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^-10m which is0.2 nm. Masses of atoms range from 10^-27to 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^-24g=1.661x10^-27kg

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,C₁₂H₂₂O₁₁.

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^-, 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⁺,Ca^2+, Mg ²⁺, A1³⁺, Sn⁴⁺, NH , H₃O ⁺

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, Cr₂O ^-, 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 (1/12th)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 ofC is 12,000 amu and relative atomic mass of H is 1.0078 amu .

Remember that: 1amu=1.66x10 ^-24g.

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

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	¹²C, ¹³ C	98.893, 1.107
Nitrogen	¹⁴ N , ¹⁵ N	99.634, 0.366
Oxygen	¹⁶ O , ¹⁷O ,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 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

“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^-6to 10^-7torr, 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 ²⁰Ne)+(fractional abundance of ²¹ Ne ) (isotopic mass of ²¹ Ne)+(fractional abundance of ²²Ne )(isotopic mass of ²²Ne).

=(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 = Mass of element in compound x100

Formula mass of the compound

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 , A_xB_y, 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 = C³H⁴O³Answer

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 CO²and 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 CO²and hydrogen to H²O vapors. These gases are passed through two pre-weighted absorbent tubes. One of the tubes contain Mf(CIO₄)²which 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 CO₂and 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			C₂H₆O
O	34.79	16.0

Empirical Formula = C₂H₆O

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: H₂O₂(hydrogen peroxide) , C₆H₆(benzene ) , C₆H₁₂O₆( glucose ).

The empirical formulas of hydrogen peroxide, benzene and glucose are HO, CH and CH₂O respectively. Some of the examples of the compounds having the same empirical and molecular formulas are: H₂O, CO ₂, NH₃, CH ₄and C₁₂H₂₂O₁₁.

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			C₂H₆O
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)

=C₆H₆O₂

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 (H₂SO ₄)=98.0g

1 gram molecule of sucrose (C₁₂H₂₂O 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 Na₂CO₃ =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^-1mol

(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^-3moles of Mg SO ₄.

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

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), N_A)

“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 =6.02x10²³atoms of H

1 mole of sodium(Na) =23g of Na =6.02x10²³atoms of Na

1 mole of water (H₂O) =18g of H₂O =6.02x10²³molecules of H₂O

1 mole of glucose (C₆H₁₂O₆)=180g of C₆H₁₂O₆ = 6.02x1023molecules of C₆H₁₂O₆

1 mole of SO^- = 96 of SO^- =6.02x1023 ions of SO^-

1 mole of NO =62g of NO =6.02x10²³ions 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 H₂O =2+16=18 g mol^-1

No of water molecules = ?

Number of molecules of H₂O = x N_A

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 H₂O contains H atoms =2atom

3.34x10²³molecules of H₂O contains H atoms =2x3.34x10²³atoms of H

=6.68x10²³atoms of H

Now, 1 Molecules of H₂O contains O atoms =1.atom

3.34x10²³Molecules of H₂O contains O atoms =1x3.34x10²³atoms of O

=3.34x10²³atoms of O

Total number of atoms =6.68x 10²³+3.34x10²³

=(6.68+3.34) x10²³

10.02x10²³atoms

(iii) Calculation for number of covalent bonds:

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

3.34x10²³molecules of H₂O contains, the number of covalent bonds =2x3.34x10²³

=6.68x10²³

Examples 9: 10.0grams of H₃PO₄have 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.

(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 =xN_A

=6.14x10²²molecules

(b) Calculation for the number of positive and negative ions in H₃PO₄:

H₃PO₄3H+ +PO^-

Now,1 molecule of H₃PO₄ contains positive H⁺ions =3

6.14x10²³molecule of contains negative PO^-ions =3x6.14x10²²+ve H⁺ions =1.842x10²³+ve H⁺ions

Now, 1 molecule of H₃PO₄contains negative PO^- ions=1

6.14x10²³molecule 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 =1.842x10²³ions

Ionic mass of H⁺=1.0008 g mol^-1

N_A =6.02x10²³ions mol^-1

Mass of H+ ions =

=0.308 g

No of PO^-ions =6.14x10²²ions

Ionic mass of PO^-ion =31+64=95g mol ^-1

N_A =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 H₃PO₄ gives positive charges in solution =3

6.14x1022 molecule of H₃PO₄ 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 dm³occupied 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

2.6.02x10²³molecules 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 dm³of 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 cm³at 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 dm³of ideal gas at STP =

0.5dm³of 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 K₃PO₄and 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) + H₂SO_4(aq) K₂SO_4(aq)+2H₂O₍₁₎

Now, 2moles of KOH =1 mole of K₂SO₄

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) K₂SO_4(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 10²³molecules mol^-1

=1.51x10²³molecules

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 C1_2(aq)+H_2(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 C1_2(aq)+H_2(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: NH₃gas can by prepared by heating together two solids NH₄C1 and C_a(OH)₂. If a mixture containing 100g of each solid is heated then.

(a) Calculate the number of grams of NH₃produced.

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

2NH₄C1_(s)+ Ca(OH) _2(s)CaC1_2(s)+2NH_3(s)+ 2H2O₍₁₎

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

Mass of NH₄C1 =100g

Mass of Ca(OH)₂ =100g

Molar mass of NH₄C1 =14+4+35.5=53.5g mol^-1

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

No of moles of NH₄C1 =

No of moles of Ca(OH)₂ =

_{1.87 moles

.35 moles}

2NH₄C1_(s)+ Ca(OH) _2(s)CaC1_2(s)+2NH_3(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 NH₃produced =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)₂is 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 NH_3..

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,CaCO₃is 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.

CaCO_3(s) CaC_(s)+CO_2(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

CaCO_3(s) CaO_(s)+CO_2(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

(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

(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

(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 O₂to produce A₁2O₃

(a) 8 g go oxygen (b) 16g 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 H₂O (b) 4.8g of C₂H₅OH

(viii) One mole of SO₂contains

(a) 6.02x10²³atoms of oxygen

(b) 18.1x10²³Molecules of SO₂

(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

(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

(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 CH₄at 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)₂is 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 dN₂are 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.

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, MnOand Cr₂O

Ans: In PO, MnOand Cr₂O 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?

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

¹⁰⁷Ag 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 ¹⁰B and ¹¹B from the following information.

Average atomic mass of boron =10.81 amu

Isotopic mass of ¹⁰B =10.0129 amu

Isotopic mass of ¹¹B =11.0093

Solution: 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)

=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 ¹⁰B =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 =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) Avogadro’s 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 H₂SO₄when 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 K₂C_rO₄has 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 CO₂occupy 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 =6.02x10²³atoms of Na

238g of U =1 mole of U =6.02x10²³atoms 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 ,6.02x10²³atoms.

(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 =6.02x10²³molecules of glucose 342 g og sucrose=1mole of sucrose =6.02x10²³molecules 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 , C₆H₁₂O₆contains 45 atoms whereas one molecules of glucose, C₁₂H₂₂O₁₁contains 24 atoms. Therefore , 6.02x10²³molecules of glucose contain different atoms as compound to6.02x10₂₃molecules 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) H₂SO₄ 2H⁺+ SO

When one molecules of H₂SO₄completely ionizes in water it produces two H⁺ion and one SOion ,.Hydrogen ion carries a unit positive charge whereas SOion 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 H₂SO₄in 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) H₂SO₄ 2H⁺+ SO

K₂C_rO₄when ionizes in water produces two k⁺ions one C O ion. Thus each formula unit of K₂C_rO₄produces three ions in solution .Hence one mg of K₂C_rO₄ has thrice the number of ion than the number of formula units ionized in water.

(f) 2g of H₂=1 mole of H₂=6.02x10₂₃molecules of H₂at STP =22.414dm³16g of CH₄=1mole of CH₄=6.02x10₂₃molecules of CH₄at STP =22.414dm³144 g of CO₂=1mole of CO₂=6.02x10₂₃molecules of CO₂at STP =22.144dm³

Although H₂, CH₄and CO₂have 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 CH₄and 44 g of CO₂occupy the same volume 22.414 dm³at 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.

(d) Mass in kilograms of 2.6x 10²⁰molecules 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 2.78x10²¹molecules of CrO₂C1₂.

(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=158gmol ^-1

Mass of KMnO₄=?

Formula used:

Mass of KMnO_{4
=}no of mole of KMnO₄x 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

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 =6.02x10²³O atoms

0.36 mole of O atoms contains =6.02x10²³x0.36 oxygen atoms

=2.17x10²³oxygen atoms

=2.17x10²³atoms Answer

(d) No of molecules of SO₂. =2.6x10²⁰molecules

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

Now, Avogadro’s number , N_A =6.02x10²³molecules of SO₂

Mass of SO₂molecules =

=27.64x10-3 g

=27.64x10^-6kg

=2.764x10^-3kg 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^-3mole of atom

=16.6x10^-3

=0.0166mole of C1 atom

=0.017 mole Answer

(f) No of mole of Ag₂CO₃ =5.136moles

Formula mass of Ag₂CO₃=215.736+12+48=275.736 g mol^-1

Mass of Ag₂CO₃=No of moles of Ag₂CO₃xformula mass of Ag₂CO₃

=5.136molx275.736 g mol^-1

=416.18g

=1416.2 g Answer

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

N_A=6.02x10²³molecules mol^-1

Molecules of CrO₂C1₂₌=2.78x10²¹molecules

Now, mass of CrO₂C1₂ =

=71.578x10^-2g

=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 6.02x10²³formula units

=4.91x10²³formula units

(i) No of K⁺ions =4.91x10²³Answer

No of CIO ions =4.91x10²³Answer

No of O atoms = 4.91x10²³x3

=14.73x10²³=1.473x10²⁴Answer

Q 11 Aspartame he artificial sweetener, has a molecular formula of C₁₄H₁₈N₂O₅.

(a) What is the mass of one mole of aspartame?

(b) How many moles are present in 52g 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^-1Answer

(b) Mass of aspartame =52g

Molecular mass of aspartame =294g mol^-1

No of moles of aspartame =

=0.1768 mol

=0.177 mol Answer

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=xN_A

=4.98x10²¹molecules.

Now,1 molecule of aspartame contains=18 H atoms

4.98x 1022 molecules =18x4.98x10²¹H 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 MF₂that 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

(d) Individual particles: 4.0g N₂O₄or 3.3g SO₂

(e) Total ions: 2.3 moles of NaC1O₃or 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: 6.02x10²³atoms of ²³⁵U or 6.02x1023 atoms of ²³⁸U

Ans:

(a) Number of molecules =moles x N_A

Number of O₂molecules =0.4x6.02x10²³=2.408x10²³molecules

No of O atoms=0.4x6.02x10²³=2.108x10²³atoms

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, O₃molecules =0.4x48=19.2g

0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms.

Ozone Answer

Mass of 1₂ =0.6x127=254g

0.6mole of 12 have larger mass than 0.6 mole of C₂H₄

1₂Answers

(d) No of molecules =

No of molecules in N₂O₄=x6.02x10²³ =2.62 x10²³molecules

No of molecules in SO₂=x6.02x10²³ =3.1x10²²molecules

3.3g of SO₂have larger number of individual particles than 4.0 g of N₂O₄.

SO₂Answer

(e) No of formula units =Moles x N_A

No of formula units of NaC1O³ =2.3x6.02x10²³=1.38x10²⁴formula units

No of ions in 1 formula units of NaC1O³=2

Total no of ions in MgC1₂ =2x1.38x10²³=2.76x10²⁴ions

No of formula units of MgC1₂ =2.0x6.02x10²³x3=3.6x10^{24 ions}

No .of ions in one formula unit of MgC1₂ =3

Total no of ions in MgC1₂ =1.20x10²⁴x3=3.6x10²⁴ions

2.0moles of MgC1₂ contain lager number of total ions than 2.3 moles of NaC1o^3-

MgC1Answer

(f) No of molecules =N_A

No of molecules in H₂O₂ =x6.02x10²³=3.68x10²³molecules

No of molecules in H₂O_{2
=}x6.02x10²³=1.95x10²³molecules

11.0g of H₂O₂contains larger number of molecules than 11.0g of H₂O₂

H₂O₂Answer

(g) No of formula units =moles xN_A

No of formula units NaBr =0.5x6.02x10²³=3.01x10²³formula units

One formula units o NaBr contain Na⁺ions =1

3.01 x10²³formula unit of NaBr contains Na ⁺ions =3.01x10²³Na⁺ions

No of formula units of NaC1 =x6.02x10²³=1.49x10²³formula units

One formula unit of NaC1 contains Na⁺ions =1

1.49x10²³formula units of NaC1 contains =1.49x10²³Na⁺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 ²³⁵Uatoms =x6.02x10²³=235g

Mass of ²³⁸U atoms =x6.02x10²³=238g

²³⁸U 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)NH₄NO₃

(b) Calculate the percentage of nitrogen and phosphorus in each of the following:

(i) NH₄H₂PO₄ (ii) (NH₄₎) PO₄(iii) (NH₄)₄PO₄

Solution:

(a) Mol-mass of NH₃ =14+4=17g

Mass of N =14g

% of N =x100

=82.35% Answer

(b) Mol-mass of NH₂CONH₂ =28+4+12+16=60g

Mass of N =28g

%of N =x100

=46.35% Answer

Mass of N =28g

% of N =x100

=21.21% Answer

(d) Mol-mass of (NH₂)₂SO₄=28+4+48=80g

Mass of N =28g

%of N =x100

=35% Answer

(I) Mol-mass of (NH₂)₂SO₄=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)₂SO₄=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)₂SO₄=42+12+31+64=149g

Mass of N =42g

Mass of P =31g

%of N =

%of P =

Q 14: Glucose C₆H₁₂O₆is 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 C₆H₁₂O₆ =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 C₆H₁₂O₆=10.5g

Mol-mass of C₆H₁₂O₆=180g

Mol-mass of =180g mol^-1

No of moles of C₆H₁₂O₆=

No of molecules of glucose =No of moles x N_A

=0.058 molx 6.02x10²³molecules mol^-1

=0.35x10²³molecules

=3.5x10²²molecules

Now, 1 molecule of glucose contains =6C-atoms

3.4x10²²molecules of glucose contains =6x3.5x10²²C-atoms

=21x10²²=2.1x10²³C atoms Answer

1 molecules of glucose contains =12H-atoms

3.5x10²²molecules glucose contains =12x3.5x10²²

=4.2x10²³H- atoms Answer

1 molecule of glucose contains =6 O –atoms

3.5 x 10²²molecules of glucose contains =6x3.5x10²²

=2.1x10²³O-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^-1At. Mass of H=1.008g mol^-1At. Mass of O =16g mol^-1

No of moles of C =

No of moles of H =

No of moles of O =

Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest ratio.

C :H :O

1 :3 :1

Empirical formula =CH³O

Empirical formula mass =31

Molecular formula =2x CH₃O

=C₂H₆O₂Answer

Q 16: Serotonin (Molecular mass= 176g mol^-1) 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

10 : 12 : 2 : 1

Empirical formula =C₁₀H₁₂N₂O

Empirical formula mass =120+12+28+16=176g mol^-1

Molecular mass =176g mol^-1

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 M₂S₃.

Solution:

Formula of compound = M₂S₃

Mass of M =3.12g

Mass of S =2.88g

Atomic mass of S =32g mol^-1

No of moles of S =

The molar ratio of M: S in the compound is :

No of moles of M =

=0.06 mole

Now, No of moles of M =

At. Mass M =

The mass of M used in the formation of M₂S₃is 3.12g. The product M₂S₃therefore 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:

At. Mass of M =

=52

Atomic number, Z =52

Q19: The octane present in gasoline burns according to the following equation.