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1st year chemistry notes
Chapter No. 1
Chapter No. 1
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Basic Concepts
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BASIC CONCEPTS
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Atom:
The term atom is derived from the Greek word “atoms” meaning indivisible.
The smallest particle of an element which may or may not have independent
existence is called an atom.
For example ,the atoms of He,Ne and A r exist independently while the
atoms of hydrogen ,nitrogen and oxygen do not have independent existence .An
atom is composed of more than 100 subatomic particles such as electron, proton ,
neutron , hyperons , neutrino, antineutrino, etc .However ,electron ,proton and
neutron are the fundamental particles of atoms. The atoms are the smallest particle
of an element which can take part in a chemical reaction.
Evidence of Atoms:
Atoms are extremely small. It is not possible actually to see them even with
a powerful optical microscope However ,the direct evidence for their existence
comes from an electron microscope. It uses beams of electrons instead of visible
light. The wavelength of electron is much shorter than that of visible light. With
optical microscopes, a clear and accurate image of an object that is smaller than the
wavelength of visible light cannot be obtained. It can only measure the size of an
object up to or above 500 nm. However, objects of the size of an atom be observed
in an electron microscope. Like light, the characteristics of an electron beam
change when it passes through or reflects off atoms in the thin layers of solids. The
electron beam takes a picture of atoms layers which can be magnified about 15
millions of times. An electron microscope photograph of a piece of graphite is
shown in the figure. The bright bands in the image are layers of carbon atoms.
(Picture)
Fig Electron microscope photograph of graphite
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X-ray work has shown that the diameters of atoms are of the order 2x10 -10 m which
is0.2 nm. Masses of atoms range from 10-27 to 10-25 kg. We can get an idea about
the small size of an atom from the fact that a full stop may have two million atoms
present in it. They are often expressed in atomic mass units (a.m.u).
amu= 1.661x 10-24 g=1.661x10-27 kg
Molecule:
“The smallest particle of a pure substance which can exist independently is
called a molecule.”
A molecule may contain one or more atoms. The number of atoms present in
a single molecule of an element is called atomicity. The molecules of elements can
be monatomic, diatomic,Triatomic and polyatomic etc, if they contain one, two
and three atoms respectively. A molecule of an element consists of one or more
similar atoms . For example , He, Ar, O2,CL2, O3, P4, S8. A molecule of a
compound consists of two or more different atoms. For example, HCI, H 2S, CO2,
NH3, H2SO4,C12H22 O11.
The sizes of molecules are bigger than atoms. Their sizes depend upon the
number of atoms present in molecules and their shapes. A molecule having a very
high molar mass is called a macromolecule. For example, hemoglobin is a
macromolecule which is found in blood. Hemoglobin carries oxygen from lungs to
all parts of the body. Each molecule of hemoglobin is made up of nearly 10,000
atoms. Hemoglobin molecule is 68,000 times heavier than a hydrogen atom.
Ions:
“The species which carry either positive or negative charge are called ions.”
An ion may be a charged atom, group of atoms or molecules. Ions are
formed by the gain or loss of electrons by neutral atoms or molecules. The number
of protons in the nucleus never changes in the formation of ions.
Examples:
Na+, Ca2+ , NH , Cl-, O2- , NO ,CO -, N ,CO+,CH
Cation
“An ion that has a positive charge is called a “Cation”.
They are formed when an atom of an element loses one or more electrons.
A A + + eThe charge on a cation may be +1, +2 or +3 . The charge present on an ion
depends upon the number of electrons lost by an atom. Energy is always required
to form positive ions. The Formation of the positive ion is an endothermic process.
The most common positive ions are formed by the metal atoms. The positive ions
having group atoms are less common.
Examples:
Anion
Na + , K + ,Ca2+, Mg 2+ , A13+ , Sn 4+ , NH , H3 O +
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“An ion that has a negative charge is called an anion.”
They are formed when a neutral atom of an element gains one or more
electrons.
B+ e - B –
Usually, energy is liberated when an electron is added to the isolated neutral atom.
The formation of a uninegetive ion is an exothermic process. The most common
negative ions are formed by the non-metal atoms.
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Examples:
F- ,CI - ,Br - ,I - , O 2- ,OH -, CO -,SO -, PO ,MnO , Cr2 O - , etc
Molecular Ion:“An ion which is formed when a molecule loses or gains an electron is
called a molecular ion.”
Positive molecular ions are formed by removing electrons from neutral molecules.
Negative molecular ions are formed when extra electrons are attached to neutral
molecules. Cationic molecular ions are more abundant than anionic ions.
Molecular ions can be generated by passing a beam of high-energy electrons ,
alpha particles or X-rays through molecules in gaseous state. The break down of
molecular ions obtained from the natural products can give important information
about their structure.
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Examples:
N , CO + , CH , N , etc
Positive ions of molecules can be generated by bombarding the gas, or vapour of
the substance with electrons. The molecular ions produced often break into
fragments, giving several different kinds of positive ions.
Thus the original molecule can give rise to a number of ions .
Relative Atomic mass:“The mass of an atom of an element as compared to the mass of an atom of
carbon-12 is called relative atomic mass.”
An atom is an extremely small particle . The mass of an individual atom is
extremely small in quantity . It is not possible to weigh individual atoms or even
small number of atoms directly . We do not have any balance to weigh such an
extremely small mass. That is why for atoms, the unit of mass used is the atomic
mass unit (amu) and not measurement I.e, grams , kilograms ,pounds and so on.
Atomic Mass Unit (amu):
“A mass unit equal to exactly one-twelfth ( th)the mass of a carbon -12 atom is
called atomic mass unit.”
For atoms , the atomic mass unit (amu) is used to express the relative atomic
because its mass of 12 units has been determined very accurately by using mass
spectrometer . The relative atomic mass of C is 12,000 amu and relative atomic
mass of H is 1.0078 amu .
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H
N
O
Na
Mg
1amu=1.66x10 -24 g.
Relative Atomic Element
mass
1.0078 amu
A1
14.0067 amu
S
15.9994 amu
C1
22.9897 amu
Cu
24.3050 amu
U
Relative
Atomic
mass
26.9815 amu
32.066 amu
35.453 amu
63.546 amu
238.0289 amu
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Remember that:
Element
Table: Relative atomic masses of some elements
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Isotopes {Greek Isotopes means → same place}
The atoms of the same element having the same atomic number but different
atomic mass are called isotopes.”
Isotopes of the same element have the same number of protons and electrons but
different number of neutrons in their nuclei. They are different kind of atoms of the
same element. Isotopes of the same element have the same chemical properties but
slightly different physical properties; they have the same position in the periodic
table because they have the same atomic number. For example, hydrogen has three
isotopes
H, Hand H called proteome, deuterium and tritium. Carbon has three isotopes
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written as C , C, C and expressed as C-12, C-13 and C-14 . Chlorine has two ,
oxygen has three nickel has five , calcium has six ,palladium has six, cadmium has
nine and tin has eleven isotopes .
Relative Abundance of Isotopes:
The isotopes of the elements have their own natural abundance. The
properties of a particular element mostly correspond to the most abundant isotopes
of that element.
The relative abundance of the isotopes of elements can be determined by mass
spectrometry .At present above 280 different isotopes of elements occur in nature.
They include 40 radioactive isotopes. About 300 unstable radioactive isotopes have
been produced artificially.
Table: Natural abundance of some common Isotopes
Element
Hydrogen
Carbon
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Isotopes
1
H, 2 H
12
C, 13 C
Abundance(%)
99.985, 0.015
98.893, 1.107
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14
Nitrogen
N , 15 N
99.634, 0.366
16
17
18
Oxygen
O, O, O
99759. 0.037, 0.204
32
33
34
36
Sulphur
S , S , S, S
95.0,0.76,4.22.0.014
35
37
Chlorine
C1, C1
75.53, 24.47
19
18
Bromine
Br , Br
50.54,49.49
Odd- Even Relationships:
1.
The elements with even atomic number usually have larger
number of stable isotopes.
2.
The elements with odd atomic number almost never possess more
than two stable isotopes. For example, the elements
As,
I and
Au have only single isotopes. These elements are known as monoisotopic elements.
The isotopes whose mass number is multiples of four are most
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3.
F,
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abundant. For example, O, Mg, Si, Ca and Fe. They form
nearly 50% of the earth‟s crust.
4.
The isotopes with even mass number and even atomic number are
more abundant and more stable. Out of 280 naturally occurring isotopes,
154 isotopes belong to this type.
Remember that: most of the elements with even atomic number have even mass
number whereas most of the elements with odd atomic number have odd mass
number.
Determination of relative atomic masses of isotopes by mass spectrometry:
Mass Spectrometer:
“An instrument which is used to measure the relative atomic masses and relative
abundances of different isotopes present in a sample of an elements is called a
mass spectrometer.”
It measures the mass to charge ratio of atoms in the form of positive ions.
Types of mass spectrometers
1.
Aston’s mass spectrograph
The first mass spectrometer known as mass spectrograph was invented by
Aston in 1919. It was designed to identify the isotopes of an element on
the basis of their atomic masses. The mass spectrograph operates on the
same principle as a mass spectrometer. The main difference is that a mass
spectrograph uses a photographic plate to detect ions instead of an
electrical device.
2.
Dumpster’s mass Spectrometer
It was designed for the identification of elements which were available in
solid state.
Mass spectrometry
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“The use of mass spectrometer to identify different isotopes of an element
by measuring their masses is called mass spectrometry.”
The method involves analysis of the path of a charged particle in a
magnetic field .
Principle of mass spectrometry
In this technique, a substance is first vaporized. It is then converted to
gaseous positive ions with the help of high energy electrons. The gaseous positive
ions are separately on the basis of their mass to charge (m/e) ratios. The results are
recorded alacrity in the form of peaks. The relative heights of the peaks give the
relative isotopes abundances.
Working of mass spectrometer
The solid substance under examination for the separation of isotopes is
converted into vapors. Under a very low pressure 10 -6 to 10-7 torr, these vapors are
allowed to enter the ionization chamber .
In ionization chamber, the vapors are bombarded with fast moving
electrons from an electron gun .The atoms present in the form of vapours collide
with electrons. The force of collision knocked out electrons from atoms. Usually,
one electron is removed form an atom. The atoms turn into positive ions. These
positive ions have different masses depending upon the nature of the isotopes
present in them. The positive ion of each isotope has its own m/e value.
When a potential difference (E) of 500-2000 volts is applied between
perforated accelerating plates, then these positive ions are strongly attracted
towards the negative plate. In this way the ions are accelerated.
These ions are then allowed to pass through a strong magnetic field of
strength (H), which will separate them on the basis of their
values. On entering
the magnetic field the ions begin to move in a circular path. The path they take
depends on the mass to charge
ratios. The ions of definite
value will move
in the form of groups one after the other and fall on the on the electrometer. The
electrometer is also called an ion collector. The electrometer develops the electric
current. The mathematical relationships for
is:
=
Where H is the strength of magnetic field, E is the strength of electrical field , r is
the radius of circular path .
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If E is increased, by keeping H constant then radius will increase and
positive ion of a particular
value will fall at a different palace as compared to
the first place. This can also be done by changing the magnetic field, Each ion sets
up a minute electrical current. The strength of the current thus measured gives the
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relative abundance of ions of ions of a definite
value.
Similarly the ions of other isotopes having different masses are made to fall
on the collector and the current strength is measured. The current strength in each
case gives the relative abundance of each of the isotopes. The same experiment is
performed with C-12 isotopes and the current strength is compared. This
comparison allows us to measure the exact mass number of the isotope. The
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following figure shows the separation of isotopes of Ne. Smaller the
value of
an isotope, the smaller the radius of curvature produces b0 the magnetic field
according to above equation.
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(Picture)
Fig: A simple Mass Spectrometer
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In modern Spectrometers, each ion hits a detector; the ionic current is
amplified and is fed to the recorder. The recorder makes a graph showing the
relative abundance of isotopes plotted against the mass number. A computer
plotted graph for the isotopes of neon is shown in the following figure.
(picture)
Fig : (Computer plotted graph for the isotopes of neon)
The separation of isotopes can be done by methods based on their
properties. Some important methods are: gaseous diffusion, thermal diffusion,
distillation, ultracentrifuge, electromagnetic separation and laser separation.
Fractional Atomic mass:
Atomic masses of elements are not exact numbers. Almost all elements
have fractional values of atomic masses. This is because the atomic mass of an
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Fractional abundance =Percent abundance x
Or Percent abundance = Fractional abundance x 100
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3.
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element is an average mass based on the number of isotopes of the element and
their natural abundance. Natural abundances of atoms are given as atomic
percentages. The mass contributed by each isotope is equal to fractional
abundance multiplied by the isotopic mass. The average or fractional atomic mass
for the element is obtained by taking the sum of the masses contributed by each
isotope.
In general.
1.
Fractional atomic mass of an element =
(fractional abundance)(
Isotopic mass) .
2.
By the symbol sigma ,
means “take the sum of the quantities
.
20
21Ne
22
90.92x
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Ne
0.9092x20=18.1840
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Example 1:
A sample of neon is found to consist of
the
percentage of 90.92%,0.26% and 8.82% respectively . Calculate the fractional
atomic mass of neon.
Solution: The mass contribution for neon isotopes are:
Isotope
Fractional abundance
Isotopic mass
Mass
contribution
=0.9092
21
20
0.0026x21=0.0546
Ne
22
0.0882x22=1.9404
Average or fractional atomic mass of neon =20.179
=20.18amu : Answer
Hence the fractional atomic mass of neon is 20.18 amu. Remember that no
individual neon atom in the ordinary isotopic mixture has mass of 20.18amu
.However Alternatively, the problem may by solved by applying the formula:
Fractional atomic mass
=
(fractional abundance)(isotopic mass)
=(fractional abundance of 20 Ne )
(isotopic mass of
20
21
Ne)+(fractional abundance of Ne ) (isotopic mass of
21
Ne)+(fractional abundance of 22Ne )(isotopic mass of
22
Ne).
=(0.9092)(20)+(0.0026)(21+(0.0882)(22)
=18.1840+0.0546+1.9404
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=20.179
= 20.18 amu Answer
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Analysis of a compound _Empirical and molecular Formulas
Both the empirical and molecular for a compound are determined from the
percentage compositions of the compound and molecular mass of the compound
obtained experimentally . The percentage of an element in a compound is the
number of grams of the element present in 100 grams of the compound.
1.
Percentage composition of a compound whose chemical
formula is not known.
When a new compound is prepared all the elements present in the
compound are first identified by qualitative analysis. After that, the mass of
each element in a given mass of the compound is determined by quantitative
analysis. From this data, the percentage of each element in the compound is
obtained by dividing the mass of the element present in the compound by the
total mass ot the compound and when multiplying to 100.
%of an element =
x100
Once the percentage composition is determined experimentally the empirical
formula can be calculated . The molecular mass of the compound is determined by
experimental methods. From empirical formula and molecular mass , the
molecular formula for the compound is determined.
2.
Percentage Composition of a Compound whose chemical
formula is known.
The percentage composition of a compound can be determined
theoretically,that is , without doing an experiment if we know the chemical
formula of the compound .The relation which can be used for this purpose is:
%of an element =
x100
Remember that the percentage composition of a pure compound does not change.
Example2: 8.657 g of a compound were decomposed into its elements and gave
5.217 g of carbon, 0.962 of hydrogen, 2.478 g of oxygen . Calculate the percentage
composition of the compound under study.
Solution:
Given:
Mass of compound = 8.657 g
Mass of carbon
=5.217 g
Mass of hydrogen =0.962 g
Mass of oxygen =2.478 g
Formula used:
(i)
% of carbon=
=
=60.26% Answer
(ii)
% of hydrogen=
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=
=11.11% Answer
(iii)
%of oxygen =
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=
= 28.62% Answer
Hence in 100 grams of the compound ,there are 60.26 grams of carbon, 11.11
grams of hydrogen and 28.62 grams of oxygen.
Empirical Formula
“ A chemical formula that gives the smallest whole number ratio of atoms
of each elements present in a compound is called an empirical formula .”
For example, in an empirical formula of a compound , A x By, there are x
atoms of element A and y atoms of element B. The empirical formula can be
determined from the percentage composition of the compound or from the
experimentally determined mass relationships of elements that make up the
compound.
Calculation of Empiriacal Formula
Empirical formula of a compound can be calculated by using the following
steps:
1.
Find the percentage composition of the compound.
2.
Find the number of gram-atoms of each element .For this purpose divide
the percentage of each element by its atoms mass.
3.
Find the atomic ratio of each element. To get this, divide the number of
gram-atoms (Moles) of each element by the smallest number of gram-atoms
(moles).
4.
Make the atomic ratio a simple whole number atomic ratio of not so
multiply it with a suitable number.
5.
Write the empirical formula having various atoms present in the above
ratio.
Example3: Ascorbic acid (vitamin C) contains 40.92% carbon, 4.58% hydrogen
and 54.5% oxygen by mass. What is the empirical formula of the ascorbic acid?
Solution:
Calculation of empirical formula:
On writing various steps in tabular from, we have
Element %
age
Atomic
mass
C
40.92 12.0
Whole
number
ratio
1x3=3
H
4.58
1.33x3=4
1.008
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No of gramatoms
Atomic
ratio
1st year chemistry notes
O
54.5
16.0
1x3=3
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Empirical Formula = C3 H4 O3 Answer
Empirical Formula From Combustion analysis
The empirical formula of organic compounds which only contain carbon,
hydrogen and oxygen can be determined by combustion, the two products of
combustion will be CO2 and H2O.These products of combustion are separately
collected and their masses are determined.
Combustion Analysis
A weighed sample of the organic compound is placed in the combustion
tube fitted in a furnace. An excess of pure oxygen is supplied to burn the
compound. The carbon in the compound is converted to CO 2 and hydrogen to H2O
vapors. These gases are passed through two pre-weighted absorbent tubes. One of
the tubes contain Mf(CIO4)2 which absorbs water and the other contains 50% KOH
which absorbs CO2. The increase in mass of potassium hydroxide tube gives the
mass of CO2. From these masses of CO2 and H2O and the mass of the organic
compound. the percentages of carbon and hydrogen in the compound can be
calculated by using the following formulas:
% of C=
%pf H=
The percentage of oxygen is obtained by the method of difference
% of oxygen = 100-(%pf carbon +%of hydrogen)
(Picture)
Fig Combustion Analysis
Example 4: A sample of liquid consisting of carbon, hydrogen, and oxygen was
subjected to combustion analysis .0.5439 g of the compound gave 1.039g of H 2O.
Determine the empirical formula of the compound.
Solution:
(i) Calculations of percentage composition:
Mass of organic compound
=0.5439 g
Mass of CO2
=1.039g
Mass of H2O
0.6369g
% of C=
=
% of H =
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=
% age of O= 100-(52.10+13.11)=34.79%
(ii)
Calculation of empirical formula:
On writing the various steps in a tabular form, we have,
H
13.11 1.008
O
34.79 16.0
atoms
ratio
formula
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52.10 12.0
Empirical
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C
Atomic
C2 H6 O
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mass
No of gram
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Element %age Atomic
Molecular Formula
Empirical Formula = C2 H6 O
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“A chemical formula of a substance that shows the actual number of atoms
of different elements present in the molecule is called a molecular formula”
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The molecular formula of a compound can only be determined if the
empirical formula and the molecular mass of the compound are known.
Examples: H2 O2 (hydrogen peroxide) , C6 H6 (benzene ) , C6 H12 O6 ( glucose ).
The empirical formulas of hydrogen peroxide, benzene and glucose are
HO, CH and CH2 O respectively. Some of the examples of the compounds having
the same empirical and molecular formulas are: H2O, CO 2 , NH3 , CH 4 and C12 H22
O11.
The empirical formula and molecular formula for a covalent compound are
related in this way:
Molecular formula =n x (Empirical formula)
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The value of „n‟ must be a whole number. Actually the value of “n” is the
ratio of molecular mass and empirical formula mass of a substance.
n=
When „n‟ is unity, the empirical formula becomes the molecular formula.
Example 4: The combustion analysis of an organic compound shows it to contain
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65.44% carbon , 5.50% hydrogen and 29.06% oxygen . What is the empirical
formula of the compound? If the molecular mass of this compound is 110.15
Solution:
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.Calculate the molecular formula of the compound.
(i)
Calculation of empirical formula:
C
65.44 12.0
H
5.50
O
29.06 16.0
Empirical
atoms
ratio
formula
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1.008
Atomic
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mass
No of gram
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Element %age Atomic
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On writing the various steps in a tabular form , we have ,
(ii)
C2 H6 O
Calculation of molecular formula :
Empirical formula mass = 36+3.024+16=55.04
Molecular mass=110.15
N==
Molecular formula
= n x (empirical formula)
= 2 ( C3H3O)
=C6 H6 O2
Concept of mole
Gram atom (mole)
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“The atomic mass of an element expressed in grams is called a gram atom.”
It is also known as a gram mole or a mole of element .
Number of gram atoms (moles) of am element =
Example :
=1.008g
1 gram atom of carbon (C)
=12.000g
1 gram atom of uranium (U)
=238.0g
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1 gram atom of hydrogen (H)
1 gram atom magnesium (Mg)
=24.000g
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It means that one gram atoms of different elements have different mass in
them . It also shown that one gram atom of magnesium is twice as heavy as an
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atom of carbon
Gram molecule (Gram mole or mole)
“The molecular mass of a substance expressed in grams is called a gram
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molecule.”
Examples :
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No. of gram molecules (moles) of a molecular substance =
1 gram molecule of oxygen (O2)
= 32g
1 gram molecule of hydrogen (H2)
=2.016g
1 gram molecule of water (H2O)
=18.0g
1 gram molecule of sulphuric acid (H2 SO 4)=98.0g
1 gram molecule of sucrose (C12 H22 O 11)
=342.0g
It means that one gram molecules of different molecular substances have different
masses.
Gram-formula (gram – mole or mole)
“The formula mass of an ionic compound expressed in grams is called a
gram formula of the ionic compound”
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Since ionic compounds do not exist in molecular form , therefore , the sum
of atomic masses individual ions gives the formula mass.
No .of gram –formula (moles ) of a substance =
=58.5g
1 gram-formula of KOH
=56.0g
1 gram-formula of Na2 CO3 =106g
1 gram-formula of Ag NO3
=170g
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Gram-Ion (Mole)
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1 gram-formula of Na C1
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Examples:
“ The atomic mass , molecular mass formula mass or ionic mass of the
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substance expressed in grams is called a mole.”
Number of moles=
Examples 6: Calculate the gram atoms (moles)in
0.1g of sodium
(b)
0.1 kg of silicon
(a)
Given: Mass of sodium
=0.1g
Atomic mass of sodium
=23g mole -1
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Solution:
(b)
=0.1g
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(a)
No of gram atoms (moles ) of Na
=
=4.3x10-1 mol
Given: Mass of silicon
=0.1kg
Atomic mass of silicon
=28.086 g mol-1
=0.1x1000=100 g
No of gram atoms (moles)of silicon =
= 3.56 mol
Example 7: Calculate the mass of 10-3 moles of Mg SO 4.
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1st year chemistry notes
Solutions:
= 10-3 mol
Given: No of moles of MgSO4
=24+32+64=120g mol -1
Formula mass of MgSO4
Formula Used:
Mass of substance
=No of moles of substance x Molar
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mass
= 10-3 mol x 120 g mol -1
Mass of Mg SO 4
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= 0.12g
Avogadro ,s Number (Avogadro Constant), NA)
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“The number of atoms, molecules and ions present in one gram – atom ,
one gram-molecule and one gram –ion respectively is called Avogadro ,s number
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Avogadro, s number is 6.02x1023.It is a constant .One mole of any
Examples:
1 mole of hydrogen (H)
1 mole of water (H2O)
of H2O
=1.008g of H
=23g of Na
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1 mole of sodium(Na)
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substance always contains 6.02x10 23 Particles.
=18g of H2O
1 mole of glucose (C6 H12 O6)=180g of C6 H12 O6
H12 O6
1 mole of SO
-
1 mole of NO
=6.02x1023 atoms of H
=6.02x1023 atoms of Na
=6.02x1023 molecules
= 6.02x1023molecules of C6
-
=6.02x1023 ions of SO
=62g of NO
=6.02x1023 ions of NO
= 96 of SO
-
One mole of different compounds has different masses but the same
number of particles .
Important Relationships
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1st year chemistry notes
The following are some useful relationships between the amounts of
substances mass and the number of particles present in them .
1.
No of atoms of an element =
2.
No of molecules of an compound =
3.
No of ions in an ionic specie
4.
Number of particles
om
=Number of moles x Avogadro
Mass of atoms
6.
=
Mass of molecules
=
e.c
number
5.
=
fre
Examples 8: How many molecules of water are there in 10.0 g ice ? Also
calculate the number of atoms of hydrogen and oxygen separately , the total
line
number of atoms and the covalent bonds present in the sample.
Solution: (i) Calculation for the number of molecules of water
Mass of water (ice)
= 10.0g
llon
Molar mass of H2 O =2+16=18 g mol-1
No of water molecules = ?
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Number of molecules of H2 O = x NA
=
(ii)
3.34x1023 molecules
Calculation for the number of atoms of hydrogen and oxygen and
total number of atoms:
No of water molecules
Now
= 3.34x1023
1 Molecule of H2 O contains H atoms =2atom
3.34x1023 molecules of H2O contains H atoms
=2x3.34x1023 atoms of
H
=6.68x1023 atoms of H
Now,
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1 Molecules of H2 O contains O atoms =1.atom
1st year chemistry notes
3.34x1023 Molecules of H2O contains O atoms
=1x3.34x1023 atoms
of O
=3.34x1023 atoms of
O
Total number of atoms
e.c
(iii)
om
+3.34x1023
=6.68x 1023
=(6.68+3.34) x1023
10.02x1023 atoms
Calculation for number of covalent bonds:
=2
fre
I Molecule of H2O contains the number of covalent bonds
3.34x1023 molecules of H2O contains, the number of covalent bonds
line
=2x3.34x1023
=6.68x1023
Examples 9: 10.0grams of H3PO4 have been dissolved in excess of water to
llon
dissociate it completely into ions. Calculate.
(a) Number of molecules in 10.0g of H3 PO4
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(b) Number of positive and negative ions in case of complete dissociation
in water.
(c) Masses of individual ions.
(d) Number of positive and negative charges dispersed in the solution.
Solution:
(a)
Calculation for the number of molecules in H3PO4:
Mass of H3PO4
=10g
Molar mass of H3PO4 = 3+31+64=98g mol-1
No . of molecules of H3PO4 =xNA
=
=6.14x1022 molecules
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1st year chemistry notes
(b)
Calculation for the number of positive and negative ions in
H3PO4 :
H3PO4
3H+ +PO
-
Now,1 molecule of H3PO4 contains positive H+ ions =3
=3x6.14x1022+ve H+
om
6.14x1023 molecule of contains negative PO - ions
ions
e.c
ions
=1.842x1023+ve H+
Now, 1 molecule of H3 PO4 contains negative PO
-
ions=1
=6.14x1022 –ve PO
Calculation for the masses of individual ions:
No, of H+ ions
=1.842x1023 ions
Ionic mass of H+
=1.0008 g mol-1
NA
=6.02x1023 ions mol-1
Mass of H+ ions
=
=
=0.308 g
-
ions
ions
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(c)
fre
6.14x1023 molecule of contains negative PO - ions =1x6.14x1022-ve PO
No of PO - ions
ww
w.a
Ionic mass of PO - ion
NA
=6.14x1022 ions
=31+64=95g mol -1
=6.02x1023 ions mol-1
Mass of PO - =
(d)
Calculation for the number of positive and negative charges
dispersed in the solution:
1 molecule of H3 PO4 gives positive charges in solution
=3
6.14x1022 molecule of H3 PO4 gives positive charges in solution =3x6.14x1023
=1.842x10 23 +ve
charges
Since the solution is always electrically neutral, therefore, number of
positive and negative charges in solution is always equal
Thus in the solution:
No. of positive charges
=No of negative
charges
Hence, the number of negative charges in the solution
= 1.842x1023
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1st year chemistry notes
(i)
Calculation for the number of moles of an ideal gas at
fre
Solution:
STP:
e.c
om
Molar Volume
“The volume , 22.414 dm3 occupied by one mole of an ideal gas at STP is
called molar volume”.
With the help of this information, we can convert the mass of a gas at STP
into its volume and vice versa, Hence.
1.
1 mole of a gas at STP
=22.414 dm3
2.6.02x1023 molecules of a gas at STP
=22.414 dm3
3.
22.414 dm3 of a gas at STP
=1 Mole
3
It should be remember that 22.414 dm of two gases has a different mass but the
same number of molecules. The reason is that the masses and the sizes of the
molecules do not affect the volumes.
Example 10: A well known ideal gas is enclosed in a container having volume 500
cm3 at STP. Its mass comes out to be 0.72 g .What is the molar mass of this gas.
ww
w.a
llon
line
Volume of ideal gas at STP =500 cm3
=0.5dm3
Now, 22.414 dm3 of ideal gas at STP
=
3
0.5dm of ideal gas at STP
=0.0223moles
(ii)
Calculation for the molar mass of the gas:
Mass of gas
=0.72g
Number of moles of gas
=0.0223 moles
Molar mass of gas
=?
Molar mass of gas
=
=
=32.28 g mol -1
Stoichiometry:
“ The study of the quantities relationships between reactants and products
in a balanced chemical equation is called Stoichiometry.”
It is based on the chemical equation and on the relationship between mass
and moles.
Stoichiometry Amount
“The amount of any reactant or product as given by the balanced chemical
equation is called stoichiometric amount.”
Assumptions
All Stoichiometry calculations are based on the following three
assumptions:
1.
Reactants are completely converted into products.
2.
No side reaction accurse.
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1st year chemistry notes
While doing Stoichiometry calculations, the law of conservation
of mass and the law of definite proportions are obeyed.
Types of Stoichiometric Relationships
The various types is useful in determining an unknown mass of reactant or
product from the given mass of one substance in a chemical reaction.
2.
Mass-mole relationship or mole-mass relationship
Such relationship is useful in determining the number of moles of a reactant
or product from the given mass of one substance and vice-versa
Number of moles=
Mole-mass relation:
om
3.
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w.a
llon
line
fre
e.c
Remember that m is mass and MM is molar mass
3.
Mass volume relationship
Such relationship is useful in determining the volume of a gas from the
given mass of another substance and vice-versa . This relationship allows us to
calculate the volume of any number of moles of a gas at STP.
Mole-volume relation:
Number of moles=
Example 11: Calculate the number of grams of K3 PO4 and water produced when
14g of KOH are reacted with excess of H2SO4 . Also, calculate the number of
molecules of water produced.
Solution:
(i)
Calculation for the number of grams of K2SO4:
Mass of KOH =14 g
Molar mass of KOH =39+16+1=56g mol-1
No . of moles KOH =
0.25mol
Equation:
2KOH(eq) + H2SO4(aq)
2moles
Now,
(ii)
2KOH (aq)
+ H2SO4(aq) K2SO4(aq) +2H2O(1)
2moles of KOH
=1 mole of K2 SO4
0.25 mole of KOH
=
=
0.125 moles of K2SO4
Molar mass of K2SO4
=78+32+64=174g Mol-1
Mass of K2SO4 Produced
=No of moles x molar mass
Mass of K2SO4 Produced
=0.125molx 174 g mol-1
Mass of K2SO4 Produced
=21.75g
Calculation for the number of molecules of water:
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1st year chemistry notes
0.25mol
Equation:
2KOH(eq) + H2SO4(aq)
2moles
K2SO4(aq) +2H2O(1)
2 moles of KOH
=2moles of H2O
0.25moles of HOH
=
Mass of H2O produced
=0.25mol x 18g mol-1=4.5g
Number of molecules of H2O =No of moles x NA
=0.25mol x6.02x 1023 molecules mol-1
=1.51x1023 molecules
Examples 12: Mg metal reacts with HCI to give hydrogen gas. What is the
minimum volume of HCI solution (27%by weight ) required to produce 12.1 g of
H2.The density of HCI solution if 1.14g cm-3
Mg(s) + 2HCI(aq)
Mg C12(aq) +H2(g)
Solution:
Mass of H2
=12.1g
Molar mass of H2
=2.016g mol-1
Mg(s) + 2HCI(aq)
1 mole of H2
Moles of H2
=2 mole of HCI
=
=12moles
Molar mass of HCI =1+35.5=36.5g mol-1
Mass of HCI
= No. of moles x molar mass
=12mol x 36.5g mol-1
=438 g
%age of HCI solution
=27
ww
w.a
Now,
llon
2moles
=
=x 6moles
Mg C12(aq) +H2(g)
line
No. of moles of H2
fre
e.c
om
Now ,
27 g of HCI are present in a mass of solution
=100g
438g of HCI are present in a mass of solution=
=1622.2g
Mass of HCI solution
Density of HCI solution
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=1622.2g
=1.14g cm -3
1st year chemistry notes
Volume of HCI solution
=
=
=1423 cm3
Limiting Reactant
“A reactant that controls the amount of the product formed in a chemical
om
reaction is called a limiting reactant.”
A limiting reactant gives the least number of moles of the product.
e.c
Generally, in carrying out chemical reactions m one of the reactants is deliberately
used in excess quantity . This quantity exceeds the amount theoretically required
fre
by the balanced chemical wquation.This is done to ensure that the other expensive
eractant is completely used up in the reaction .Sometimes, this strategy is applied
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to increase the speedof reactions. In this way excess reactant is left behind at the
end of reaction and the other reactant in completely consumed .This reactant which
is completely used up in the reaction is Known as the limiting reactant .Once this
llon
reactant is used up , the reactant stops and no additional product is formed .Hence
the limiting reactant controls the amount of the product formed in a chemical
Example:
ww
w.a
reaction .
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1st year chemistry notes
Identification of Limiting Reactant
To identify a limiting reactant, the following three steps are performed.
1.
Convert the given amount of each reactant to moles.
2.
Calculate the number of moles of product that could be produced form each
reactant by using a balanced chemical equation.
om
Example 13: NH3 gas can by prepared by heating together two solids NH4C1 and
Ca (OH)2. If a mixture containing 100g of each solid is heated then.
Calculate the number of grams of NH3 produced.
(b)
Calculate the excess amount of reagent left unrelated.
Solution:
(a)
CaC12(s) +2NH3(s) + 2H2O(1)
fre
2NH4C1(s) + Ca(OH) 2(s)
e.c
(a)
Calculation for the number of grams of NH3
=100g
line
Mass of NH4 C1
Mass of Ca(OH)2
=100g
=14+4+35.5=53.5g mol-1
Molar mass of NH4 C1
=40+34=74g mol -1
llon
Molar mass of Ca(OH)2
=
ww
w.a
No of moles of NH4 C1
No of moles of Ca(OH)2
1.87 moles
.35 moles
2NH4C1(s) + Ca(OH) 2(s)
CaC12(s) +2NH3(s) + 2H2O(1)
2moles
1mole
Now,
2molesof NH4CI
1 mole of Ca(OH)2
1.35
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moles
2moles
=2moles of NH3
1.87 moles of NH4CI
Also,
=
of
=
=2moles of NH3
Ca(OH)2
=
1st year chemistry notes
=2.70 moles of NH3
Since NH4C1 produces the least number of moles of NH3 , therefore, it is
limitation reactant.
No of moles of NH3 produced
=1.87moles
Molar mass of NH3
=14+3=17g mol-1
=No of moles NH3xmolar mass of
om
Mass of NH3produced
NH3
e.c
=1.87mol x17g mol-1
=31.79g
Calculation for the excess amount of reagent left un reacted
fre
(b)
calculation as follows:
Now,
2moles of NH4C1
=1 mole of Ca(OH)2
=
llon
1.87 moles of NH4C1
line
The reactant, Ca(OH)2 is used in excess , its amount left un reacted can be
=0.935moles of Ca(OH)2
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w.a
Amount of excess Ca(OH)2=Amount of Ca(OH)2 taken-amount of Ca(OH)2reacted
=1.35-0.935=0.415moles
Mass of uncreated Ca(OH)2 =No of moles x Formula mass
=0.415 mol x74 g mol -1
=30.71g
It means we should mix 100g of NH4C1with 69.29g of Ca(OH)2to get 1.87
moles of NH3..
Yield
“The amount of the product formed in a chemical reaction is called the
yield.
Theoretical Yield
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1st year chemistry notes
“The amount of the product calculated from the balanced chemical
equation is called the theoretical yield of the product .’’
It is the maximum amount of the product that can be produced by a given
amount of a reactant according to balanced chemical equation . In most chemical
reactions the amount of the product is less than the theoretical yield.
om
Actual yield
“The amount of the product actually abtained in a chemical reaction is
e.c
called the actual yield of the product .”
The actual yield of the product is always less than the theoretical yield of
fre
the product.
Causes of less actual yield
line
In most chemical reactions, the actual yield is always less than the
theoretical yield of the product due to the following reasons:
1.
A practically inexperienced worker cannot get the expected yield
2.
llon
because of many short comings.
Product may be lost during the processes like filtration, separation
ww
w.a
by distillation , separation by a separating funnel , washing ,drying and
crystallization if not properly carried out.
3.
4.
5.
Side reaction may occur which reduce the amount of the product.
The reaction may not go to completion.
There may have been impurities in one or more of the reactants.
Percentage yield of product
A chemist is usually interested in the efficiency of a reaction. The
efficiency of a reaction is expressed by comparing the actual and theoretical yields
in the form of the percentage yield.
%age yield of product=
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1st year chemistry notes
Example 14: When lime stone,CaCO3 is roasted , quicklime, CaO is produced
according to the following equation. The actual yield of CaO is 2.5kg, when 4.5gk
of lime stone in roasted .What is the percentage yield of this reaction.
CaC(s) +CO2(g)
Solution:
Actual yield of CaO
=2.5kg =2500g
Mass of lime stone CaCO3
=4.5kg =4500g
=40+12+48=100g mol-1
Molar mass of CaCO3
= 40+16=56g mol -1
e.c
Molar mass of CaO
45000g
CaCO3(s)
CaO(s) +CO2(s)
56g
fre
100g
100g of CaCO3
4500g of CaCO3
=56g of CaO
line
Now ,
om
CaCO3(s)
llon
Theoretical yield of CaO
=
=2520g pf CaO
=2520
Q1.
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w.a
%yield =
=
=99.21%
EXERCISE
Select the most suitable answer from the given ones in each question.
(i)
The mass of one mole of electrons is
(a)
Properties which depend upon mass
(b)
Arrangement of electrons in orbital
(c)
Chemical properties
(d)
The extent to which they may be affected in
electromagnetic field
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1st year chemistry notes
(ii)
which of the following statements is not true?
(a)
isotopes with even atomic masses are comparatively
abundant
(b)
isotopes with odd atomic masses and even atomic number
are comparatively abundant
atomic masses are average masses of isotopes.
om
(c)
(d)
Atomic masses are average masses of isotopes
Many elements have fractional atomic masses, this is because
The mass of the atom is itself fractional
(b)
Atomic masses are average masses of isobars
(c)
Atomic masses are average masses of isotopes.
fre
(a)
line
(iii)
e.c
proportional to their relative abundance
(d)
Atomic masses are average masses of isotopes
proportional to their relative abundance
llon
(iv) The mass of one mole of electrons is
a 008mg(b)
0.55mg
(c)
0.184mg
(d)
1.673mg
ww
w.a
(v) 27g of Al will react completely with how much mass of O2 to
produce A12O3
(a)
8 g go oxygen
(c)
(vi)
32g of oxygen
(viii)
(d)
16g of oxygen
24g of oxygen
The number of moles of CO2 which contain 8.0 g of oxygen .
(a)
(vii)
(b)
0.25
(b)
0.50
(c)
1.0
(d)
The largest number of molecules are present in
(a)
3.6g of H2 O (b)
4.8g of C2H5 OH
(c)
2.8 g of CO
5.4g of N2O5
(d)
One mole of SO2 contains
(a)
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6.02x1023 atoms of oxygen
1.50
1st year chemistry notes
(b)
18.1x1023 Molecules of SO2
(c)
6.02x1023 atoms of sulphur
(d)
4 gram atoms of SO2
(a)
2.24 dm3
(c)
1.12 dm3
22.4dm3
(b)
112 cm3
(d)
A limiting reactant is the one which
(a)
is taken in lesser quantity in grams as compared to other
(b)
is taken in lesser quantity in volume as compared to the
e.c
(x)
The volume occupied by 1.4 g of N2at STP is
om
(ix)
fre
reactants
reactants
©
give the maximum amount of the product which is required
(e)
give the minimum amount of the product under
Q2:
(ii)d
(iii)d
(iv)b
(v)d
(vi)a
(vii)a
(viii)c (ix)c
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w.a
(x)d
(i)a
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consideration
Ans:
line
other
Fill in the blanks :
(i)
(ii)
The unit of relative atomic mass is-----------
The exact masses of isotopes can be determined by ------------
spectrograph.
(iii)
The phenomenon of isotopes was first discovered by --------------
(iv)
Empirical formula can be determined by combustion analysis for
those compound which have-----------and -----------in them.
(v)
A limiting reagent is that which controls the quantities of ------------
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1st year chemistry notes
(vi)
I mole of glucose has-----------atoms of carbon ---------------of
oxygen and ----------of hydrogen.
(vii)
4g of CH4 at Oo C and I am pressure has ---------molecules of
CH4 .
(viii)
Stoichiometry calculations can by performed only when -----------
amu
(ii)
mass (iii)
(v)
1.505x1023
(vi)
fre
(viii) conservation
Indicate true or false as the case my be:
(i)
Neon has three isotopes and the fourth one with atomic mass
line
Q3:
carbon, hydrogen
Products
6x6.02x1023,6x6.02x1023,12x6.02x1023
(vii)
Soddy (iv)
e.c
Ans: (i)
om
--law is obeyed.
20.18 amu.
(ii)
Empirical formula gives the information about he total number
(iii)
During combustion analysis Mg(CIO4)2 is employed to absorb
ww
w.a
water vapors.
(iv)
llon
of atoms present in the molecule
Molecular formula is the integral multiple of empirical formula
and the integral multiple can never be unity.
(v)
The number of atoms in 1.79 g of gold and 0.023g of sodium are
equal.
(vi)
The number of electrons in the molecules of CO an dN2 are 14
each, so 1 mg go each gas will have same number of electrons.
(vii)
Avogadro‟s hypothesis is applicable to all types of gases, i.e.,
ideal and non-ideal .
(viii)
Actual yield of a chemical reaction may by greater than the
theoretical yield.
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1st year chemistry notes
Ans. (i)
False (ii)
False (iii)
True
(v)
False (vi)
true
False (viii)
Q4:
(vii)
(iv)
false
False
What are ions? Under What condition are they produced ? can you
explain the places of negative charge in PO , MnO and Cr2 O
the negative charge resides on singly covalent
om
Ans: In PO , MnO and Cr2 O
bonded oxygen because it contains seven electrons three electron pairs and one
e.c
electron from covalent bond in its cuter most shell.
(Picture)
(a)
What are isotopes? How do you deduce the fractional atomic
fre
Q4:
masses of
line
Elements form the relative isotopes abundance? Give two examples in
support of your answer.
How does a mass spectrograph show the relative aboundace of
llon
(b)
isotopes of an element?
©
What is the justification of two strong peaks in the mass spectrum
Ans
ww
w.a
for bromine; while for iodine only one peak at 127 amu , is indicated?
The two strong peak in the mass spectrum for bromine represent two
different isotopes of bromine having nearly equal natural abundances. Only one
peak at 127 amu in the mass spectrum for iodine indicates that it has only one
isotope of atomic mass 127 amu.
Remember that the peak heights are proportional to the natural abundances of the
isotopes in the given sample , the larger the height of the peak, the greater is the
natural abundance of the isotopes in the sample.
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1st year chemistry notes
Q5:
Silver has atomic number 47 and has 16 known isotopes but two occur
naturally I,e, Ag _____107 . and Ag _____109 . Given the following mass
spectrometric data, calculated the average atomic mass of silver,
Isotopes mass (amu) percentage abundance
107
Isotopes
48.16
The mass contribution for silver are:
Fractional abundance
isotopic mass
Ag
109Ag
mass
107
0.5184x107=55.4688
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contribution
107
108.90476
e.c
Solution:
Ag
51.84
om
109
Ag 106.90509
0.4816x109=52.4944
107
Fractional atomic mass of silver
=107.9632
Q6:
llon
Hence the fractional atomic mass of silver is =107.9632 Ans
Boron with atomic number 5 has two naturally occurring isotopes.
Calculate the percentage abundance of
Solution:
ww
w.a
information.
10
B and
11
B from the following
Average atomic mass of boron
=10.81 amu
Isotopic mass of 10B
=10.0129 amu
Isotopic mass of 11B
=11.0093
Let, the fractional abundance of 10B =x
The fractional abundance of 11B
=1-x
Remember that the sum of the fractional abundances of isotopes must be
equal to one, now, The equation to determine the atomic mass of element is
(fractional abundance ) (isotopic mass ) (fractional abundance of 10B)(isotopic
mass of 10B )+(fractional abundance of 11B) (isotopic mass of 11B)
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1st year chemistry notes
=Average atomic mass of Boron
(x)(10.0129)+(1-x)(11.0093) =10.81
10.0129x+11.00093x
=10.81
10.0129x-11.00093x =10.81-11.0093
-0.9964x
=-0.1993
e.c
om
x
=
10
Fractional abundance of B =0.2000
Fractional abundance of 11B =(1-0.2000)=0.8000
By percentage the fractional abundance of isotope is
%of 10B
=0.2000x100 =20% Answer
% of 11B
(i)
Gram atom
molecular mass
(iv)
volume
(v)
Avogadro‟s number
Gram
molar
(vii)
Percentage yield
(c)
23 g of sodium and 238g of uranium have equal number of atoms in
Mg atom is twice heavier than that of carbon
ww
w.a
(b)
llon
Justify the following statements:
(a)
the
(viii)
Gram molecular mass (iii)
Gram ion
(vi)
Stoichiometry
Q8:
(ii)
fre
Define the following terms and give three examples of each.
line
Q7:
=0.8000x100 =80%Answer
180g of glucose and 342 g of sucrose have the same number of
molecules but different number of atoms present in them.
(d)
4.9g of H2 SO4 when completely ionized in water , have equal number
of positive and negative charges but the number of positively charged
ions are twice the number of negatively charged ions.
(e)
One mg of K2 Cr O4 has thrice the number of ions than the number of
formula units when ionized in water.
(f)
Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the
volumes of 22.414 dm3 , although the sizes and masses of molecules
of three gases are very different from each other.
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Solution:
(a)
23g of Na
=1 mole of Na
=6.02x1023 atoms of Na
238g of U
=1 mole of U
=6.02x1023 atoms of U.
Since equal number of gram atoms(moles) of different elements contain
om
equal number of atoms. Hence , 1 mole (23g ) of sodium and 1 mole (238)g of
uranium contain equal number of atoms , i , e ,6.02x1023 atoms.
Since the atomic mass of Mg (24) is twice the atomic mass of carbon
e.c
(b)
(12) therefore, Mg atom is twice heavier than that of carbon. Or
fre
Mass of 1 atom of Mg=
Mass of 1 atom of C =
line
Since the mass of one atom of Mg is twice the mass of one atom of C ,
therefore, Mg atom is twice heavier than that of carbon.
(c)
180 g of glucose = 1 mole of glucose =6.02x1023 molecules of
sucrose
llon
glucose 342 g og sucrose=1mole of sucrose
=6.02x1023 molecules of
ww
w.a
Since one mole of different compounds has the same number of molecules.
Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the
same number (6.02x1023)of molecules. Because one molecule of glucose , C6H12O6
contains 45 atoms whereas one molecules of glucose, C 12 H22 O11 contains 24
atoms. Therefore , 6.02x1023 molecules of glucose contain different atoms as
compound to6.02x1023 molecules of sucrose. Hence , 180 g of glucose and 342g
og sucrose have the same number of molecules but different number of atoms
present in them.
(d)
H2 SO4
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2H+ + SO
1st year chemistry notes
When one molecules of H2 SO4 completely ionizes in water it produces two
H+ ion and one SO ion ,.Hydrogen ion carries a unit positive charge whereas SO
ion carries a double negative charge. To keep the neutrality , the number of
hydrogen are twice than the number of soleplate ions. Similarly the ions produced
om
by complete ionization of 4.8g of H2 SO4 in water will have equal number of
positive and negative but the number of positively charged ions are twice the
(e)
e.c
number of negatively charged ions.
2H+ + SO
H2 SO4
ion. Thus
fre
K2 Cr O4 when ionizes in water produces two k+ ions one C O
each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2
line
Cr O4 has thrice the number of ion than the number of formula units ionized in
water.
(f)
2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP
llon
=22.414dm3 16g of CH4 =1mole of CH4 =6.02x1023 molecules of CH4
at STP =22.414dm3 144 g of CO2 =1mole of CO2 =6.02x1023
ww
w.a
molecules of CO2 at STP =22.144dm3
Although H2 , CH4 and CO2 have different masses but they have the same
number of moles and molecules . Hence the same mumber of moles or the same
number of molecules of different gases occupy the same volume at STP . Hence 2
g of H2 ,16g of CH4 and 44 g of CO2 occupy the same volume 22.414 dm3 at STP.
The masses and the sizes of the molecules do not affect the volumes.
Q10: Calculate each of the following quantities
(a)
mass in grams of 2.74 moles of KMnO4 .
(b)
Moles of O atoms in 9.0g of Mg (NO3)2 .
(c)
Number of O atoms in 10.037g of Cu SO4 .5H2 O.
(d)
Mass in kilograms of 2.6x 1020 molecules of SO2 .
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1st year chemistry notes
Moles of C1 atoms in 0.822g C2H4C12 .
(f)
Mass in grams of 5.136 moles of silver carbonate .
(g)
Mass in grams of 2.78x1021 molecules of CrO2 C12 .
(h)
Number of moles and formula units in 100g of KC1O3 .
(i)
Number of K+ ions C1O ions, C1 atoms, and O atoms in (h)
Solution:
(a)
No of moles of KMnO4
Mass of KMnO4
= no
=39+55+64=158g mol -1
=?
fre
Mass of KMnO4
Formula used:
=2.74moles
e.c
formula mass of KMnO4
om
(e)
of mole of KMnO4 x formula mass of KMnO4
line
=2.74 mol x 158 g mol-1
=432.92g Answer
Mass of Mg (NO3)2 =9g
llon
(b)
Formula mass of Mg (NO3)2 =24+28+96=148g mol -1
No of moles of O atoms
=?
ww
w.a
Formula used:
No of mole of Mg (NO3)2
=
Now,
I mole of Mg (NO3)2 contains =6moles of O atoms
0..06 moles of Mg (NO3)2contains
=6x0.6
=0.36 moles of O atoms
Alternatively ,
148g of Mg (NO3)2 contains =6moles of O atoms
g of Mg (NO3)2contains
(c)
=
=0.36 mole Answer
4
Mass of CuSO . 5H2O=10.037g
Formula mass of CuSO4. 5H2O=63.54+32+64+90
=249.546g mol -1
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1st year chemistry notes
No of moles of CuSO4. 5H2O
No of moles of CuSO4. 5H2O
Now,
1 mole of CuSO4 .5H2O contains
9moles of O atoms
0.04 mole of CuSO4 .5H2O contains=9x0.04
=0.36 moles of O atoms
I mole of O atoms contains
=6.02x1023 O atoms
0.36 mole of O atoms contains
=6.02x1023 x0.36 oxygen
om
=
Now,
=?
=
Now,
No of molecules of SO2 .
Molecular mass of SO2 .
Avogadro‟s number , NA
=
Now,
(f)
mol-1
(g)
llon
=27.64x10-6 kg
=2.764x10-3 kg Answer
Mass of C2 H4C1
= 0.822g
Molecular mass of C2 H4C1
=24+4+71=99 g mol-1
No of moles of C2 H4C1
=
1 mole of C2 H4C1 contains
=2moles of C1 atoms
8.3x10-3mole of C2 H4C1 contains =2x8.3x10-3 mole of atom
=16.6x10-3
=0.0166mole of C1 atom
=0.017 mole Answer
No of mole of Ag2 CO3
=5.136moles
Formula mass of Ag2 CO3
=215.736+12+48=275.736 g
ww
w.a
(e)
=
=
=27.64x10-3 g
line
Mass of SO2 molecules
fre
(d)
=2.17x1023 oxygen atoms
=2.17x1023 atoms Answer
=2.6x1020 molecules
=32+32=64 g mol-1
=6.02x1023 molecules of SO2
e.c
atoms
Mass of Ag2 CO3=No of moles of Ag2 CO3xformula mass of Ag2 CO3
=5.136molx275.736 g mol-1
=416.18g
=1416.2 g Answer
Molecular mass of CrO2C12
=52+32+71=155g mol-1
NA
=6.02x1023 molecules mol-1
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1st year chemistry notes
Molecules of CrO2C12==2.78x1021 molecules
Now,
mass of CrO2C12
=
=
om
Mass of KCIO3
Formula mass of KCIO3
No of moles of KCIO3
No of moles of KCIO3
=
=0.816mole Answer
No of moles x Avogadro,s No
=0.816mole x 6.02x1023 formula units
=4.91x1023 formula units
=4.91x1023 Answer
fre
=
No of K+ ions
line
No of formula units
(i)
e.c
(h)
=71.578x10-2 g
=0.71578
=0.716 g Answer
=100g
=39x35.5+48=122g mol-1
=?
=4.91x1023 Answer
=4.91x1023 Answer
= 4.91x1023 x3
=14.73x1023 =1.473x1024 Answer
Q 11 Aspartame he artificial sweetener, has a molecular formula of C14 H18 N2O5
.
(a)
What is the mass of one mole of aspartame?
(b)
How many moles are present in 52g of aspartame?
(c)
What is the mass in grams of 10.122 moles of aspartame?
(d)
How many hydrogen atoms are present in 2.34g of aspartame?
(a)
Molecular mass of aspartame =168+18+28+80=295g mol-1
Mass of 1 mole of aspartame =294g mol-1 Answer
(b)
Mass of aspartame =52g
Molecular mass of aspartame =294g mol-1
ww
w.a
llon
No of CIO ions
No of CIO ions
No of O atoms
No of moles of aspartame
=
=
=0.1768 mol
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1st year chemistry notes
(c)
om
(d)
=0.177 mol Answer
No moles of aspartame
= 10.122 moles
Molecular mass of aspartame =294g mol-1
Mass of aspartame =No of moles x Molar mass
=10.122mol x 294g mol-1
=2975.87 g Answer
Mass of aspartame =243g
Molar mass of aspartame
=294g mol -1
No of molecules of aspartame=?
ww
w.a
llon
line
fre
e.c
No of molecules of aspartame=
xNA
=
=
=4.98x1021 molecules.
Now,1 molecule of aspartame contains=18 H atoms
4.98x 1022 molecules =18x4.98x1021 H atoms
=89.64x1021H atoms
=8.964x1022 H atoms Answer
Q 12: A sample of 0.600 mole of a metal M reacts completely with excess of
fluorine to from 46.8g MF2 .
(a)
How many moles of F are present in the sample of MF2 that forms.
(b)
which elements is represented by the symbol M ?
Solution:
(a)
Formula of compound
=MF2
No of moles of M
=0.6 mol
Mass of MF2 =46.8g
The molar of M:F in the compounds;
No of moles of F
Mass of F
=0.6x2=1.2mol Answer
=No of moles of Fx At . mass
Mass of compound
=1.2x19=22.8g
=46.8g
of F
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1st year chemistry notes
Mass of metal, M
=46.8-22.8
=24
At mass of M
=
ww
w.a
llon
line
fre
e.c
om
=
(b)
The atomic mass of the elements, M
=40
The metal is calcium, Ca Answer
Q 12 : In each pair , choose the larger of the indicated quantity ,or state if the
samples are equal.
(a)
Individual particles: 0.4 mole of oxygen molecules or0.4mole of
oxygen atom.
(b)
Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms
(c)
Mass: 0.6 mole of C2 H4 or 0.6mole of 12
(d)
Individual particles: 4.0g N2O4 or 3.3g SO2
(e)
Total ions: 2.3 moles of NaC1O3 or 2.0mole of MgC12
(f)
Molecules: 11.0g of H2Oor 11.0g H2O2
(g)
Na+ ion: 0.500 moles of NaBr or 0.0145kg NaC1
(h)
Mass: 6.02x1023 atoms of 235U or 6.02x1023 atoms of 238U
Ans:
(a)
Number of molecules
=moles x NA
Number of O2 molecules
=0.4x6.02x1023 =2.408x1023 molecules
No of O atoms=0.4x6.02x1023=2.108x1023 atoms
There are equal number of individual particles in 0.4 mole of oxygen
molecules and 0.4 mole of oxygen atom. In general, equal number of moles of
different substances contains equal number of particles.
Both are equal
Answer
(b)
Mass of substance
= moles x molar mass
Mass of oxygen atoms
=0.4x16=64g
Mass of ozone, O3 molecules =0.4x48=19.2g
0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen
atoms.
Ozone Answer
(c)
Mass of C2H4
=0.6x28=1.68g
Mass of 12
=0.6x127=254g
0.6mole of 12 have larger mass than 0.6 mole of C2H4
12 Answers
(d)
No of molecules
=
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1st year chemistry notes
No of molecules
=
NA
x6.02x1023=3.68x1023
llon
No of molecules in H2 O2
molecules
=
line
(f)
fre
e.c
om
No of molecules in N2 O4 = x6.02x1023 =2.62 x1023 molecules
No of molecules in SO2 =x6.02x1023
=3.1x1022 molecules
3.3g of SO2 have larger number of individual particles than 4.0 g of N2 O4 .
SO2 Answer
(e)
No of formula units
=Moles x NA
3
No of formula units of NaC1O
=2.3x6.02x1023=1.38x1024 formula
units
No of ions in 1 formula units of NaC1O3=2
Total no of ions in MgC12
=2x1.38x1023=2.76x1024 ions
No of formula units of MgC12
=2.0x6.02x1023 x3=3.6x1024 ions
No .of ions in one formula unit of MgC12 =3
Total no of ions in MgC12 =1.20x1024 x3=3.6x1024 ions
2.0moles of MgC12 contain lager number of total ions than 2.3 moles of
3NaC1o
MgC1 Answer
units
ww
w.a
(g)
No of molecules in H2 O2
x6.02x1023=1.95x1023 molecules
=
11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2
H2 O2Answer
No of formula units
=moles xNA
No of formula units NaBr
=0.5x6.02x1023=3.01x1023 formula
One formula units o NaBr contain Na+ ions =1
3.01 x1023 formula unit of NaBr contains Na +ions =3.01x1023 Na+ ions
No of formula units of NaC1
=
x6.02x1023
=1.49x1023formula units
One formula unit of NaC1 contains Na+ ions
=1
23
1.49x10 formula units of NaC1 contains
=1.49x1023 Na+ ions
0.500 moles of NaBr contains lager number of Na+ ions than 0.0145kg
ofNaC1.
NaBr Answer
(h)
Mass of atoms of an element =
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1st year chemistry notes
Mass of 235Uatoms
Mass of 238U atoms
=x6.02x1023 =235g
=x6.02x1023=238g
238
U Answer
Calculate the percentage of nitrogen in the four important fertilizer
ww
w.a
llon
line
fre
e.c
om
Q 13: (a)
i.e.,
(i)NH3
(ii)NH2CONH2(Urea) (iii)(NH4)2SO4
(iv)NH4 NO3
(b)
Calculate the percentage of nitrogen and phosphorus in each of the
following:
(i)
NH4H2PO4
(ii)
(NH4)) PO4 (iii)
(NH4)4 PO4
Solution:
(a)
Mol-mass of NH3
=14+4=17g
Mass of N
=14g
% of N
=x100
=82.35% Answer
(b)
Mol-mass of NH2 CONH2 =28+4+12+16=60g
Mass of N
=28g
%of N
=x100
=46.35% Answer
(c)
Mol-mass of (NH2 )2 SO4
=28+8+32+64=132g
Mass of N
=28g
% of N
=x100
=21.21% Answer
(d)
Mol-mass of (NH2 )2 SO4 =28+4+48=80g
Mass of N
=28g
%of N
=x100
=35% Answer
(I)
Mol-mass of (NH2 )2 SO4
=14+6+31+64=115g
Mass of N
=14g
Mass of P
=31g
%of N
=x100=12.17% Answer
%of P
==26.96% Answer
(II)
Mol-mass of ((NH2 )2 SO4 =28+9+31+64=132g
Mass of N
=28g
Mass of P
=
%of N
= =21.21% Answer
%of P
= =23.48% Answer
(III) Mol-mass of (NH2 )2 SO4
=42+12+31+64=149g
Mass of N
=42g
Mass of P
=31g
%of N
=
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1st year chemistry notes
ww
w.a
llon
line
fre
e.c
om
%of P
=
Q 14: Glucose C6 H12 O6 is the most important nutrient in the cell for generating
chemical potential energy. Calculate the mass% of each element in glucose and
determine the number of C,H and O atoms in 10.5g go the sample.
Solution:
Mol-mass of glucose C6 H12 O6
=72+12+96=180g
Mass of C
=72
Mass of H
=12
Mass of O
=96
% of C
= =40% Answer
% of H
= =6.66% Answer
% of O
= =53.33% Answer
Mass of C6 H12 O6 =10.5g
Mol-mass of C6 H12 O6
=180g
Mol-mass of
=180g mol-1
No of moles of C6 H12 O6
=
No of molecules of glucose
=No of moles x NA
=0.058 molx 6.02x1023 molecules mol-1
=0.35x1023 molecules
=3.5x1022 molecules
Now, 1 molecule of glucose contains
=6C-atoms
22
3.4x10 molecules of glucose contains
=6x3.5x1022 C-atoms
=21x1022 =2.1x1023 C atoms Answer
1 molecules of glucose contains
=12H-atoms
22
3.5x10 molecules glucose contains
=12x3.5x1022
=4.2x1023 H- atoms Answer
1 molecule of glucose contains
=6 O –atoms
22
3.5 x 10 molecules of glucose contains
=6x3.5x1022
=2.1x1023 O-atoms Answer
Q 16: Ethylene glycol is used as automobile antifreeze .It has 38.7% carbon, 9.7%
hydrogen and 51.6% oxygen. Its molar mass is 62.1 grams mol-1 .Determine its
molecular formula.
Solution:
% of C=38.37 g
% of H
=9.7g
% of
O=51.6g
At. Mass of C=12g mol-1
At. Mass of H=1.008g mol-1 At. Mass of O =16g
mol-1
No of moles of C
=
No of moles of H
=
No of moles of O
=
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1st year chemistry notes
1
:3
:1
3
Empirical formula =CH O
Empirical formula mass
=2x CH3 O
=C2 H6 O2 Answer
-1
Q 16: Serotonin (Molecular mass= 176g mol ) is a compound that conducts
nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08%
O. What is its molecular formula?
Solution:
No of moles of C
=
No of moles of H
=
No of moles of N
=
No of moles of O
=
C
:
H
:
N
:
O
Atomic ratio
ww
w.a
llon
line
fre
Molecular formula
=31
e.c
n=
om
Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest
ratio.
C
:H
:O
10
:
12
:
2
:
1
Empirical formula
=C10 H12 N2 O
Empirical formula mass
=120+12+28+16=176g mol-1
Molecular mass
=176g mol-1
n=
Q17: An unknown metal M reacts with S to from a compound with a formula
M2S3 .If 3.12 g of M reacts with exactly 2.88 g of sulphur ,what are the names of
metal M and the compound M2 S3 .
Solution:
Formula of compound
= M2 S3
Mass of M
=3.12g
Mass of S
=2.88g
Atomic mass of S
=32g mol-1
No of moles of S
=
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1st year chemistry notes
No of moles of M
om
No of moles of S
=
The molar ratio of M: S in the compound is :
=
Now,
No of moles of M
e.c
=0.06 mole
=
line
fre
At. Mass M =
The mass of M used in the formation of M2S3 is 3.12g. The product M2S3 therefore
also contains 3.12g of M, because mass is conserved . The amount of M before and
after reaction must be the same. Since we know both the number of moles of M
and the mass of M , we can cal calculate the atomic mass of M as follows:
=
=52
Atomic number, Z
=52
Q19: The octane present in gasoline burns according to the following equation.
2C8 H18 (i)
+ 2502(g)
16CO 2(g) + 18H2O (i)
(a)
How many moles of O2 are needed to react fully with 4 moles
of actane?
(b)
How many moles of CO2 can be produced from one mole of
actane?
(c)
How many moles of water are produced by the combustion of 6
moles of octane?
(d)
If this reaction is to be used to synthesize 8 moles of CO 2 how
many grams of oxygen are needed? How many grams of octane will
be used?
Solution:
ww
w.a
llon
At. Mass of M
4 moles
(a)
2C8 H18 (i)
2 moles
+ 2502(g)
25 moles
2 moles of C8 H18
4 moles of C8 H18
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16CO 2(g) + 18H2O (i)
=25 moles of O2
=
1st year chemistry notes
=50moles of O2 Answer
(b)
1 moles
2C8 H18 (i)
2 moles
+ 2502(g)
Now, 2 moles of C8 H18
1 mole of C8 H18
16CO 2(g) + 18H2O (i)
=16 moles of CO2
=
=8 moles of CO2 Answer
+ 2502(g)
Now, 2 moles of C8 H18
6 moles of C8 H18
(c)
=18 moles of H2 O(i)
=
=54 moles of H2 O
6 moles
2C8 H18 (i)
2 moles
+ 2502(g)
16CO 2(g) + 18H2O (i)
1800moles
line
=25 moles of O2
=
=12.5 moles of CO2
Mol-mass of O2
=32g mol-1
=12.5 molx 32g mol-1
=400g of O2
Now, 16moles of CO2
=2moles of C8 H18
8 moles of CO2
=
=1 mole of C8 H18
Mol-mass of C8 H18
=96+18=114g mol-1
Mass of C8 H18
=No of moles of C8 H18xMol.mass ofC8
H18
=1 molx 114 g mol-1
114g Answer
Q19: Calculate the number of grams of A12 S3 which can be prepared by the
reaction of 20 g of A1 and 30 g of sulphur. How much the non-limiting reaction is
in excess ?
Solution:
Mass of A1
=20g
Molar mass of A1
=27g mol-1
No of moles of A1
=
Mass of S
= 30g
Molar mass of S
=32g mol-1
No of moles of S
=
ww
w.a
llon
Now, 16 moles of CO2
8 moles of CO2
16CO 2(g) + 18H2O (i)
e.c
2C8 H18 (i)
2 moles
om
6 moles
fre
(c)
0.74 mole 0.94 mole
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1st year chemistry notes
2A1
+
2 mole
3S
A12 S3
1 mole
3 mole
Now, 2 moles of A1
0.74 moles of A1
Now,
3 moles of S
0.94 moles of S
=1 mole of A12 S3
=
=0.37 mole of A12 S3
=1 moles of A12 S3
=
0.94 mole
2A1
+
2 mole
A12 S3
3 mole
=2 moles of A1
=
=
Mass of A1
=No of moles of A1 x molar mass of A1
=0.63x 27
=17g of A1
Mass of A1available =20g
Mass of A1 which reacts completely =17g with available S
Excess of A1
=20-17=3g
Q20: A mixture of two liquids, hydrazine N2H4 and N2 O4 are used as a fuel in
rockets. They produce N2 and water vapors. How many grams of N2 gas will be
formed by reacting 100g of N2 O4 and 200g g of N2 O4.
2N2H4 + N2O2
3N2 +4 H2O
Solution:
Mass of2N2H4
=100g
Mass of N2O2
=200g
Molar mass of 2N2H4
=28+4=32g mol-1
Molar mass of N2O2
=28+64=92g mol-1
No of moles of N2H4
=
llon
3 moles of S
0.94 moles of S
ww
w.a
Now ,
3S
line
fre
e.c
om
=0.313 mole of A12 S3
Since S give the least number of moles of A12 S3 therefore, it is the limiting
reactant.
No of moles of A12 S3
=0.313 mole
Molar mass of A12 S3
=150g mol-1
Mass of A12 S3=No of moles of A12 S3xMolar mass of A12
S3
=0.313molx 150 g mol-1
=46.95 g of A12 S3 Answer
The non-limiting reactant is A1 which is in excess. Now mass of A1
required reacting completely with 0.94 moles of S can be calculated as:
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1st year chemistry notes
No of moles of N2O2
=
3.125moles 2.174 moles
2N2H4 + N2O2
2 moles
1mole
3N2 +4 H2O
3moles
=3moles of N2
=
=4.69 mole of N2
Now , 1 mole of N2O2
=3moles of N2
2.174 moles of N2O4
=
=6.52 mole of N2O2
Since N2H4gives the least number of moles of N2, hence it is the limiting
reactant.
Amount of N2 produced
=4.69 moles
2
Molar mass of N
=28g mol-1
Mass of N2
=4.69g molx 28g mol-1
=131032 g Answer
Q21: Silicon carbide (SiC) is an important ceramic material . It is produced by
allowing sand (SiO2 )to react with carbon at high temperature.
SiO2 +
3C
SiC +
2CO
When 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is
produced.
Solution:
Mass of SiO2
=100 kg=100000g
Mass of SiC produced
=5.14 kg =51400g
SiO2
60g
Now,
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Now , 2moles of N2H4
3.125moles of N2H4
+
3C
SiC +
2CO
40g
60g of SiO2
100000g of SiO2
=40g of SiC
=
=66666.67 g
Actual yield of Sic =51400 g
Theoretical yield of SiC
=66666.67g
% yield =
=
=77.1%
Q22: (a)
What is Stoichiometry? Give its assumptions? Mention two
important law , which help to perform the Stoichiometry calculations.
(b)
What is a limiting reactant? How does it control the quantity of the
product formed? Explain with three examples
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1st year chemistry notes
(a)
Define yield. How do we calculate the percentage yield of a
chemical reaction?
(b)
What are the factors which are mostly responsible for the low yield
of the products in chemical reactions.
Q24: Explain the following with reasons.
(j)
Law of conservation of mass has to be obeyed during
Stoichiometric calculations.
(ii)
Many chemical reactions taking place in our surrounding
involves the limit reactants.
(iii)
No individual neon atom in the sample of the element has a mass
of 20.18amu.
(iv)
One mole of H2 SO4 should completely react with two moles of
NaOH. How does Avogadro, s number help to explain it.
(v)
One mole H2 O has two moles of bonds , three moles of atoms ,
ten moles of electrons and twenty eight moles of the total fundamental
particles present in it.
(vi)
N2 and CO have the same number of electrons, protons and
neutrons.
Ans. (i)
According to law of conservation of mass, the amount of
each element is conserved in a chemical reaction. Chemical equations
are written and balanced on the basis of law of conversation of mass.
Stoichiometry calculations are related with the amounts of reactants
and products in a balanced chemical equation. Hence, law of
conservation of mass has to be obeyed during stoichiometric
calculations.
(ii)
In our surrounding many chemical reactions are taking place
which involve oxygen. In these reactions oxygen in always in excess
quantity while other reactant are in lesser amount. Thus other
reactants act as limiting reactants.
(iii)
Since the overall atomic mass of neon in the average of the
determined atomic masses of individual isotopes present in the sample
of isotopic mixture .Hence, no individual neon atom in the sample has
a mass of 20.18amu.
(iv)
H2 SO4
+2NaOH
Na2 SO4
+ 2H2 O
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Q 23:
1 mole
2moles
2 moles of H+ ions
2x6.02x1023 H+ ions
2 moles of OH ions
2x6.02x1023 OH ions
Once mole of H2 SO4 consists of 2 moles of H+ ions that contains
twice the Avogadro‟s number of H+ ions. For complete neutralization
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1st year chemistry notes
(v)
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(vi)
it needs 2 moles of one mole of H2 SO4 should completely react with
two moles of NA OH.
Since one molecule of H2 O has two covalent bonds between H
and O atoms. Three atoms, ten electrons and twenty eight total
fundamental particles present in it. Hence, one mole of H 2 O has two
moles of bond, three moles of atoms, ten moles of electrons and
twenty eight moles of total fundamental particle present in it.
In N2 there are 2 N atoms which contain 14 electrons (2x7),14
protons (2x7) and 14 neutrons (2x7) . In CO, there are one carbon and
one oxygen atoms. It contains 14 electrons (6carbon e +8 oxygen e),
14 protons (6 C proton +8 O proton ) and 14 neutrons (6 neutrons +8
O neutrons).Hence , N2 and CO have the same number of electrons,
protons and neutrons. Remember that electrons, protons and neutrons
of atoms remain conserved during the formation of molecules in a
chemical reaction.
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