Revision work
1) Define the term isotopes.
2) How many protons,
neutrons and electrons are
present in each of the
following atoms?
In Modern Definition, C-12 is chosen as
the standard reference
Atomic Mass Unit ( amu)
• Mass of one atom of Carbon – 12 = 12 amu
Modern Definition of relative atomic mass
Old definition of relative atomic mass
•
Relative atomic mass (Numerical)
• 1 Chlorine consists of two isotopes with mass numbers 35 and 37,
with their natural abundances 76% and 24% respectively.
• Calculate a) average mass of one atom of Chlorine
•
b) relative atomic mass of Chlorine
Isotopes of Chlorine
(numerical’s answer)
• a) Let us consider 100 Cl atoms found in nature.
• Total mass of 100 Cl atoms = 76 x 35 amu + 24 x 37 amu
= 3548 amu
Therefore, average mass of one atom of Cl = 3548 amu/100
= 35.5 amu
b) Relative atomic mass of Chlorine
•
Problem no 2
• Magnesium consists of three isotopes with mass numbers
24, 25 and 26 , and with natural abundances 79%, 10% and
11% respectively.
• Calculate a) average mass of one atom of Magnesium
•
b) relative atomic mass of Magnesium
Problem No 3
• Boron consists of two isotopes with nucleon number 10 and
11, and with natural abundances 20% and 80% respectively.
• Calculate relative atomic mass of Boron.
Isotopes of Magnesium
Isotopes of Chlorine
Summary about Chlorine
• i) Atomic number of Chlorine is 17.
• ii) Mass numbers for first Cl atom is 35 and for 2nd Cl
atom is 37.
• iii) Average mass for one atom of Chlorine = 35.5 amu.
• iv) Average mass of one atom of element in amu is
atomic mass.
• v) Atomic mass of Chlorine is 35.5 amu.
• vi) Relative atomic mass of Chlorine is 35.5.
Isotopes of Magnesium
Summary about Magnesium
• i) Atomic number of Magnesium is 12.
• ii) Mass numbers for first Mg atom is 24 , for 2nd Mg atom is
25 and for third Mg atom is 26.
• iii) Average mass for one atom of Magnesium = 24.3 amu.
• iv) Average mass of one atom of element in amu is atomic
mass.
• v) Atomic mass of Magnesium is 24.3 amu.
• vi) Relative atomic mass of Magnesium is 24.3.
•
https://youtu.be/0N_CIdmI5Cg
Problem No 4
Most elements exist naturally as a mixture of isotopes, each
with their own relative isotopic mass.
Lead has four stable isotopes. Information about three of these
isotopes is given.
Isotope
Pb-204
Pb-206
Pb-208
Relative isotopic Percentage abundance
mass
204.0
1.4
206.0
24.1
208.0
52.4
a) Define the term
relative isotopic mass.
For example
Relative isotopic mass of
Pb-208 is equal to ratio of
mass of one atom of
Pb-208 to 1/12 the mass
of one atom of C-12
b) The relative atomic mass of Lead is 207.21.
Calculate the percentage abundance and hence the
relative isotopic mass of the fourth isotope of Lead.
• Percentage abundance of the fourth isotope of Lead
= 100 – (1.4 + 24.1 + 52.4)
= 100 – 77.9
= 22.1
Relative atomic mass of Lead is 207.12.
Let relative isotopic mass of the fourth isotope of
Lead is a.
• Learners can build and test isotopes using this simulation:
phet.colorado.edu/en/simulation/isotopes-and-atomic-mass
1 mole = 6.02 x 1023 particles(Avogadro’s
constant)
For example
1 mole of e - = 6.02 x 1023 electrons
1 mole of H = 6.02 x 1023 hydrogen atoms
1 mole of H2 = 6.02 x 1023 hydrogen molecules
1 mole of CO2 = 6.02 x 1023 CO2 molecules
Comparison
Species
He
C-12
Mg
Relative atomic
mass/relative
isotopic mass
4
12
24.3
Atomic
mass/isotopic
mass
4 amu
12 amu
24.3 amu
1 mole
4g
12 g
24.3 g
No of particles
6.02 x 1023
Helium atoms
6.02 x 1023
Carbon -12
atoms
6.02 x 1023
Magnesium
atoms
Molar mass
• Mass of 1 mole of substance is called molar mass.
• Molar mass of Helium is 4 g mol-1
• Molar mass of Magnesium is 24.3 g mol-1
Molar mass of Calcium is 40.1 g mol-1
Copy and complete the following table
Element
Molar mass
Na
23 g mol-1
Cu
Ar
Fe
Ba
Answer
Element
Molar mass
Na
23 g mol-1
Cu
63.5 g mol-1
Ar
39.9 g mol-1
Fe
55.8 g mol-1
Ba
137.3 g mol-1
Revision
• Relative atomic mass of Lithium (Ar) = 6.9
• Atomic mass of Lithium = 6.9 amu
• Molar mass of Lithium = 6.9 g mol-1
• 6.9 g of Lithium contains 6.02 x 1023 Lithium atoms
Calculate the no of moles present in each of the following
samples.
a) 30.0 g carbon b) 24.0 g Sulphur c) 16.0 g Magnesium
• a) Mass = 30 g
• Molar mass of Carbon = 12 g mol-1
• No of moles = Mass/Molar mass
•
= 30 g/12 g mol-1
•
= 30 g/12 g mol-1
•
= 2.5 mol
Mole or Mol ?
•
Rupees in short is written as Rs
•mole in short is written as mol
•5 mole or 5 mol
How many atoms are present in each of the
following samples?
i)
1.5 mol of H
ii)
3.99 gram of Argon
iii) 48.6 gram of Magnesium
Answer
• i) 1 mol of H contains 6.02 x 10 23 H atoms
Therefore 1.5 mol of H contains 1.5 x 6.02 x 10 23 H atoms
= 9.03 x 1023 H atoms
ii) No of mol = mass/molar mass
= 3.99 g/ 39.9 g mol-1 = 0.1 mol
Therefore No of Argon atoms = 0.1 x 6.02 x 10 23
= 6.02 x 1022
What is the mass (in gram) in each of
following samples?
i) 0.1 mol of Mg
ii) 10 mol of Li
iii) 6.02 x 1021 atoms of Iron
iv) 6.02 x 1025 atoms of Helium
Answer
• i) No of mol (n) = 0.1 mol
• Molar mass of Magnesium = 24.3 g mol-1
No of mole (n) = mass (m)/Molar mass(M)
Therefore, m = n x M
= 0.1 mol x 24.3 g mol-1
= 0.1 mol x 24.3 g mol-1
= 2.43 g
Answer of q. no. iii)
No of mol of iron = 6.02 x 1021 / 6.02 x 1023
n = 0.01 mol
molar mass of iron (M) = 55.8 g mol-1
mass (m) = n x M
= 0.01 mol x 55.8 g mol-1
= 0.558 g
Which one of the following samples contains
highest no of atoms?
• a) 2 g of hydrogen.
• b) 9.03 x 1023 atoms of oxygen
• C) 16.0 g of oxygen
• d) 7 g of Helium gas
Definition
Calculate relative molecular mass (Mr) of
following. (Relative atomic mass (Ar) of H= 1, C =
12, N= 14, O = 16)
i) CO2
• ii) NH3
• iii) C3H7OH
• Iv) C6H12O6
Answer
• iii) Relative molecular mass (Mr) of C3H7OH
= 3 x 12 + 8 x 1 + 1 x 16
= 36 + 8 + 16 = 60
Note: Like Ar , Mr also does not have any unit.
Answers i) 44 ii) 17 iii) 60 iv) 180
Calculate relative formula mass (Mr) of following.
• i) KCl
• iii) BaCl2. 2H2O
ii) MgSO4
iv) CuSO4. 5 H2O
• v) (NH4)2SO4
• Note: Compounds given above are not made up of molecules.
•
They are made up of ions.
• (Ar of K = 39.1, Mg = 24.3, Ba = 137.3 , Cl = 35.5 , Cu = 63.5 ,
S = 32.1)
Answer
• i) Mr of KCl = 1x 39.1 + 1 x 35.5
= 39.1 + 35.5
= 74.6
iv) Mr of CuSO4. 5 H2O = 1 x 63.5 + 1 x 32.1 + 4 x 16 + 5 x 18
= 63.5 + 32.1 + 64 + 90
= 249.6
Other answers: ii) 120.4 iii) 244.3 iii) 132.1
Revision
• i) Relative molecular mass (Mr) of CO2 = 44
• Molecular mass of CO2 = 44 amu
• Molar mass of CO2 = 44 g mol-1
• 44 g of CO2 contains 6.02 x 1023 molecules of CO2
Calculate the no of moles present in each of the following
samples.
a) 90 g of H2O(l) b) 36 g of C6H12O6 (s) c) 51 g of NH3 (g)
Answers
• a) 5 mol of H2O
• b) 0.2 mol of glucose
• C) 3 mol of NH3
A drop of water weighs 0.04 g.
Calculate the number of H2O molecules present in
the drop of water.
Mass (m) = 0.04 g
Molar mass of water (M) = 18 g mol-1
No of mol of H2O (n) = m/M
= 0.04 g/18 g mol-1
= 2.22 x 10-3 mol
Therefore, number of H2O molecules = 2.22 x 10-3 x 6.02 x 1023
= 1.34 x 1021
What is the mass of each of the following
samples?
• a) 1.5 moles of MgSO4
• b) 0.333 moles of AlCl3
• c ) 0.25 mole of CaCO3
d) 4 mol of CH4
Answer
i) relative molecular mass (Mr ) of MgSO4
= 1 x 24.3 + 1 x 32.1 + 4 x 16 = 120.4
1.5 mol of MgSO4 = 180.6 g
ii) Mr of AlCl3 = 133.5
mass of 0.333 mol of AlCl3 = 44.45 g
iii) Mr of CaCO3 = 100.1
0.25 mol of CaCO3 = 25.025 g
iv) Mr of CH4 = 16
4 mol of CH4 = 64 g
A gas syringe contains 64 g of Oxygen gas.
Calculate the number of moles of Oxygen.
Relative atomic mass of O (Ar)
= 16
Mass (m) = 64 g
Molar mass(M) = 16 g mol-1
n = m/M
= 64 g/16 g mol-1
= 4 mol of O
Relative molecular mass of O2 (Mr)
= 32
Mass (m) = 64 g
Molar mass(M) = 32 g mol-1
n = m/M
= 64 g/32 g mol-1
= 2 mol of O2
A mole of ?
Be specific
• 1 molecule of H2O is made up of 2 atoms of H and 1 atom of
O
• 1 mol of H2O contains 2 mol of H and 1 mol of O.
• 1 mol of H2O contains 2 x 6.02 x 10 23 atoms of H and 6.02 x
1023 atoms of O.
• 1 mole of CaCl2 contains 1 mol of Ca++ ion and 2 mol of Clion
Fill in the blanks about ethane (C2H6 )
• 1 molecule of ethane is made up of ….. atoms of C and …..
atoms of H.
• 1 mol of ethane is made up of …… mol of C and ….. mol of H.
1 mol of C2H6 contains ………………… atoms of C and ……………..
atoms of H.
Relative molecular mass (Mr ) of C2H6 is ……………..
Molar mass of ethane is …………..
Answer for fill in the blanks about ethane (C2H6 )
• 1 molecule of ethane is made up of 2 atoms of C and 6
atoms of H.
• 1 mol of ethane is made up of 2 mol of C and 6 mol of H.
1 mol of C2H6 contains 2 x 6.02 x 1023 atoms of C and
6 x 6.02 x 1023 atoms of H.
Relative molecular mass (Mr ) of C2H6 is 30
Molar mass of ethane is 30 g mol-1
What is the mass of each of the following samples
in gram?
• a) 1 mol of H
• b) 1 mol of H2
c) 1 mol of C2H5OH
d) 5 mol of CaCl2
Answer
• a) 1 g of Hydrogen
• b) 2 g of Hydrogen
• c) 46 g
• d) 555.5 g
Suppose you wrote your name with a stick of chalk
(CaCO3) on the ground. 0.05 g of calcium carbonate is
deposited on the ground.
• a) Calculate how many mol of calcium carbonate is
deposited on the ground after writing your name with
a stick of chalk.
• b) Calculate how many particles of calcium carbonate
are deposited on the ground
c) Calculate how many individual atoms of oxygen are
deposited.
First 20 elements and their normal form
S.No
Name of element
Symbol/
formula
Mass of
1 mol
S.no
Name of element
Symbol/
formula
Mass of
1 mol
1
Hydrogen
H2 (g)
2g
11
Sodium
Na (s)
23 g
2
Helium
He (g)
4g
12
Magnesium
Mg (s)
24.3 g
3
Lithium
Li (s)
6.9 g
13
Aluminium
Al (s)
27 g
4
Beryllium
Be (s)
9g
14
Silicon
Si (s)
28.1 g
5
Boron
B (s)
10.8 g
15
Phosphorous
P (s)
31 g
6
Carbon
C (s)
12 g
16
Sulphur
S (s)
32.1 g
7
Nitrogen
N2 (g)
28 g
17
Chlorine
Cl2 (g)
71 g
8
Oxygen
O2 (g)
32 g
18
Argon
Ar (g)
39.9 g
9
Fluorine
F2 (g)
38 g
19
Potassium
K (s)
39.1 g
10
Neon
Ne (g)
20.2 g
20
Calcium
Ca (s)
40.1
Ions formed by the first 20 elements
S.
No.
Name of
element
Electron
arrangement
Formula of
Ion/s formed
S.No Name of
.
element
Electron
arrangemen
t
Formula of
Ion/s formed
1
Hydrogen
1
H + or H -
11
Sodium
2,8,1
Na +
2
Helium
2
None
12
Magnesium
2,8,2
Mg ++
3
Lithium
2,1
Li +
13
Aluminium
2,8,3
Al +++
4
Beryllium
2,2
Be ++
14
Silicon
2,8,4
Si ++++ or Si ----
5
Boron
2,3
B +++
15
Phosphorous 2,8,5
6
Carbon
2,4
C ++++ or C ----
16
Sulphur
2,8,6
7
Nitrogen
2,5
Chlorine
2,8,7
Cl -
8
Oxygen
2,6
N
O --
17
18
Argon
2,8,8
None
9
Fluorine
2,7
F-
19
Potassium
2,8,8,1
K+
---
P --S --
Monoatomic ion contains only one atom in the ion.
Polyatomic ion contains 2 or more atoms in the ion.
S.No
Name of ion
1
hydroxide
Formula S.No Name of ion Formula
of ion
of ion
OH 6
sulphite
SO3 --
2
carbonate
CO3 --
3
4
Bicarbonate/
HCO3 hydrogencarbonate
sulphate
SO4 --
5
Phosphate
--PO4
7
nitrate
NO3 -
8
nitrite
NO2 -
9
ammonium
NH4 +
Rule for writing formula of ionic compound
• i) Positive ion is written in the left side of the formula
• ii) Polyatomic ion is enclosed in a bracket if the number of the
polyatomic ion in the compound is more than 1.
• Write down molecular formula of aluminium sulphate
3
2
Al
SO 4
Al2(SO4)3
Examples of ionic compounds
Compound name
formula
Cation
anion
Calcium chloride
CaCl2
Ca++
Cl -
Sodium chloride
NaCl
Na+
Cl -
Lithium fluoride
LiF
Li+
F-
Iron(II) oxide
FeO
Fe++
O --
Aluminium sulphide
Al2S3
Al+++
S --
Fe2(SO4)3
Fe+++
SO4 --
Iron(III) sulphate
Correct following molecular formulae if
necessary
• i) MgOH2
ii) (NH4)Cl
• iii) Cl3Al
iv) Cu(II)SO4
Water is a covalent compound
• Water is a covalent
compound.
• If you take water in a
glass, there will be large
no of seperate seperate
H2O molecules
Water is made up of large number of separate
H2O molecules.
Sodium chloride is an ionic compound
Sodium chloride is made up of ions. It does
not contain separate NaCl molecules
Na +Cl Na
+Cl-
Na +Cl -
Na +Cl Na
+Cl -
Na +Cl-
Na +ClNa
+Cl-
X
Na +Cl-
Ionic compound like sodium chloride are not made up of molecules.
Hence we use the term relative formula mass for them.
Formula of sodium chloride is NaCl because ratio of Na+ ion and Cl- ion
is 1:1 in the compound.
Sodium Chloride is made up of ions.
A precipitation reaction is a reaction that
yields an insoluble product—a precipitate—
when two solutions are mixed.
• i) NaCl(aq) + AgNO3 (aq)
AgCl(s) + NaNO3(aq)
•
White ppt.
• Here, Silver Chloride produced is insoluble in water
and its colour is white.
Convert a molecular equation into net ionic
equation
• i) NaCl (aq) + AgNO3 (aq)
Na+(aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)
Omit spectator ions.
Na+(aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)
Ag + (aq) + Cl – (aq)
AgCl (s)
This is the net ionic equation.
AgCl(s) + NaNO3 (aq)
AgCl (s) + Na+ (aq) + NO3- (aq)
AgCl (s) + Na+ (aq) + NO3- (aq)
Convert following molecular reaction into net ionic
equation.
ii) BaCl2 (aq) + Na2SO4 (aq)
iii) FeCl3 (aq) + 3 NaOH (aq)
iv) KOH (aq) + HNO3 (aq)
v) Zn (s) + H2SO4 (aq)
BaSO4 (s) + 2 NaCl (aq)
White ppt
Fe(OH)3 (s) + 3 NaCl (aq)
reddish brown ppt
KNO3 (aq) + H2O (l)
ZnSO4 (aq) + H2 (g)
https://youtu.be/uaBHB5GSgHU
• While converting molecular equations into ionic
equations, do not convert solid, liquid and
gaseous substances into ions.
Some more examples
H2SO4 (aq) + 2 KOH (aq)
K2SO4 (aq) + 2 H2O (l)
2H+ (aq) + SO4- - (aq) + 2K+ (aq) + 2OH– (aq)
2K+ (aq) + SO4- - (aq) + 2 H2O (l)
Omit spectator ions.
2H+ (aq) + SO4- - (aq) + 2K+ (aq) + 2OH– (aq)
2K+ (aq) + SO4- - (aq) + 2 H2O (l)
2H+ (aq) + 2OH– (aq)
2 H2O (l)
H+ (aq) + OH– (aq)
H2O (l)
Some more examples
Na2CO3 (s) + 2 HCl (aq)
2 NaCl (aq) + H2O (l) + CO2 (g)
Na2CO3 (s) + 2H+ (aq) + 2Cl- (aq)
Omit spectator ions.
Na2CO3 (s) + 2H+ (aq) + 2Cl- (aq)
2Na + (aq) + 2Cl -(aq) +
H2O(l) + CO2(g)
2Na + (aq) + 2Cl -(aq) +
H2O(l) + CO2(g)
Na2CO3 (s) + 2H+ (aq)
2Na + (aq) + H2O(l) + CO2(g)