What is a mole in chemistry ?
The mole (symbol mol) is the SI of measurement for amount of substance.
The mole concept is very basic concept of chemistry and has several key significances:
Mole as counting particles of substance :
Mole concept provides a way to count atoms, molecules, ions, and other entities in chemistry .
One mole of a substance contains Avogadro's number of entities, which is equal to 6.23x10^23 entities. These entities could be atoms, molecules, ions, electrons, protons, neutrons etc.
Example:
A mole of Na means 6.023x10^23 atoms of Na.
Two mole of Ca means 2X6.023x10^23 atoms of Ca.
Three mole of Mg means 3X6.023x10^23 atoms of Mg.
1 mole of H2O means 6.023x10^23 molecules of H2O.
1 mole of Cl- ion means 6.023x10^23 of Cl- ion.
Mole as quantitative relationships with weight & volume of substance :
The Mole concept is also used to establish relationships between the masses of substances and the number of particles they contain, and their volumes.
Molar Mass
A mole is also used to express the amount of a substance in grams. This is called the molar mass or molecular weight.
One mole of any substance is equal to its molar mass in grams.
For example,
The molar mass of carbon (C) =12 grams per mole, means one mole of carbon weighs 12 grams.
The molar mass of oxygen (O) =16 grams per mole, means one mole of oxygen weighs 16 grams.
Molar mass of any element is equal to atomic mass of element and molar mass of any compound is equal to molecular mass of compound.
Molar Volume
For gases, the concept of the mole is linked to the molar volume at standard temperature and pressure, which is useful for calculating gas volumes involved in reactions. At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters of volume.
This is known as the molar volume of a gas at STP. In case of solids and liquids, the volume occupied by one mole varies depending on the substance's density and molecular structure.
Frequently Asked Questions
Q: How many particles are equal to 1 mole?
Ans : 6.22X10^23 Particles .Q: How many moles are in 32 gm of oxygen?
Ans : 2 moles .Q: What is Avogadro's number used for?
Ans : Avogadro's number 60.22x10^23 is equal to the number of particles in 1 mole (or mol) of a substance .Q: How many mole are in 18 gm of water?
Ans : 1 moles .Problem Based on Mole Concept
Q1 : How many atoms of oxygen are there in 2 mole of H2O ?
Solution:
2 mole of H2O = 2x6.023x1023 molecules of H2O
1 molecule of H2O = 1 oxygen atom
total oxygen atom = 2x6.023x1023 atoms
Q2 : How many electrons are there in 2 mole of Na+ ion?
Solution:
1 mole of Na = 6.023x1023 atoms of Na
2 mole of Na = 2x6.023x1023 atoms of Na
Na+ has = 10 electrons
Therefore, 2 moles of Na+ has = 2x10x6.023x1023 electrons of Na+
Q3 : How many atoms are there in 2 mole of Na2So4?
Solution:
1 mole of Na2SO4 = 6.023x1023 molecules of Na2SO4
2 mole of Na = 2x6.023x1023 molecules of Na2SO4
Na2SO4 has = 2+1+4 atoms = 7atoms
Therefore, 2 moles of Na2SO4 has = 2x7x6.023x1023 atoms of Na2SO4
Q4 : How many molecules of water are there in 54 g of H2O?
Solution:
1 mole of H2O has = 18 g = Molecular Weight
In 54 g of H2O has = 54/18 g of molecules of H2O = 3 molecules
Q5 : Calculate the mass of 12.044×10^23 Oxygen atoms?
Solution:
Given, O atom = 12.044 x 1023 atoms = 2x6.022x1023
Since, 1 mole = 6.022 X 1023= atmoic weight
Therefore, Mass = 2 x 16 = 32g
Q6 :How many atoms of hydrogen are there in 36 g of NH4?
Solution:
Molecular mass of NH4 = 14 + 4 = 18g
Therefore, 36g has = 36/18 = 2 molecules
1 mole = 6.022 x 1023 molecules of NH4
1 mole = 4 x 6.022 x 1023 atoms of H
2 mole will have = 2 x 4 x 6.022 x 1023 atoms of H
Q7 :An atom of some element Y weighs 6.644 × 10-23 g. Calculate the number of atoms in 40 g of it?
Solution:
Given, Mass of an atom = 6.644 x 10-23 g
Mass of 6.023 x 1023 = 6.023 x 1023 x 6.644 x 10-23 = 6.023 x 6.644 = 40 (approx)
Number of moles of the element = 40/40 = 1 mol
Number of atoms in 40g of the element = 6.023 x 1023
Q8 :Find the number of moles and number of atoms of H and S in 10 mole of H2S?
Number of moles of H in H2S = 2 mol
Number of atoms of S in H2S = 10 x 6.023 x 1023
Number of moles of S in H2S = 1 mol
Solution:
Given, Number of moles of H2S = 10 mol
Number of molecules in 10 moles of H2S = 10 x 6.023 x 1023
Number of atoms of H in H2S = 20 x 6.023 x 1023
Number of moles of H in H2S = 2 mol
Number of atoms of S in H2S = 10 x 6.023 x 1023
Number of moles of S in H2S = 1 mol
Q9 : Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3).
Solution:
Given, Molecules of CaCO3 = 6.022 x 1023
Molecular mass of CaCO3 = 40(1) + 12(1) + 16(3) = 100 amu
Number of moles in 6.022 x 1023 molecules of CaCO3 = (6.022 x 1023)/(6.022 x 1023) = 1
So, the mass of 6.022 × 1023 molecule of CaCO3 = 100g
Q10 : Calculate the mass of 12.044 × 1023 carbon atoms.
Solution:
Given, numbers of carbon atoms = 12.044 x 1023
We known that, 6.023 x 1023 of carbon have 12g of carbon.
So, 1 atom of carbon will have 12/(6.022 x 1023)g of carbon.
Therefor, 12.044 x 1023 atoms of carbon will have 12(12.044 x 1023)/(6.022 x 1023) = 12 x 2 = 24g of carbon.
Q11 :Calculate the mass of 2 moles of N2.
Solution:
Molecular mass of N2 = 14(2) = 28 amu
1 mole of N2 = 28 amu
2 mole of N2 = 2 x 28 = 56 amu
Q12 : If the mass of proton is doubled and that of neutron is halved, the molecular weight of CO2, consisting only C12 and O16 atoms, will
(a) not change(b) increases by 25%
(c) decrease by 25%
(d) increase by 50%
Solution:
Q13 : 1mole of atom X has mass, 0.444 times the mass of 1 mole of atom Y. Atomic mass of X is 2.96 times the mass of one atom of C-12.what is the atomic weight of Y.
(a) 15.77 amu(b) 80 amu
(c) 46.66 amu
(d) 40 amu
Solution:
Q14 : Twenty molecules of SO3 will weigh as much as...molecules of oxygen
(a) 100(b) 50
(c) 15
(d) 8
Solution:
Q15 : A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical formula and molecular formula?
(a) CH2Cl(b)C2H2Cl
(c) C2HCl
(d)C2H4Cl2
Solution:
Consider sample mass = 100 g of the above compound,which contains 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.
Moles of hydrogen = 4.07 g/ 1.008g = 4.04
Moles of carbon = 24.27 g /g12.01 = 2.021
Moles of chlorine = 71.65g/ g35.453 = 2.021
Now, dividing each of the mole values obtained above by the smallest number amongst them
2.021 is the smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl .
Now, write down the empirical formula of compound is CH2Cl .
Molecular formula will be
Empirical formula mass calculation
For CH2Cl, empirical formula mass is 12.01 + (2 × 1.008) + 35.453 = 49.48g
Dividing Molar mass by empirical formula mass
Molar mass/ empirical formula mass = 98.96g / 49.48 g
=> 2 = (n)
Empirical formula = CH2Cl
n=2
molecular formula = C2H4Cl2
