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 fomula of the molecule. It can be deduced by knowing the weight %of allthe constituent element
with their atomic masses for the given compound.
For example: CsH2O6, CH3COOH, HCHO
All have same empirical formula CH,0, but they are different.
The empirical fomula of a compound can be determined by the following steps:
Write the name of detected elements in column-1 present in the compound.
Write the corresponding atomic mass in column-2.
Wite the experimentally determined percentage composition by weight of each element present
in the compound in column-3.
number of atoms of
Divide the percentage of each element by its atomic weight to get the relative
each element in column-4.
the smallest number
Divide each number obtained for the respective elements in step (3) by
among those numbers so as to get the simplest ratio in column-5.
then multiply all the numbers by a
If any number obtained in step (4) is not a whole number
simplest ratio of the atoms of
suitable integer to get whole number ratio. This ratio will be the
the compound can be written
different elements present in the compound. Empirical formula of
with the help of this ratio in column-6.
Molecular formulla = Empirical formula x n
Molecular mass
Empirical formula mass
Molecular mass =2xVapour density
pressure and temperature.)
(Vapour density: Ratio of density of vapour to the hydrogen at similar
15.81%. What is the empirical
Illustration 19. Acompound contains C = 71.23%, H= 12.95% and O =
formula of the compound?
Element's Atomic Weight % Relative Simplest Empirical
Solution: number of atomic formula
name with mass in
compound atom ratio
their symbol
12 71.23 71.23 5.936
Carbon (C) =5.936 =6
12 0.988
1 12.95 12.95 12.95
Hydrogen (H) =12.95 = 13 CH,,0
1 0.988
Oxygen (O) 16 15.81 15.81 0.988
=0.988 =1
16 0.988
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Illustration 20. The simplest formula of acompound containing50% of element X (Atomic mass 10)
and 50% of the element Y (Atomic mass = 20) is:
(A) XY (B) X,Y
(C) XY; (D) X>Y,
Solution: Element's Atomic Weight %in Relative Simplest Empirical
name with number of atomic ratio formula
mass compound
their atom
Symbol
X 10 50 50/10 = 5 5/2.5 =2
X,Y
|Y 20 50 50/20 = 2.5 2.5/2.5 =1
Hence (B) is correct.
IlMustration 21. Acompound of carbon, hydrogen and nitrogen contains three elements in the ratio
9:1:3.5. Calculate the empirical formula. f its molecular mass is 108, what is the
molecular formula?
Solution:
Element Element Atomic mass Relative number of atoms Simplest ratio
ratio
0.75
Carbon 9 12 -=0.75 =3
12 0.25
Hydrogen 1 1 1 1
-=1 =4
1 0.25
14 3.5 0.25
Nitrogen 3.5 =0.25 =1
14 0.25
Empirical formula = CgH4N
Empirical formula mass = (3 x 12) + (4 x 1) + 14 = 54
mol. mass 108
=2
Emp. formula mass 54
Thus, molecular formula of the compound
=2x empirical formula
=2x CzsH,N = CeHgN
llustration 22. 2.38 gm of uranium was heated strongly in acurrernt of air. The resuiting 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.
2.806 g of the oxide contain uranium = 2.38 g
2.38
Percentage of uranium = -x 100= 84.82
2.806
Hence, the percentage of oxygen in the oxide
= 100.00 - 84.82 = 15.18
Step 2: To calculate the empirical formula
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Percentage
8ofelements
of
mass
At. ratio
atonmic no.atomic
ratio
Element Symbol elements
Relative no. of Simplest Simplest
atoms whole
Percentage
At. mass
Uranium U 84.82 238 84.82 0.3562 3
=0.3562
238 0.3562
Oxygen 15.18 16 15.18 0.94875 8
=0.94875
16 0.3562
=2.666
Hence the empirical formula of the oxide is U,0&.
llustration 23. 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
Carbon Hydrogen Chlorine
10.06 0.84 89.10
Step 2: Dividing the percentage compositions by the respective atomic weights of the
elements
10.06 0.84 89.10
12 35.5
0.84 0.84 2.51
Step 3: Dividing each value in step 2 by the smallest number among them to get simple
atomic ratio
0.84 0.84 2.51
0.84 0.84 0.84
Step 4: Ratio of the atoms present in the molecule C :H:CI
1:1: 3
.:.The empirical formulaof the compound C,H,CI, or CHCI,.
Illustration 24. 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: Step1: Percentage composition of the elements present in the compound.
C C
14.5 1.8 64.46 19.24
Step 2: Dividing by the respective atomic weights
14.5 1.8 64.46 19.24
12 1 35.5 16
1.21 1.8 1.81 1.2
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Step 3: Dividing the values in step 2 among them by the smallest number.
1.21 1.8 1.81 1.2
1.20 1.2 1.2 1.2
Step 4: Multiplication by a suitable integer to get whole number ratio.
(1x2) (1.5x2) (1.5x 2) (1×2)
2 3 3 2
..The simplest ratio of the atoms of different elements in the compound.
C:H:CI:0 =2:3:3:2
.Empirical formula of the compound C,H,CI,O,.
illustration 25. The empirical formula of a compound is CH,o. Its molecular weight is 90. Calculate the
molecular formula of the compound. (Atomic weights C= 12, H=1, O= 16)
Solution: Empirical formula = CH,0
Empirical formula weight = (12 +2 + 16) = 30
molecular weight 90
n= ..n= =3
empirical formula weight 30
The molecular fornula (CH,0), = C,H,0,
EXERCISE -2
1. Chlorine has isotopes CI and CI, There are three "CI isotopes for every one Cl isotopes in
asample of chlorine. Calculate the atomic mass of chlorine.
2. Natural hydrogen gas is a mixture of H and H in the ratio of 5000:1. Calculate the atomic mass
of the hydrogen.
3. Hydrogen and oxygen are combined in the ratio 1:16 by mass in hydrogen peroxide. What is the
percentage composition of hydrogen peroxide?
4 A mineral contained MgO= 31.88%, SiO, = 63.37% and H,O = 4.75%. Show that the
fomula for the mineral is HzMg;Si,O12 (H = 1, Mg =24, Si = 28, O = 16) simplest
5. A colouless 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 sulphate.
6. Detenine the percentage of water of crystallization in pure sample of green vitriol
(FeSO47H,0).
(A) 45.32% (B) 42.23%
(C) 50% (D)48.32%
7. A gaseous hydrocartbon contains 85.7% carbon and 14.3%
hydrogen.
1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the molecular fomula of the
hydrocartbon
(A) CaHe (B) C,Hs
(C) CzHe (D) CAH0
An automobile antifreeze consists of 38.7% C, 9.7% H and remaining oxygen by
weight. When
0.93g of it are vapourized at STP 336 ml of vapour are formed. Find the molecular formula
artífreeze
of the
(A)CH,O, (B) C,H,0
(C) CaHO, (D) C,H,O4
