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 C_{6}H_{6 }.
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
in 100g |
Atomic weight (At.wt) (from PT) |
# moles = (W/ At.wt) |
Simplest ratio(SR)= (# moles/lowest quotient) |
Rounding off to a Whole Number. |
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 C_{2}H_{2}Cl_{2}O_{2}.
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 CH_{2}O.It’s vapour density is found to be 30, calculate its molecular formula.
Solution –
Empirical formula = CH_{2}O.
∴ 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( CH_{2}O).
=C_{2}H_{4}O_{2} .
Thus, the molecular formula of the compound is C_{2}H_{4}O_{2} .
Another useful formula for calculation of molecular weight is –
Molecular weight = [w(in g) × 22.4 dm^{3}] / V_{o}.
w= weight of the substance in grams.
V_{o}= Volume of the gas at STP(standard temperature and pressure).
22.4 dm^{3}= 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 – V_{o}= 2.0 liters = 2 dm^{3}…. [liters and dm^{3}(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 dm^{3} )] / 2 dm^{3}
= 28g,
To find the empirical formula we carry the following calculations as seen earlier –
Element |
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. |
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 CH_{2}.
empirical formula weight = 12=2 = 14g.
Molecular weight = 28g.
∴ n = 28/14 = 2.
Thus, the molecular formula of the compound is C_{2}H_{4 .}
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