you should take into account the stoichiometry of the

-31-

reaction. Do not be tempted to apply the dilution equation to solve

the problem because it fails to take into account the stoichiometry

of the reaction.

e.g.

A flask contains a solution with an unknown amount of HCl. This

solution is titrated with 0.207 M NaOH. It takes 4.47 mL NaOH to

complete the reaction. What is the mass of HCl ?

A , HCl + NaOH NaCl + H2O 33.77 mg 4.47 mL

mw 36.5 0.207 M

0.925 mmol 0.925 mmol