Class 9 Molecular mass, Formula unit mass and Mole concept
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Class -9 chapter - 3 for cbse - ncert

Molecular mass, Formula unit mass and Mole concept

Topics to be covered

`=>` Molecular mass
`=>` Formula unit mass
`=>` Mole concept

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The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is therefore the relative mass of a molecule expressed in atomic mass units `color{red}((u))`.
Q 3217323280

(a) Calculate the relative molecular mass of water `(H_2O)`.
(b) Calculate the molecular mass of `HNO_3`.

Solution:

(a) Atomic mass of hydrogen ` = 1 u`

oxygen = 16 u

So the molecular mass of water, which
contains two atoms of hydrogen and
one atom of oxygen is `= 2 Γ— 1+ 1Γ—16 = 18 u`

(b) The molecular mass of `HNO_3` = the atomic mass of `H` + the atomic mass of `"N+ 3 Γ— the atomic mass of O"`


`= 1 + 14 + 48 = 63 u`

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`color{green}(β€’)` The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound.

`color{green}(β€’)` The only difference is that we use the word formula unit for those substances whose constituent particles are ions.

`color{green}(β€’)` For example, sodium chloride has a formula unit `color{red}(NaCl)`. Its formula unit mass can be calculated as– `color{red}(1 Γ— 23 + 1 Γ— 35.5 =58.5u)`
Q 3237323282

Calculate the formula unit mass of `CaCl_2.`

Solution:

Atomic mass of `"Ca + (2 Γ— atomic mass of Cl)"`
`= 40 + 2 Γ— 35.5 = 40 + 71 = 111 u`

πŒπŽπ‹π„ π‚πŽππ‚π„ππ“

`color{green}(β€’)` One mole of any species (atoms, molecules, ions or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.

`color{green}(β€’)` The number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed, with a value of `color{red}(6.022 Γ— 10^(23))`. This number is called the Avogadro Constant or Avogadro Number (represented by `color{red}(N_0)`),named in honour of the Italian scientist, Amedeo Avogadro.

`color{green}(β€’)` The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units `color{red}((u))`.

`color{green}(β€’)` Molar mass of atoms is also known as gram atomic mass. For example, atomic mass of hydrogen `color{red}(=1u)`. So, gram atomic mass of hydrogen `color{red}(= 1 g)` (To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from `color{red}(β€˜u’)` to `color{red}(β€˜g’)` .)
Q 3267323285

Calculate the number of moles for the following:
(i) `52 g` of `He` (finding mole from mass)
(ii) `12.044 Γ— 10^(23)` number of `He` atoms (finding mole from number of particles).

Solution:

No. of moles `= n`
Given mass `= m`
Molar mass `= M`
Given number of particles `= N`
Avogadro number of particles `= N_0`

(i) Atomic mass of `He = 4 u`
Molar mass of `He = 4g`

Thus, the number of moles `= " given mass"/ " molar mass "`

`=> n = m/M = 52/4 = 13`

(ii) we know,

1 mole `= 6.022 Γ— 10^(23)`

The number of moles

`= " given number of particles " / " Avogadro number "`

`=> N/N_0 = (12.044 xx 10^(23))/(6.022xx10^(23)) = 2`
Q 3207323288

Calculate the mass of the following:
(i) `0.5` mole of `N_2` gas (mass from mole of molecule)
(ii) `0.5` mole of `N` atoms (mass from mole of atom)
(iii) `3.011 Γ— 10^(23)` number of `N` atoms (mass from number)
(iv) `6.022 Γ— 10^(23)` number of `N_2` molecules (mass from number)

Solution:

(i) mass = molar mass Γ— number of moles

`β‡’ m = M Γ— n = 28 Γ— 0.5 = 14 g`

(ii) mass = molar mass Γ— number of moles

`β‡’ m = M Γ— n = 14 Γ— 0.5 = 7 g`

(iii) The number of moles, `n`

`= " given number of particles"/"Avogadro number" = N/ N_0`

` = ( 3.011 xx 10^(23))/(6.022xx10^(23))`

`=> m = M xx n = 14 xx ( 3.011 xx 10^(23))/(6.022 xx 10^(23))`

`=14 Γ— 0.5 = 7 g`

(iv) `n = N/N_0`


`=> m = M xx N/ N_0 = 28 xx ( 6.022 xx 10^(23))/(6.022 xx 10^(23))`

` = 28 xx 1 = 28 g`
Q 3227423381

Calculate the number of particles in each of the following:
(i) `46 g` of `Na` atoms (number from mass)
(ii) `8 g` `O_2` molecules (number of molecules from mass)
(iii) 0.1 mole of carbon atoms (number from given moles)

Solution:

(i) The number of atoms

` = " given mass "/"molar mass " xx "Avogadro number"`

`=> N = m/M xx N_0`

`=> N = 46/23 xx 6.022xx10^(23)`

`=> N = 12.044 xx 10^(23)`

(ii) The number of molecules

` = " given mass " / " molar mass " xx " Avogadro number"`

`=> N = m/M xx N_0`

atomic mass of oxygen ` = 16 u`

`therefore` molar mass of `O_2` molecules

` = 16 xx 2 = 32 g`

`=> N = 8/32 xx 6.022 xx 10^(23)`

`=> N = 1.5055 xx 10^(23)`

`1.51 xx 10^(23)`

(iii) The number of particles (atom) = number of moles of particles Γ— Avogadro number

`N = n xx N_0 = 0.1 xx 6.022xx10^(23)`

` = 6.022 xx 10^(22)`
 
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