What is a Mole? Learn Formula and Easy Ways to Calculate Moles Yashwant Parihar, April 16, 2026April 16, 2026 The mole is one of the most important concepts in chemistry, but it can seem confusing at first. In simple words, a mole helps us count very tiny particles like atoms and molecules easily. Just like we use a dozen to count 12 items, chemists use the mole to count particles. In this post, you will learn what a mole is and explore simple methods to calculate moles using easy formulas and examples.What is a Mole?A mole is the amount of substance that contains as many species (atoms, molecules, ions and other species) as there are atoms in exactly 12g of C-12.Mass of one atom of C-12 = 1.99×10-23gNo of atoms in 12g of C-12 = 12/1.99×10-23g = 6.022×1023 g (Avagadron’s Number NA)One mole is the collection of 6.022×1023 g particles. (atoms, molecules, ions, etc.)1 mol of Na = 6.022×1023 g atoms of Na1 mol of H2 gas = 6.022×1023 g molcules of H21 mol of NaCl = 6.022×1023 g formula unit of NaCl1 mol of Cl– = 6.022×1023 g ions of Cl–Molar MassIt is the mass of one mole of substance.For atoms: Molar Mass = Gram Atomic MassFor Molecules: Molar Mass = Gram Molecular MassMolar Mass of He = 4g/molMolar Mass of Oxygen = 16g/molMethods to Calculate MolesIn Terms of Mass:Numberofmoles=GivenmassofsubstanceMolarmassofsubstanceNumber\:of\:moles\; = \frac {Given\:mass\:of\:substance}{Molar\:mass\:of\:substance}For example:Convert 36g of Carbon into moles?Numberofmoles=3612=3molNumber\:of\:moles\; = \frac {36}{12} = 3\:molIn Terms of the number of Particles:Numberofmoles=Givennumberofparticles6.022×1023gNANumber\:of\:moles\; = \frac {Given\:number\:of\:particles}{6.022\times10^{23} g\:N_A}For example:Convert 6.022×1021 atom of H into moles?\Numberofmoles=6.022×10216.022×1023gNA=10−2Number\:of\:moles\; = \frac {6.022\times10^{21}}{6.022\times10^{23} g\:N_A} = 10^{-2}In terms of Volume of Gas:At STP(Standard Temperature and Pressure):Old STP = T = 0o C = 273.15K P = 1 atmNew STP = T = 0o C = 273.15K P = 1 bar1 atm = 1.012 bar1 bar = 0.987 atmFrom the ideal gas equation, PV = NRT 0.987 x V = 1 x 0.087 x 273.15K V = 22.7LOld STP, PV = NRT 1 x V = 1 x 0.087 x 273.15K V = 22.4LNumberofmoles=volumeofgasatSTPcondtions22.7/22.4Number\:of\:moles\; = \frac {volume\:of\:gas\:at\:STP\:condtions}{22.7/22.4}Y-Map to Calculate the Moles Chemistry class 11 Chemistry