52.TYPES OF ORGANIC FORMULAE – (2)- Molecular formula.

After having studied the empirical formula in the previous post , let us now move forward and learn what is a molecular formula for a molecule and how to calculate it.We shall solve some numerical problems later , before concluding this post.

Molecular Formula –

A chemical formula that shows the total number and kinds of atoms in a molecule, but not their structural arrangement.
e.g.- The molecular formula for benzene is C6H6 .

Molecular formula can be found out from empirical formula by the following equation –

Molecular formula = n × Empirical formula ,
n= Integer 1,2,3 ..etc
&
n=(molecular weight) / (empirical formula weight)

Let us solve a few problems to completely understand these calculations.


Problem 1 –  An organic monobasic acid contains 18.6% carbon,1.55% hydrogen, 55.04% chlorine.Its molecular weight is 129.Calculate the molecular formula of the acid.
Solution
We first add up all the percentages given in the problem, to ensure that they sum upto  a value of 100.If their sum is less than 100 , then we can be sure that oxygen is present too.In this case, we can be very sure that Oxygen is present as it is clearly mentioned that the compound is a monobasic acid (i.e it has -COOH group in it).

18.6% + 1.55% +55.04% = 75.19% . Thus, 100 – 75.19 = 24.81 % is oxygen.

Element

%

composition

(Given)

Weight

(W)

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

18.6

18.6

12

(18.6/12) =1.55

(1.55/1.55) = 1

1

H

1.55

1.55

1

(1.55/1) = 1.55

(1.55/1.55) = 1

1

Cl

55.04

55.04

35.5

(55.04/35.5)=1.55

(1.55/1.55) = 1

1

O

24.81

24.81

16

(24.81/16)=

1.55

(1.55/1.55) = 1

1

Empirical formula weight = (Weight of C+Weight of H+Weight of Cl+Weight of oxygen)
= 12+1+ 35.5+ 16
                                                = 64.5

Molecular weight is given = 129.

∴n=(molecular weight) / (empirical formula weight)
n= 129/64.5
n=2

Molecular formula  = n × Empirical formula
Molecular formula = 2 × (CHClO)

∴The molecular formula of the compound  is C2H2Cl2O2.


Sometimes the problems one encounters provide the vapour density values. Though we have not yet discussed this term, it will be sufficient to just know that,

Molecular weight = 2 × Vapour Density

Problem 2 – The empirical formula of a compound is CH2O.It’s vapour density is found to be 30, calculate its molecular formula.

Solution – 

Empirical formula =  CH2O.
∴ Empirical formula weight = 12 +2+16 = 30.
Molecular weight = 2 × vapour density
= 2 × 30
= 60.

n= Molecular weight ÷ Empirivcal formula weight = 60/30 = 2.

Molecular formula = n (empirical formula)
=2( CH2O).
=
C2H4O2 .


Thus, the molecular formula of the compound is C2H4O2 .


Another useful formula for calculation of molecular weight is –

Molecular weight = [w(in g) × 22.4 dm3] / Vo.
w= weight of the substance in grams.
 Vo= Volume of the gas at STP(standard temperature and pressure).
 22.4 dm3= 1 mole of gas at STP conditions.

We shall study the exact meaning of these terms in later posts.For now, we just learn the formula , plug values in and solve the problems.

Problem – A sample of gas occupies 2.0 liters at STP. The sample contains 2.143g of carbon and 0.358g of hydrogen. Find the empirical and molecular formulas of the gas.

Solution – 

Given –  Vo= 2.0 liters  = 2 dm3…. [liters and  dm3(decimeter cube) are equivalent terms].
w
 = (2.143g + 0.358g)  = 2.501g ≈    2.5g

using the formula above,

Molecular weight = [(2.5 g ) (22.4 dm3 )] / 2 dm3
= 28g,

To find the empirical formula we carry the following calculations as seen earlier –

Element

Weight

(W)

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

2.143

12

(2.143/12)= 0.179

(0.179/0.179) = 1

1

H

0.358

1

(0.358/1) = 0.358

(0.358/0.179) = 2

2

Thus, the empirical formula is CH2.

empirical formula weight = 12=2 = 14g.
Molecular weight = 28g.

∴ n = 28/14 = 2.

Thus, the molecular formula of the compound is C2H4 .


In old days , empirical formula calculations lead us to molecular formula.However, now with the advent of many advanced techniques, the computers give us the desired results quickly. Techniques like mass spectrometry , give us molecular weight (largest m/z ratio) directly, most of the times. We shall explore all these spectrometric parameters in our posts on spectroscopy soon. 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!

 

References and further reading-

1.http://www.dictionary.com/
2.Precise Chemistry textbook by Sheth Prakashan Kendra.
3.http://www2.hawaii.edu/~voyce/Chap27/Voyce_Chap27.htm

 

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