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  Analysis of Molecular Formulas Homework Help - K-12 Grade Level, College Level Chemistry

Introduction to Analysis of Molecular Formulas

Even though structural formulas are important to unique description of organic compounds, it is interesting and instructive to evaluate the information that may be achieved from a molecular formula alone. The Three valuable rules are listed below:

1. The quantity of hydrogen atoms that may be bonded to a particular given number of carbon atoms is limited by the valence of carbon. The carbon and hydrogen's (hydrocarbons) compounds the maximum number of hydrogen atoms that can be bonded to n carbons is 2n + 2 (n is an integer). In case of methane, CH4, n=1 & 2n + 2 = 4. Origin of this formula is manifest by considering that a hydrocarbon made up of a chain of carbon atoms. Here each middle carbon has two hydrogens and two end carbons have three hydrogens each. Therefore, a six-carbon chain (n = 6) may be written as H-(CH2)6-H and the entire hydrogen count is (2 x 6) + 2 = 14. This relationship does not change the presence of oxygen (valence = 2), so previously defined C4H10O isomers follow the rule n=4 &2n + 2 = 10. The Halogen atoms (valence = 1) should be counted equal to hydrogen, as defined by C3H5Cl3, n = 3 & 2n + 2 = 8 = (5 + 3). In the presence of nitrogen, each nitrogen atom (valence = 3) increases the maximum number of hydrogens by one. 

Some Plausible                         C7H16O3, C9H18, C15H28O3, C6H16N2
Molecular Formulas

Some Impossible                     C8H20O6, C23H50, C5H10Cl4, C4H12NO
Molecular Formulas

2. The total number of odd-valenced atoms is even for stable organic compounds. So, when even-valenced atoms like carbon and oxygen are bonded jointly in any number and in any manner, number of remaining unoccupied bonding sites must be even. The univalent atoms like H, F, Cl, etc are occupied these sites.  Total number of their will must be even. Nitrogen is also an odd-valenced atom (3) and if nitrogen occupies a bonding site on carbon it adds two additional bonding sites, so maintaining even/odd parity.

Some Plausible                         C4H4Cl2, C5H9OBr, C5H11NO2, C12H18N2FCl
Molecular Formulas

Some Impossible                     C5H9O2, C4H5ClBr, C6H11N2O, C10H18NCl2
Molecular Formulas

3. In stable compounds of the oxygen, hydrogen & carbon, the number of hydrogen atoms reflects the number of the double bonds and rings in their structural formulas. Consider hydrocarbon with a molecular structure containing a simple chain of four carbon atoms that is CH3CH2CH2CH3. Molecular formula this chain is C4H10 (the highest number of bonded hydrogens by the 2n + 2 rules). If a ring is formed by four carbon atoms, the two hydrogens must be lost. Similarly, the introduction of a double bond involves the loss of two hydrogens and a triple bond the loss of four hydrogens.

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After the discussion and examples it should be clarified that molecular formula of a hydrocarbon (CnHm) provides information about double bonds and/or number of rings that must be shown in its structural formula. By the 2nd rule m must be an even number so if m < (2n + 2) the variation is also an even number that reflects double bonds and any rings. The triple bond is counted as two double bonds.

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It requires a modified analysis in Presence of the one or more nitrogen atoms or halogen substituents. The formula which is described above may be extended to such compounds by a few simple principles

  1.  
    • The relationship does not alter due to the Presence of oxygen.
    • In molecular formula all halogens must be replaced by hydrogen.
    • In formula each nitrogen molecule must be replaced by a CH moiety.

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