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8/13/2019 74. Mole Concept and Stoichiometry-1
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Chemistry
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Mole concept and stoichiometry
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Session Objectives
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Session Objective
Problems related to
1. Mole concept
2. Stoichiometry
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Concept of equivalence
weightNo. of equivalents
equivalent weight
Equivalent weight can be defined as
gm atomic weight /Molar massEquivalent weight (E)
n factor
Therefore, no. of equivalents = no. of moles x n – factor
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Illustrativeproblem 1
3 2 2 4 3 4 3 30.5 mole 0.1 mole
3Pb(NO ) Cr (SO ) 3PbSO 2Cr(NO )
2 4 3
2 4 3 4
Since Cr (SO ) is the limiting reagent
Molar ratio of Cr (SO ) and PbSO is 1 : 3
4Hence, moles of PbSO 0.3
0.5 mole of lead nitrate is mixed with 0.1mole of chromium sulphate in water. Themaximum number of moles of lead sulphatethat can be obtained is
(a) 0.6 mole (b) 0.5 mole
(c) 0.3 mole (d) 0.1 mole
Solution:
Hence, the answer is (c).
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Illustrative problem 2A compound of iron and chlorine is solublein water. An excess of silver nitrate wasadded to precipitate chloride ion assilver chloride. If a 134.8 mg of the compoundgave 304. 8 mg of AgCl,
what is the formula of the compound(Fe = 56, Ag = 108, Cl = 35.5)
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SolutionxLet the molecular formula FeCl
x 3 3 xFeCl AgNO Fe(NO ) xAgCl
x0.1348Moles of FeCl
56 35.5x
300.3 48Moles of AgCl 2.12 10143.5
x0.1348 0.0021256 35.5x
2
0.1348x 0.1187 0.0753xx 1.99 2Formula is FeCl
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Illustrative example 3
Let molecular mass of Haemoglobin M
0.33M 4 56
100M 67878.8 gm
Blood haemoglobin contains 0.33%iron. Assuming that there are fouratoms of iron per molecule ofhaemoglobin, its approximate molecularmass is found to be
(a) 34,000 (b) 17,000
(c) 67,879 (d) 85,000
Solution:
Hence, the answer is (c).
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Illustrative example 4
2 4 4 2Mg H SO MgSO H
2448
Moles of H liberated 0.0222400
Moles of pure Mg in the sample
Amount of pure Mg 0.024 24 0.48g0.48
%purity 100 96%0.5
0.5 g of an impure sample of magnesium
contains its own oxide as an impurity,when heated with dil. H 2SO 4 it gave 448 mlof hydrogen at N.T.P. Calculate thepercentage purity of magnesium. At wt. OfMg = 24.
Solution:
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Illustrative example 5Determine the percentage composition ofa mixture of anhydrous sodium carbonateand sodium bicarbonate from thefollowing data:
Weight of the mixture taken = 2 g
Loss in weight on heating = 0.124 g
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Solution
2 3x gm
Na CO no effect
3 2 3 2 22 x (2 x)mole mole84 2 84
2NaHCO Na CO CO H O
Let mass of Na 2CO 3=x gmMass of NaHCO
3=(2 – x) gm
Loss in weight on heating is dueto the decomposition of NaHCO 3.After decomposition, weight of the remaining substance =(2 – 0.124)g=1.876 g
In the mixture,
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Solution
(2 x)x 106 1.876
84 2
168 x 212 106 x 315.17
103.168x62
1.664g
2 31.664
%Na CO 100283.2
3%NaHCO 16.8
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Illustrative example 6
Let the ratio of metal and oxide is x : yx 50
For 1st oxide, 1 :1y 50
Formula MO (given)
2 3
x 40For 2nd oxide , 2 : 3y 60
Formula M O
Two oxides of a metal contain 50%and 40% of the metal by mass respectively.The formula of the first oxide is MO.Then the formula of the second oxide is
(a) MO 2 (b) M 2 O3
(c) M 2 O (d) M 2 O5
Solution:
Hence, the answer is (b).
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Illustrative example 7
6.25:251 : 4
Let the ratio of carbon and hydrogen in the hydrocarbon is75 25C : H :12 1
4Empirical formula CH Molecular formula
A hydrocarbon contains 75% of carbon.
Then its molecular formula is.(a) CH 4 (b) C 2H4
(c) C 2H6 (d) C 2H2
Solution:
Hence, the answer is (a).
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ustrat veproblem 8
To meet the hourly requirement of energyof an astronaut moles of sucrose required
34.2 0.1342
moles of sucrose required in one day 2.4
12 22 11 2 2 2C H O 12O 12CO 11H O
2For 2.4 moles sucrose, amount of O needed
2.4 12 32g921.6 g
An hourly requirement of an astronanut canbe satisfied by the energy released when
34.2 g of sucrose (C 12 H22 O11 ) are burnt inhis body. How many grams of oxygen wouldbe needed in a space capsule to meet hisrequirement for one day ?
Solution:
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Illustrative problem 960 g of a compound on analysis
gave 24 g C, 4 g H and 32 g O.The empirical formula of thecompound is:
(a) C 2H4O2 (b) C 2H2O2 (c) CH 2O2 (d) CH 2O
Solution:
% of C = 24
100 40%60
4
100 6.6660% of H =
% of O = 32
100 53.3360
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Solution
Element Percentage Atomic ratio Simplest ratio
C 40 40 3.3312
3.33 13.33
H 6.66 6.666.66
16.66
23.33
O 53.3353.33 3.33
163.33 13.33
Hence, answer is (d).Hence empirical formula CH 2O.
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Simple Titrations
Find out the concentration of a solution
with the help of a solution of knownconcentration.
1 1 2 2N V N V
For mixture of two or more substances
N1V1 + N 2V2 + ……= NV Where V=(V 1 + V 2 + …..)
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Normality of mixing two acids
1 1 2 2
1 2
N V +N VN =
V + V
Normality of mixing acid and bases
1 1 2 2
1 2
N V -N VN =
V + V
2 2 1 1
1 2
N V -N Vor N= V + V
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Questions
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Illustrative example 10Find the molality of H 2SO 4 solutionwhose specific gravity(density) is1.98 g/ml and 95% mass by volume H 2SO 4 .
100 ml solution contains 95 g H 2SO 4 .
95
98Moles of H 2SO 4 =
Mass of solution = 100 × 1.98 = 198 g
Mass of water = 198 – 95 = 103 gMolality =
95 100098 103
= 9.412 m
Solution:
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Illustrative example 11A sample of H 2 SO 4 (density 1.787 g/ml)is 86% by mass. What is molarity of acid?What volume of this acid has to beused to make 1 L of 0.2 M H 2SO 4?
d×10×xM = Molecular mass
1.787×10×86= =15.68 molar98
Let V 1 ml of this H 2SO 4 are used to prepare 1 L of 0.2 MH2SO 4.M1V1 = M 2V2
15.68 × V 1 = 0.2 × 1000
10.2×1000
V = =12.75 ml15.68
Solution:
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Illustrative example 12A mixture is obtained by mixing 500ml0.1M H 2 SO 4 and 200ml 0.2M HCl at 25 0C.
Find the normality of the mixture.
2 2 1 1
1 2
N V + N VWe know, N =
V + VFor the mixture, 500 0.1 2 200 0.2 1
N 0.2700
Solution:
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Illustrative example 13500 ml 0.2 N HCl is neutralized with250 ml 0.2 N NaOH. What is thestrength of the resulting solution?
HCl + NaOH + NaCl + H 2O
Equivalents of HCl 3
500 0.2 10 egvEquivalents of NaOH -3= 250×0.2×10 egv
Equivalence of excess HCl 3 3(500 0.2 10 250 0.2 10 egv)
Normality of HCl (excess)-3 3
500×10 ×10= = 0.067 N750
Strength of HCl = .067 × 36.5 g/litre
= 2.44 g/litre
Solution:
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Solution
1 1 2 2
1 2
N V -N VN =
V + V
0.2 × 1 × 500 - 0.2 × 1 × 250N =
500 + 250
N = 2.44 NStrength of HCl = .067 × 36.5 grams/litre
= 2.44 grams/litre
Normality of HCl (excess),
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Alkali metals: very energetic
They readily form oxides and hydroxideswhich are strongly alkaline.
They do not occur free in nature.
Li
Na
K
Rb
Cs
Helen kabre se farar
Group 1 elements(Alkali metals)
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Alkaline earth metals:
Oxides of Ca, Sr and Ba form alkaline hydroxideswhen dissolved in water and occur in the earth’scrust.
Be
MgCa
Sr
Ba
Ra
Why? IE>IE of IBear mugs can serve bar rats
roup e emen s(Alkaline earth
metals)
G 13 l t
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Except Boron, all others are metals.
B
Al
Ga
In
Tl
Bob allen gone indrains jennis lessons
Al is longer than Ga
Group 13 elements(Boron family)
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ns 2np 2
Carbon is a typical non-metal.
Si and Ge are metalloids.
Sn and Pb are metals.
C
Si
Ge
Sn
Pb
Can sily or Gervans snatch lead
Group 14 elements(Carbon family)
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N and P are non-metals.
As and Sb are metalloids.
Bi is a true metal.
N
P
As
Sb
Bi
Never put arsence in silver bullet bear
Group 15 elements(Nitrogen family)
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ns2
np4
First four elements are calledchalcogen meaningore forming.
O
S
Se
Te
Po
Oh, she sells tie moles
Group-16 elements(Oxygen family)
Group 17 elements
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ns 2np 5
2. Diatomic molecule in theelemental form.
1. Astatine is radioactive withvery short half-life period.
F
Cl
Br
I
At
Fat Clyde bribed Innocents
Sea salt producer
Group-17 elements(Halogen family)
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