chemistry. session opener mole concept and stoichiometry

Chemistry

Session Opener

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

Illustrative problem 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 0.1 mole of chromium sulphate in water. The maximum number of moles of lead sulphate that 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 2

A compound of iron and chlorine is solublein water. An excess of silver nitrate was added to precipitate chloride ion as silver chloride. If a 134.8 mg of the compound gave 304. 8 mg of AgCl, what is the formula of the compound

(Fe = 56, Ag = 108, Cl = 35.5)

Solution

xLet the molecular formula FeCl

x 3 3 xFeCl AgNO Fe(NO ) xAgCl

x

0.1348Moles of FeCl

56 35.5x

300.3 48

Moles of AgCl 2.12 10143.5

x0.13480.00212

56 35.5x

2

0.1348x 0.1187 0.0753x

x 1.99 2

Formula is FeCl

Illustrative example 3

Let molecular mass of Haemoglobin M

0.33M 4 56

100M 67878.8gm

Blood haemoglobin contains 0.33% iron. Assuming that there are four atoms of iron per molecule of haemoglobin, its approximate molecular mass is found to be

(a) 34,000 (b) 17,000

(c) 67,879 (d) 85,000

Solution:

Hence, the answer is (c).

Illustrative example 4

2 4 4 2Mg H SO MgSO H

2

448Moles of H liberated 0.02

22400Moles of pure Mg in the sample

Amount of pure Mg 0.024 24 0.48g

0.48% 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. H2SO4 it gave 448 ml of hydrogen at N.T.P. Calculate the percentage purity of magnesium. At wt. Of Mg = 24.Solution:

Illustrative example 5

Determine the percentage composition of a mixture of anhydrous sodium carbonate and sodium bicarbonate from the following 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 Na2CO3=x gmMass of NaHCO3=(2–x) gmLoss in weight on heating is due to the decomposition of NaHCO3.

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

103.168x

621.664g

2 3

1.664% Na CO 100

283.2

3% NaHCO 16.8

Illustrative example 6

Let the ratio of metal and oxide is x : y

x 50For 1st oxide, 1 : 1

y 50

Formula MO (given)

2 3

x 40For 2nd oxide, 2 : 3

y 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) MO2 (b) M2O3

(c) M2O (d) M2O5

Solution:

Hence, the answer is (b).

Illustrative example 7

6.25 : 25

1: 4

Let the ratio of carbon and hydrogen in the hydrocarbon is

75 25C : H :

12 1

4Empirical formula CH Molecular formula

A hydrocarbon contains 75% of carbon. Then its molecular formula is.

(a) CH4 (b) C2H4

(c) C2H6 (d) C2H2

Solution:

Hence, the answer is (a).

Illustrative problem 8

To meet the hourly requirement of energy

of an astronaut moles of sucrose required

34.20.1

342

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 32g

921.6 g

An hourly requirement of an astronanut can be satisfied by the energy released when 34.2 g of sucrose (C12H22O11) are burnt in his body. How many grams of oxygen would be needed in a space capsule to meet his requirement for one day ?Solution:

Illustrative problem 9

60 g of a compound on analysisgave 24 g C, 4 g H and 32 g O.The empirical formula of thecompound is:

(a) C2H4O2 (b) C2H2O2 (c) CH2O2 (d) CH2O

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 6.66

6.661

6.66

23.33

O 53.33 53.33

3.3316

3.33

13.33

Hence, answer is (d).Hence empirical formula CH2O.

Simple Titrations

Find out the concentration of a solutionwith the help of a solution of knownconcentration.

1 1 2 2N V N V

For mixture of two or more substances

N1V1 + N2V2 + ……= NVWhere V=(V1 + V2 + …..)

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 10

Find the molality of H2SO4 solution whose specific gravity(density) is 1.98 g/ml and 95% mass by volume H2SO4.

100 ml solution contains 95 g H2SO4.

95

98Moles of H2SO4 =

Mass of solution = 100 × 1.98 = 198 g

Mass of water = 198 – 95 = 103 g

Molality =

95 1000

98 103= 9.412 m

Solution:

Illustrative example 11A sample of H2SO4 (density 1.787 g/ml) is 86% by mass. What is molarity of acid? What volume of this acid has to be used to make 1 L of 0.2 M H2SO4?

d×10×xM=

Molecular mass

1.787×10×86= =15.68 molar

98

Let V1 ml of this H2SO4 are used to prepare 1 L of 0.2 M H2SO4.M1V1 = M2V2 15.68 × V1 = 0.2 × 1000

10.2×1000

V = =12.75 ml15.68

Solution:

Illustrative example 12A mixture is obtained by mixing 500ml 0.1M H2SO4 and 200ml 0.2M HCl at 250C. Find the normality of the mixture.

2 2 1 1

1 2

N V + N VWe know, N =

V + V

For the mixture, 500 0.1 2 200 0.2 1N 0.2

700

Solution:

Illustrative example 13500 ml 0.2 N HCl is neutralized with 250 ml 0.2 N NaOH. What is the strength of the resulting solution?

HCl + NaOH + NaCl + H2O

Equivalents of HCl 3500 0.2 10 egv

Equivalents 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 3500×10 ×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 N

Strength of HCl = .067 × 36.5 grams/litre

= 2.44 grams/litre

Normality of HCl (excess),

Alkali metals: very energetic

They readily form oxides and hydroxides which 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 hydroxides when dissolved in water and occur in the earth’s crust.

Be

Mg

Ca

Sr

Ba

Ra

Why? IE>IE of I

Bear mugs can serve bar rats

Group 1 elements(Alkaline earth metals)

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)

ns2np2

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)

ns2np4

First four elements are calledchalcogen meaningore forming.

O

S

Se

Te

Po

Oh, she sells tie moles

Group-16 elements (Oxygen family)

ns2np5

2. Diatomic molecule in the elemental form.

1. Astatine is radioactive with very short half-life period.

F

Cl

Br

I

At

Fat Clyde bribed Innocents

Sea salt producer

Group-17 elements (Halogen family)

Thank you