Concentration is the word used frequently in the calculation of material balance, energy balance, analyzing the purity of the substance. In physics and chemistry, it is a regular word. Without gaining enough knowledge on concentration concept 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 as per unit volume. We should be knowing, that mass, weight and mole are three physical quantities which tell us about the matter content but in various phenomena’s 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 which represents the number of molecules present in available mass.
So, these differences are clearly distinguished and now let a substance called ‘X’ it take of unit volume which means that 1cm of length, breadth, and height respectively.
Concentration thus can be calculated in terms of
Concentration
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Units
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Mass
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Mass/ volume
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Kg/m3(gc)
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Weight
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Weight/ volume
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Kg/m3
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Mole
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Mole/volume
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Kmole/m3
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The density of solid is the term used for 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 a mass in gram of an element which is equal numerically to its atomic weight.
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.
Mostly gram mole is commonly termed as 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. By summing all the atomic weight multiplied with their respective no of molecules or elements participate in the formation of the compound.
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Example 1: Find the molecular weight of sodium hydroxide NaOH
atomic weight of Na = 23
atomic weight of O = 16
atomic weight of H = 1
Now ,
= 23 X 1
=16 X 1
=1 X 1
one atom of each element are present in this compound so they multiplied with it, now add up
= 23 X 1 + 16 X 1 + 1 X 1
= 23 + 16 + 1
= 40
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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
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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 ion (OH-) donated for each atomic weight or molecular 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 gram of that compound / compound molecular weight
Mostly gram mole is commonly termed as 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. By summing all the atomic weight multiplied with their respective no of molecules or elements participate in the formation of the compound.
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Example 1: Find the molecular weight of sodium hydroxide NaOH
atomic weight of Na = 23
atomic weight of O = 16
atomic weight of H = 1
Now ,
= 23 X 1
=16 X 1
=1 X 1
one atom of each element are present in this compound so they multiplied with it, now add up
= 23 X 1 + 16 X 1 + 1 X 1
= 23 + 16 + 1
= 40
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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
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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 ion (OH-) donated for each atomic weight or molecular weight.
Equivalent weight = Molecular weight / valence
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Ex: (1)
Chemicals: K+ Mg2+ Na2CO3
Ionic / molecular weight: 39 24 106
For K+ valence is +1 so, 39/1 = 39
similarly,
Mg2+ valence is +2, so equivalent weight is 24/2 = 12
Na2CO3 valence is +2, so equivalent weight is 106/ 2 = 53
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Equivalent weight does not have signs so if you find the valence of any sign, use the numerical value and molecular weight, divide the molecular weight with the numerical value of the valence.
It is defined as the number of gram -equivalents of solute dissolved in one litre of solution. and designated by symbol 'N'
where,
Gram equivalent is the ratio of the weight of the compound to equivalent weight of the compound
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 is defined as gram moles of solute dissolved in one kilogram of solvent
Note: to find the concentration of solute in g/l we can use the formula as
= Normality X equivalent weight
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Ex: (2)
HCl:
molecular weight of HCl : 1 X 1 + 1 X 35.5 = 36.5
valence of HCl : 1
there fore equivalent weight of HCl = 36.5 / 1 = 36.5
NaOH:
molecular weight of NaOH: 1 X 23 + 1X 16 + 1 X 1 =40
valence of NaOH: 1
there fore equivalent weight of NaOH: 40 / 1 = 40
H2SO4:
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
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Ex: (3)
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
similarly,
Mg2+ valence is +2, so equivalent weight is 24/2 = 12
Na2CO3 valence is +2, so equivalent weight is 106/ 2 = 53
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Equivalent weight does not have signs so if you find the valence of any sign, use the numerical value and molecular weight, divide the molecular weight with the numerical value of the valence.
Normality:
It is defined as the number of gram -equivalents of solute dissolved in one litre of solution. and designated by symbol 'N'
Normality (N) = Gram equivalent of solute / Volume of solution in liter
Gram equivalent is the ratio of the weight of the compound to equivalent weight of the compound
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.
Molarity (M) = Gram moles of solute / Volume of solution in a liter
Molality:
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
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Ex: (2)
Find the Equivalent weight of HCl, NaOH, H2SO4
HCl:
molecular weight of HCl : 1 X 1 + 1 X 35.5 = 36.5
valence of HCl : 1
there fore equivalent weight of HCl = 36.5 / 1 = 36.5
NaOH:
molecular weight of NaOH: 1 X 23 + 1X 16 + 1 X 1 =40
valence of NaOH: 1
there fore equivalent weight of NaOH: 40 / 1 = 40
H2SO4:
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
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Ex: (3)
Find the Equivalent weight of H3PO4, CaCl2, FeCL3, Al2(SO4)3, KMnO4
H3PO4:
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
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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 solution given as 1.196 kg/l
Now find 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 convert 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
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Ex: (5)
To prepare 1 N of the Na2CO3 solution, how many grams of Na2CO3 must be dissolved in 1.0 L of water?
Ans:
By using Normality formula = gram equivalent of solute / volume of solution in liter
Given:
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 g
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Specific gravity is used for indirect measurement of concentrations of aqueous solutions.which is the ratio of density of solution at T1 (temperature)to that of density of water at T2 (temperature)and specific gravity of a gas is the ratio of average molecular weight of the gas to the molecular weight of air as same temperature.
In water analysis, the impurities are in traces and their concentration is expressed in mg/ litre or Parts per million.
