What are the empirical and molecular formulas of nicotine?

Jun 25, 2016

Empirical formula$\setminus {C}_{5} {H}_{7} N$
Molecular Formula$\setminus {C}_{10} {H}_{14} {N}_{2}$

Explanation:

The empirical and the molecular formulas differ in the level of information that each one of them could provide.

The empirical formula of a compound is the simplest formula that gives the ratio of atoms of each of the elements present in the compound, however it does not provide the actual number of atoms of each of the elements present.

On the other hand, the molecular formula provides, in addition to the ratio of atoms, the actual number of atoms of each of the elements present in the compound. It does not show the way atoms are connected and therefore a structural formula is needed.

The following illustrates the difference between an empirical formula, a molecular formula and a structural formula using nicotine as an example.

${C}_{5} {H}_{7} N$ is the empirical formula of nicotine. It tells that in one molecule of nicotine there are 5 atoms of carbon for each 7 atoms hydrogen and 1 atom of nitrogen.

${C}_{10} {H}_{14} {N}_{2}$ is the molecular formula of nicotine. It provides the ratio of atoms of each of the elements present $5 : 7 : 1$ it also provides the actual number of atoms. It tells that in one molecule of nicotine there are 10 atoms of carbon, 14 atoms of hydrogen and 2 atoms of nitrogen.

Note that a higher level of information is provided by the structural formula. 