# 8.06: Molecular Formulas

- Page ID
- 178156

## Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the *relative* numbers of a compound’s elements. Determining the *absolute* numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules).

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule.

The table below lists families of compounds with the same empirical formula

Empirical Formula | Empirical Formula Molar Mass | Molecular Formula | Molecular Formula Molar Mass | Molecular Formula Mass/Empirical Formula Mass |
---|---|---|---|---|

CH_{3} |
15.035 g/mole | CH_{3} |
15.035 g/mole | \[\mathrm{\dfrac{15.035\,\dfrac{g}{mole}}{15.035\,\dfrac{g}{mole}} = 1}\] |

C_{2}H_{6} |
30.07 g/mole | \[\mathrm{\dfrac{30.07\,\dfrac{g}{mole}}{15.035\,\dfrac{g}{mole}} = 2}\] | ||

C_{3}H_{9} |
45.105 g/mole | \[\mathrm{\dfrac{45.104\,\dfrac{g}{mole}}{15.035\,\dfrac{g}{mole}} = 3}\] | ||

CH_{4}O |
33.04 g/mole | CH_{4}O |
32.04 g/mole | \[\mathrm{\dfrac{32.04\,\dfrac{g}{mole}}{32.04\,\dfrac{g}{mole}} = 1}\] |

C_{2}H_{8}O_{2} |
64.08 g/mole | \[\mathrm{\dfrac{64.08\,\dfrac{g}{mole}}{32.04\,\dfrac{g}{mole}} = 2}\] | ||

C_{3}H_{12}O_{3} |
96.13 g/mole | \[\mathrm{\dfrac{96.13\,\dfrac{g}{mole}}{32.04\,\dfrac{g}{mole}} \approx 3}\] |

\[\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}= \mathit n\: formula\: units/molecule}\]

The molecular formula is then obtained by multiplying each subscript in the empirical formula by *n*, as shown by the generic empirical formula A_{x}B_{y}:

\[\mathrm{(A_xB_y)_n=A_{nx}B_{nx}}\]

For example, consider a covalent compound whose empirical formula is determined to be CH_{2}O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

\[\mathrm{\dfrac{180\:amu/molecule}{30\:\dfrac{amu}{formula\: unit}}=6\:formula\: units/molecule}\]

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

\[\ce{(CH2O)6}=\ce{C6H12O6}\]

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.

Example \(\PageIndex{5}\): Determination of the Molecular Formula for Nicotine

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

**Solution **

Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

\[\begin{alignat}{2}

&\mathrm{(74.02\:g\: C)\left(\dfrac{1\:mol\: C}{12.01\:g\: C}\right)}&&= \:\mathrm{6.163\:mol\: C}\\

&\mathrm{(8.710\:g\: H)\left(\dfrac{1\:mol\: H}{1.01\:g\: H}\right)}&&= \:\mathrm{8.624\:mol\: H}\\

&\mathrm{(17.27\:g\: N)\left(\dfrac{1\:mol\: N}{14.01\:g\: N}\right)}&&= \:\mathrm{1.233\:mol\: N}

\end{alignat}\]

Next, we calculate the molar ratios of these elements relative to the least abundant element, \(\ce{N}\).

\[\begin{alignat}{2}

&\mathrm{\left(\dfrac{6.163\:mol\: C}{1.233\:mol\: N}\right)}= \:\mathrm{\left(\dfrac{4.998\:mol\: C}{1\:mol\: N}\right)}=\:\mathrm{\left(\dfrac{5\: mol\:C}{1\:mol\:N}\right)}\\

&\mathrm{\left(\dfrac{8.624\:mol\: H}{1.233\:mol\: N}\right)}= \:\mathrm{\left(\dfrac{6.994\:mol\: H}{1\:mol\: N}\right)}=\:\mathrm{\left(\dfrac{7\: mol\:H}{1\:mol\:N}\right)}\\

&\mathrm{\left(\dfrac{1.233\:mol\: N}{1.233\:mol\: N}\right)}= \:\mathrm{\left(\dfrac{1\:mol\: N}{1\:mol\: N}\right)}

\end{alignat}\]

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C_{5}H_{7}N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

\[\mathrm{\dfrac{40.57\:g\: nicotine}{0.2500\:mol\: nicotine}=\dfrac{162.3\:g}{mol}}\]

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

\[\mathrm{\dfrac{162.3\:g/mol}{81.13\:\dfrac{g}{formula\: unit}}=2\:formula\: units/molecule}\]

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

\[\ce{(C5H7N)2}=\ce{C10H14N2}\]

Exercise \(\PageIndex{5}\)

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

**Answer:**

C_{8}H_{10}N_{4}O_{2}

## Summary

The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

## Key Equations

- \(\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}=\mathit n\: formula\: units/molecule}\)
- (A
_{x}B_{y})_{n}= A_{nx}B_{ny}

## Glossary

- empirical formula mass
- sum of average atomic masses for all atoms represented in an empirical formula

## Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).