Empirical formula

In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.^[1] A simple example of this concept is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S₂O₂. This means that sulfur monoxide and disulfur dioxide, both compounds of Sulfur and Oxygen will have the same empirical formula.

An empirical formula makes no mention of the arrangement or number of atoms. It is standard for many ionic compounds, like CaCl₂, and for macromolecules, such as SiO₂.

The molecular formula, on the other hand, shows the number of each type of atom in a molecule. Also the structural formula shows the arrangement of the molecule. It is possible for different types of compounds to have equal empirical formulas.

Examples

• Glucose (C₆H₁₂O₆), ribose (C₅H₁₀O₅), acetic acid (C₂H₄O₂), and formaldehyde (CH₂O) all have different molecular formulas but the same empirical formula: CH₂O. This is the actual molecular formula for formaldehyde, but acetic acid has double the number of atoms, ribose has five times the number of atoms, and glucose has six times the number of atoms.
• The chemical compound n-hexane has the structural formula CH₃CH₂CH₂CH₂CH₂CH₃, which shows that it has 6 carbon atoms arranged in a chain, and 14 hydrogen atoms. Hexane's molecular formula is C₆H₁₄, and its empirical formula is C₃H₇, showing a C:H ratio of 3:7.

Calculation

Suppose you are given a compound such as methyl acetate, a solvent commonly used in paints, inks, and adhesives. When methyl acetate was chemically analyzed, it was discovered to have 48.64% carbon (C), 8.16% hydrogen (H), and 43.20% oxygen (O). For the purposes of determining empirical formulas, we assume that we have 100 g of the compound. If this is the case, the percentages will be equal to the mass of each element in grams.

Step 1

Change each percentage to an expression of the mass of each element in grams. That is, 48.64% C becomes 48.64 g C, 8.16% H becomes 8.16 g H, and 43.20% O becomes 43.20 g O.

Step 2

Convert the amount of each element in grams to its amount in moles.

$\left(\frac{48.64 \mbox{ g C}}{1}\right)\left(\frac{1 \mbox{ mol }}{12.01 \mbox{ g C}}\right) = 4.049\ \text{mol}$

$\left(\frac{8.16 \mbox{ g H}}{1}\right)\left(\frac{1 \mbox{ mol }}{1.008 \mbox{ g H}}\right) = 8.095\ \text{mol}$

$\left(\frac{43.20 \mbox{ g O}}{1}\right)\left(\frac{1 \mbox{ mol }}{16.00 \mbox{ g O}}\right) = 2.7\ \text{mol}$

Step 3

Divide each of the found values by the smallest of these values (2.7)

$\frac{4.049 \mbox{ mol }}{2.7 \mbox{ mol }} = 1.5$

$\frac{8.095 \mbox{ mol }}{2.7 \mbox{ mol }} = 3$

$\frac{2.7 \mbox{ mol }}{2.7 \mbox{ mol }} = 1$

Step 4

If necessary, multiply these numbers by integers in order to get whole numbers; if an operation is done to one of the numbers, it must be done to all of them.

$1.5 \times 2 = 3$

$3 \times 2 = 6$

$1 \times 2 = 2$

Thus, the empirical formula of methyl acetate is C₃H₆O₂. This formula also happens to be methyl acetate's molecular formula.

References

1. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "Empirical formula".