Page 149 - Chemistry ICSE Class X

Mole Concept and Stoichiometry 135
Thus, the simplest ratio of Na, C and O in the given compound is 2 : 1 : 3.
So, empirical formula of the given compound is Na CO .
2
3
Then,
Empirical formula mass of the compound = (2 × 23 u) + (1 × 12 u) + (3 × 16 u)
= 46 u + 12 u + 48 u = 106 u
Molecular mass 106 u
So, n = = = 1
Empirical formula mass 106 u
Then, the molecular formula of the given compound is
Molecular formula = 1 × Empirical formula
= 1 × Na CO = Na CO 3
2
3
2
EXAMPLE 5.22. A compound is found to contain 11.2% nitrogen (N), 3.2%
hydrogen (H), 41.2% chromium (Cr) and 44.4% oxygen (O). Determine the
stoichiometric (empirical) formula. Atomic masses (u) are: N = 14, H = 1, Cr = 52
and O = 16.
Solution: The above results are written as follows:

Simplest
Percentage Atomic Simplest whole
Elements Atomic ratio
composition mass atomic ratio number
ratio
N 11.2 14 u 11.2/14 = 0.8 0.8/0.8 = 1 2
H 3.2 1 u 3.2/1 = 3.2 3.2/0.8 = 4 8
Cr 41.2 52 u 41.2/52 = 0.8 0.8/0.8 = 1 2
O 44.4 16 u 44.4/16 = 2.8 2.8/0.8 = 3.5 7
Since, the simplest atomic ratios cannot have fractional values, hence the simplest
whole number atomic ratio of N, H, Cr and O are 2, 8, 2, 7 respectively. Therefore,
Stoichiometric (empirical) formula of the compound = N H Cr O .
2
2
8
7
(The actual formula of this compound is (NH ) Cr O . This compound is named as
2
4 2
7
+
2–
ammonium dichromate. Ammonium dichromate contains NH and Cr O ions).
4
7
2
Calculations Based on Chemical Equations
A balanced chemical equation provides information regarding the relative
amount, mass or volume of the reactants and products. ͻ DŽůĞͲDŽůĞ ƌĞůĂƟŽŶƐŚŝƉ
The fundamental behind all these relationships is the mole concept. ͻ DŽůĞͲDĂƐƐ ƌĞůĂƟŽŶƐŚŝƉ
The number of moles of any substance is directly related to its mass and ͻ DĂƐƐͲDĂƐƐ ƌĞůĂƟŽŶƐŚŝƉ
its number of molecules or atoms. Therefore, different relationships can be ͻ DŽůĞͲsŽůƵŵĞ ƌĞůĂƟŽŶƐŚŝƉ
written between the reactants and products. Some typical relationships are ͻ DĂƐƐͲsŽůƵŵĞ ƌĞůĂƟŽŶƐŚŝƉ
given alongside and below. ͻ sŽůƵŵĞͲsŽůƵŵĞ ƌĞůĂƟŽŶƐŚŝƉ
Some useful relationships are,
23
ƒ 1 mol { 6.02 × 10 chemical species
{ 1 g-mol { 1 molar mass = 22.4 litres at NTP
ƒ No. of moles of a substance = Mass of the substance
Molar mass of the substance
Volume (in litres) of the substance at NTP
ƒ No. of moles of a gaseous substance =
22.4

23
ƒ No. of atoms or molecules in a sample of substance = Mass of the substance × (6.02 × 10 )
Molar mass of the substance