Fundamentals of mole concept and chemical calculations — lesson. Science TNSB Mentoring, Class 10 Crash course.

Difference between atoms and molecules:

Atom	Molecule
An atom is the smallest particle of an element.	A molecule is the smallest particle of an element or compound.
Except in noble gases, atoms do not exist in free state.	Molecule exists in a free state.
Except for some noble gases, other atoms are highly reactive.	Molecules are less reactive.
Chemical bonds do not exist between atoms.	Atoms in a molecule are held by chemical bonds. Example: Water molecule.

Mole:

In the SI system, the mole (or mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly $12\ g$ (or $0.012\ kg$) of the carbon - $12$ isotope.
Experimentally, the number of atoms in $12$ $g$ of carbon-$12$ is determined. This is called 'Avogadro's number' ($N_A$), named after an Italian scientist Amedeo Avogadro who proposed its significance. Its value is $6.023\times10^{23}$.
Hence, one mole of a substance contains $6.023\times10^{23}$ entities. Thus, $5$ moles of oxygen molecules contain $5\times6.023\times10^{23}$ molecules.

PYQ - Mole

Mole concept:

The study of the collection of particles by using mole as the counting unit, in order to express the mass and volume of such unit particles in bulk of matter is known as mole concept. The number of moles of a substance can be determined in several of ways depending on the data available, as follows:

$Number of moles = \frac{Mass}{Atomic mass}$

$Number of moles = \frac{Mass}{Molecular mass}$

$Number of moles = \frac{Number of atoms}{{6.023 \times 10}^{23}}$

$Number of moles = \frac{Number of molecules}{{6.023 \times 10}^{23}}$

Mole concept

Calculating number of moles from mass and volume:

1.	Calculate the number of moles in $46$ $g$ of sodium.

Data:

Given mass of sodium $= 46$ $g$.

Atomic mass or mass number of sodium $= 23$

The formula to find the number of moles from the mass is,

Number of moles = \frac{Mass of the element}{Atomic mass of the element}

\begin{array}{l} Number of moles & = & \frac{46}{23} \\ = & 2 \end{array}

Hence, the number of moles in $46$ $g$ of sodium $=2$ $moles$

2.	Calculate the number of moles in $5.6$ $L$ of oxygen at STP.

Data:

Given volume of oxygen $= 5.6$ $L$

Molar volume of oxygen at STP $= 22.4$ $L$

The formula to find the number of moles from the volume is,

Number of moles = \frac{Given volume of O_{2} at STP}{Molar volume at STP}

\begin{array}{l} Number of moles & = & \frac{5.6}{22.4} \\ = & 0.25 moles \end{array}

Hence, the number of moles in $5.6$ $L$ of oxygen at STP$=0.25$ $moles$

3.	Calculate the number of moles of a sample that contains $12.046\times10^{23}$ atoms of iron.

Data:

Given the number of atoms of iron $=12.046\times10^{23}$

Avogadro's number $=6.023\times10^{23}$

$Number of moles = \frac{Number of atoms of iron}{Avogadro number}$

$\begin{array}{l} Number of moles & = & \frac{12.046 \times 10^{23}}{6.023 \times 10^{23}} \\ = & 2 \end{array}$

Hence, the number of moles of a sample that contains $12.046\times10^{23}$ atoms of iron$=2$ $moles$

Calculation of mass from Avogadro number:

1.	Calculate the mass of $1.51\times10^{23}$ molecules of water.

Data:

Given molecules of $H_2O=1.51\times10^{23}$

Molecular mass of $H_2O=18$

Avogadro's number $=6.023\times10^{23}$

\begin{array}{l} Number of moles & = & \frac{Number of molecules of water}{Avogadro number} \\ = & \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} \\ = & \frac{1}{4} \\ = & 0.25 mole \end{array}

\begin{array}{l} Number of moles & = & \frac{Mass}{Molecular mass} \\ 0.25 & = & \frac{Mass}{18} \\ Mass & = & 0.25 \times 18 \\ Mass & = & 4.5 g \end{array}

Hence, the mass of $1.51\times10^{23}$ molecules of water $=4.5$ $g$

2.	Calculate the mass of $5\times10^{23}$ molecules of glucose.

Data:

Given molecules of glucose $=1.51\times10^{23}$

Molecular mass of glucose $=180$

Avogadro's number $=6.023\times10^{23}$

\begin{array}{l} Mass of glucose & = & \frac{Molecular mass \times Number of particles}{Avogadro number} \\ = & \frac{180 \times 5 \times 10^{23}}{6.023 {\times 10}^{23}} \\ = & 149.43 g \end{array}

Hence, the mass of $5\times10^{23}$ molecules of glucose $=149.43$ $g$

PYQ - Calculation of moles

Avogadro's law:

Avogadro’s law states that “equal volumes of all gases under similar conditions of temperature and pressure contain the equal number of molecules”.

As a result, the volume of any gas must be proportional to the number of molecules present in it. If '$V$' is the volume and '$n$' is the number of molecules in a gas, then Avogadro law is written as follows:

\begin{array}{l} Volume of gas \propto Number of molecules in it \\ V \propto n \\ V = Constant \times n \end{array}

PYQ - Avogadro’s law

Applications of Avogadro's law:

It describes the law of Gay-Lussac.
It helps in identifying the atomicity of gases.
Avogadro's law can be used to find the molecular formula of gases.
The relation between molecular mass and vapour density is determined by it.
It helps to calculate the gram molar volume of all gases (i.e., $22.4$ litre at STP).

PYQ - Application of Avogadro's law

Percent composition:

The percentage composition of a compound represents the mass of each element present in $100$ $g$ of the compound.

Mass percent of an element = \frac{Mass of the element in the compound}{Molecular mass of the compound} \times 100

PYQ - Percent composition

Relationship between vapour density and relative molecular mass:

Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
The relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to the mass of one atom of hydrogen.

Vapour density (VD) = \frac{Mass of a given volume of gas or vapour at STP}{Mass of the same volume of hydrogen}

According to Avogadro's law, all gases have the same number of molecules in equal volumes. Thus, let the number of molecules in one volume $= n$, then cancelling '$n$' which is common and since hydrogen is a diatomic,

VD = \frac{Mass of 1 mole of a gas or vapour at STP}{Mass of 2 atoms of hydrogen}

When you compare the formula for vapour density and relative molecular mass, and by substituting the values, we get,

2 \times Vapour density = Relative molecular mass of a gas

Relative molecular mass of a gas = 2 \times Vapour density

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2. Fundamentals of mole concept and chemical calculations

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