Concentration Calculation and Methods of Determining Concentration of Compounds, Molarity, Molality, and Normality with solved example

Concentration is the word used frequently in the calculation of material balance, energy balance, and analyzing the purity of the substance. In physics and chemistry, it is a regular word. Without gaining enough knowledge on concentration concepts you are not able to understand the depth concepts of environmental science, mass transfer and reaction engineering, etc.

Concentration is a term used to know the amount of matter present in the measurable volume, mostly followed by unit volume. We should be knowing, that mass, weight, and mole are three physical quantities that tell us about the matter content but in various phenomena let us discuss it.

MASS:  It is a physical matter that exists in the universe.

WEIGHT: It is the force exerted by that mass by the influence of gravity.

MOLE:  It is the number that represents the number of molecules present in available mass.

So, these differences are clearly distinguished and now let a substance called β€˜X’ it take of unit volume which means 1cm of length, breadth, and height respectively.

Concentration thus can be calculated in terms of

MassMass/ volumeKg/m3(GC)
WeightWeight/ VolumeKg/m3

The density of solid is the term used for the mass/volume ratio

Liquid density is molecular mass/ molar volume

Vapour density is given the ideal gas equation PV= nRT.

The Gram Atom: It is used to specify the amounts of chemical elements. It is defined as the mass in grams of an element which is equal numerically to its atomic weight.

Gram atom of element =weight in grams of the element / atomic weight of the element

Mole: a mole is defined as the amount of substance equal to molecular weight (formula weight)

The Gram Mole: It is used to specify the amounts of chemical compounds. It is defined as the mass in grams of the substance that is equal numerically to its molecular weight.

The gram mole of compound = weight in grams of that compound  / compound molecular weight

Mostly gram mole is commonly termed a mole in textbooks. If you see a mole consider it to be a gram mole.

The molecular weight of the compound is found from the atomic weights of elements involved in the formation of compounds. Summing all the atomic weight multiplied by their respective no of molecules or elements participate in the formation of the compound.

Example 1: Find the molecular weight of sodium hydroxide NaOH
                 the atomic weight of Na = 23
                 the atomic weight of O   = 16
                 the atomic weight of H   =  1
= 23 X 1
=16 X 1
=1 X 1
 one atom of each element is present in this compound so they multiplied with it, now add up

= 23 X 1 + 16 X 1 + 1 X 1
=  23 + 16 + 1 
Ex 2: Find Molecular weights of H2SO4, Na2CO3, KMnO4

Remember the atomic weight of H= 1, S= 32, O = 16

molecular weight of H2SO4 = 2 X 1 + 1 X 32 + 4 X 16 = 98

Atomic weight of Na= 23, C = 12 , O = 16
molecular weight ofNa2CO3 = 2 X 23 + 1 X  12 + 3 X 16 = 106

Atomic weight of K = 39 , Mn = 55,  O = 16
molecular weight of KMnO4 = 1 X 39 + 1 X 55 + 4 X 16 = 158

Molarity, Molality, and Normality are important terms used in the determination of concentration when concerned with chemical material balance or chemical reactions studies.

The equivalent weight of an element or compound: is defined as the ratio of the atomic weight or molecular weight to its valence. The valence of an element or a compound does depend on the number of hydrogen ions (H+) accepted or hydroxyl ions (OH-) donated for each atomic weight or molecular weight.

Equivalent weight = Molecular weight/valence


Ex: (1)

Calculate the equivalent weight (EW) of the following ions or molecules:

Chemicals:                      K+      Mg2+        Na2CO3
Ionic/molecular weight:  39          24           106

For K+ valence is +1 so, 39/1 = 39
Mg2+ valence is +2, so the equivalent weight is 24/2 = 12
Na2CO3 valence is +2, so the equivalent weight is 106/ 2 = 53
Equivalent weight does not have signs so if you find the valence of any sign, use the numerical value and molecular weight, and divide the molecular weight with the numerical value of the valence.


It is defined as the number of grams -equivalents of solute dissolved in one liter of solution. and designated by the symbol ‘N’

Normality (N) =  Gram equivalent of solute / Volume of solution in a liter

Gram equivalent  is the ratio of the weight of the compound to the equivalent weight of the compound


It is defined as the number of gram moles of the solute dissolved in one liter of solution. It is designated by the symbol M.

Molarity (M) =  Gram moles of solute / Volume of solution in a liter


It is defined as gram moles of solute dissolved in one kilogram of solvent

Molality = gram moles of solute/mass of solvent in Kg

Note: to find the concentration of solute in g/l we can use the formula as
= Normality  X equivalent weight
Ex: (2)

Find the Equivalent weight of HCl, NaOH, H2SO4

the molecular weight of HCl: 1 X 1 +  1 X 35.5 = 36.5
the valence of HCl: 1
therefore equivalent weight of HCl = 36.5 / 1 = 36.5


molecular weight of NaOH: 1 X 23 + 1X 16 +  1 X 1 =40
the valence of NaOH: 1
therefore the equivalent weight of NaOH: 40 / 1 = 40

molecular weight of  H2SO4: 2 X 1 + 1X 32 +  4 X 16 =98
the valence of  H2SO4: 2
therefore equivalent weight of  H2SO4: 98 / 2 = 49
Ex: (3)

Find the Equivalent weight of H3PO4, CaCl2, FeCL3, Al2(SO4)3, KMnO4

molecular weight of  H3PO4: 3 X 1 + 1X 31 +  4 X 16 =98
valence of H3PO4: 3
equivalent weight of  H2SO4: 98 / 3 = 32.67

CaCl2 :
molecular weight of CaCl2 : 1 X 40 + 2 X 35.5 = 111
valence of CaCl2 :2
equivalent weight of  CaCl2 : 111 / 2 = 55.5

FeCl3 :
molecular weight of FeCl3 : 1 X 56 + 3 X 35.5 = 162.5
valence of FeCl3: 3
equivalent weight of  FeCl3 : 162.5 / 3 = 54.17

Al2(SO4)3 :
molecular weight of Al2(SO4)3 : 2 X 27 + 3 X 32 + 2X 16 =342
valence of Al2(SO4)3: 6

equivalent weight of  Al2(SO4)3 : 342 / 6 = 57

KMnO4 :
molecular weight of KMnO4 : 1 X 39 + 1 X 55 + 4X 16 =158
valence of KMnO4: 5

equivalent weight of  KMnO4 : 158 / 5 = 31.6
Ex: (4)

Find the normality, molarity, and molality of the caustic solution containing 20% NaOH by weight, the density of the solution is 1.196 kg/l

Basis: 100 kg of solution 
so, the solution contains 20 kg of NaOH (20X 100/100) and 80 kg of water(solvent)
The density of the solution given as 1.196 kg/l

Now find the volume of the solution by using mass and density as volume = mass / density
= 100/ 1.196 = 83.62 liters

Moles of NaOH in solution = weight of NaOH in the solution / molecular weight of NaOH
                                            = 20/40 = 0.5 kmol = 500 mol (also called as gram mole or g mole)

Molarity =  gram mole of NaOH / volume of solution in a liter
              =  500 / 83.52
              =  5.98 mol/ liter

For NaOH the valence = 1
Equivalent weight = molecular weight/valence
                           =  40 / 1 = 40 
Gram equivalent weight = weight of NaOH / equivalent weight
                                    =  20000 gram / 40   (since we have 20 kg converted to gram it becomes 20X1000 g)
                                    =    500

Normality =  gram equivalent of NaOH / volume of solution in a liter
                 = 500 / 83.52
                 = 5.98

Molality= gram moles of NaOH /  kg of solvent
               = 500 / 80 
               = 6.25 mol/kg

Ex: (5)

To prepare 1 N of the Na2CO3 solution, how many grams of Na2CO3 must be dissolved in 1.0 L of water?

By using Normality formula = gram equivalent of solute/volume of solution in a liter
The 1.0-liter solution, 1 N of Na2CO3
We know the molecular weight Na2CO3 = 106, valence = 2
Gram equivalent = weight of Na2CO3 / equivalent weight of Na2CO3

1N = weight of Na2CO3/ 53/1 liter solution

weight of Na2CO3 = 53 g

Specific gravity is used for the indirect measurement of concentrations of aqueous solutions. which is the ratio of the density of solution at T1 (temperature)to that of the density of water at T2 (temperature)and the specific gravity of a gas is the ratio of the average molecular weight of the gas to the molecular weight of air at the same temperature.

In water analysis, the impurities are in traces and their concentration is expressed in mg/liter or Parts per million.

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