Class 11 Chemistry


Atomic Mass

One atomic mass unit is defined as a mass exactly equal to one twelfth of the mass of one carbon-12 atom.

1 amu = 1.66056 × 10–24 g

Mass of an atom of hydrogen = 1.6736 × 10–24 g

Thus, in terms of amu, the mass of hydrogen atom

= (1.66056 × 10–24 g) ÷ (1.6736 × 10–24 g)

= 1.0078 amu = 1.0080 amu

At present, ‘amu’ has been replaced by ‘u’, which is known as unified mass.

Average Atomic Mass: Many elements exist in nature as more than one isotope. Average atomic mass is calculated after taking into account the existence of these isotopes and their relative abundance in nature. In the Periodic Table, average atomic masses of elements are mentioned.

Molecular Mass: The sum of atomic masses of the elements present in a molecule gives the molecular mass of that molecule.

Example: Molecular mass of methane (CH4) can be calculated as follows:

= (12.011 u) + 4 × (1.008 u)

= 12.011 + 4.032 = 16.043 u

Example: Molecular mass of water (H2O) can be calculated as follows:

= 2 × (1.008) + (16.00 u)

= 2.016 + 16.00 = 18.016 u = 18.02 u

Formula Mass: Some substances do not contain discrete molecules as their constituent units. In such compounds, positive and negative entities are arranged in a three-dimensional structure. In such cases, the formula is used to calculate the formula mass instead of molecular mass.

Example: Formula mass of sodium chloride

Atomic mass of sodium + Atomic mass of chlorine

= 23.0 u + 35.5 u = 58.5 u

Mole Concept

One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.

Mass of a C-12 atom = 1.992648 × 10-23 g

Mass of one mole of carbon = 12 g

So, number of atoms in 1 mole carbon atom

`= (12g)/(1.992648×10^(-23)g)`

= 6.0221367 × 1023 atoms/mol

We can also say that 1 mole of a particular atom contains 6.022 × 1023 atoms

This number is also called Avogadro’s constant and is denoted by NA.

Molar Mass: The mass of one mole of a substance in grams is called its molar mass. Molar mass in grams is numerically equal to atomic/molecular/formula mass in u.

Molar mass of water = 18.02 g mol-1

Molar mass of sodium chloride = 58.5 g mol-1

Percentage Composition

Percentage composition of a compound gives mass percentage of different constituents of a compound.

Mass % of an element
= (Mass of that element in compound × 100) ÷ Molar mass of compound

Example: Let us find mass percentage of oxygen and hydrogen in water molecule.

Molar mass of water (H2O) = 18.02 g

Mass % of hydrogen


Mass % of oxygen


Empirical Formula or Molecular Formula: Empirical formula represents the simplest whole number ratio of various atoms present in a compound. On the other hand, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. Following example illustrates the application of mass percentage to find empirical formula and molecular formula of a compound.

Example: A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Solution: Let us first convert mass percentages into grams

Mass of hydrogen `=(4.07)/(100)×98.96=4.03`

Mass of carbon `=(24.27)/(100)×98.96=24.03`

Mass of chlorine `=(71.65)/(100)×98.96=70.90=71.0`

Let us now convert each into number of mole

Moles of hydrogen = 4.03 ÷ 1.008 = 4

Moles of carbon = 24.03 ÷ 12.01 = 2

Moles of chlorine = 71.0 ÷ 35.5 =2

Let us now find the ratio for different elements, which is as follows:

H : C : Cl = 2 : 1 : 1

So, empirical formula is: CH2Cl

Writing Molecular Formula:

Determining empirical formula mass

= 12.01 + (2 × 1.008) + 35.453

= 49.48 g

Divide molar mass by empirical formula mass

= 98.96 ÷ 49.48 = 2

Multiply the numbers in empirical formula by 2 (quotient found above) to get molecular formula.

So, molecular formula: C2H4Cl2

Stoichiometric Calculations

Mass Percent
= (Mass of Solute ÷ Mass of Solution) × 100

Mole Fraction:

The ratio of number of moles of a particular component to the total number of moles of the solution gives mole fraction of that component. If a substance A dissolves in substance B and their number of moles respectively are; nA and nB then mole fractions of A and B can be given by following equations:

Mole fraction of A `=(n_A)/(n_A+n_B)`

Mole fraction B `=(n_B)/(n_A+n_B)`

Molarity: The number of moles of the solute in 1 liter of solution is called molarity. It is denoted by M.

Molarity (M) = No. of moles of solute ÷ Volume of solution in liter

Molality: The number of moles of solute present in 1 kg of solvent is called molality. It is denoted by m.

Molality (m) = No. of moles of solute ÷ Mass of solvent in kg