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CHAPTER
1
Atoms, Molecules, Mol Concept
and Principles of Volumetric
Analytical Methods
A. BASIC CONCEPTS AND DETAILS
1. Element: Basic (or fundamental) inseparable elemental material (such as Fe, Mg,
etc.).
2. Compound: A pure substance that can be separated from a mixture (For example,
Separation of NaCl from sea water).
3. Dalton’s model of an atom and molecule:
(i) Atom: The smallest fundamental unit of an element. Atoms of a given element
are all alike but differ from atoms of other elements. An atom of an element
has a definite mass. Atoms are indestructible.
(ii) Molecule: A group of atoms capable of independent existence. A compound is
composed of group of atoms of different elements.
4. Avogadro’s hypothesis: Equal volumes of all gases at the same temperature and
pressure contain equal number of molecules.
5. Atomic weight of an element: It is defined relative to the weight of oxygen atom
taken as 16. It is the average weight of its atoms (which may have different masses
due to different isotopes present in it) relative to that of oxygen atom. The gram
atomic weight is the atomic weight of an element in grams and is numerically equal
to its atomic weight in atomic mass units (see. a.m.u. below).
6. Atomic mass unit (at mass): It is based on carbon-12 standard i.e., it assigns a
mass of 12 a.m.u’s to the mass of 6C12 isotope. Thus
12 a.m.u. =
12.000
6.023 × 10
23
g atom–1
The average atomic mass of carbon atom = 12.011 a.m.u.
1
1 a.m.w. =
= 1.66 × 10–24 g
6.023 × 10 23
7. Mol: It is the amount of any substance that contains as many elementary particles
(may be atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12.
The number of elementary particles found by experiment is 6.023 × 1023 (known as
1
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OBJECTIVE PHYSICAL CHEMISTRY
Avogadros’ number). Thus mol is the amount of any substance containing
6.023 × 1023 entities.
8. Molar mass: It is the mass of one mol of any substance. For example, in the case
of carbon-12.
12 a.m.u. =
12.000
6.023 × 10
23
× 6.023 × 10 23
= Molar mass of carbon-12 in grams.
9. Empirical formula: It is the relative number of different types of atoms present
in a given compound. It is obtained from the analytical chemical data on the compound
and the atomic masses of various elements present in the compound.
10. Molecular formula: It gives the exact number of atoms in a given compound.
11. Chemical equation: It is a manifestation of a chemical reaction both qualitatively
and quantitatively. The quantitative aspects of mass and volume relationships between
the reactants and products is known as stoichiometry.
In writing a chemical equation, the following points are important:
(i) It must be balanced correctly to maintain conservation of mass of all species
taking part in the reaction.
(ii) The physical state of all reactants and products, whether liquid, gas or solid
must be indicated. It is especially important in thermochemical equations. A
balanced chemical equation is helpful in calculation of masses of reactants and
products.
12. Law of conservations: It states “when chemical reactions occur, matter is neither
created nor destroyed.”
13. Laws of chemical combination: There are three laws:
(i) Law of definite proportions,
(ii) Law of multiple proportions,
(iii) Law of reciprocal or combining proportions.
From law of definite proportions, one may state that a given compound is always
composed of the same elements combined together in fixed proportions by weight.
Law of multiple proportions states that if two elements combine together in more
than one proportion of weight to form a number of compounds, then the weights of
one of the elements which combine with a fixed weight of the other are in a simple
ratio of whole numbers.
The law of reciprocal proportions states that different weights of various elements
that combine with a certain constant weight of some other element (taken as
standard) are also the weights in which they combine with each other or as a simple
multiple of these weights.
14. pH: In aqueous solutions, the H+ concentration varies over 14 powers of ten (i.e.,
0 to 10–14). A compact presentation of such large variations in [H+] is made possible
by Sorenson who introduced pH concept given by
pH = –log10[H+]
ATOMS, MOLECULES, MOL CONCEPT, PRINCIPLES OF VOLUMETRIC ANALYTICAL METHODS
3
A neutral solution has a pH of 7 while an acid solution has a pH less than 7. A basic
solution has a pH greater than 7.
15. Neutralisation curve: It is a graph showing variation of pH with volume of the
reagent during acid-base titration. In the case of a strong acid vs strong base
titration, the pH changes over a large range (about 7 pH units) while the change
in pH is only about 3 to 4 units in the case of a titration of a weak acid or a weak
base by the other reagent.
16. Indicators in acid-base titrations: Some indicators, their pH range and colour
changes in acidic and basic media are listed below:
Indicator
Colour change
Acid
Alkaline
Useful pH range
Methyl orange
→
red
yellow
3.1–4.4
Methyl red
→
red
yellow
4.2–6.3
Phenol red
→
yellow
red
6.8–8.4
Phenolphthalein
→
colourless
red
8.3–10.0
17. Different volumetric analytical methods: (i) Neutralisation method,
(ii) Precipitation analytical methods, (iii) oxidation – reduction method,
(iv) Complexometric method.
18. Basic terminology used in volumetric methods:
(i) Standard solution: It implies a solution of known concentration. It is often
used as a titrant.
(ii) Equivalent weight of an acid: It is the number of parts by weight of the
acid that contains one part by weight of replaceable hydrogen. The number of
replaceable H atoms in an acid molecule is termed as “basicity”.
Equivalent weight of an acid =
Molecular weight of acid
Basicity
(iii) Equivalent weight of a base: It is the number of parts by weight of the base
that completely neutralises a gram equivalent of acid. Analogous to basicity,
the “acidity” of a base refers to the number of –OH groups in the molecule of
the base.
Equivalent weight of a base =
Molecular weight of base
Acidity
(iv) Equivalent weight of an oxidising or reducing agent: It is equal to the
molecular weight of the reagent divided by the number of electrons required
in the particular reaction.
19. Equation in volumetric analytical calculations: The equation is
V1N1 = V2N2
where V1, N1, V2, N2 are the volumes (V) and normalities of reactants 1 and 2.
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OBJECTIVE PHYSICAL CHEMISTRY
20. Primary and Secondary standard solutions: A primary standard solution is
prepared by directly weighing a known amount of the substance and dissolving it
in a known amount of the solvent. The substanced used as a primary standard must
be obtainable in a pure and stable form, must be non-hygroscopic, must have a high
equivalent weight and must be capable of reacting stoichiometrically with the other
reagent.
A secondary standard solution needs to be standardised against a primary standard.
21. Useful relations in volumetric analysis:
(i) When two substances A and B are titrated against each other, the identity
milliequivalents of A = milliequivalents of B holds.
(ii) In a solution of substanced A, the product
VA × NA = milliequivalents A
The milliequivalents of A may also be written as
milliequivalents of A =
Mgs of A
Equivalent weight of A
22. Methods of expressing the concentration of a solution:
(i) grams per dm3 (or litre) of solution (grams dm–3)
(ii) weight per cent of solute in the given solution: weight of solute is 100 grams
of solution.
(iii) Molarity (C): It is defined as mols of solute in one dm3 (or litre) of solution
(mol dm–3 (or litre–1). Molarity depends on temperature.
(iv) Molality (m): It is defined as mols of solute in one kilogram of solvent
(mol kg–1), molality is independent of temperature.
The following relations involving molarity, molality with the weights WA, WB,
the weights of solvent and solute and “d” the density of solution are worth
noting. MB is the molecular weight of solute.
C=
1000 dWB
M B W A + WB
m=
1000 WB
WA × M B
b
g
FG W × d IJ
HW + W K
C= m
A
A
(v) mol fraction (Xs) of solute =Xs =
B
moles of solute
moles of solute + moles of solvent
=
WB / M B
nB
=
WB / M B + W A / M A
nB + n A
=
WB M A
n
≈ B (when nA >> nB)
W A M B + WB M A
nA
ATOMS, MOLECULES, MOL CONCEPT, PRINCIPLES OF VOLUMETRIC ANALYTICAL METHODS
5
Mole fraction is a dimensionless quantity. The sum of molfractions of solute
and solvent is equal to unity.
(vi) Normality (N) of a solution: It is defined as the equivalents of solute in one
dm3 (or litre) of soution (equivalent dm–3 or litre–1).
23. Oxidation-reduction reactions:
(i) Oxidation: It is a process that involves loss of electrons by a neutral or
charged species.
Example, Fe2+ → Fe3+ + e
It is a process in which the oxidation number of an element increases.
(ii) reduction: It is a process that involves gain of elections by a mental species
by or charged species.
Example, Cu2+ + e → Cu+
It is a process in which the oxidation number of an element decreases.
(iii) Oxidation-reduction reaction: It is a reaction in which some species
undergoes oxidation while other species undergoes reduction (Example: reaction
between KMnO4 and FeSO4 in acid solution)
(iv) Oxidation number: It refers to the number of electrons gained or lost in
producing an anion or cation during a reaction. A cation has a positive oxidation
number whereas an anion has a negative oxidation number.
The following table shows the oxidation numbers of different elements in their
compounds.
Species
Value of oxidation
number
Exclusions
Element
0
—
Alkali metal ion in its compounds
+1
—
Alkaline earth metal ion in its
compounds
+2
—
H atom in its compounds
+1
In hydrides like NaH or CaH2, it
is negative
“O” atom in its compounds
– 2
In compounds like F2O it is +2
“Cl” in its compounds
– 1
Except in its compounds with
other elements like “O” or other
halogen.
In calculating the oxidation number—(i) of an element in a compound, the net
charge on the compound is zero, (ii) of an atom in an ion, the sum of oxidation
numbers of all atoms is equal to the net charge on the ion.
Oxidation number of an atom in a compound or ion may be +ve, −ve, zero or even
fractional. It is distinguished from valency in that the valency of an atom in a
compound is a whole number and does not carry any sign.
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OBJECTIVE PHYSICAL CHEMISTRY
24. Oxidising and reducing agents: Oxidising agent is a compound which increases
the oxidation number of an element of the oxidised species. It is a species getting
reduced.
Reducing agent is a compound which decreases the oxidation number of an element
of the reduced species. It is a species getting oxidised.
25. Balancing of redox equations: Two methods are adopted: (i) ion electron method,
(ii) oxidation state method.
(i) Ion electron method: The various steps in using this method are: (a) Partial
equations are written for oxidation and reduction steps, (b) Oxygen atoms are
balanced by adding the required number of H2O molecules and H+ ions to one side
or the other of the equation, (c) The charge on the partial equations is balanced by
adding required number of electrons, (d) The number of electrons in the two partial
equations are equalised by multiplication with suitable numbers (e) The two equations
are now added and balanced, (f) In reaction in basic media: (i) each excess ‘‘O’’ atom
is balanced by adding one H2O molecule to the same side and two OH− ions to the
other side; (ii) Each excess H; atom is balanced by adding one OH− ion to the same
side and one H2O molecule to the other side.
(ii) Oxidisation state method:
(a) The atoms undergoing change in oxidation state are found out.
(b) Electron balancing is done by adding the required number of electrons to the
appropriate side.
(c) The number of electrons is equalised by using suitable multiplication factor.
(d) The complete redox equation is obtained by adjusting the coefficients on
both sides of the equation.
26. Equivalent weights of some common oxidants and reductants: The table
below contains some well-known oxidising and reducing agents along with their
equivalent weights based on partial ionic equations.
Oxidising agent
1. KMnO4
Partial ionic equation
MnO4– + 8H+ + 5e → Mn2+ + 4H2O
(in acid medium)
2. KMnO4
MnO4– + e → MnO42–
(in strongly basic medium)
3. K2Cr2O7
4. I2
Equivalent weight = Molecular
weight/number of e– in
the reaction
Molecular weight
5
Molecular weight
1
Cr2O72– + 14H+ + 6e → 2Cr3+ + 7H2O
Molecular weight
6
I2 + 2e → 2I–
Molecular weight
2
ATOMS, MOLECULES, MOL CONCEPT, PRINCIPLES OF VOLUMETRIC ANALYTICAL METHODS
BrO3– + 6H+ + 6e → Br– + 3H2O
Molecular weight
6
6. KIO3
IO–3 + 6H+ + 5e → I2 + 3H2O
(in acid medium)
Molecular weight
5
7. KIO3
IO3– + 6H+ + 4e → I2 + 3H2O
(in strong acid medium)
Molecular weight
4
1. H2C2O4·2H2O
C2O42– → 2CO2 + 2e
(acid medium)
Molecular weight
2
2. Na2S2O3·5H2O
2S2O32– → S4O62– + 2e
Molecular weight
2
3. H2O2
2H2O2 → 2H2O + O2
Molecular weight
2
5. KBrO3
Reducing agents
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B. OBJECTIVE TYPE QUESTIONS
(i) The multiple choice questions given below contain one correct answer out of four
choices. Indicate the correct answer.
1. Which of the following statements is not true of Dalton’s theory?
(a) An atom has a definite fixed mass.
(b) Atoms can’t be created or destroyed.
(c) Atoms of all elements are similar and indistinguishable.
(d) A molecule of an element consists of a group of atoms.
2. The atomic mass of an element is
(a) the mass of the most abundant isotope of the element in a.m.u.
(b) the average atomic mass of the naturally occurring isotopes of the elements in
a.m.w.
(c) the weight of one mole of the atoms of the most abundant isotope of the element
in grams.
(d) the weight of one atom of the element in grams.
3. If the atomic mass unit is based on O16 isotope, its value in grams is
16
1
(a)
(b)
6.023 × 10 23
16 × 6.023 × 1023
16 × 2
16
1
×
(c)
(d) 6.023 × 10 23
23
16
6.023 × 10
4. The number of atoms present in 9.6 × 1011 a.m.u. of oxygen is
(a) 8 × 1012
(b) 4 × 1010
(c) 2 × 1015
(d) 6 × 1010
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OBJECTIVE PHYSICAL CHEMISTRY
5. The number of grams of oxygen (O2) gas present in 6 × 1010 of oxygen atoms is
(a) 1.59 × 10–12
(b) 2.12 × 10–10
(c) 8.48 × 10–9
(d) 4.24 × 10–11
6. The weight of water containing 6 × 1010 atoms of oxygen is
(a) 7.16 × 10−13
(b) 1.79 × 10−12
(c) 1.43 × 10−10
(d) 3.58 × 10−11
7. Which of the following statements rigorously defines the ‘‘mol’’ of any substance?
(a) It is the number of atoms in one gram atom of any substance
(b) It refers to the number of molecules in a gram molecule of any substances
(c) It refers to the number of ions in a gram ion of any ionic species.
(d) It refers to the amount of a substance that contains as many elementary particles
(atoms, molecules or ions) as there are atoms in 12 grams of carbon-12.
8. The gram molecular weight of a substance is
(a) the weight of a molecule of the substance in grams.
(b) the sum of the weights of all the atoms is the molecule in grams.
(c) the weight of the substance in grams, numerically equal to its molecular weight
in atomic mass units.
(d) none of the above.
9. The equivalent weight of a substance is defined as
(a) the weight of the substance that can combine with one part by weight of hydrogen.
(b) the weight of the substance that can combine with 16 parts by weight of oxygen.
(c) the weight of the substance that can combine with one mole of chlorine gas.
(d) the weight of the substance that can combine with hydrogen, oxygen or chlorine
in the ratios indicated under (a), (b), or (c).
10. Oxidation of (an element or) a compound by a reagent in modern terminology is
(a) the removal of hydrogen gas from the compound.
(b) the loss of electrons from the oxidising agent (oxidant).
(c) the loss of electrons from the reducing agent (reductant).
(d) the addition of an electronegative element to the compound.
11. Which of the following statements regarding the oxidation number of an atom in a
given species is correct?
(a) It is synonymous with valency.
(b) It cannot have a fractional value.
(c) It may be positive or negative.
(d) It increases in a reduction process.
12. Analysis of a compound gave the following per cent composition for various elements:
K = 37.0, Fe = 13.3, C = 17.0, N = 19.9, H = 1.4 and the rest oxygen. The entire
hydrogen in the compound is combined with oxygen to form water. The molar mass
of the compound is 422 a.m.w. The ratio of its molecular to empirical formula is
(Given atomic masses in a.m.u. as, K = 39.0, Fe = 56.0, C = 12.0, O = 16.0, H = 1.0)
(a) 3
(b) 2
(c) 1
(d) 4
ATOMS, MOLECULES, MOL CONCEPT, PRINCIPLES OF VOLUMETRIC ANALYTICAL METHODS
9
13. A carbohydrate compound contains the three elements carbon, hydrogen and oxygen
in the weight ratio of 8.1.3: 10.6. The molar mass of the compound is 180. The ratio
of molecular to empirical formula is (Given atomic masses in a.m.u. as C = 12.0,
H = 1.0, O = 16.0)
(a) 2
(b) 4
(c) 6
(d) 1
14. 5 grams of sodium carbonate hydrate, on strong heating, loses 3.15 grams of water
leaving anhydrous salt. The number of water molecules in the hydrate is
(a) 8
(b) 7
(c) 4
(d) 10
15. A gaseous compound contains carbon = 12.12%, oxygen = 16.16% and chlorine =
71.72% by weight. 9.90 × 10-4 kgs of the compound occupies a volume of 4.55 × 10–4 m3
at a pressure of 55.42 kPa at 303K. The ratio of molar mass to the empirical formula
weight of the compound is
(a) 1
(b) 6
(c) 4
(d) 2
16. The oxidation number of S in Na2S4O6 is
(a) −2.0
(b) 2.0
(c) −1.5
(d) 2.5
17. The oxidation numbers of S2− and S2−
2 are respectively
(a) −2 and −1
(b) −1 and + 2
(c) −1 and −2
(d) 1 and −2
18. The oxidation numbers of Zn and Al in K2 ZnO2 and NaAlO2 are respectively
(a) –1 and 3
(b) 2 and 3
(c) 1 and –3
(d) 2 and –2.
19. 25.2 gms of ammonium dichromate on heating decomposes according to
(NH4)2 Cr2O7 = Cr2O3 + N2 + H2O
to give 15.2 grams of Cr2O3 and 2.24 × 10–3 m3 of N2 at S.T.P.
The amount of water produced in the reaction (in gms) is (given atomic masses as:
Cr = 52.0, N = 14.0, O = 16.0, H = 1.0)
(a) 3.6
(b) 4.8
(c) 7.2
(d) 5.4
20. 13.2 grams of (NH4)2 SO4 was heated with requisite amount of sodium hydroxide
solution. After removing the ammonia and water formed in the reaction, the remaining
sodium sulphate formed, weighed 14.2 grams. Using the law of conservation of mass,
the amounts of ammonia and water formed in the reaction (in gms) are respectively,
21.
(a) 3.6 and 3.4
(b) 3.4 and 3.6
(c) 7.5 and 3.8
(d) 3.8 and 7.6
Two experiments were carried to prepare calcium carbonate. In one, 0.74 gms of
Ca(OH)2 reacted with 0.44 gms of CO2 to give one gram CaCO3. In a second experiment,
0.555 gms CaCl2 was treated with 0.53 grams of sodium carbonate in solution to give
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OBJECTIVE PHYSICAL CHEMISTRY
0.5 grams CaCO3. Using law of definite proportions, the weight ratios of Ca, C and
O in both experiments to yield CaCO3 are
(Molar masses: Ca(OH)2 = 74, CaCO3 = 100, Na2CO3 = 106, CaCl2 = 111, atomic
masses: Ca = 40, C = 12, O = 16)
(a) 4.0 : 3.33 : 1
(b) 1 : 4 : 3.33
(c) 3.33 : 4 : 1
(d) 4 : 1 : 3.33
22. Nitrogen forms three oxides. In the first oxide, 1.4 grams of nitrogen combines with
1.6 grams of oxygen, in the second one, 0.7 grams of nitrogen combines with
1.6 grams of O2 and in the third one, 2.8 grams of nitrogen combines with 8.0 grams
of oxygen. From law of multiple proportions, the ratios of O2 combining with a fixed
weight of nitrogen in the three oxides are
(a) 2 : 4 : 5
(b) 5 : 4 : 2
(c) 4 : 2 : 5
(d) 2 : 5 : 4
23. Calcium hydride contains 95.24 wt. % of calcium and the rest hydrogen. Hydrogen
chloride contains 97.26 wt. % of chlorine and the rest hydrogen. Calcium and chlorine
combine together to form calcium chloride. From law of reciprocal proportions, the
ratio of calcium to chlorine in CaCl2 is
(Show details of calculation)
(a) 3.1 : 2.0
(b) 7.1 : 4.0
(c) 2.0 : 3.1
(d) 4.0 : 7.1
24. Silicon dioxide contains 46.75 wt. % silicon and carbon dioxide contains 27.27 wt. %
carbon. From law of reciprocal proportions the percent ratio of silicon to carbon in
silicon carbide is
(a) 3 : 7
(b) 4 : 9
(c) 7 : 3
(d) 9 : 4
25. A certain glass contains 11.70 wt. % CaO, 75.34 wt. % SiO2 and the rest as sodium
oxide. The ratio of silicon to sodium atoms in the glass is
(Molar mass of SiO2 = 60.1, Molar mass of Na2O = 62)
(a) 1
(b) 3
(c) 2
(d) 4
26. 5 grams of a sample of sodium chloride containing potassium chloride as impurity
was treated with dilute perchloric acid solution. The resulting precipitate of potassium
chlorate was filtered, dried carefully and weighed 0.465grams. The per cent purity of
NaCl is
(Atomic masses: K = 39.0, Cl = 35.5, O = 16.0, Na = 23.0)
(a) 86.9
(b) 75.00
(c) 95.0
(d) 69.8
27. Ethylene is polymerised to polythylene (C2H2)n using an organic aluminium compound
as catalyst. The polymer was found to contain 0.473 per cent aluminium for every
polymer molecule. The value of ‘‘n’’ in the polymer is
(a) 250
(b) 185
(c) 158
(d) 204