MOLE CONCEPT
Mole is a unit of amount of substance.
The amount of a substance that contains Avogadro’s number of particles is called one mole.
The number 6,02.10²³ is called Avogadro’s number and shown as N_A.
One mole represents 6,02.10²³ (often taken as 6.10²³ ) units.
For example:

• 6,02.10²³ pencils=1 mol pencil
• 6,02.10²³ apple = 1 mol apple
• 6,02.10²³ atoms = 1 mol atom
• 6,02.10²³ molecules =1mol molecule
• 1 mole particles= 6,02.10²³ particles
• 1 mole oxygen atoms= 6,02.10²³ O atoms
• 1 mole hydrogen molecules =6,02.10²³ H₂ molecules

Mole number of an element or a compound is shown with symbol n.
Mole number of a substance can be found from number of particles by the following formula:

  

Example: Calculate the number of moles of each of the following:
a) 3.10²³ N₂ molecules
b) 12.10²³ Fe atoms
c) 12.10²³  H₂O molecules

Molar Mass

• The mass of 1 mole of a substance is called molar mass.
• Molar mass is shown as M.
• Its unit is g/mole.
• Molar masses of the elements can be found from periodic table.
• Molar masses of the compounds can be found from the molar masses of the elements.

Example: Find the molar mass of water, H₂O
M(H₂O)=2M(H)+M(O)
                   =2.1+1.16
                   =2+16
                   =18 g/mole
Example: Find the molar mass of diphosphorus trioxide, P₂O₃.
Example: Find the molar mass of ammonium sulfate, (NH₄)₂SO₄.

Molar Volume:

• At STP (standard temperature and pressure: 1 atm pressure and 0^oC)  1 mole of each gas occupies 22.4 liter volume.
• Molar volume is shown as V_m.
• Mole number of a gas can be found from molar volume by proportion or by the following formula:

                   

Example: Find the mole number of hydrogen gas having a volume of 5.6 liter
Example: Find the mole number of oxygen gas having a volume of 11.2liter
Example: Find the mole number of nitrogen gas having a volume of 22.4 liter

DETERMINING EMPIRICAL AND MOLECULAR FORMULAE

The empirical formula is the simplest formula for a compound.
A molecular formula is the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound.
For example, if the empirical formula of a compound is C₃H₈ , its molecular formula may be C₃H₈ , C₆H1₆ , etc.
An empirical formula is often calculated from elemental composition data. The weight percentage of each of the elements present in the compound is given by this elemental composition.
Let's determine the empirical formula for a compound with the following elemental composition:
40.00% C, 6.72% H, 53.29% O.
The first step will be to assume exactly 100 g of this substance. This means in 100 g of this compound, 40.00 g will be due to carbon, 6.72 g will be due to hydrogen, and 53.29 g will be due to oxygen. We will need to compare these elements to each other stoichiometrically. In order to compare these quantities, they must be expressed in terms of moles. So the next task will be to convert each of these masses to moles, using their respective atomic weights:

        

Take notice that since the composition data was given to four significant figures, the atomic weights used in the calculation were to at least four significant figures. Using fewer significant figures may actually lead to an erroneous formula.

Now that the moles of each element are known, a stoichiometric comparison between the elements can be made to determine the empirical formula. This is achieved by dividing through each of the mole quantities by whichever mole quantity is the smallest number of moles. In this example, the smallest mole quantity is either the moles of carbon or moles of oxygen (3.331 mol):

The ratio of C:H:O has been found to be 1:2:1, thus the empirical formula is: CH₂O. Again, as a reminder, this is the simplest formula for the compound, and not necessarily the molecular formula. Suppose we know that the molecular weight of this compound is 180 g/mol. With this information, the molecular formula may be determined. The formula weight of the empirical formula is 30 g/mol. Divide the molecular weight by the empirical formula weight to find a multiple:

The molecular formula is a multiple of 6 times the empirical formula:

C(1 x 6) H(2 x 6) O(1 x 6) which becomes C₆H₁₂O₆

Alternatively, the empirical and molecular formula may be determined from experimental data. Suppose we have a compound containing the elements, C, H and S. A 7.96 mg sample of this compound is burned in oxygen and found to form 16.65 mg of CO₂. The sulfur in 4.31 mg of the compound is converted into sulfate by a series of reactions, and precipitated as BaSO₄. The BaSO₄ was found to have a mass of 11.96 mg. The molecular weight of the compound was found to be 168 g/mol. Using this data, what is the molecular formula of the compound?

The strategy will be to use stoichiometry to determine the mass percent of each of the elements in the compound, and then use the mass percentages to determine the empirical formula. Notice that since all the data is in milligrams, we may carry out the calculations using milli- units throughout.

The only source of carbon for the CO₂ formed came from the compound, thus, determine the milligrams of carbon found in 16.65 mg of CO₂:

Now that the "part" of the sample due to carbon is known, one may calculate the percent carbon in the compound, using the mass the sample as the "whole":

The only source of sulfur for the precipitate of BaSO₄ , came from the compound, thus, determine the milligrams of sulfur in 11.96 mg of BaSO₄:

Similarly, determine the percent sulfur in the compound, using the mass of sulfur as the "part" and the mass of compound as the "whole":

The percentage of hydrogen may determined by difference:

%H = 100.0% - 57.1 %C - 38.1 %S = 4.8 %H

From the elemental composition, we may determine the empirical formula, in the same manner as used in the first example. First, assume exactly 100 g of the compound. In 100 grams of the compound, 57.1 g would be due to carbon, 38.0 g would be due to sulfur and 4.9 g would be due to hydrogen. Convert each of these masses into moles using the corresponding atomic weight for each element:

Now that the moles of each element are known, the empirical formula may be determined by dividing the moles of each element by the smallest number of moles. This yields a ratio of the number of each element in the empirical formula.

 

The ratio of C:H:S has been found to be 4:4:1, thus the empirical formula is: C₄H₄S. The molar mass of the empirical formula is 84 g/mol. Since the molecular weight of the actual compound is 168 g/mol, and is double the molar mass of the empirical formula, the molecular formula must be twice the empirical formula:

C(4 x 2) H(4 x 2) S(1 x 2) which becomes C₈H₈S₂.