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Some Basic Concepts of Chemistry - Mole Concept and Molar Masses

Grade 11ICSEChemistry

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.). This number is known as Avogadro's Number (NAN_A).

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Molar Mass (MM) is the mass of one mole of a substance, usually expressed in g/molg/mol. It is numerically equal to the atomic or molecular mass in amuamu.

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Gram Atomic Mass is the mass of 11 mole of atoms of an element. For example, Gram Atomic Mass of O=16 gO = 16\,g.

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Gram Molecular Mass is the mass of 11 mole of molecules of a substance. For example, Gram Molecular Mass of O2=32 gO_2 = 32\,g.

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Molar Volume: At STP (Standard Temperature and Pressure: 0∘C0^\circ C and 1 atm1\,atm), one mole of any ideal gas occupies a volume of 22.4 L22.4\,L (or 22.4 dm322.4\,dm^3).

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Percentage Composition calculates the relative mass of each element in a compound using the formula: Mass of element in 1 mole of compoundMolar mass of compound×100\frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100.

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Empirical Formula represents the simplest whole-number ratio of various atoms present in a compound, while the Molecular Formula shows the exact number of atoms of each element present in a molecule.

📐Formulae

Number of moles (n)=Given Mass (m)Molar Mass (M)\text{Number of moles } (n) = \frac{\text{Given Mass } (m)}{\text{Molar Mass } (M)}

Number of moles (n)=Number of particles (N)NA\text{Number of moles } (n) = \frac{\text{Number of particles } (N)}{N_A}

Number of moles for gases at STP (n)=Volume in Litres22.4\text{Number of moles for gases at STP } (n) = \frac{\text{Volume in Litres}}{22.4}

Molecular Mass=2×Vapour Density\text{Molecular Mass} = 2 \times \text{Vapour Density}

Molecular Formula=n×(Empirical Formula)\text{Molecular Formula} = n \times (\text{Empirical Formula})

n=Molar MassEmpirical Formula Massn = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}

💡Examples

Problem 1:

Calculate the number of moles and molecules present in 4.4 g4.4\,g of CO2CO_2.

Solution:

Molar mass of CO2=12+(2×16)=44 g/molCO_2 = 12 + (2 \times 16) = 44\,g/mol. n=mM=4.4 g44 g/mol=0.1 moln = \frac{m}{M} = \frac{4.4\,g}{44\,g/mol} = 0.1\,mol. Number of molecules N=n×NA=0.1×6.022×1023=6.022×1022N = n \times N_A = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} molecules.

Explanation:

First, find the molar mass by summing atomic masses. Then, use the mass-to-mole ratio, and finally multiply by Avogadro's number to find the count of molecules.

Problem 2:

What is the volume occupied by 14 g14\,g of N2N_2 gas at STP?

Solution:

Molar mass of N2=2×14=28 g/molN_2 = 2 \times 14 = 28\,g/mol. n=14 g28 g/mol=0.5 moln = \frac{14\,g}{28\,g/mol} = 0.5\,mol. Volume at STP =n×22.4 L=0.5×22.4=11.2 L= n \times 22.4\,L = 0.5 \times 22.4 = 11.2\,L.

Explanation:

Convert the given mass of Nitrogen gas to moles. Since it is at STP, multiply the moles by the molar volume (22.4 L22.4\,L) to find the total volume.

Problem 3:

A compound has an empirical formula CH2CH_2 and a molar mass of 42 g/mol42\,g/mol. Find its molecular formula.

Solution:

Empirical formula mass of CH2=12+(2×1)=14 uCH_2 = 12 + (2 \times 1) = 14\,u. n=Molar MassEmpirical Mass=4214=3n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{42}{14} = 3. Molecular Formula=3×(CH2)=C3H6\text{Molecular Formula} = 3 \times (CH_2) = C_3H_6.

Explanation:

Determine the mass of the empirical unit. Divide the total molar mass by this empirical mass to find the multiplier nn, then multiply the subscripts of the empirical formula by nn.

Mole Concept and Molar Masses - Revision Notes & Key Formulas | ICSE Class 11 Chemistry