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Saturday, 26 September 2015

FSc Notes Chemistry Part 1 Chapter 1 Basic Concepts Lecture 2

FSc Notes Chemistry Part 1 Chapter 1 Basic Concepts

Determination of Relative Atomic Masses of Isotopes by Mass

Spectrometry:

Mass spectrometry is a technique by means of which the relative abundance and atomic mass of the isotopes of an element can be determined. The instrument used in this technique is called Mass Spectrometer Although there are several methods for the determination of atomic mass and molecular mass, but mass spectrometry is the most direct and accurate method. In this component is called vaporizer. Here the sample (element or molecule ) is vaporized. This vaporized sample is then passed into another component known as filament. Here high speed electrons are passed through the vaporized sample. These high speed electrons collide with the electrons of the atoms of sample and get them out of these atoms. Thus tiredly charged ions. These tiredly change ions of mass " m" and change "e" are passed through tiredly charged plates. Thus the +ve ions are accelerated. Then the +ve ions are padded through the poles of magnet (magnetic field) which forces them into a circular path. The radius of the path depend upon the charge to mass ration ( e/m) of the tire ions. The ions with larger e/ m have smaller radius and the ions (+ ve ) with smaller e/m have larger radius. The ions with different masses and same charges are separated. + ve ions with different e/m are thus separated and the detector receive them of different points producing different signals. Thus we easily come to know about the no of isotopes, their atomic masses and their relative abundance. One of the first application of mass spectrometer was determined by F. W. Aston. It was about Neon ( Ne ). It was told or explained that the naturally occurring on has B isotopes ie Ne-20 ( 90.92 %) Ne- 21 ( 0.26 %) and Ne-22 ( 8.82% ). All this information is got from the graph as shown in the following diagram.

Chemical Formula:

The symbolic representation of a compound is known as its chemical formula.
Or
The representation of a compound in terms of symbols of its elements is known as chemical formula of that compound.

Advantages Or Characteristics of Chemical Formula:

The chemical formula of a compound has the following advantages i.e.
  • It provides us the following information about the compound.
  • It tells us about the elements present in the compound.
  • It tells us about the no of atoms of each element.
  • It tells us about the chemical composition of compound i.e. tells us the ratio b/w the atoms of elements of the compound.
  • It shows us about the no of moles of each element present in one mole of the compound.
For example the chemical formula of water is H2O. Thus H2O provides us the following information about water .
There are H and O element in water.
There are 2 atoms of H element.
The ratio b/w the atoms of H & O is 2:1
There are 2 moles of H and One mole of a elements in one mole of H20

Types Of Chemical Formula:

There are two types of chemical formula.
  1. Molecular formula
  2. Empirical or simplest formula
Molecular Formula:
The formula of a compound which shows the actual no of atoms of its element is called as molecular formula of that compound. Molecular Formula Benzene C6H6 Glucose C6H12O6 Water H2O

Empirical or Simplest Formula:
The formula of a compound which shows the simplest whole ratio b/w the atoms of its elements is known as empirical formula of that compound.
For example: Empirical Formula Benzene CH, Glucose CH2O, Water H2O

Relationship between Molecular Formula & Empirical Formula

The molecular and empirical formula of a compound ( molecule ) have the following relationship. There are some molecules for which both molecular and empirical formula are same. E.g. : water has H2O as molecular as well as empirical formula Molecular Formula is an integral multiple of the Empirical Formula i.e.
M. F = N X E. F
Where "N" is an integer and its value may be 1, 2,3…………… The value of "N" is determined as
N = Molecular Mass / Formula Mass
Examples: The E.F of Acetic Acid is CH2O. Its Molecular Mass is 60g/mol. What is its M,f
Data
E.g. Acetic Acid = CH2O
There fore Mass of Acetic Acid = 12+ 1+2+16 = 30g
Molecular Mass of Acetic acid = 60 g/mol
M. F = ?
M. F = n E.F
= Mole mass/ For Mass (CH2O)
= 60/30 (CH2O)
= 2 (CH2O)
M.F = C2H4O (CH3 COOH)

Percentage Composition:

The components ( Part ) of each element ( by mass ) is known as percentage composition of the compound. The percentage of element of a compound may be determined in two cases. If the chemical formula of the compound is known, then the percentage of any element is determined as: Element = Mass of that element* 100

Molecular mass of compound

For example : what is the % age of C & O in CO2 If we don’t know about the chemical formula of the compound. Whose % age composition is to be determined, then first of all, we determine, the chemical composition of compound by a process known as chemical analysis and then find out the % age of each element.

Chemical Analysis:

The process by which the chemical composition of a compound is determined is called chemical analysis:
Let we have an organic compound containing carbon, hydrogen and oxygen elements. During chemical this compound a weighted quantity ( known mass ) of this sample compound is placed in the combustion tube which is fitted with a furnace. Oxygen is supplied at one end of the combustion tube .
At the other end of the combustion tube two consecutive jars are placed one containing a water absorbing substance like 50% KOH and the other jar containing a CO2 absorber like. Magnesium per chlorate [ Mg ( ClO4 )2 ]. These two jars are pre- weight. As a result of combustion of the organic compound, all the hydrogen converts into H2O and carbon into CO2.
These two gases ( H2O (g) & CO2 ) pass into the jars where they are absorbed and as a result the weights of both the jars increased. From the increase in the weight of jars, we can easily, determine the masses of H2O and CO2 formed from a known mass of the organic compound .Now the percentage of each element is determined as follow:
% C = Mass of CO2 / Mass of org. compound * Mass of C (12)/ M. Mass of CO2 * 100
% H = Mass of H2O / Mass of org. compound * Mass of H (2)/ M. Mass of H2O (18) * 100

The %age of oxygen element is determined indirectly as:
%O = 100 – ( %C + %H)

Example : During combustion analysis of an organic compound containing C,H and O, 1.039 g
of CO2 and 0.636 g of H2O are produced from 0.5439 g of the org, compound.
Determined the %age composition of compound.
Mass of Data / org. compound = 0.5439 g
Mass of Co2 = 1.039g
Mass of H2o = 0.6369 g
Solution:
%C = 1.039/0.5439 * 12/44 * 100 = 52.108%
%H = 0.6369/0.5439 * 2/18*100 = 13.11%
%O = 100 – (52.108 + 13.11) = 34.77%

Empirical Formula of a Compound

The empirical formula is the simplest formula that gives information about the simple ratio of atoms present in a compound.

Determination of Molecular Formula from Empirical Formula:

The molecular formula of molecule can be determined from its empirical formula as follow
Mol . Formula = n* Emp. formula
Here “n” is an integer and its value can be find as:
N = Molecular Mass
Formula Mass
N= 1, 2, 3 ………………………

Molecular Mass:
It is the sum of the atomic masses of all the elements present in the molecular formula of a molecule.
e.g : The molecular formula of benzene is C6H6. Thus its molecular mass is 12*6+1*6===== 78a. m. u or 78g/ mole.

Formula Mass:
It is the sum of the atomic masses of all the elements present in the empirical formula of a molecule or formula unit of a compound. e.g The emp. Formula of benzene is CH, so its formula mass will be 12*1+1*1 ===13a. m. u or 13g/mol

Example :
The combustion analysis of an organic compound shows that it contains 65. 44%C, 5.50% H and 29.06 % O. what is its emp. Formula. If the molecular mass of the compound is 110.15 g/mol, then what will be the molecular formula of the compound Data.

Written by: Asad Hussain

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