Chemistry Atomic and Molecular Masses, Mole Concept and Percentage Composition
Please Wait... While Loading Full Video

Class 11 Chapter 1 - Basic Concepts of Chemistry

Atomic and Molecular Masses, Mole Concept and Percentage Composition

Click for Only Video

Topics Covered :

`●` Atomic Mass Unit ( amu) or unified mass(u)
`●` Atomic and Molecular Masses
`●` Average Atomic Mass
`●` Mole Concept and Molar Masses
`●` Percentage Composition

Atomic Mass Unit ( amu) or unified mass(u)

`=>` The atomic mass or the mass of an atom is actually very-very small because atoms are extremely small.

`=>` in the nineteenth century, it was not possible to measure such small masses

`=>` scientists could determine mass of one atom relative to another by experimental means

`=>` Hydrogen, being lightest atom was arbitrarily assigned a mass of 1 (without any units) and other elements were assigned masses

relative to it.

`=>` Today, we have sophisticated techniques e.g., mass spectrometry for determining the atomic masses fairly accurately.

` => ` So today technically we can write the accurate mass of an atom in grams but ` text ( for the sake of convenience ) ` we write the atomic mass not in gm or kg ( SI unit of mass) but in atomic mass unit.

1 amu `= 1/12 xx` mass of one `C-12` (`text()^12C`) atom


`= 1/12 xx 1.9927 xx 10^(-23)` g = `1.66056xx10^(-24)` g

` => ` 1 amu `= 1.66056xx10^(-24)`g

Atomic and Molecular Masses :

(i) Atomic Mass :

Atomic mass of an atom is mass of an atom in amu or u.


i.e. The atomic mass of atom is given relative to `C-12`. `text()12^C` is assigned a mass of exactly `12` atomic mass unit (amu) and masses of all other are given relative to this standard.


Similarly, mass of oxygen`-16`(`text()16^O`) atom is `15.995` amu.

The unit ‘ amu ‘ is replaced by ‘u’ known as unified mass.

Some important masses:

mass of a proton = 1.007276 u

mass of neutron = 1.008664 u

mass of an electron = 0.00054858 u

Q 1973856746

Calculate the atomic mass (average) of chlorine
using the following data:
Class 11 Exercise 1 Q.No. 9
Solution:

Fractional abundance of `text()^35 Cl =0. 7577`,
Molar mass `= 34 .9689`
Fractional abundance of `text()^37 Cl = 0.2423`,
Molar mass `= 36 .9659`
Average atomic mass
`= (0 · 7577) (34·9689 am u) + (0.2423) (36.9659 arn u)`
`=26.4959 + 8.9568 = 35.4527`

Average Atomic Mass

Many elements exist as more than one isotope. So, when we consider their abundance we calculate average atomic mass.
e.g. From the data given in table, the average atomic mass of carbon is.

`(98.892xx12+1.108xx13.00335+2xx10^(-10) xx14.00317)/(100)=12.011u`

When we use atomic masses of elements in calculations, we actually use average atomic masses of elements

In the periodic table of elements, the atomic masses mentioned for different elements actually represented their average atomic masses.

Average mass is same as atomic weight.

Average Atomic mass of some elements to remember.

Average Atomic mass of `C= 12.01` u
Average Atomic mass of `H=1.008` u
Average Atomic mass of `O=16` u
Average Atomic mass of `Cl=35.5` u
Average Atomic mass of `N= 14.007` u
Q 1973856746

Calculate the atomic mass (average) of chlorine
using the following data:
Class 11 Exercise 1 Q.No. 9
Solution:

Fractional abundance of `text()^35 Cl =0. 7577`,
Molar mass `= 34 .9689`
Fractional abundance of `text()^37 Cl = 0.2423`,
Molar mass `= 36 .9659`
Average atomic mass
`= (0 · 7577) (34·9689 am u) + (0.2423) (36.9659 arn u)`
`=26.4959 + 8.9568 = 35.4527`

Molecular Mass :

Definition : It is the sum of atomic masses of the elements present in a molecule

It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding together.

e.g (i) `CH_4` (methane)

So, molecular mass of methane` = 1xx (12.011u) +4xx (1.008u) =16.0434u`

(ii) `H_2O` so, molecular mass of water `= 2xx (1.008u) +16.00u) =18.02u`

Formula Mass :


Some compounds such as `NaCl` does not contain discrete molecules as their constituents.

In `NaCl`, positive(`Na^+`) and negative (`Cl^-`) change are arranged in 3D structure. One `Na^+` is surrounded by 6 `Cl^-`and vice-versa.

So, formula such as `NaCl` is used to calculate the formula mass as `NaCl` does not exist as a single entity.

So, formula mass of `NaCl` = atomic mass of `Na` + atomic mass of `Cl` `= 23.0 u + 35.5 u = 58.5 u`

Mole Concept and Molar Masses :

Since the number of atoms and molecular is very large even in very small amount, a unit is required to handle such a large number.

Definition of Mole : One mole is the amount of a substance that contains as many particle or entities (Atoms, molecules, ions or other
particles) as there are atoms in exactly 12 g (or 0.012kg) of the `text()^12C` isotope.

1 mole `= 6.0221367xx10^23`

Note: (i) In SI system, mole is the seventh base quantity for the amount of a substance.
(ii) One mole of a substance always contains the same number of entities irrespective of substance.

Since Mass of `text()^12C = 1.992648 xx10 ^(-23) g`

Mass of `1` mole of `text()^12C = 12g`

So, the number of atoms in `text()^12C = (text(12g/mol) text()^12C) /(1.992648 xx 10^(-23) g//text()^12C)`

`= 6.0221367xx10^23` atoms/mol

This number of entities is 1 mole and called as Avogadro constant denoted by `N_A`.

Definition of Molar Mass : The mass of one mole of a substance in grams is called its molar mass.

Note : The molar mass in grams is numerically equal to atomic/molecular/formula mass in u.

e.g. Molar mass of `H_2O = 18.02 gmol^-1`

Molar mass of `NaCl` `= 58.5g mol^-1`


Q 1963767645

In three moles of ethane `(C_2H_6)` calculate the
following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Class 11 Exercise 1 Q.No. 10
Solution:

`=> 1` molecule of `C_2H_6` contains `2` atoms of `C` and `6` atoms of `H`
After multiplying by `N_A` (Avagadro's number)
`N_A` molecules of `C_2H_6` will contain `2N_A` atoms of `C` and `6 N_A` atoms of `H`

This can be further written as

`1` mole of `C_2H_6` will contain `2` mole of `C` atoms and `6` mole of `H` atoms.

(i) `1` mole of `C_2H_6` contains `2` moles of carbon
atoms. Hence `3` moles of `C_2H_6` will contain
C-atoms `= 6` moles.

(ii) `1` mole of `C_2H_6` contains `6` moles of
hydrogen atoms. Hence `3` moles of `C_2H_6`
will contain H-atoms `= 18` moles.

(iii) `1` mole of `C_2H_6` contains `6. 02 xx 10^2`
molecules. Hence `3` moles of `C_2H_6` will
contain ethane molecules `= 3 xx 6.02 xx 1 0^2`
`= 18.06 xx 10^23 ` molecules.

Percentage Composition :

Percentage composition gives the idea about the purity of a given sample by analyzing the given data.

Mass `%` of an element `= text(mass of that element in the compound)/text(molar mass of the compound)`

e.g. For `H_2O`

Molar mass of water = `18.02`g

Mass `%` of `H` `= (2xx1.008)/18.02 xx100`

Mass `%` of `O = 16.00/18.02 xx100= 88.79`
Q 2644445353

What is the percentage of carbon, hydrogen and oxygen in ethanol?

Solution:

Molecular formula of ethanol is : `C_2H_5OH`

Molar mass of ethanol is : `(2xx12.01 + 6xx1.008 + 1xx 16.00)g = 46.068 g`

Mass per cent of carbon `= (24.02g)/(46.068g) xx 100 = 52.14%`

Mass per cent of hydrogen `= (6.048g)/(46.068g) xx 100 = 13.13%`

Mass per cent of Oxygen `= (16.00g)/(46.068g) xx 100 = 34.73%`
Q 1218634500

One litre of `CO_2` is passed over hot coke.The volume becomes `1.4` litre.The percent composition of products is

(A)

`0.6` litre

(B)

`0.8` litre `CO_2`

(C)

`0.6`L `CO_2` and `0.8`L `CO`

(D)

none of these

Solution:

`CO_2(g) +C rightarrow 2CO(g)`

`(1-x)+2x rightarrow 1.4`

`x=0.4`

The percent composition of `CO_2` = `1-x =1-0.4 =0.6`

The percent composition of product `CO` `= 2x = 2xx 0.4 = 0.8`

`therefore` The percent composition of products is `0.8` litre

Hence `3` is the correct answer.
Correct Answer is `=>` (C) `0.6`L `CO_2` and `0.8`L `CO`
Q 1953345244

Calculate the mass per cent of different
elements present in sodium sulphate (`Na_2SO_4`)
Class 11 Exercise 1 Q.No. 2
Solution:

Mass % of an element

`=(text(Mass of:hat element in the compound))/(text (Molar mass of the compound)) xx 100`

Now, molar mass of

`Na_2SO_4 =2 (23.0)+ 32 .0+4 xx 16 .0= 142 gmol^(-1)`

Mass percent of sodium `=46/142 xx 100 =32.39 %`

Mass percent of sulphur `=32/142 xx100=22.54%`

Mass percent of oxygen `=64/142 xx 100 =45.07%`

How to calculate molar mass and number of moles

Method to calculate molar mass :

`=>` Write the formula of substance

`=>` Calculate formula mass in amu or `u`

`=>` Replace `u` or amu by `g//mol`

This will give molar mass in `g//mol`

E.g. Calculate the molar mass of the following substance:
(a) Ethyne `C_2H_2`
(b) Sulpur molecule , `S_8`
(c) Phosphorus molecule , `P_4` (Atomic mass of phosphorus `=31`
(d) Hydrochloric acid `HCl`
(e) Nitric acid, `HNO_3`

Sol.
(a) Molar mass of ethyne `(C_2H_2)`

`= 2 xx` atomic mass of `C+2 xx` atomic mass of `H`

`=2 xx 12 + 2 xx 1= 26 u`

`=2 xx 12 + 2 xx 1= 26 g//mol`

(b) Molar mass of sulphur `(S_8)`

`=8 xx `atomic mass of `S`

`=8 xx 32 = 256 u`

`= 8 xx 32 = 256 g//mol`

(c) Molar mass of phosphorus `(P_4)`

`4 xx ` atomic mass of `P`

`4 xx 31= 124 u`

`4 xx 31= 124 g//mol`

(d) Molar mass of hydrochloric acid `(HCl)`

`=1 xx ` atomic mass of `H+1 xx` atomic mass of `Cl`

`= 1 xx 1 +35.5 = 36.5 u`

`= 1 xx 1 +35.5 = 36.5 g//mol`

(e) Molar mass of nitric acid `(HNO_3)`

`= 1 xx ` atomic mass of `H+1 xx `atomic mass of `N+3 xx ` atomic mass of `O`

`=1 xx 1 +1 xx 14+3 xx 16= 63 u`

`=1 xx 1 +1 xx 14+3 xx 16= 63 g`

How to calculate number of moles

1. If mass of substance is given , `"No. of moles" = ("mass of substance in g")/("molar mass")`

E.g. 1. What will be the mass of `5` mole of `SO_2`?

`=>` Molecular mass of `SO_2= 64` gm

`5= ("mass (gm)")/64`

`:. ` mass `= 320` gm


E.g. 2. `(a)` How many mole of `O` atoms are present in `88` gm `CO_2` ?
`(b)` What will be the mass of `10` mole of `H_3PO_4`?

Ans. `(a) 4` mole `(b) 980` gm

2. If no. of particles (Atoms/ molecules/ ions) is given , `"No of moles" = ( " given no of particles (Atoms/ molecules/ions")/(N_A)`

Eg. 1. A piece of `Cu` contains `6.022 xx 10^24` atoms. How many mole of `Cu` atoms does it contain?

`=>` No. of mole `=(6.022 xx 10^(24))/(N_A)= (6.022 xx 10^(24))/(6.022 xx 10^(23))=10` mole

Eg. 2. `5` mole of `CO_2` are present in a gaseous sample . How many molecules of `CO_2` are present in the sample?

`=> 5 N_A`

3. If the volume of gas is given , `"No of moles " = ("volume of gas at 1 bar pressure and" 273 K " in lit")/(22.7)`

or No. of moles `= ("volume of gas at " 1 "atm and" 273 K("in Lit"))/(22.4)`


E.g. 1. A sample of `He` gas occupies `5.6` litre volume at `1` atm and `273 K`. How many mole of `He` are present in the sample?

`=>` No. of mole `=(5.6)/(22.4)=0.25`

E.g. 2. How many volume will be occupied by `2` mole `CO_2` gas at STP?

`=>45.4 L`
 
SiteLock