Rapid Scan of Chapter # 1 (Stoichiometry)
The number of atoms present
in a molecule is called its atomicity.
1 amu = 1/NA
= 1 Avogram = 1/6.02 × 1023 g = 1.66 × 10-24 g = 1.66 ×
10-27 kg = 1 dalton
Reason of Choosing C-12 as standard
Average Atomic Mass
An element’s atomic mass is a weighted average of the isotopic masses of the element’s naturally occurring isotopes as compared to the mass of one atom of C-12. OR The atomic weight or relative atomic mass (Ar) of an element is a statistical mean of the atomic masses of different isotopes of the element i.e. an element’s relative atomic mass is a weighted average of the isotopic masses of the element’s naturally occurring isotopes.
Examples
1. Naturally occurring carbon is composed of two isotopes, 12C with 98.892 % (98.889%) abundance (0.98892 fraction) and 13C with exact mass 13.0034 amu with 1.108 % (1.111%) abundance (0.01108 fraction). Thus the average isotopic mass i.e. the atomic weight of a large collection of carbon atoms is 12.011 amu.
Average atomic weight of C =(0.98892) (12) + (0.01108) (13.0034) = 11.867+ 0.144 = 12.011 amu
2. Naturally
occurring chlorine is a mixture of two isotopes; 35Cl which has an
abundance of 75.53% or roughly 75% (0.7553 fraction)
and an isotopic mass of 34.969 (35) amu and 37Cl which has an
abundance of 24.47% or roughly 25% (0.2447 fraction)
and an isotopic mass of 36.966 (37) amu. Thus the average isotopic mass i.e.
the atomic weight of a large collection
of chlorine atoms is 35.46 amu.
Atomic Mass, Molecular Mass, Formula Mass, Average Atomic Mass and Gram Particle Masses
1. The
average of masses of all isotopes of an element is called average or fractional atomic mass of that
element
2. The
fractional atomic mass of an element depends on the natural
abundance of isotopes and number of
isotopes of that
element.
3. Atomic mass = Mass of one atom = Mass of
Proton + Mass of neutron (P+n) = Mass number (A)
4. Gram atomic mass = mass of one mole of atoms = mass in g of 6.02 × 1023 atoms = Atomic
mass in gram
5. Atomic mass of O is 16 amu, so gram atomic
mass of O is 16 g.
6. 16 g O contains 6.02 × 1023 oxygen atoms
7. Atomic mass = Average mass of an atom/1/12 mass of on atom of C-12
8. Mass of an atom = atomic mass/Avogadro’s
number
9. Mass of an atom = atomic mass x 1.661 × 10-24
g.
10. Average
atomic mass = Relative abundance ×
mass no. + Relative abundance × mass no./total abundance (100)
11. Molecular
mass (weight) = Mass (weight) of one molecule = Sum of atomic masses
of all atoms
12. Gram
molecular mass = mass of one mole of molecules = mass in g of 6.02 ×
1023 molecules = molecular mass in g.
13. Molecular
mass = 2 × vapour density (V.P).
14. Vapour
density = (Mass in g × 22400)/volume in mL at STP
15. Mass of a
molecule = molecular mass/Avogadro’s number
16. Mass of one
Mg atom = 24/6.02 × 1023 = 3.9867 × 10-23 g
17. Formula mass
(weight) = Mass (weight) of one formula unit = Sum of atomic masses
of all atoms in ionic compound
18. Gram Formula
mass = mass of one mole of formula units = mass in g of 6.02 × 1023
formula unit = formula mass in g.
19. Gram formula is the gram formula mass of ionic
compounds.
20. One gram formula of NaCl = 58.5 g = 6.02 × 1023
formula unit = 2 × 6.02 × 1023 ions.
21. 35.5 amu is the average atomic mass of chlorine (showing the composition of isotopic mixture of chlorine)
1. Avogadro’s number is the number of
particles present in one mole of a
substance.
2. Avogadro’s number is denoted by NA
3. 6.02 × 1023
represents Avogadro’s number.
4. Number of
particles = mole (n) × NA
5. Number of
particles = (mass/molar mass) × NA
6. Number of
particles =(volume at STP)/molar volume × NA
7. Total
number of atoms = mole (n) × NA × atomicity (no. of atoms
in one molecule).
8. Total
number of electrons = mole (n) × NA × total number of
electrons in one molecule.
9. 64 g of
SO2 contains 6.023 × 1023 number of molecules.
10. The number
of carbon atoms in 1 mole of sugar (C12H22O11)
are approximately 72 × 1023 (12NA).
11. The number
of carbon atoms in 180 g of glucose (C6H12O6)
are approximately 36 × 1023 (6NA).
12. The number
of carbon atoms in 0.5 mole of sugar (C12H22O11)
are approximately 36 × 1023 (6NA).
13. 20g
calcium contains the same number of atoms as that of 16 g of S.
14. 20g
calcium contains the same number of atoms as that of 6 g of C.
15. 20g
calcium contains the same number of atoms as that of 12 g of Mg.
16. 20g calcium contains the same number of atoms as that of 28 g of Fe.
17. Number of hydrogen atoms present in one mole of water is 1.204 × 1024.
18. Number of ionizable hydrogen ions given by 1 mole of H2SO4 in water = 2 × 6.02 × 1023 or 1.204 × 1024.
19. 1 mole of FeCl3 contains total ions = 4NA ions = 24.08 × 1023 ions = 2.408 × 1024 ions.
Mole Concept (Mole, Avogadro’s no., molar volume, molar mass and other basics topics)
1. 1 mole = NA × number of species = 6.02 × 1023
species (atoms, molecules, ions, formula units etc.)
2. Mole is
derived from Latin means heap or pile
3. Mole is the SI unit for the amount of
substance of specific number of particles.
4. A mole is the gram atomic mass or gram molecular
or gram formula mass of any substance (element or covalent compound or ionic compound) which
contains 6.02 × 1023 particles (atoms, molecules, ions etc.).
5. Mole represents the number of chemical
entities in a fixed mass.
6. One moles is the amount of substance which
contains as many elementary particles as there are in 0.012 kg of 12 g of C-12.
7. Number of
moles (n) or gram atom or gram molecule or gram formula = mass in gram/molar mass
8. Number of
moles (n) = Number of particles (NP)/Avogadro’s
number (NA)
9. Number of
moles (n) = Volume of gas at STP (Vg)/Molar
volume (VM)
10. 1 mole of hydrogen atom (H) weighs 1 g which contains 6.02 × 1023 atoms.
11. 1 mole of hydrogen molecule (H2) weighs 2
g which
contains 6.02 × 1023
molecules
12. 1 mole of sodium chloride (Na+Cl−) weighs 58.5
g which
contains 6.02 × 1023
+ 6.02 × 1023 ions
13. one mole of carbon and one mole of
magnesium contains same number of
atoms.
14. Ionic
compound is assumed to consist of formula units.
15. One of
mole MgCl2 contains 6.02 × 1023 formula
units.
16. one mole of MgCl2 contains one mole of Mg2+ ion and two moles of Cl− ions.
17. one mole of MgCl2 contains 6.02 × 1023 Mg2+ ions
and 12.04 × 1023 (2 x 6.02 × 1023)
Cl− ions.
18. one mole of FeCl3
contains 6.02 × 1023 FeCl3
formula units
19. one mole of FeCl3
contains 6.02 × 1023 Fe3+
ions
20. one mole of FeCl3
contains 3(6.02 × 1023) Cl−
ions
21. one mole of H2O contains 6.02 × 1023 H2O molecules
22. one mole of H2O
contains 6.02 × 1023 O atoms
23. one mole of H2O
contains 2(6.02 × 1023) H atoms
24. Number of moles = mass (g)/molar mass (gmol-1)
= Np/NA = volume of gas at STP or RTP/molar volume
(22.4Lmol-1)
25. Number of moles = Molarity × volume in L
26. Millimoles = Molarity × volume in mL
27. Number
of hydrogen atoms in one mole of water = 2(6.02 × 1023) or 1.204 × 1024 H atoms H
atoms
28. Number
of electrons in one coulomb of charge = 6.25 ×
1018 (1/1.6022 × 10-19).
28. Charge
on one electron = 1.6022 × 10-19
C.
29. Charge
on one mole electrons = 96485 C (1 faraday or
1F).
30. Moles = % age of element/atomic mass
Different Amount shown by 1 mole of water (H2O)
1. 18 g of water
2. 6.02 × 1023 water molecules
3. 18.06 × 1023 atoms or 1.806 ×
1024 atoms or 3 × 6.02 × 1023 atoms or 3NA
atoms
4. 2 mole of H-atoms
5. 1 mole of O-atoms
6. 2 g of H
7. 16 g of O
8. 12.04 × 1023 H-atoms or 2 ×
6.02 × 1023 H-atoms or 2NA hydrogen-atoms
9. 6.02 × 1023 O-atoms or 1 ×
6.02 × 1023 O-atoms or 1NA oxygen-atoms
10. 12.04 × 1023 covalent bonds or 2
× 6.02 × 1023 covalent bonds or 2NA covalent bonds
11. 10 mole electrons
12. 60.2 × 1023 electrons or 6.02 ×
1024 electrons or 10 × 6.02 × 1023 electrons or 10NA
electrons
13. 1 mole H+ ions on
auto-ionization
14. 6.02 × 1023 H+ ions
or 1 × 6.02 × 1023 H+ ions or 1NA H+
ions on auto-ionization
15. 1g mole of H+ ions on
auto-ionization
16. 1 mole of OH- ions on
auto-ionization
17. 17 g of OH- ions on
auto-ionization
18. 6.02 × 1023 OH- ions
or 1 × 6.02 × 1023 OH- ions or 1NA OH-
ions on auto-ionization
19. 2 mole of total ions on auto-ionization (1
mole H+ and 1 mole OH- ions)
20. 12.04 × 1023 or 2 × 6.02 × 1023
or 2NA of total ions on auto-ionization (1 mole H+ and 1
mole OH- ions)
Different Amount shown by 1 mole of Glucose (C6H12O6)
1. 180 g of glucose
2. 6.02 × 1023 molecules
3. 144.48 × 1023 atoms or 1.4448
× 1025 atoms or 24 x 6.02 × 1023 atoms
4. 6 mole of C-atoms
5. 12 mole of H-atoms
6. 6 mole of O-atoms
7. 72 g of C
8. 12 g of H
9. 72 g of O
10. 36.12 × 1023 C-atoms or 6 x 6.02
× 1023 C-atoms or 6NA carbon-atoms
11. 72.24 × 1023 H-atoms or 12 x
6.02 × 1023 H-atoms or 12NA hydrogen-atoms
12. 36.12 × 1023 C-atoms or 6 x 6.02 × 1023 C-atoms or 6NA oxygen-atoms
Different Amount Shown By 1 Mole Of Carbon dioxide (Co2)
1. 18 g of Co2
2. 6.02 × 1023 Co2 molecules
3. 18.06 × 1023 atoms or 1.806 ×
1024 atoms or 3 x 6.02 × 1023 atoms or 3NA
atoms
4. 1 mole of C-atoms
5. 2 mole of O-atoms
6. 12 g of C
7. 32 g of O
8. 6.02 × 1023 C-atoms or 1 ×
6.02 × 1023 C-atoms or 1NA carbon-atoms
9. 12.04 × 1023 O-atoms or 2 ×
6.02 × 1023 O-atoms or 2NA oxygen-atoms
10. 12.04 × 1023
double covalent bonds or 2 × 6.02 × 1023 double covalent bonds or 2NA
double covalent bonds
11. 22 (6+8+8) mole electrons
12. 132.44 × 1023 electrons or 22 ×
6.02 × 1023 electrons or 22NA electrons
13. 22.4 dm3 at STP (22400 cm3
or 0.0224 m3)
14. 24 dm3 at RTP (24000 cm3 or 0.024 m3)
1. The mass
in grams of one mole of any pure substance is known as molar mass.
2. Molar
mass is expressed in grams per mole (g/mol).
3. The molar
mass of Na2CO3 is 106 g/mol.
4. The molar
mass of glucose (C6H12O6) is 180 g/mol.
5. The molar
mass of sugar (C12H22O11) is 342 g/mol.
6. The molar
mass of acetic acid (CH3COOH) is 60 g/mol.
7. The molar
mass of ethanol (C2H5OH) is 46 g/mol.
8. The molar
mass of KMnO4 is 158 g/mol.
9. The molar
mass of FeSO4.7H2O is 278 g/mol.
10. The molar
mass of Mohr’s salt (FeSO4.(NH4)2SO4.6H2O
is 392 g/mol.
11. The molar
mass of oxalic acid (H2C2O4.2H2O is
392 g/mol.
12. The molar
mass of CaCl2 is 111 g/mol.
13. The molar
mass of Al2O3 is 102 g/mol.
14. The molar
mass of Al2(SO4)3 is 342 g/mol.
15. The molar
mass of Fe2(SO4)3 is 400 g/mol.
16. 75.0 g of
table salt contains number of moles of NaCl 1.28.
17.Mass of 1 mole of electrons = 9.11 × 10-31 × 6.02 × 1023 = 5.48 × 10-7 kg = 0.548 mg
1. Molar volume is the volume of one mole of an
ideal gas at STP.
2. STP stands for standar temperature which is 0oC (273K) and standard pressure which is 1
atm (760 torr).
3. At
STP, volume of one mole of an idealgas is 22.414
dm3 or 22414 cm3 or 0.0224 m3.
4. At
Room temperature and presssure (RTP), molar volume is 24 dm4 or 24000cm3.
5. Molar
volume = volume of 1 mole of any gas at STP or NTP = 22.4 dm3 or
22400 cm3 or 0.0224 m3
6. Molar volume = V/n = RT/P
7. Molar volume = Molar mass/density at STP
8. 32
g O2 = 1 mol of O2 = 6.02 ×1023 molecules = 2
× NA atoms of oxygen = 22.414 dm3 of O2 at STP
9. Molar volume is obtained by dividing molar
mass of gas with its mass density.
10. The volume
of 1 gm of H2 gas at STP is 11.2 dm3.
11. Volume of
1 mole of chlorine gas at STP is 22.4 dm3.
12. 5.6 dm3
of N2 gas at STP weighs 7 g.
13. 1 liter of
H2 gas at STP weighs 0.089 g.
14. 3.2 g of O2
gas at STP has a volume 2.24 dm3.
15. 35.5 g of
Cl2 gas at STP has a volume 11.2 dm3.
16. The volume occupied by 2.8 g of nitrogen gas at STP is 2.24 dm3
stoichiometry (Stoichiometric relations, Limiting reactant, Yield and its types)
1. The
word stoichiometry derives from two Greek words; stoicheion meaning “element” & metron meaning “measurement”.
2. For
stoichiometric calculations, it is assumed that all reactants must be converted into products.
3. For
stoichiometric calculations, it is assumed that no
side reaction takes place.
4. For
stoichiometric calculations, it is assumed that law
of conservation of mass and law of constant composition must be obeyed.
5. The reactant which is entirely consumed
first during chemical reaction is called limiting
reactant/limiting reagent.
6. limiting reactant is
that which gives least number of moles of product
7. Limiting
reactant is present in smaller
amount than stoichiometric amount.
8. Limiting
reactant controls the quantity of product and controls the reaction.
9. The
reactant that is not completely consumed is called as excess reactant.
10. The amount of the product obtained as a
result of the chemical reaction is called yield.
11. There are three
types of yields namely actual yield, theoretical yield and percent yield.
12.The maximum amount of the product calculated
from the balanced chemical equation by using its limiting reactant (given amount of
reactant) is known as theoretical /
stoichiometric yield/calculated/expected yield.
13. theoretical
yield of a reaction is always greater
than the actual yield of the same reaction.
14. The actual
amount of product which is formed in an experiment is called practical/experimental/actual yield.
15. The actual yield of a reaction is always less than the theoretical yield
16.The efficiency of a chemical reaction can be
checked by calculating its percentage yield
17. percentage yield is
expressed by comparing the actual yield and theoretical yields.
18. The ratio of practical yield to theoretical
yield is referred as percent yield
19. Stoichiometric
calculation based on chemical equation provides us estimation about theoretical yield.
20. %age yield =
Efficiency = actual yield/theoretical yield × 100
Uncertainty in Measurement (SF, Exponential notation, basics topics and other topics of measurement)
1. The digits in a number which show
reliability in measurement are known as Significant
Figures.
2. The
certain digits of a measured quantity plus one uncertain last digit are called Significant Figures.
3. Zeroes lying in between non-zero digits are significant.
4. Final
zeros to the right of the decimal point are significant.
5. Zeroes
locating the decimal point in number less than one are not significant.
6. Zeros
locating the decimal point in number greater than one are not significant.
7. In
exponential notation, the numbers are expressed as the product of two numbers; N × 10±x
8. Standard
Scientific Notation is one in which decimal point is after one digit of co-efficient number.
9. If the number to be expressed in exponent
is greater than 1, the exponent is positive
integer (x >1).
10. If the number to be expressed in exponent is
less than 1, the exponent is negative integer (x
<1).
11. There are 2 significant figures in 0.0021.
12. 1dm3 of volume is equal to 1 liter.
13. The decimal fractional part of logarithm is
called mantissa.
14. In log system the characteristic of 1000 is 3.
Empirical
and Molecular Formula
1. The type
of formula which gives simplest whole number ratio of different combining atoms
in a molecule of a compound
is called empirical or experimental formula.
2. It is
obtained from %age composition of
elements.
3. Ionic compounds are represented by their
empirical formula.
4. Giant covalent compounds are also represented
by their empirical formula e.g. diamond, sand (SiO2) etc.
5. Empirical
formula of glucose (C6H12O6), acetic acid (CH3COOH)
and formaldehyde (HCHO) is CH2O.
6. Empirical
formula of benzene (C6H6) & acetylene (C2H2)
is CH.
7. Empirical
formula of oxalic acid (C2H2O4) is CHO2.
8. Empirical
formula of hydrogen peroxide (H2O2) is HO.
9. The empirical formula of acetylene is
CH. Its molecular formula will be C2H2
if its molecular mass is 26.
10. The chemical formula of glucose is C6H12O6
11. The formula of lime water is Ca(OH)2
12. In
combustion analysis, water is absorbed in Mg(ClO4)
and CO2 is absorbed in 50%
KOH.
13. Molecular
formula = (empirical formula)n
14. Integer (n) = molar mass/empirical formula
mass
15. Percentage of element in compound = (Mass of element × number of atoms/total mass
of compound) × 100
16. Molecular formula gives
total number of atoms in a molecule.
17.Covalent compounds are represented by their
molecular formulas.
18. Molecular formula is same as empirical formula
or its some integral multiple.
19.If value
of integer (n) is unity, then
empirical and molecular formula is same.
20.Water,
carbon dioxide, ammonia, methane, methanol, sucrose have same EF and MF.
21. Combustion
analysis is also called elemental analysis.
22. Combustion
analysis is only for organic compounds having C,
H and O.
23. The sole
products of combustion analysis are CO2
and H2O.
24.Percentage
of C and H is calculated from the masses of CO2
and H2O.
25. Percentage
of O is calculated by difference method
26. Absorption
of water in Mg(ClO4)2 is a physical
change.
27. Absorption
of CO2 in 50% aqueous KOH is a chemical
change which is in fact an
acid-base neutralization reaction between
acidic CO2 and base KOH.
28.%age of C = (mass of CO2/Mass of
organic compound) × 12/44 × 100
29.%age of H = (mass of H2O/Mass of
organic compound) × 2/18 × 100
30. %age of O = 100 – (%age of C + %age of H)
31. Mole fraction of element = % of element/atomic
mass
32. Simplest ratio of element = mole ratio/least mole ratio
Isotopes
1. J. J. Berzelius determined atomic masses and
symbols of elements.
2. The
existence of isotopes of elements was first discovered by J.J. Thomson in 1913.
3. The
name of isotope was introduced by Soddy
4. Isotopes have same atomic number (Z), same
number of protons and electrons, same electronic configuration, same chemical properties and same position in
the periodic table.
5. Isotopes have different mass number (A),
different atomic mass, different number of neutrons, different physical properties, half-lives and different
rate of reaction.
6. Calcium and palladium have 6
isotopes each.
7. Cadmium
has 9 isotopes
8. Tin has 11 isotopes.
9. Total number of isotopes = 580.
10. Total number of natural isotopes = 280.
11. Natural radioactive isotopes (out of 280) = 40.
12. non-radioactive isotopes
(out of 280) = 240.
13. Synthetic
unstable
radioactive isotopes produced through artificial disintegration = 300
14. Isotopes with
even atomic mass and even atomic number = 154
15. The isotopes whose mass
number is multiples
of four are
most abundant e.g. O, Mg, Si, Ca,Fe
form nearly 50%.
16. The elements with even atomic number usually have larger number of stable isotopes.
Calculating PEN (Protons-Electrons-Neutrons)
Numbers
Number of protons = Atomic number
Number of neutrons = Mass number (A) – atomic number (Z)
Number of electrons = Number of protons = Atomic number (for neutral atom)
Number of electrons= Atomic number – Charge (for cation or positive ion)
Number of electrons= Atomic number + Charge (for anion or negative ion)
Different Types of Atomic Species
Classification of elements on the basis of number of isotopes
