MOLE CONCEPT
The mole is a fundamental unit in chemistry used to quantify the amount of a substance. Historically, it was defined as the mass of a substance containing the same number of particles as present in 12 grams of the carbon-12 isotope. One mole of a substance is equivalent to 6.022 x 1023 particles (This number is known as Avogadro’s number and represented by NA), which include atoms, molecules, or ions. Mass of one mole any substance is equal to its atomic mass or molecular mass in grams. The mass of one mole of H2O is 18 g. Therefore, gram molecular mass of H2O is 18 g/mol.
Number of moles can be calculated by atomic mass, molecular mass, number of particles(atoms, molecules, ions) and also can be calculated from the volume occupied by an ideal gas at STP. One mole of any ideal gas at STP(standard temperature and pressure) occupies 22.4 liters of volume.
Calculation of number of moles from atomic and molecular masses
$$\text{Number of molecules} = \dfrac{\text{Given mass}}{\text{Atomic mass (or) Molecular mass}}$$
Calculation of number of moles the given number of particles
$$\text{Number of moles} = \dfrac{\text{Number of particles}}{6.022 \times 10^{23}}$$
In the formula number particles can be atoms, molecules or ions. If the number of moles are given then multiply with Avogadro’s number to get the number of particles.
Note: To calculate the total number of atoms, multiply the total number of molecules by the atomicity of the molecule. The atomicity represents the total number of atoms present in a single molecule.
.Example: Number of atoms in CO2 = molecules of CO2 x 3
= Moles of CO2 x 6.022 x 1023 x 3
Example: Number of oxygen atoms in CO2 = molecules of CO2 x 2 (∵Atomicity of oxygen in CO2 is 2)
Calculation of number of moles of ideal gas at STP
$$\text{Number of moles} = \dfrac{\text{Volume of gas in Liters}}{22.4}$$
SOLVED EXAMPLES
Q.1 Calculate number of moles present in 69 g of Na
Ans: Given mass = 69g; Atomic mass of Na = 23 g/mol
Number of moles = 69/23 = 3 moles.
Q.2 What is the mass (in mg) of 2.3 moles of Ca?
Ans: Given number of moles = 2.3; Atomic mass of Ca = 40 g/mol
Given mass = Moles x Atomic mass = 2.3 x 40 = 92 g = 92 x 10-3 mg = 9.2 x 10-2 mg
Q.3 What is the mass of 0.9 moles of water?
Ans: Given moles of H2O = 0.9 g; Molecular mass of H2O = 2(1) + 16 = 18 g/mol.
Mass of H2O = Moles x molecular mass = 0.9 x 18 = 16.2 g
Q.4 How many moles of C6H12O6 present in 18 g of glucose(C6H12O6). Given atomic mass of C = 12 u, H = 1 u, O = 16 u?
Ans: Given mass of glucose = 18 g; Molecular mass of C6H12O6 = 6(12) + 12(1) + 6(16)=180 g/mol
Number of moles of glucose = 18/180 =0.1 moles.
Q.5 Calculate number of moles of 1.055 x 1024 water molecules.
Ans: Given number of water molecules = 1.055 x 1024
$$\text{Number of moles} = \dfrac{1.055 \times 10^{24}}{6.022 \times 10^{23}} = 1.75$$
Q.6 How many CO2 molecules are present in 1.5 moles of carbon dioxide gas?
Ans: Number of CO2 molecules = Number of moles of CO2 x 6.022 x 1023
$$1.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$$
= 1.5 x 6.022 x 1023
= 9.033 x 1023
Q.7 Calculate number of hydrogen atoms, number of oxygen atoms and total number of atoms present in 1.5 g of CH3COOH.[Atomic mass of C = 12 u, H = 1 u, O = 16 u]
Ans: Given mass of CH3COOH = 1.5 g; Molecular mass of CH3COOH = 2(12) + 4(1) + 2(16)=60 g/mol.
Atomicity of oxygen in CH3COOH = 2, Atomicity of H = 4 and Total number of atoms in the molecule = 8.
$$\text{Number of moles} = \dfrac{1.5}{60} = 0.025$$
Number of Hydrogen atoms = Number of moles of CH3COOH x NA x Atomicity of H in CH3COOH
= 0.025 x 6.022 x 1023 x 4 = 6.022 x 1022
Number of Hydrogen atoms = 0.025 x 6.022 x 1023 x 2 = 3.011 x 1022
Total number of atoms = 0.025 x 6.022 x 1023 x 8 = 12.044 x 1022 = 1.2044 x 1023
Q.8 Calculate mass of CH4 which contain 4.8176 x 1023 number of hydrogen atoms.
Ans: Given number of hydrogen atoms = 4.8176 x 1023; Molecular mass of CH4 = 12 + 4 = 16 g/mol. Atomicity of hydrogen in CH4 = 4
Number of Hydrogen atoms = Number of moles of CH4 x NA x Atomicity of H in CH4
4.8176 x 1023 = Moles of CH4 x 6.022 x 1023 x 4
Moles of CH4 = 0.2
$$\dfrac{\text{Mass of CH4}}{\text{Molecular mass of CH4}} = 0.2$$
$$\Rightarrow \text{Mass of CH4} = 0.2 \times 16 = 3.2\, g$$
Q.9 What mass of CH4 contain same number of hydrogen atoms as present in 15 g of C2H6?
Ans: Given Mass of C2H6 =15 g ; Molecular mass of C2H6 = 2(12) + 6(1) =30 g/mol
Moles of C2H6 =15/30 =0.5 moles
Number of hydrogens in 15 g of C2H6 = Moles of C2H6 x 6.022 x 1023 x Atomicity of H in C2H6
= 0.5 x 6.022 x 1023 x 6 = 18.066 x 1023
Let the mass of CH4 = x g; Molecular mass of CH4 = 12+4 = 16 g/mol
Moles of CH4 =x/16 moles
Number of hydrogens in x g of CH4 = Moles of CH4 x 6.022 x 1023 x Atomicity of H in CH4
= (x/16) x 6.022 x 1023 x 4 = (x/16) x 6.022 x 1023
Given: Number of hydrogen atoms in 15 g of C2H6 = Number of hydrogens in ‘x’ g of CH4
$$18.066 \times 10^{23} = \dfrac{x}{4} \times 6.022 \times 10^{23}$$
x = 12 g
Q.10 What is the volume occupied by 1.5 moles of O2 gas at STP?
Ans: Given Moles of O2 =1.5
Volume occupied by O2 = Number of moles x 22.4 = 1.5 x 22.4 = 33.6 L
Q.11 How many moles of CO2 is present in 5600 ml of CO2 at STP?
Ans: Volume of CO2 at STP = 5600 ml = 5.4 L
Number of moles = 5.6/22.4=0.25
Q.12 How many CH4 molecules are present in 250 mg go CH4 gas at STP?
Ans: Mass of CH4 = 250 mg = 0.25 g; Molecular mass of CH4 = 16 g/mol
Number of moles of CH4 = 0.25/16 =0.0156
Number of molecules = Moles x 6.022 x 1023
= 0.0156 x 6.022 x 1023 = 0.094 x 1023 = 9.4 x 1021
Q.13 What mass of CO2 occupies same volume occupied by 9 g of water vapor at STP?
Ans: Given mass of H2O = 9 g, Molecular mass of H2O = 18 g/mol
Moles of H2O = 9/18 =0.5 moles
1 mole of H2O occupies 22.4 liters of volume at STP. Therefore, Volume of 0.5 moles of H2O 11.2 liters of volume at STP.
Volume occupied by CO2 = 11.2 liters.
1 mole of CO2 occupies 22.4 liters. Therefore, 11.2 liters of CO2 = 0.5 moles
1 mole of CO2 = 44 g
Therefore, mass of 0.5 mole of CO2 = 22g