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Some Basic Concepts of Chemistry - Mole Concept and Molar Mass

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The mole is the SI unit for the amount of substance. One mole contains exactly 6.02214076×10236.02214076 \times 10^{23} elementary entities (atoms, molecules, ions, etc.).

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The constant 6.022×10236.022 \times 10^{23} is known as the Avogadro Constant, denoted by NAN_A.

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Molar mass is the mass of one mole of a substance in grams. It is numerically equal to the atomic or molecular mass expressed in uu (atomic mass units), but the units are g mol−1g \text{ mol}^{-1}.

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Gram Atomic Mass is the mass of 1 mole1 \text{ mole} of atoms (e.g., O=16 gO = 16 \text{ g}), while Gram Molecular Mass is the mass of 1 mole1 \text{ mole} of molecules (e.g., O2=32 gO_2 = 32 \text{ g}).

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At STP (Standard Temperature and Pressure: 273.15 K273.15 \text{ K} and 1 atm1 \text{ atm}), 1 mole1 \text{ mole} of any ideal gas occupies a volume of 22.4 L22.4 \text{ L}, known as Molar Volume.

📐Formulae

n=mMwhere n=moles, m=given mass, M=molar massn = \frac{m}{M} \quad \text{where } n = \text{moles, } m = \text{given mass, } M = \text{molar mass}

n=NNAwhere N=number of particles, NA=6.022×1023n = \frac{N}{N_A} \quad \text{where } N = \text{number of particles, } N_A = 6.022 \times 10^{23}

n=V22.4where V=volume of gas at STP in Litresn = \frac{V}{22.4} \quad \text{where } V = \text{volume of gas at STP in Litres}

Mass Percentage of an element=Mass of element in 1 mole of compoundMolar mass of compound×100\text{Mass Percentage of an element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100

Molecular Mass=n×Empirical Formula Mass\text{Molecular Mass} = n \times \text{Empirical Formula Mass}

💡Examples

Problem 1:

Calculate the number of moles in 12.044×102212.044 \times 10^{22} atoms of Helium (HeHe).

Solution:

Given N=12.044×1022N = 12.044 \times 10^{22} and NA=6.022×1023N_A = 6.022 \times 10^{23}. Using the formula n=NNAn = \frac{N}{N_A}: n=12.044×10226.022×1023=210=0.2 molesn = \frac{12.044 \times 10^{22}}{6.022 \times 10^{23}} = \frac{2}{10} = 0.2 \text{ moles}

Explanation:

To find the number of moles from the number of particles, we divide the given count of atoms by Avogadro's number.

Problem 2:

Calculate the mass of 0.5 moles0.5 \text{ moles} of CO2CO_2 gas.

Solution:

Molar mass of CO2=12+(2×16)=44 g/molCO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}. Using the formula m=n×Mm = n \times M: m=0.5 mol×44 g/mol=22 gm = 0.5 \text{ mol} \times 44 \text{ g/mol} = 22 \text{ g}

Explanation:

First, calculate the molar mass by summing the atomic masses of Carbon and Oxygen. Then, multiply the moles by this molar mass to get the total mass.

Problem 3:

Find the volume occupied by 8 g8 \text{ g} of Oxygen gas (O2O_2) at STP.

Solution:

Molar mass of O2=32 g/molO_2 = 32 \text{ g/mol}. First, find moles: n=8 g32 g/mol=0.25 molesn = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles}. Now, calculate volume: V=n×22.4 L=0.25×22.4=5.6 LV = n \times 22.4 \text{ L} = 0.25 \times 22.4 = 5.6 \text{ L}

Explanation:

Convert the mass of the gas to moles first, then multiply by the molar volume (22.4 L22.4 \text{ L}) to find the volume at STP.

Mole Concept and Molar Mass - Revision Notes & Key Formulas | CBSE Class 11 Chemistry