## Determination of Empirical Formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative *numbers*, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound.

To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

\(1.17\text{ g C} × \frac{1\text{ mol C}}{12.01\text{ g C}} = 0.142\text{ mol C}\)

\(0.287\text{ g H} × \frac{1\text{ mol H}}{1.008\text{ g H}} = 0.284\text{ mol H}\)

Thus, we can accurately represent this compound with the formula C_{0.142}H_{0.248}. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

\(\text{C}_{\frac{0.142}{0.142}}\text{ H}_{\frac{0.248}{0.142}}\) or \(\text{CH}_2\)

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH_{2}. This may or not be the compound’s *molecular formula* as well; however, we would need additional information to make that determination (as discussed later in this section).

Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

\(\text{Cl}_{0.150}\text{O}_{0.525} \)\(= \text{Cl}_{\frac{0.150}{0.150}}\text{ O}_{\frac{0.525}{0.150}} \)\(= \text{ClO}_{3.5}\)

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl_{2}O_{7} as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

- Deriving the number of moles of each element from its mass
- Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
- Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

The figure below outlines this procedure in flowchart fashion for a substance containing elements A and X.

### Determining a Compound’s Empirical Formula from the Masses of its Elements

A sample of the black mineral hematite (see image below), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

**Solution**

For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

\(34.97\text{ g Fe} \left ( \frac{\text{mol Fe}}{55.85\text{ g}} \right ) = 0.6261\text{ mol Fe}\)

\(15.03\text{ g O} \left ( \frac{\text{mol O}}{16.00\text{ g}} \right ) = 0.9394\text{ mol O}\)

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

\(\frac{0.6261}{0.6261} = 1.000\text{ mol Fe}\)

\(\frac{0.9394}{0.6261} = 1.500\text{ mol O}\)

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe_{1}O_{1.5}). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

\(2 (\text{Fe}_1\text{O}_{1.5}) = \text{Fe}_2\text{O}_3\)

The empirical formula is Fe_{2}O_{3}.

## Video Illustrating Derivation of Empirical Formulas

For additional worked examples illustrating the derivation of empirical formulas, you might watch the video below by Mr. Causey.