Page 148 - Chemistry ICSE Class X

134 ICSE Chemistry – 10

Step 2. If the sum of the percentage of all these elements is not 100, then the
difference gives the percentage of oxygen. Calculate the percentage of
oxygen in the given compound by using the following relationship:
Percentage of oxygen = 100 – (Sum of the percentage of all other
elements present in the compound).
Step 3. Divide the percentage of each element by the atomic mass of the
respective element. The ratio so obtained is called atomic ratio.
Step 4. Divide the atomic ratios by the lowest value, and convert these into the
nearest whole numbers. These whole numbers give the simplest ratio
between the number of atoms of the various elements present in the
compound.
Step 5. Write down the empirical formula of the compound.
Step 6. Calculate the empirical formula mass by adding the atomic masses of
all the atoms present in the empirical formula.
Step 7. Obtain molecular mass (or molar mass) either from experiment or
from the vapour density of the compound by using the relationship,
Molecular mass = 2 × Vapour density
Step 8. Obtain the value of n through the relation,
Molecular mass
n =
Empirical formula mass
‘n’ is an integer, such as 1, 2, etc.
Step 9. The molecular formula is then obtained by
Molecular formula = n × Empirical formula
The above method of writing the molecular formula of a compound is
illustrated through the following numerical problems:

EXAMPLE 5.20. A compound contains 75% carbon and 25% hydrogen.
Determine its empirical formula. The molecular mass of this compound is
16 amu. Determine its molecular formula also. The atomic masses are: C = 12 u,
H = 1 u.
Solution: Percentage Atomic Atomic Simplest
Elements
composition mass ratio atomic ratio
C 75 12 75/12 = 6.25 6.25/6.25 = 1
H 25 1 25/1 = 25 25/6.25 = 4
So,
Empirical formula of the compound = C H or CH 4
4
1
Then, Empirical formula mass = (1 × 12 u) + (4 × 1 u) = 12 u + 4 u = 16 u
Molecular mass (given) = 16 u
Molecular mass 16 u
So, n = = = 1
Empirical formula mass 16 u
Therefore, Molecular formula = 1 × Empirical formula = 1 × CH = CH 4
4
EXAMPLE 5.21. A substance on analysis gave the following percentage
composition : Na = 43.4%, C = 11.3% and O = 45.3%. Determine its empirical and
molecular formulae. Given the relative molecular mass of the compound is 106.
Solution: Percentage Atomic Simplest
Elements Atomic ratio
composition mass atomic ratio
Na 43.4 23 43.4/23 = 1.89 1.89/0.94 = 2
C 11.3 12 11.3/12 = 0.94 0.94/0.94 = 1
O 45.3 16 45.3/16 = 2.83 2.83/0.94 = 3