Showing posts with label Molecular mass. Show all posts
Showing posts with label Molecular mass. Show all posts

Monday, 11 February 2019


Empirical Formula


 “Empirical formula (or stoichiometric formula) of a substance is the simplest formula which gives the lowest whole-number ratio between the number of atoms of different elements present in the substance.”

Molecular Formula


“A shorthand notion for the molecule of a substance in terms of symbols and number of atoms of each element present in it is called its molecular formula.”

Molecular and empirical formulae of any compound can be determined from the given composition of that compound. This is illustrated below.

Formula of an unknown compound

The chemical formula of a compound can be determined from the composition of the compound. The composition of a compound is commonly expressed in terms of percentage of each element present in it.

To determine the formula of an unknown compound, follow the following steps:
Step 1. Determine the percentage of each element present in the compound from the mass of each element present in a certain known mass of the compound.

Step 2. If the sum of the percentage of all these elements is not 100, then the difference gives the percentage of oxygen. Calculate the percentage of oxygen in the given compound by using the following relationship:

Percentage of oxygen = 100 – (Sum of the percentage of all other elements
present in the compound).

Step 3. Divide the percentage of each element by the atomic mass of the respective element. The ratio so obtained is called atomic ratio.

Step 4. Divide the atomic ratios by the lowest value, and convert these into the nearest whole numbers. These whole numbers give the simplest ratio between the number of atoms of the various elements present in the compound.

Step 5. Write down the empirical formula of the compound.

Step 6. Calculate the empirical formula mass by adding the atomic masses of all the atoms present in the empirical formula.

Step 7. Obtain molecular mass (or molar mass) either from experiment or from the vapour density of the compound by using the relationship,

Molecular mass = 2 × Vapour density

Step 8. Obtain the value of n through the relation,
n = Molecular mass / Empirical formula mass
 ‘n’ is an integer, such as 1, 2, etc.

Step 9. The molecular formula is then obtained by
Molecular formula = n × Empirical formula

The above method of writing the molecular formula of a compound is illustrated through the following numerical problems:




EXAMPLE 1.
A compound contains 75% carbon and 25% hydrogen. Determine its empirical formula. The molecular mass of this compound is 16 amu. Determine its molecular formula also. The atomic masses are: C = 12 u, H = 1 u.

Solution: 








So, Empirical formula of the compound = C1H4 or CH4

Then, Empirical formula mass = (1 × 12 u) + (4 × 1 u) = 12 u + 4 u = 16 u
 Molecular mass (given) = 16 u
So, n = Molecular mass / Empirical formula mass = 16 u / 16 u = 1

Therefore, Molecular formula = 1 × Empirical formula = 1 × CH4 = CH4



EXAMPLE 5.2.
A substance on analysis gave the following percentage composition : Na = 43.4%, C = 11.3% and O = 45.3%. Determine its empirical and molecular formulae. Given the relative molecular mass of the compound is 106.

Solution: 




Thus, the simplest ratio of Na, C and O in the given compound is 2 : 1 : 3.
So, empirical formula of the given compound is Na2CO3.

Then,

Empirical formula mass of the compound   = (2 × 23 u) + (1 × 12 u) + (3 × 16 u)

= 46 u + 12 u + 48 u = 106 u

So, n = Molecular mass / Empirical formula mass = 106 u / 106 u = 1
Then, the molecular formula of the given compound is
 Molecular formula = 1 × Empirical formula = 1 × Na2CO3 = Na2CO3


EXAMPLE 5.3.

A compound is found to contain 11.2% nitrogen (N), 3.2% hydrogen (H), 41.2% chromium (Cr) and 44.4% oxygen (O). Determine the stoichiometric (empirical) formula. Atomic masses (u) are: N = 14, H = 1, Cr = 52 and O = 16.

Solution: The above results are written as follows:















Since, the simplest atomic ratios cannot have fractional values, hence the simplest whole number atomic ratio of N, H, Cr and O are 2, 8, 2, 7 respectively.

Therefore,
Stoichiometric (empirical) formula of the compound = N2H8Cr2O7.

(The actual formula of this compound is (NH4)2Cr2O7. This compound is named as ammonium dichromate. Ammonium dichromate contains NH4+ and Cr2O72– ions).