Chemistry Atomic and Molecular Masses, Mole Concept and Percentage Composition
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### 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 = 18.02g

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)

0.6L CO_2 and 0.8L 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) 0.6L CO_2 and 0.8L 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