# 51. TYPES OF ORGANIC FORMULAE-(1) – Empirical formula.

Let us dedicate this post in understanding the empirical formula in organic chemistry. In successive posts, we shall study other types of formulae.It is imperative that one understands these correctly as they form the basis of understanding the structures of organic molecules.

### Empirical Formula –

This is the simplest formula that shows the simplest whole number ratio of atoms of different elements in a molecule.

e.g. Benzene empirical formula is CH.This formula shows that in benzene molecule, the ratio of carbon atoms to hydrogen atoms is 1:1 , i.e the # of C-atoms = # of H-atoms. But the finding the ratios is not always this simple.The correct understanding of the mathematical and chemical aspect of this term is necessary.

So what does simplest whole number ratio mean?

As we have already studied in mathematics,a whole number means a number without a decimal point or without fractions. 0,1,2,3,4,5….etc, are all whole numbers.

Now,consider a molecule, C6H14 – Hexane.The ratio of no.of carbon atoms to hydrogen atoms in this molecule is – Undoubtedly, ‘6/14’ is a whole number ratio but  is NOT the simplest whole number ratio.Mathematically, we can simplify this fraction further by dividing both the numerator and denominator by 2 to yeild 3/7. However, if we further try to simplify this fraction we end up getting a decimal figure.Thus, 3/7 is the simplest whole number ratio for C6H14 . Thus, the empirical formula for hexane is C3H7.

Why do we need whole numbers only?

It is because we are talking about atoms here. Number of atoms in a molecule is always a whole number.We cannot have 2.33 atoms of hydrogen participating in a reaction.Atom as an entity is a whole and cannot be subdivided, in molecule formation.Thus, the number of atoms is always a whole number.

### Procedure to calculate empirical formula (3steps)-

Step 1 Knowing the mass of each element in the sample.

• Generally, the percentage composition(% composition) of elements in a compound are given. Percentage composition refers to the amount/mass of that element present in the given sample.Generally, we assume the sample to be 100 g. Thus, for 100g of sample the % composition is equal to the mass of the element in 100 g of sample(as % refers to amount in each hundred).
e.g.–  In the above example (benzene,C6H6) , C= 50% and H= 50%.
So , if we take 100g of benzene as the sample,
50% of 100g of benzene = 50g will be that of carbon and
50% of 100g of benzene=50g of hydrogen.

If the sample is not exactly 100g,  then , % composition is NOT equal to the mass of the element.
Thus, if we have 160g of benzene as the sample, the mass of C and H present in it will be –
50% of 160g of benzene = 80g of Carbon and
50% of  160g of benzene = 80g of Hydrogen. • If the sum of masses of the given elements does not add up to 100, it indicates that oxygen  is present.
e.g.- If we are asked to calculate the empirical formula of a substance which contains 40%carbon and 6.66% hydrogen, we should immediately conclude that oxygen is present.How much oxygen is present?
40+6.66 = 46.66% of Carbon and hydrogen in 100 g of sample.
∴ amount of oxygen = 100 – 46.66 = 53.34%.

Step 2 – Converting atomic mass to moles.

We haven’t learned the concept of a mole yet.We shall learn the mole concept in detail in our future posts. For now, understanding the following shall suffice –

For most stoichiometric calculations(calculations involving measurement of weight/amount of substances), we work with moles. To convert the amount from grams to moles, we just have to divide the amount of substance in grams by the atomic weight of that element(We get this from periodic table, PT). The quantity thus obtained gives no of moles of that element present. When we write an empirical formula, we are talking about the ratio of no.of atoms of these elements. In problems , we are given %composition of these elements , which refers to the mass of the element. These two are different quantities.
e.g. In the hexane example, when we say C3H7 is the empirical formula for hexane, we mean that for every 3 atoms of Carbon, there will be 7 atoms of hydrogen.This DOESNOT MEAN THAT FOR EVERY 3 gms OF CARBON THERE WILL BE 7gms OF HYDROGEN.

1mole of any substance = 6.023 × 1023 atoms.
Thus, we need the mole concept to convert the mass ratios into atom ratios!

Step 3Finding the simplest ratio and rounding off to get a whole number.

In step 2 , we obtained number of moles of each element present in the compound.However, for empirical formula consideration, we need to find out the simplest ratio of moles.

How to find the simplest ratio of moles ?

This can be done by simply dividing all the mole quantities by a number smallest among them/lowest quotient.As we need whole numbers, we later round the figures off (use significant number rules) to obtain simple whole numbers.

Lets us solve some examples to figure out how to do the calculations.

Problem – The percentage composition of an aromatic hydrocarbon is – C= 92.4% and H= 7.6% .Find the empirical formula of this compound.

Solution –  We find the solution in a tabular form as it makes things look easy. Here, the amount of sample is not given, so we assume it to be 100g.

 Element % composition (Given) Weight in 100g Atomic weight (At.wt) (from PT) # moles = (W/ At.wt) Simplest ratio(SR)= (# moles/lowest quotient) Rounding off to a Whole Number. (WN) C 92.4% 92.4g 12 (92.4/12) =7.7 (7.7/7.6) = 1.01 1 H 7.6% 7.6g 1 (7.6/1) = 7.6 (7.6/7.6)= 1 1

Empirical formula is CH.

In the next post , we shall study how to calculate molecular formula from empirical formula and solve many more problems to deepen our understanding of these concepts.Till then,

Be a perpetual student of life and keep learning..

Good day !

NOTE
THERE MIGHT BE A PROBLEM IN OBSERVING THE EMPIRICAL CALCULATION TABLE ABOVE, ON YOUR PHONE.KINDLY SWITCH TO A LAPTOP /MAC BOOK TO SEE THE ENTIRE TABLE.UNFORTUNATELY, I AM NOT ABLE TO FIX THIS PROBLEM.
THANKS!