We studied the concept of empirical formula in the previous post. The empirical formula gives us the relative number of elements present in the compound. However, we need to know the absolute number of atoms, of each element, present in the molecule. The molecular formula provides us with information. Let us proceed to learn all about the molecular formula in this post.
The Molecular Formula
The molecular formula is a chemical formula that shows the total number and kinds of atoms in a molecule. However, it does not show the structural arrangement of the atoms in it.
e.g.- The molecular formula for benzene is C6H6 .This formula tells us that there are six carbon atoms and six hydrogen atoms in a molecule of benzene. However, we cannot know the exact structure of benzene by just looking at this formula.
The molecular formula can be derived by comparing the molecular/molar mass with its empirical formula mass.
Molecular weight /Molecular mass is the mass of one molecule of a compound.
e.g.Carbon dioxide (CO2) molecule.
This molecule has one carbon atom and two oxygen atoms in it. The sum of the atomic weights of one carbon atom and two oxygen atoms will be the molecular weight of this compound. The atomic weight of any element can be found by referring to the periodic table. It is mentioned below the symbol for that element.
Molecular weight of CO2 = 1(At.wt of C) + 2(At.wt of O)
= 12.011 + 2(15.99 )
= 12.011 + 31.98
= 43.991
≈ 44 amu
Molar mass is the mass of one mole(6.023 × 1023 atoms) of a substance. The numeric value of the molecular weight and molar mass is the same. Only the units are different. While the molecular mass is expressed in amu (atomic mass unit), the molar mass is expressed in g/mol.
Thus, for CO2,
The molecular weight = 44 amu,
Molar mass = 44g/mol.
The empirical formula weight is the sum of the average atomic masses of all the atoms represented in an empirical formula.
e.g.- In the previous post, we mentioned that the empirical formula for benzene is CH.
Empirical formula weight= 1(At. wt of carbon) + 1( At.wt. of hydrogen) = 12 + 1 = 13.
The molecular formula can be found out from the empirical formula by the following equation –
Molecular formula = n × Empirical formula ,
n= Integer 1,2,3 ..etc
&
n=(molecular weight) / (empirical formula weight)
If we know the empirical formula, we can easily find the empirical formula weight. The molecular weight of any compound can be found in various ways (e.g.- through Mass Spectrometry). Once we know these two quantities, we can use the above relation to find ‘n’.
Let us solve a few problems to completely understand these calculations.
Problem 1 – An organic monobasic acid contains 18.6% carbon,1.55% hydrogen, and 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 up to 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 |
The empirical formula is CHCLO.
Therefore, the empirical formula weight = (Weight of C+ Weight of H+ Weight of Cl+ Weight of oxygen)
= 12+1+ 35.5+ 16
= 64.5
Molecular weight (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 vapor 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.Its vapor 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 ÷ Empirical 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 the 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
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 CH2.
empirical formula weight = 12=2 = 14g.
Molecular weight = 28g.
∴ n = 28/14 = 2.
Thus, the molecular formula of the compound is C2H4.
In the old days, empirical formula calculations lead us to molecular formulas. However, now with the advent of many advanced techniques, computers give us the desired results quickly. Techniques like mass spectrometry, give us molecular weight (the largest m/z ratio) directly, most of the time. 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!
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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