DALTON’S ATOMIC THEORY:

(1) Matter is made up of extremely small, indivisible particles called atoms.

(2) Atom of same substance are identical in all respect i.e. they posses same size, shape, mass, chemical properties etc.

(3) Atom of different substances are different in all respect i.e. they posses different shape, size, mass and chemical properties etc.

(4) Atom is the smallest particle that takes part in chemical reactions.

(5) Atom of different elements may combine with each other in a fixed, simple, whole number ratio to form compound atoms.

(6) Atom can neither be created nor destroyed i.e. atoms are indestructible.

Limitations of Dalton’s theory:

The main failures of Dalton’s atomic theory are:

(1) Atom was no more indivisible. It is made up of various sub-atomic particles like electrons, proton and neutron etc.

(2) It failed to explain how atoms of different elements differ from each other.

(3) It failed to explain how and why atoms of elements combine with each other to form compound atoms or molecules.

(4) It failed to explain the nature of forces that bind together different atoms in molecules.

(5) It failed to explain Gay Lussac’s law of combining volumes.

(6) It did not make any distinction between ultimate particle of an element that takes part in reaction (atoms) and the ultimate particle that has independent existence (molecules).

Modern Atomic theory:

(1) Atom is no longer supposed to be indivisible. Atom has a complex structure and is composed of sub-atomic particles such as electrons protons and neutrons.

(2) Atom of the same element may not be similar in all respects e.g. isotopes.

(3) Atom of different elements may be similar in one or more respects e.g. isobars.

(4) Atom is the smallest unit which takes part in chemical reactions.

(5) The ratio in which atoms unite may be fixed and integral but may not be simple. e.g. In sugar molecules C₁₂H₂₂O₁₁ the ratio of C, H and O atoms is 12:22:11 which is not simple.

(6) Atom of one element can be changed into atoms of other element for e.g. transmutation.

(7) Mass of atom can be changed in energy. 

     (E=MC²) According to Einstein mass energy relationship, mass and energy are inter-convertible. Thus atom is no longer indestructible.

ILLUSTRATIVE EXAMPLE: An important postulate of Dalton’s atomic theory is:

                        (A) an atom contains electrons, protons and neutrons

                        (B) atom can neither be created nor destroyed nor divisible

                        (C) all the atoms of an element are not identical

                        (D) all the elements are available in nature in the form of atoms

SOLUTION: The statement written in (B) is about law of mass conservation which is true for all chemical reaction. Hence (B) is correct.

PERCENTAGE COMPOSITION OF COMPOUNDS:

Law of Definite Proportions:  Compounds are consistent chemical combinations of atoms that can be expressed as:

(i)  Ratio of masses

(ii) Ratio of atoms

(iii) Ratio of moles of atoms

(iv) Percent composition

ILLUSTRATIVE EXAMPLE (1): Calculate the percent Nitrogen in Dinitrogen (N₂O) Monoxide?

SOLUTION:

ILLUSTRATIVE EXAMPLE (2): What is the % composition of each element in (Mg(OH)₂) magnesium hydroxide?

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Haemoglobin contains 0.33% of Iron by weight. The molecular weight of it is approx. 67200. The numbers of iron atoms (Atomic wt of Fe=56 u) present in one molecule of Haemoglobin are?

SOLUTION:

ILLUSTRATIVE EXAMPLE (4): The hydrated salt, Na₂SO4.nH₂O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be?

SOLUTION:

ILLUSTRATIVE EXAMPLE (5): Air contain 20 % O₂ by volume. How many cm³ of air will be required for oxidation of 100 cm³(ml) of acetylene?

SOLUTION:

Since air contain 20 % oxygen by volume then amount of air required to react with 100cc/ml of C₂H₂

Try yourself:

Exercise (4): Two oxides of metal contain 72.4 and 70 of metal respectively if formula of 2nd oxide is M₂O₃ find that of the first.

Ans Key : M₃O_{4
Determination of Empirical and Molecular Formula:}

DETERMINATION OF EMPIRICAL AND MOLECULAR FORMULA:

(1)  PERCENTAGE COMPOTION OF COMPOUNDS:
(2) EMPIRICAL FORMULA:

It is the formula which expresses the smallest whole number ratio of the constituent atom within the molecule. Empirical formula of different compound may be same. So it may or may not represent the actual formula of the molecule. It can be deduced by knowing the weight % of the entire constituent element with their atomic masses for the given compound.

For example: C₆H₁₂O₆, CH₃COOH, HCHO All have same empirical formula CH₂O, but they are different.

The empirical formula of a compound can be determined by the following steps:

Step-(1) Write the name of detected elements in column-1 present in the compound.

Step-(2) Write the corresponding atomic mass in column-2.

Step-(2) Write the experimentally determined percentage composition by weight of each element present in the compound in column-3.

Step-(2) Divide the percentage of each element by its atomic weight to get the relative number of atoms of each element in column-4.

Step-(2) Divide each number obtained for the respective elements in step (3) by the smallest number among those numbers so as to get the simplest ratio in column-5.

Step-(2) If any number obtained in step (4) is not a whole number then multiply all the numbers by a suitable integer to get whole number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. Empirical formula of the compound can be written with the help of this ratio in column-6.

ILLUSTRATIVE EXAMPLE (1): A compound contains C =71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?

SOLUTION:

ILLUSTRATIVE EXAMPLE (2): The simplest formula of a compound containing 50% of element X (Atomic mass = 10) and 50% of the element Y (Atomic mass = 20) is:

                        (A) XY                                                  (B) X₂Y

                        (C) XY₂                                                 (D) X₂Y₃

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): A compound of carbon, hydrogen and nitrogen contains these elements in the ratio 9:1:3.5. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula?

SOLUTION

Empirical formula = C₃H₄N

Empirical formula mass = (3 ´ 12) + (4 ´ 1) + 14 = 54

Thus, molecular formula of the compound

 = 2 ´ empirical formula = 2 ´ (C₃H₄N )= C₆H₈N₂

 ILLUSTRATIVE EXAMPLE (4): 2.38 gm of uranium was heated strongly in a current of air. The resulting oxide weighed 2.806 g. Determine the empirical formula of the oxide. (At. mass U = 238; O = 16).

SOLUTION:

Step 1: To calculate the percentage of uranium and oxygen in the oxide.

Step 2: To calculate the empirical formula

ILLUSTRATIVE EXAMPLE (5): Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present. Carbon 10.06%, hydrogen 0.84%, chlorine 89.10%. Calculate the empirical formula of the compound.

SOLUTION:

Step 1: Percentage of the elements present

Step 2:      Dividing the percentage compositions by the respective atomic weights of the elements

Step 3: Dividing each value in step 2 by the smallest number among them to get simple atomic ratio  

Step 4:      Ratio of the atoms present in the molecule  C  : H  : Cl

                                                                                     1  :  1  :   3

      The empirical formula of the compound is  C₁H₁Cl₃ o r CHCl₃.

ILLUSTRATIVE EXAMPLE (6): A carbon compound on analysis gave the following percentage composition. Carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound

SOLUTION:

Step 1: Percentage composition of the elements present in the compound.

Step 2:      Dividing by the respective atomic weights

Step 3:      Dividing the values in step 2 among them by the smallest number.

Step 4:      Multiplication by a suitable integer to get whole number ratio.

       Thus the simplest ratio of the atoms of different elements in the compound.

                                            C : H : Cl : O = 2 : 3 : 3 : 2

 (1) MOLECULAR FORMULA:

The formula which represents the actual number of each individual atom in any molecule is known as molecular formula. For certain compounds the molecular formula and the empirical formula may be same.

Molecular formula= (Empirical formula) n

Molecular Weight = Empirical Weight * n

If the vapour density of the substance is known, its molecular weight can be calculated by using the equation.

Molecular Weight= 2* Vapour Density

ILLUSTRATIVE EXAMPLE (6): The empirical formula of a compound is . Its molecular weight is 90. Calculate the molecular formula of the compound. (Atomic weights C = 12, H = 1, O = 16)

SOLUTION:

Empirical formula =CH₂O

Empirical formula weight = (12 + 2 + 16) = 30

         The molecular formula = (CH₂O)₃=C₃H₆O₃

ILLUSTRATIVE EXAMPLE (7): Certain non metal X forms two oxides I and II. The mass % of oxygen in 1st X₄O₅ is 43.7, which is same as that of X in 2nd oxide. Find the formula of 2nd oxide.

SOLUTION:

Try yourself:

Exercise (1):  A crystalline hydrated salt, on being rendered anhydrous, loses 45.6% of its mass. The percentage composition of the anhydrous salt is: Al = 10.5%, K = 15.1%, S =24.8% and oxygen = 49.6%. Find the empirical formula of the anhydrous and crystalline hydrated salt.        [K = 39; Al = 27; S = 32; O = 16; H = 1]

Exercise (2): A colourless crystalline compound has the following percentage composition: Sulphur 24.24%, nitrogen 21.21%, hydrogen 6.06% and the rest is oxygen. Determine the empirical formula of the compound. If the molecular mass is 132, what is the molecular formula of the compound? Name the compound if it is found to be sulphite.

Exercise (3) A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen.
1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the molecular formula of the hydrocarbon.

Answers Keys:     

Exercise (1) Empirical formula of anhydrous salt = KAS₂O₈ Hydrated salt composition; % anhydrous part = 54.4% and % H2O = 45.6%; Empirical formula of hydrated salt = KAIS₂O₈.12H₂O.

Exercise (2) Empirical formula SN₂H₈O₄; Molecular formula = SN₂H₈O₄; Name – Ammonium sulphite (NH₄)₂SO₃.

Exercise (3) C₂H_{4
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GAY LUSSAC'S LAW OF COMBINING VOLUME:

At given temperature and pressure the volumes of all gaseous reactants and products bear a simple whole number ratio to each other.

For example:

I.e. one volume of hydrogen reacts with one volume of chlorine to form two volumes of HCl gas. I.e. the ratio by volume which gases bears is 1:1:2 which is a simple whole number ratio.

Similarly other examples are:

ILLUSTRATIVE EXAMPLE (14):  In the reaction

the ratio of volumes of nitrogen, hydrogen and ammonia is 1: 3 : 2. These figures illustrate the law of:

                         (A) Constant proportions                               (B) Gay-Lussac

                        (C) Multiple proportions                                 (D) reciprocal proportions

SOLUTION:  The above ratio of 1: 3: 2 illustrates the Gay-Lussac law of combining volume. 

                          Hence (B) is correct.

ILLUSTRATIVE EXAMPLE (15): How much volume of oxygen will be required for complete combustion of 40 ml of acetylene (C₂H₂) and how much volume of carbon dioxide will be formed? All volumes are measured at NTP.

SOLUTION:

So for complete combustion of 40 ml of acetylene, 100 ml of oxygen are required and 80 ml of CO₂ is formed.

LAW OF RECIPROCAL PROPORTION:

If two elements B and C react with the same mass of a third element (A), the ratio in which they do so will be the same or simple multiple if B and C reacts with each other.

The above law is the basis of law of equivalent masses, which will be described latter in details.

For example: 

Ratio of masses of carbon and sulphur which combine with fixed mass (32 parts) of oxygen is

                                                12:32 or 3:8                                                                … (1)

In CS₂ ratio of masses of carbon and sulphur is 12: 64 or 3:16                                 … (2)

The two ratios (1) and (2) are related to each other by 3/8:3/18 or 2:1

ILLUSTRATIVE EXAMPLE (11): One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combines together, the ratio of their masses will be governed by:

                        (A) Law of definite proportions                      (B) law of multiple proportions

                        (C) Law of reciprocal proportions                  (D) law of conservation of mass

SOLUTION:  

From this when A combined with C the ratio is A/C = 1/3 Hence (C) is correct.    

ILLUSTRATIVE EXAMPLE (12): Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as          

                        (A) 23: 35.5                                                     (B) 35.5: 23

                        (C) 1: 1                                                            (D) 23: 1

SOLUTION:  1 atom of hydrogen combined with 1 atom of chlorine and 1 atom of sodium. So when sodium reacts with chlorine the ratio is 1: 1 by atom. While 23: 35.5 by mass. Hence (A) is correct.

ILLUSTRATIVE EXAMPLE (13): Copper sulphide contains 66.6% Cu, copper oxide contains 79.9% copper and sulphur trioxide contains 40% Sulphur. Show that these data illustrates law of reciprocal proportions.

SOLUTION:  

In copper sulphides Cu: S mass ratio is 66.6: 33.4

In sulphur trioxide Oxygen: Sulphur (O: S) mass ratio is 60:40

Now in CuS 33.4 parts of sulphur combines with Cu = 66.6 parts

40.0 parts of sulphur combines with cu=66.68*40/33.4=79.8 parts

Now the ratio of masses of Cu and O which combines with same mass (40 parts) of sulphur separately is 79.8:60                                                                                       … (1)

                                                Cu: O ratio by mass in CuO is 79.9: 20.1        … 2)

                                                Ratio 1: Ratio 2 = 1:3

Which is simple whole number ratio; hence law of reciprocal proportion is proved.

TRY YOURSELF:

Exercise 9. Carbon combines with hydrogen to form three compounds A, B and C. The  percentage of hydrogen in A, B and C are 25, 14.3 and 7.7 respectively. Which law of chemical combination is illustrated?

Exercise 10. 61.8g of A combines with 80 g of B. 30.9g of A combines with 106.5g of C, B and C combine to form compound CB₂. Atomic weight of C and B are respectively 35.5 and 6.6. Show that the law of reciprocal proportion is obeyed.

Exercise 11. Carbon dioxide contains 27.27% carbon, carbon disulphide contains 15.97% carbon and sulphur dioxide contains 50% sulphur. Show that these figures illustrate the law of reciprocal proportions.

`Exercise 12. Aluminium oxide contains 52.9% aluminium and carbon dioxide contains 27.27% carbon. Assuming the validity of the law of reciprocal proportions, calculate the percentage of aluminium in aluminium carbide.