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### DETERMINATION OF EMPIRICAL AND MOLECULAR FORMULA:

(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
.