152096955 long questions basic concept

1st year chemistry notes

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Chapter No. 1

BASIC CONCEPTS

Chapter No. 1

Basic Concepts

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



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, H2S, 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+, Ca

2+ , NH , Cl

-, O

2- , 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 +

+ e-

The 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: Na +

, K +

,Ca2+,

Mg 2+

, A13+

, Sn 4+

, NH , H3 O +

Anion



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.

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.

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 .



Remember that: 1amu=1.66x10 -24

g.

Table: Relative atomic masses of some elements

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

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 Isotopes Abundance(%)

Hydrogen 1H,

2 H 99.985, 0.015

Carbon 12

C, 13

C 98.893, 1.107

Element Relative Atomic

mass

Element Relative Atomic

mass

H 1.0078 amu A1 26.9815 amu

N 14.0067 amu S 32.066 amu

O 15.9994 amu C1 35.453 amu

Na 22.9897 amu Cu 63.546 amu

Mg 24.3050 amu U 238.0289 amu



Nitrogen 14

N , 15

N 99.634, 0.366

Oxygen 16

O , 17

O ,18

O 99759. 0.037, 0.204

Sulphur 32

S , 33

S , 34

S, 36

S 95.0,0.76,4.22.0.014

Chlorine 35

C1, 37

C1 75.53, 24.47

Bromine 19

Br , 18

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 F, As, I and

Au have only single isotopes. These elements are known as mono-

isotopic elements.

3. The isotopes whose mass number is multiples of four are most

abundant. For example, O, Mg, Si, Ca and Fe. They form

nearly 50% of the earths 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. Astons 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. Dumpsters mass Spectrometer It was designed for the identification of elements which were available in

solid state.

Mass spectrometry



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 .



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

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

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.

(Picture)

Fig: A simple Mass Spectrometer

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



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 .

3. Fractional abundance =Percent abundance x

Or Percent abundance = Fractional abundance x 100

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

20Ne 90.92x =0.9092 20

0.9092x20=18.1840

21Ne 21 0.0026x21=0.0546

22Ne 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

Ne)+(fractional abundance of 21

Ne ) (isotopic mass of 21

Ne)+(fractional abundance of 22

Ne )(isotopic mass of 22

Ne).

=(0.9092)(20)+(0.0026)(21+(0.0882)(22)

=18.1840+0.0546+1.9404



=20.179

= 20.18 amu Answer

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=



=

=11.11% Answer

(iii) %of oxygen =

=

= 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 , Ax 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

No of gram-

atoms

Atomic

ratio

Whole

number

ratio

C 40.92 12.0

1x3=3

H 4.58 1.008

1.33x3=4



O 54.5 16.0

1x3=3

Empirical Formula = C3 H

4 O

3 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 CO2

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 H2O.

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 =



=

% 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,

Element %age Atomic

mass

No of gram

atoms

Atomic

ratio

Empirical

formula

C 52.10 12.0

H 13.11 1.008

C2 H6 O

O 34.79 16.0

Empirical Formula = C2 H6 O

Molecular Formula

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

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)



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

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

.Calculate the molecular formula of the compound.

Solution: (i) Calculation of empirical formula:

On writing the various steps in a tabular form , we have ,

Element %age Atomic

mass

No of gram

atoms

Atomic

ratio

Empirical

formula

C 65.44 12.0

H 5.50 1.008

C2 H6 O

O 29.06 16.0

(ii) 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)



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 gram atom of hydrogen (H) =1.008g

1 gram atom of carbon (C) =12.000g

1 gram atom of uranium (U) =238.0g

1 gram atom magnesium (Mg) =24.000g

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

atom of carbon

Gram molecule (Gram mole or mole)

The molecular mass of a substance expressed in grams is called a gram

molecule.

No. of gram molecules (moles) of a molecular substance =

Examples :

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



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 =

Examples:

1 gram-formula of Na C1 =58.5g

1 gram-formula of KOH =56.0g

1 gram-formula of Na2 CO3 =106g

1 gram-formula of Ag NO3 =170g

Gram-Ion (Mole)

The atomic mass , molecular mass formula mass or ionic mass of the

substance expressed in grams is called a mole.

Number of moles=

Examples 6: Calculate the gram atoms (moles)in

(a) 0.1g of sodium =0.1g

(b) 0.1 kg of silicon

Solution: (a) Given: Mass of sodium =0.1g

Atomic mass of sodium =23g mole -1

No of gram atoms (moles ) of Na =

=

=4.3x10-1

mol

(b) Given: Mass of silicon =0.1kg

=0.1x1000=100 g

Atomic mass of silicon =28.086 g mol-1

No of gram atoms (moles)of silicon =

= 3.56 mol

Example 7: Calculate the mass of 10-3

moles of Mg SO 4.



Solutions: Given: No of moles of MgSO4 = 10-3

mol

Formula mass of MgSO4 =24+32+64=120g mol -1

Formula Used: Mass of substance =No of moles of substance x Molar

mass

Mass of Mg SO 4 = 10-3 mol x 120 g mol -1

= 0.12g

Avogadro ,s Number (Avogadro Constant), NA)

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

.

Avogadro, s number is 6.02x1023

.It is a constant .One mole of any

substance always contains 6.02x10 23

Particles.

Examples:

1 mole of hydrogen (H) =1.008g of H =6.02x1023

atoms of H

1 mole of sodium(Na) =23g of Na =6.02x1023

atoms of Na

1 mole of water (H2O) =18g of H2O =6.02x1023

molecules

of H2O

1 mole of glucose (C6 H12 O6)=180g of C6 H12 O6 = 6.02x1023molecules of C6

H12 O6

1 mole of SO- = 96 of SO

- =6.02x1023 ions of SO

-

1 mole of NO =62g of NO =6.02x1023

ions of NO

One mole of different compounds has different masses but the same

number of particles .

Important Relationships



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 =Number of moles x Avogadro

number

5. Mass of atoms =

6. Mass of molecules =

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

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

Molar mass of H2 O =2+16=18 g mol-1

No of water molecules = ?

Number of molecules of H2 O = x NA

=

3.34x1023 molecules

(ii) Calculation for the number of atoms of hydrogen and oxygen and

total number of atoms:

No of water molecules = 3.34x1023

Now 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, 1 Molecules of H2 O contains O atoms =1.atom



3.34x1023

Molecules of H2O contains O atoms =1x3.34x1023

atoms

of O

=3.34x1023

atoms of

O

Total number of atoms =6.68x 1023

+3.34x1023

=(6.68+3.34) x1023

10.02x1023

atoms

(iii) Calculation for number of covalent bonds:

I Molecule of H2O contains the number of covalent bonds =2

3.34x1023

molecules of H2O contains, the number of covalent bonds

=2x3.34x1023

=6.68x1023

Examples 9: 10.0grams of H3PO4 have been dissolved in excess of water to

dissociate it completely into ions. Calculate.

(a) Number of molecules in 10.0g of H3 PO4

(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



(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

6.14x1023

molecule of contains negative PO- ions =3x6.14x10

22+ve H

+

ions =1.842x1023

+ve H+

ions

Now, 1 molecule of H3 PO4 contains negative PO- ions=1

6.14x1023

molecule of contains negative PO- ions =1x6.14x1022-ve PO

- ions

=6.14x1022 ve PO - ions (c) 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

No of PO- ions =6.14x10

22 ions

Ionic mass of PO- ion =31+64=95g mol

-1

NA =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



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

It should be remember that 22.414 dm3

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.

Solution: (i) Calculation for the number of moles of an ideal gas at

STP:

Volume of ideal gas at STP =500 cm3 =0.5dm3

Now, 22.414 dm3 of ideal gas at STP =

0.5dm3 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.



3. 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:

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

2KOH (aq) + H2SO4(aq) K2SO4(aq) +2H2O(1)

Now, 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

(ii) Calculation for the number of molecules of water:



0.25mol

Equation: 2KOH(eq) + H2SO4(aq) K2SO4(aq) +2H2O(1)

2moles

Now , 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

No. of moles of H2 =

=x 6moles

Mg(s) + 2HCI(aq) Mg C12(aq) +H2(g)

2moles

Now, 1 mole of H2 =2 mole of HCI

Moles of H2 =

=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

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 =1622.2g

Density of HCI solution =1.14g cm -3



Volume of HCI solution = =

=1423 cm3

Limiting Reactant

A reactant that controls the amount of the product formed in a chemical

reaction is called a limiting reactant.

A limiting reactant gives the least number of moles of the product.

Generally, in carrying out chemical reactions m one of the reactants is deliberately

used in excess quantity . This quantity exceeds the amount theoretically required

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

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

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

reaction .

Example:



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.

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.

(a) Calculate the number of grams of NH3 produced.

(b) Calculate the excess amount of reagent left unrelated.

2NH4C1(s) + Ca(OH) 2(s) CaC12(s) +2NH3(s) + 2H2O(1)

Solution: (a) Calculation for the number of grams of NH3

Mass of NH4 C1 =100g

Mass of Ca(OH)2 =100g

Molar mass of NH4 C1 =14+4+35.5=53.5g mol-1

Molar mass of Ca(OH)2 =40+34=74g mol -1

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

2moles

Now, 2molesof NH4CI =2moles of NH3

1.87 moles of NH4CI =

Also, 1 mole of Ca(OH)2 =2moles of NH3

1.35 moles of Ca(OH)2 =



=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

Mass of NH3produced =No of moles NH3xmolar mass of

NH3

=1.87mol x17g mol-1

=31.79g

(b) Calculation for the excess amount of reagent left un reacted

The reactant, Ca(OH)2 is used in excess , its amount left un reacted can be

calculation as follows:

Now, 2moles of NH4C1 =1 mole of Ca(OH)2

1.87 moles of NH4C1 =

=0.935moles of Ca(OH)2

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



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.

Actual yield

The amount of the product actually abtained in a chemical reaction is

called the actual yield of the product .

The actual yield of the product is always less than the theoretical yield of

the product.

Causes of less actual yield

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

because of many short comings.

2. Product may be lost during the processes like filtration, separation

by distillation , separation by a separating funnel , washing ,drying and

crystallization if not properly carried out.

3. Side reaction may occur which reduce the amount of the product.

4. The reaction may not go to completion.

5. 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=



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.

CaCO3(s) CaC(s) +CO2(g)

Solution: Actual yield of CaO =2.5kg =2500g

Mass of lime stone CaCO3 =4.5kg =4500g

Molar mass of CaCO3 =40+12+48=100g mol-1

Molar mass of CaO = 40+16=56g mol -1

45000g

CaCO3(s) CaO(s) +CO2(s)

100g 56g

Now , 100g of CaCO3 =56g of CaO

4500g of CaCO3 =

=2520g pf CaO

Theoretical yield of CaO =2520

%yield =

=

=99.21%

EXERCISE

Q1. 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



(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

(c) atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes

proportional to their relative abundance

(iii) Many elements have fractional atomic masses, this is because

(a) The mass of the atom is itself fractional

(b) Atomic masses are average masses of isobars

(c) Atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes

proportional to their relative abundance

(iv) The mass of one mole of electrons is

a 008mg(b) 0.55mg (c) 0.184mg (d) 1.673mg

(v) 27g of Al will react completely with how much mass of O2 to

produce A12O3

(a) 8 g go oxygen (b) 16g of oxygen

(c) 32g of oxygen (d) 24g of oxygen

(vi) The number of moles of CO2 which contain 8.0 g of oxygen .

(a) 0.25 (b) 0.50 (c) 1.0 (d) 1.50

(vii) 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 (d) 5.4g of N2O5

(viii) One mole of SO2 contains

(a) 6.02x1023

atoms of oxygen



(b) 18.1x1023

Molecules of SO2

(c) 6.02x1023 atoms of sulphur

(d) 4 gram atoms of SO2

(ix) The volume occupied by 1.4 g of N2at STP is

(a) 2.24 dm3

(b) 22.4dm3

(c) 1.12 dm3 (d) 112 cm

3

(x) A limiting reactant is the one which

(a) is taken in lesser quantity in grams as compared to other

reactants

(b) is taken in lesser quantity in volume as compared to the

other

reactants

give the maximum amount of the product which is required

(e) give the minimum amount of the product under

consideration

Ans: (i)a (ii)d (iii)d (iv)b (v)d (vi)a (vii)a (viii)c (ix)c

(x)d

Q2: Fill in the blanks :

(i) The unit of relative atomic mass is-----------

(ii) 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 --------

-----



(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 -----------

--law is obeyed.

Ans: (i) amu (ii) mass (iii) Soddy (iv) carbon, hydrogen

(v) Products (vi)

6x6.02x1023,6x6.02x1023,12x6.02x1023

(vii) 1.505x1023

(viii) conservation

Q3: Indicate true or false as the case my be:

(i) Neon has three isotopes and the fourth one with atomic mass

20.18 amu.

(ii) Empirical formula gives the information about he total number

of atoms present in the molecule

(iii) During combustion analysis Mg(CIO4)2 is employed to absorb

water vapors.

(iv) 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) Avogadros 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.



Ans. (i) False (ii) False (iii) True (iv) false

(v) False (vi) true (vii) False (viii) False

Q4: What are ions? Under What condition are they produced ? can you

explain the places of negative charge in PO , MnO and Cr2 O

Ans: In PO , MnO and Cr2 O the negative charge resides on singly covalent

bonded oxygen because it contains seven electrons three electron pairs and one

electron from covalent bond in its cuter most shell.

(Picture)

Q4: (a) What are isotopes? How do you deduce the fractional atomic

masses of

Elements form the relative isotopes abundance? Give two examples in

support of your answer.

(b) How does a mass spectrograph show the relative aboundace of

isotopes of an element?

What is the justification of two strong peaks in the mass spectrum

for bromine; while for iodine only one peak at 127 amu , is indicated?

Ans 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.



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

Ag 106.90509 51.84

109 Ag 108.90476 48.16

Solution: The mass contribution for silver are:

Isotopes Fractional abundance isotopic mass mass

contribution

107Ag 107 0.5184x107=55.4688

109Ag 107 0.4816x109=52.4944

Fractional atomic mass of silver =107.9632

Hence the fractional atomic mass of silver is =107.9632 Ans

Q6: Boron with atomic number 5 has two naturally occurring isotopes.

Calculate the percentage abundance of 10

B and 11

B from the following

information.

Average atomic mass of boron =10.81 amu

Isotopic mass of 10

B =10.0129 amu

Isotopic mass of 11

B =11.0093

Solution: Let, the fractional abundance of 10

B =x

The fractional abundance of 11

B =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 10

B)(isotopic

mass of 10

B )+(fractional abundance of 11

B) (isotopic mass of 11

B)



=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

x =

Fractional abundance of 10

B =0.2000

Fractional abundance of 11

B =(1-0.2000)=0.8000

By percentage the fractional abundance of isotope is

%of 10

B =0.2000x100 =20% Answer

% of 11

B =0.8000x100 =80%Answer

Q7: Define the following terms and give three examples of each.

(i) Gram atom (ii) Gram molecular mass (iii) Gram

molecular mass (iv) Gram ion (v) molar

volume (vi) Avogadros number (vii)

Stoichiometry (viii) Percentage yield

Q8: Justify the following statements:

(a) 23 g of sodium and 238g of uranium have equal number of atoms in

the (b) Mg atom is twice heavier than that of carbon

(c) 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.



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

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.

(b) Since the atomic mass of Mg (24) is twice the atomic mass of carbon

(12) therefore, Mg atom is twice heavier than that of carbon. Or

Mass of 1 atom of Mg=

Mass of 1 atom of C =

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

glucose 342 g og sucrose=1mole of sucrose =6.02x1023

molecules of

sucrose

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, C12 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 2H+

+ SO



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

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

number of negatively charged ions.

(e) H2 SO4 2H+

+ SO

K2 Cr O4 when ionizes in water produces two k+

ions one C O ion. Thus

each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2

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

=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

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 .



(e) 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 =2.74moles

formula mass of KMnO4 =39+55+64=158g mol -1

Mass of KMnO4 =?

Formula used:

Mass of KMnO4 = no of mole of KMnO4 x formula mass of KMnO4

=2.74 mol x 158 g mol-1

=432.92g Answer

(b) Mass of Mg (NO3)2 =9g

Formula mass of Mg (NO3)2 =24+28+96=148g mol -1

No of moles of O atoms =?

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 =

=0.36 mole Answer

(c) Mass of CuSO4. 5H2O=10.037g

Formula mass of CuSO4. 5H2O=63.54+32+64+90

=249.546g mol -1



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

Now, I mole of O atoms contains =6.02x1023

O atoms

0.36 mole of O atoms contains =6.02x1023

x0.36 oxygen

atoms

=2.17x1023

oxygen atoms

=2.17x1023

atoms Answer

(d) No of molecules of SO2 . =2.6x1020

molecules

Molecular mass of SO2 . =32+32=64 g mol-1

Now, Avogadros number , NA =6.02x1023

molecules of SO2

Mass of SO2 molecules =

=

=27.64x10-3 g

=

=27.64x10-6

kg

=2.764x10-3

kg Answer

(e) Mass of C2 H4C1 = 0.822g

Molecular mass of C2 H4C1 =24+4+71=99 g mol-1

No of moles of C2 H4C1 =

Now, 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

(f) No of mole of Ag2 CO3 =5.136moles

Formula mass of Ag2 CO3 =215.736+12+48=275.736 g

mol-1

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

(g) Molecular mass of CrO2C12 =52+32+71=155g mol-1

NA =6.02x1023

molecules mol-1



Molecules of CrO2C12==2.78x1021

molecules

Now, mass of CrO2C12 =

=

=71.578x10-2

g

=0.71578

=0.716 g Answer

(h) Mass of KCIO3 =100g

Formula mass of KCIO3 =39x35.5+48=122g mol-1

No of moles of KCIO3 =?

No of moles of KCIO3 =

= =0.816mole Answer

No of formula units No of moles x Avogadro,s No

=0.816mole x 6.02x1023

formula units

=4.91x1023

formula units

(i) No of K+

ions =4.91x1023

Answer

No of CIO ions =4.91x1023

Answer

No of CIO ions =4.91x1023

Answer

No of O atoms = 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

No of moles of aspartame =

=

=0.1768 mol



=0.177 mol Answer

(c) 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

(d) Mass of aspartame =243g

Molar mass of aspartame =294g mol -1

No of molecules of aspartame=?

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

=89.64x1021

H 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 =0.6x2=1.2mol Answer

Mass of F =No of moles of Fx At . mass

of F

=1.2x19=22.8g

Mass of compound =46.8g



Mass of metal, M =46.8-22.8

=24

At mass of M =

=

(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 235

U or 6.02x1023 atoms of 238

U

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 =



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

No of formula units of NaC1O3 =2.3x6.02x10

23=1.38x10

24 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

NaC1o3-

MgC1 Answer

(f) No of molecules = NA

No of molecules in H2 O2 = x6.02x1023

=3.68x1023

molecules

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

(g) No of formula units =moles xNA

No of formula units NaBr =0.5x6.02x1023

=3.01x1023

formula

units

One formula units o NaBr contain Na+

ions =1

3.01 x1023

formula unit of NaBr contains Na +ions =3.01x10

23 Na

+ ions

No of formula units of NaC1 = x6.02x1023

=1.49x1023

formula units

One formula unit of NaC1 contains Na+

ions =1

1.49x1023

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 =



Mass of 235

Uatoms =x6.02x1023

=235g

Mass of 238

U atoms =x6.02x1023

=238g 238

U Answer Q 13: (a) Calculate the percentage of nitrogen in the four important fertilizer

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 =



%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

3.4x1022

molecules of glucose contains =6x3.5x1022

C-atoms

=21x1022

=2.1x1023

152096955 long questions basic concept

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