Basic calculation chemical

1. Basic Chemical Calculations
Dr. Bimal Das

2. Before dealing with chemical process calculations, we need to be familiar with basic
chemical principles underlying the subject of discussion of the following terms.
(i) Atomic weight
(ii) Molecular weight
(iii) Basis of calculation.
Atomic weight
It may be defined as the mass of an atom
that assigns carbon a mass of twelve exactly.
Molecular weight
It may be defined as the sum of the atomic
weights of the atoms using which a molecule
of a compound is formed.

3. Determine the molecular weight of the following compounds.
(i) Na2CO3
(ii) HNO3
(iii) MgSO4
(iv) H2SO4
(v) HCl
Solution:
(i) Na2CO3
Atomic weights: Na = 23, C = 12 and O = 16
Molecular weight of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16
= 46 + 12 + 48 = 106
(ii) HNO3
Atomic weights: H = 1, N = 14 and O = 16
Molecular weight of HNO3 = 1 × 1 + 1 × 14 + 3 × 16
= 1 + 14 + 48 = 63
(iii) MgSO4
Atomic weights: Mg = 24, S = 32 and O = 16
Molecular weight of MgSO4 = 1 × 24 + 1 × 32 + 4 × 16
= 24 + 32 + 64 = 120
(iv) H2SO4
Atomic weights: H = 1, S = 32 and O = 16
Molecular weight of H2SO4 = 2 × 1 + 1 × 32 + 4
× 16
= 2 + 32 + 64 = 98
(v) HCl
Atomic weights: H = 1, and Cl = 35.5
Molecular weight of HCl = 1 × 1 + 1 × 35.5
= 1 + 35.5 = 36.5

4. Gram atom is used to specify the amounts of chemical elements. It is defined as a mass in
gram of an element which is numerically equal to its atomic weight. It can be represented
as
• Calculate the kg atoms of carbon which weigh 48 kg.
Solution: Basis: 48 kg of carbon
Atomic weight of carbon = 12

5. In general gmol or kgmol is used to specify the amounts of chemical compounds. It is
defined as the mass in grams of substance that is numerically equal to its molecular
weight. It can be represented as
Calculate the moles of oxygen present in 640 grams.
Solution: Basis: 640 g of oxygen
Molecular weight of oxygen = 32
How many kilograms of ethane are there in 220 kgmol?
Solution: Basis: 220 kgmol of ethane
Chemical formula of ethane = C2H6
Atomic weights: C = 12 and H = 1
Molecular weight of ethane = 2 × 12 + 6 × 1
= 24 + 6 = 30
kg of ethane = kgmol of ethane × Molecular weight of ethane
= 220 × 30 = 6600

6. Equivalent Weight
It is defined as the ratio of the atomic weight or molecular weight to its valency.
The valency of an element or a compound does depend on the number of hydrogen ions H+
accepted or the hydroxyl ion OH– donated for each atomic weight or molecular weight. It
can be expressed as

8. Calculate the equivalent moles of Na2SO4 in 1288 g of Na2SO4,10H2O crystals.
Solution: Basis: 1288 g of Na2SO4,10H2O crystals

9. Mole fraction is the ratio of moles of individual components to the total moles of the
system. For binary system of A and B, the mole fraction may be represented as

10. Mass Per cent:
It is the mass of any component expressed as the percentage of total mass of the system. It can
be expressed as
Volume Per cent:
It is the pure component volume of any
component expressed as a percentage of the
total volume of the system. It can be expressed
as

11. Mole Per cent:
It is the moles of any component expressed as the percentage of the total moles of the system.
It can be expressed as

12. An aqueous solution of sodium chloride is prepared by dissolving 20 kg of NaCl in 80 kg of
water. Calculate mole% composition of solution.
Solution: Basis: 20 kg of NaCl and 80 kg of water
Molecular weight of NaCl = 1 × 23 + 1 × 35.5 = 23 + 35.5 = 58.5
Molecular weight of H2O = 2 × 1 + 1 × 16 = 2 + 16 = 18

13. Concept of Normality, Molarity and Molality
There are three ways of expressing the concentration of solution containing either solid or liquid
solute, namely
1. Normality
2. Molarity
3. Molality
Normality
It is defined as the number of gram equivalents of solute
dissolved in one litre of solution. It is designated by the
symbol N. It can be expressed as
Molarity
It is defined as the number of gram moles of the solute
dissolved in one litre of solution. It is designated by the
symbol M. It can be expressed as
Molality
It is defined as the gram moles of the solute dissolved in one
kilogram of the solvent. It can beexpressed as

14. Example : 196 grams of sulphuric acid (H2SO4) are dissolved in water to prepare one litre of
solution. Determine the normality, molarity and molality.

16. Concept of PPM (Parts Per Million)
It is the short form of parts per million, i.e. parts of one substance present in million parts of
another substance, specially solvent. It is commonly used as a unit of concentration.
When a large amount of solute substance is present in a solvent, it can be comfortably represented
by the unit g/l. But, when the solute substance present in solvent is very very small, the
uncomfortness arises to represent the unit as g/l. To overcome this difficulties, ppm unit is used.
The unit as ppm is commonly used to measure the small level of pollutants present in air. For
example, Permissible Exposure Limit (PEL) of CO in air is 50 ppm. It is also used as a measure of
small level of pollutant present in drinking water. For example, permissible exposure limit of
Arsenic in drinking water is 1 ppm.

17. EXAMPLE: A sample of water contains 2000 ppm solids. Find the concentration of solids by
weight percentage.
Solution: Basis: 106 kg of water sample