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. 

LAW OF MILTIPLE PROPORTION:

When two elements combine to form two or more than two different compounds then the different masses of one element B which combine with fixed mass of the other element bear a simple ratio to one another.

For example: Carbon forms two oxides in oxygen

The ratio of masses of oxygen in CO and CO₂ for fixed mass of carbon (12)
is 16: 32 = 1: 2.

ILLUSTRATIVE EXAMPLE (9): The law of multiple proportions is illustrated by the pair of compounds:

(A) Sodium chloride and sodium bromide     

(B) Water and heavy water

(C) Sulphur dioxide and sulphur trioxide       

(D) Magnesium hydroxide and magnesium oxide

SOLUTION:  In SO₂ 32 gram of suphur react with 32 gram of oxygen. Similarly for SO₃ fixed mass of sulphur (32 gram) react with 48 gram of oxygen. The ratio of oxygen’s mass = 32:48 = 2:3 which support law of multiple proportions. Hence (C) is correct answer.

ILLUSTRATIVE EXAMPLE (10): Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show that these figures illustrate the law of multiple proportions.

SOLUTION:  

Step 1: To calculate the percentage composition of carbon and oxygen in each of the two oxides

Step 2:  To calculate the weights of carbon which combine with a fixed weight i.e., one part by weight of oxygen in each of the two oxides.

In the first oxide, 57.1 parts by weight of oxygen combine with carbon = 42.9 parts.

1 part by weight of oxygen will combine with carbon = 42.9/57.1=0.751

In the second oxide 72.7 parts by weight of oxygen combine with carbon = 27.3 parts.

1 part by weight of oxygen will combine with carbon =27.3/72=0.376

Step 3:  To compare the weights of carbon which combine with the same weight of oxygen in both the oxides-The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751: 0.376 or 2:1

Since this is a simple whole number ratio, so the above data illustrate the law of multiple proportions.

TRY YOURSELF:

Exercise (1): Metal M and chlorine combine in different proportions to form two compounds A and B. The mass ratio M: Cl is 0.895: 1 in A and 1.791: 1 in B. What law of chemical combination is illustrated?

Exercise (2): By means of the following analytical results show that law of multiple proportions is true:

Exercise (3): 1 g of a metal, having no variable valency, produces 1.67 g of its oxide when heated in air. Its carbonate contains 28.57% of the metal. How much oxide will be obtained by heating 1 g of carbonate?

Exercise (4): Phosphorous and chlorine form two compounds. The first contains 22.54% by mass of phosphorous and the second 14.88% of phosphorous. Show that these data are consistent with law of multiple proportions.

TRY YOURSELF: SOLUTION:

(1) Mass of metal which combine with 1 part of chlorine are in the ratio of 1:2, which is a simple ratio. Hence, law of multiple proportions is illustrated.

(2) The masses of mercury which combine with 1 part of chlorine are in the ratio of 2:1 which is simple ratio. Hence, law of multiple proportions is illustrated.     

(3) 0.477 g

(4) Prove yourself ….

Limitation of law of multiple proportions: Non Stoichiometric compound do not follow this law for example Fe_0.93O₁.

LAW OF DEFINITE OR CONSTANT COMPOSITION:

A chemical compound always contains same elements in definite proportion by mass and it does not depend on the source of compound.

For example:

The composition of CO₂ obtained by different means always having same hence Law of definite proportion is proving.

ILLUSTRATIVE EXAMPLE (2): Ammonia contains 82.65 % N₂ and 17.65% H₂. If the law of constant proportions is true, then the mass of zinc required to give 10 g Ammonia will be:

                        (A)       8.265 g                                    (B) 0.826 g

                        (C)       82.65 g                                     (D) 826.5 g

SOLUTION: The mass of zinc required to give 10 g ammonia will be

ILLUSTRATIVE EXAMPLE (3): Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of:

                        (A) Conservation of mass                    (B) Constant composition

                        (C) Multiple proportion                      (D) Constant volume

SOLUTION: As in water

Both values always constant. Obey law of constant composition. Hence (B) is correct.

ILLUSTRATIVE EXAMPLE (4): 6.488 g of lead combine directly with 1.002 g of oxygen to form lead peroxide PbO₂. Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition

SOLUTION:

Step 1: To calculate the percentage of oxygen in first experiment.

Step 2: To compare the percentage of oxygen in both the experiments.

            Percentage of oxygen in PbO₂ in the first experiment = 13.38

            Percentage of oxygen in PbO₂ in the second experiment = 13.38

Since the percentage composition of oxygen in both the samples of PbO₂ is identical, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (5): Copper oxide was prepared by the following methods:

(A) In one case, 1.75 g of the metal was dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.

(B) In the second case, 1.14 g of metal dissolved in nitric acid was precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.

(C) In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide. Show that the given data illustrate the law of constant composition.

SOLUTION:

Step 1:  In the first experiment.

            2.19 g of copper oxide contained 1.75 g of Cu.

            100 g of copper oxide contained = 1.75/2.19*100=79.91 %

Step 2:  In the second experiment.

            1.43 g of copper oxide contained 1.14 g of copper.

            100 g of copper oxide contained =1.14/1.43*100=79.72 %.

Step 3: In the third experiment.

            1.83 g of copper oxide contained 1.46 g of copper

            100 g of copper oxide contained =1.46/1.83*100 = 79.78 %

Thus the percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (6): 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO was produced in the process. Show that these results illustrate the law of constant proportions.

SOLUTION: The ratio of copper and oxygen is 1: 0.25. Hence, the law of constant proportions is illustrated.

ILLUSTRATIVE EXAMPLE (7): In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment 2.9 g of iron oxide give 2.03 g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions.

SOLUTION: Ratio of oxygen and iron in both the experiment is 1:2.33

ILLUSTRATIVE EXAMPLE (8): 2.8 g of calcium oxide (CaO) prepared by heating limestone were found to contain 0.8 g of oxygen. When one gram of oxygen was treated with calcium, 3.5 g of calcium oxide were obtained. Show that the results illustrate the law of definite proportions.

SOLUTION:  try yourself……

Limitation of law definite proportions: Discovery of isotopes created some limitations to this law. Isotopes of an element have different atomic mass hence they form same chemical compounds with different compositions;

For example: