Basic Concepts Notes
Chapter 1 Basic concept notes and solved examples
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.
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+, Ca2+ , NH , Cl-, O2- , NO,CO-, N,CO+,CH
“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-
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.
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 ofC is 12,000 amu and relative atomic mass of 
H is 1.0078 amu .
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
|
12C, 13 C
|
98.893, 1.107
|
Nitrogen
|
14 N , 15 N
|
99.634, 0.366
|
Oxygen
|
16 O , 17 O ,18 O
|
99759. 0.037, 0.204
|
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.
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.
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 10-6 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 .
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.
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.
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.
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
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=
% 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 H4 O3 Answer
Empirical Formula From Combustion analysis
The empirical formula of organic compounds which only contain carbon, hydrogen and oxygen can be determined by combustion, the two products of combustion will be CO2 and H2O.These products of combustion are separately collected and their masses are determined.
Combustion Analysis
A weighed sample of the organic compound is placed in the combustion tube fitted in a furnace. An excess of pure oxygen is supplied to burn the compound. The carbon in the compound is converted to 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 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 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.”
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 H12O6
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 :
H3PO43H+ +PO-
Now,1 molecule of H3PO4 contains positive H+ ions =3
6.14x1023 molecule of contains negative PO- ions =3x6.14x1022+ve H+ions =1.842x1023+ve H+ ions
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 =6.14x1022 ions
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 .
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%
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