74. Mole Concept and Stoichiometry-1

8/13/2019 74. Mole Concept and Stoichiometry-1

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Chemistry





Mole concept and stoichiometry



Session Objectives



Session Objective

Problems related to

1. Mole concept

2. Stoichiometry



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



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).



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)



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



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).




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:



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



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,



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




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).




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).



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:



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



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.



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 + …..)



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



Questions



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:



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:



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:



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:



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),



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)



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



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)



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)



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)



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



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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74. Mole Concept and Stoichiometry-1