WBCHSE Class 11 Notes For Mole Concept- Formula, Explanations, Examples

Mole Concept Formulas And Examples For WBCHSE Class 11

When we are dealing with elements and compounds and the reactions between them, it often becomes necessary (or more convenient) to know the number of atoms or molecules that a sample of an element or compound may contain.

This is because atoms or molecules are the basic units taking part in a chemical reaction. Reactions take place between atoms of reactants and molecules of products are formed. Consider one example.

“WBCHSE Class 11 Chemistry, mole concept, formula, explanations, and examples”

We know that one molecule of oxygen combines with two molecules of hydrogen to produce two molecules of water vapor. Suppose we wish to take molecules of oxygen and hydrogen in the ratio 1: 2 so that all the molecules of the reactants are used up in a reaction. How do we do so, since we cannot count molecules? The mole, a chemist’s unit for counting atoms and molecules, was introduced to solve such problems.

• Let us go back to the concept of (relative) atomic mass. The ratio of the masses of the hydrogen atom and the oxygen atom is 1.008:16, so the ratio of the masses of the hydrogen molecule and the oxygen molecule is 2.016: 32. Then the ratio of the mass of n molecules of hydrogen to that of n molecules of oxygen must also be 2.016: 32.
• In that case, if the ratio of the mass of a sample of hydrogen to the mass of a sample of oxygen is 2.016: 32, they must both contain the same number of molecules. In particular, a sample of hydrogen weighing 2.016 g and a sample of oxygen weighing 32 g must contain the same number of molecules. Let us say this number is N. By the same logic a gram-molecule of any compound (or element) must contain the same number of molecules as 2.016 g of hydrogen, i.e., N. This equals the number of molecules present in one gram-molecule of a compound (or element).
• As mentioned earlier, units like the gram-atom and gram-molecule have been replaced by the mole. A mole of a substance contains tire Avogadro number of elementary particles. These elementary particles may be atoms or molecules or any other discrete particles such as ions, electrons, and protons.
• A mole is defined as the amount of a substance that contains as many entities (atoms, molecules, or other particles like ions) as there are atoms in exactly 0.012 kg (or 12 g) of the carbon-12 isotope. In SI units, we represent mole by the symbol mol.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

To determine the number of atoms in 12 g of ¹²C isotope, the mass of a carbon-12 atom was determined by a mass spectrometer, which came out to be 1.992648 x 10^-23 g. Knowing that 1 mol of carbon weighs 12 g, the number of atoms in it is equal to

⇒ \(\frac{12 \mathrm{~g}}{1.992648 \times 10^{-23} \mathrm{~g}}=6.022137 \times 10^{23}\) = 6.022137 x 10²³

Thus, we can say that a mole has 6.022137 x 10²³ atoms, molecules, ions, etc. This number is denoted by the symbol NA and is known as the Avogadro constant as it is a physical quantity and not a pure number.

In most of the calculations, its value with four significant figures is used, i.e., 6.022 x10²³. Put simply, the mole is the chemical counting unit used by chemists for 6.022 x 10²³ particles of a substance.

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“WBCHSE Class 11, chemistry notes, on mole concept, and relation to chemical equations”

• Although, while using the term mole, the kind of particles involved are specified, this may not always be the case. If we simply refer to one mole of a substance, it means we are talking about the naturally occurring form of that substance. For example, one mole of oxygen contains 6.022 x 10 ²³ oxygen molecules. Similarly, one mole of sodium contains 6.022 x 10²³ atoms of sodium, and one mole of sodium carbonate (Na₂CO₃) contains 6.022 x 10²³ molecules of sodium carbonate.
• According to Avogadro’s hypothesis, equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules. This means 6.022 x 10²³ molecules of any gas at stp (0°C and 1 bar pressure) must have a particular volume. This volume has been found experimentally to be equal to 22.7 L and is called the molar volume. Put simply, 1 mole of any gas at stp has a volume of 22.7 L.
• It should be clear from the preceding discussion that a mole can be expressed in terms of mass, number, or volume.
• In terms of mass, a mole is defined as the amount of a substance whose mass in grams is numerically equal to its atomic mass (if the substance is atomic) or molecular mass (if the substance is molecular) in amu or u.

∴ 1 mole of oxygen (atomic) weighs 16 g.

∴ 1 mole of oxygen (molecular) weighs 32 g.

∴ 1 mole of carbon (atomic) weighs 12 g.

∴ 1 mole of H₂O (molecular) weighs 18 g.

WBCHSE Class 11 Chemistry Mole Concept Notes

Here again, we must remember that one mole of hydrogen weighs 2.0 g if it has not been specified whether we are talking of atoms or molecules. Similarly, one mole of sodium weighs 23.0 g and one mole of sodium carbonate weighs 106.0 g. The mass of one mole of any substance will be its molar mass.

∴ Molar mass of hydrogen (H₂) = 2.0 g/mol^-1

∴ Molar mass of sodium (Na) = 23.0 g mol^-1.

∴ Molar mass of sodium chloride (NaCl) = 585 g/mol^-1.

• Therefore, the molar mass is equal to the atomic mass or molecular mass expressed in grams, depending upon whether the substance contain atoms or molecules.
• In terms of volume (related to gases), a mole is the amount of a gas that occupies a volume of 22.7 L at stp. This means 1 mole of oxygen (gas) or 1 mole of CO₂ (gas) occupies 22.7 L at stp.
• Tire mole concept can also be applied to ionic compounds. Ionic compounds do not contain molecules. They are made up of positively and negatively charged ions. The formula of an ionic compound does not represent one molecule, but provides the ratio between the constituent ions.
• The formula of sodium chloride, for example, does not represent a molecule of sodium chloride—it merely expresses the ratio of sodium and chloride ions in sodium chloride. While referring to ionic compounds, we use the term formula unit, rather 6.022 x 10²³ formula units. The mass of one mole of formula units of an ionic compound is equal to its formula mass (rather than molecular mass) expressed in grams, or gram-formula mass.

“Mole concept, WBCHSE Class 11, chemistry notes, and key formulas”

Thus, a mole of NaCl weighs 58.5 g (1 gram-formula mass) and contains 6.022 x 10²³ formula units of NaCl or 6.022 x 10²³ Na⁺ ions and 6.022 X 10²³ Cl^– ions. Similarly, a mole of ferric chloride (FeCl₃)weighs 161.7 g and contains 6.022 x 10²³ FeCl₃ formula units or 6.022 x 10²³ Fe³⁺ ions and 3 x 6.022 x 10²³Cl^– ions.

As you know, the mole is a counting unit used for a huge number of particles (Avogadro constant). Thus, the mole is not a useful measure for things much larger than atoms or molecules. Further, being an SI unit, it can be used with prefixes.

“Mole concept, step-by-step explanation, and problem-solving techniques, WBCHSE notes”

For example, 1 m mol = 10^-3 mol,

∴ 1 μ mol = 10^-6 mol,

∴ 1 n mol =10^-9 mol.

We can conveniently convert a larger number of entities (atoms, molecules, etc.) into a number of moles and vice versa using the Avogadro constant.

⇒ Number of moles of entities/particles = \(\frac{\text { number of entities } / \text { particles }}{N_{\mathrm{A}}}\)

How To Solve Mole Concept Problems For WBCHSE Chemistry

The following summary should help you remember what you have learned about the mole and the Avogadro constant.

The mole concept and Avogadro constant are used in various types of chemical calculations.

“WBCHSE Class 11 Chemistry, mole concept, molar mass, and gram-mole relationship”

• Molar masses are also used as conversion factors between masses in grams and several moles. To find out the number of moles of an element we divide the mass of the element by its molar mass. To calculate the number of atoms in a given number of moles of a substance, we multiply the number of moles by the Avogadro constant. To calculate the mass of one atom we divide the atomic mass by the Avogadro constant.
• It is easier to calculate the molar volume of a liquid or solid if we know the molar mass and density at any given temperature and pressure as these do not change much with variation in temperature and pressure.

Example 1. Calculate the mass of

1. An atom of copper and
2. A molecule of sulfur dioxide.

Solution:

1. Atomic mass of copper = 64 u.

∴ mass of 1 mole of copper atoms = 64 g.

⇒ mass of 6.022 x 10²³ atoms of copper = 64 g.

∴ mass of 1 atom of Cu = \(\frac{64}{6.022 \times 10^{23}}\) = 10.63 x 10^-23 g.

2. Molecular mass of SO₂ =1×32 +2×16 = 64 u.

∴ mass of 1 mole of SO₂ = 64 g

⇒  mass of 6.022 x 10²³ molecules of SO₂ = 64 g.

∴ mass of 1 molecule of SO₂ = \(\frac{64}{6.022 \times 10^{23}}\) = 10.63 x 10^-23 g.

“WBCHSE Class 11, chemistry notes, on mole concept, Avogadro’s number, and calculations”

Example 2. Calculate the number of ozone molecules and that of oxygen atoms present in 64 g of ozone.
Solution:

Given

64 g of ozone

The molecular mass of ozone (O₃) = 16 x 3 = 48 u.

∴ mass of 1 mole of ozone molecules = 48 g.

⇒ 48 g of ozone contains of 6.022 x 10²³ molecules.

∴ 64 g of ozone will contain \(\frac{6.022 \times 10^{23}}{48}\) x 64 = 830 x 10²³ molecules.

1 molecule of ozone contains 3 atoms of oxygen.

∴ 830 x 10²³ molecules of ozone will contain 3 x 830 x 10²³ = 24.9 x 10²³ oxygen atoms.

Example 3. Assuming that the density of water is 1 g cm^-3, calculate (a) the volume of one molecule of water and (b) the radius of a water molecule, taking it to be spherical.
Solution:

Given

Assuming that the density of water is 1 g cm^-3,

1. Volume of 1 mole of H₂O = 18 cm³ (because density of H₂O = 1g cm^-3 and mass of 1 mole of H₂O = 18 g)

⇒ volume of 6.022 x 10²³ molecules of H₂O = 18 cm³.

∴ volume of 1 molecule of H₂O = \(\frac{18}{6.022 \times 10^{23}} \mathrm{~cm}^3\) = 2.99 x 10^-23 cm³.

2. Let R be the radius of a water molecule. Then

4/3 πR³ = 2.99 x 10^-23 cm³ (volume of a sphere of radius r =4/3 πr³).

∴ R³ = 7.13 x 10^-24 = (7.13)^1/3 x 10^-8= 1.92 x 10^-8 cm.

“Mole concept, definition, formulas, and numerical problems, WBCHSE syllabus”

Example 4. Calculate the number of moles in the following.

1. 460 g of H₂SO₄
2. 67,2 L of CO₂ at stp

Solution:

1. Mass of 1 mole of H₂SO₄ = 2 x 1 + 32 + 4 x 16 = 98 g.

∴ number of moles in 460 g of H₂SO₄ = 1/98 x 460 = 5.

2. 1 mole of CO₂ occupies 22.4 L at STP.

∴ 67.2 L of CO₂ at stp contains 1/22.4 x 67.2 = 3 moles of CO₂.