# 10.12: Determining Empirical Formulas

In the early days of chemistry, there were few tools for the detailed study of compounds. Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials. The "new" field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely. The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios. We did not know exactly how many of these atoms were actually in a specific molecule.

## Determining Empirical Formulas

An empirical formula is one that shows the lowest whole-number ratio of the elements in a compound. Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical. However, we can also consider the empirical formula of a molecular compound. Ethene is a small hydrocarbon compound with the formula $$\ce{C_2H_4}$$ (see figure below). While $$\ce{C_2H_4}$$ is its molecular formula and represents its true molecular structure, it has an empirical formula of $$\ce{CH_2}$$. The simplest ratio of carbon to hydrogen in ethene is 1:2. There are two ways to view that ratio. Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen. Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen. So the subscripts in a formula represent the mole ratio of the elements in that formula. Figure 10.12.1: Ball-and-stick model of ethene, $$\ce{C_2H_4}$$.

In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula. The steps to be taken are outlined below.

1. Assume a $$100 \: \text{g}$$ sample of the compound so that the given percentages can be directly converted into grams.
2. Use each element's molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.
4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.
5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.

Example 10.12.1

A compound of iron and oxygen is analyzed and found to contain $$69.94\%$$ iron and $$30.06\%$$ oxygen. Find the empirical formula of the compound.

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$\%$$ of $$\ce{Fe} = 69.94\%$$
• $$\%$$ of $$\ce{O} = 30.06\%$$

Unknown

• Empirical formula $$= \ce{Fe}_?\ce{O}_?$$

Steps to follow are outlined in the text.

Step 2: Calculate.

1. Assume a $$100 \: \text{g}$$ sample.

$69.94 \: \text{g} \: \ce{Fe}$

$30.06 \: \text{g} \: \ce{O}$

2. Convert to moles.

$69.94 \: \text{g} \: \ce{Fe} \times \frac{1 \: \text{mol} \: \ce{Fe}}{55.85 \: \text{g} \: \ce{Fe}} = 1.252 \: \text{mol} \: \ce{Fe}$

$30.06 \: \text{g} \: \ce{O} \times \frac{1 \: \text{mol} \: \ce{O}}{16.00 \: \text{g} \: \ce{O}} = 1.879 \: \text{mol} \: \ce{O}$

3. Divide both moles by the smallest of the results.

$\frac{1.252 \: \text{mol} \: \ce{Fe}}{1.252} = 1 \: \text{mol} \: \ce{Fe} \: \: \: \: \: \frac{1.879 \: \text{mol} \: \ce{O}}{1.252} = 1.501 \: \text{mol} \ce{O}$

4/5. Since the moles of $$\ce{O}$$ is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.

$1 \: \text{mol} \: \ce{Fe} \times 2 = 2 \: \text{mol} \: \ce{Fe} \: \: \: \: \: 1.501 \: \text{mol} \: \ce{O} \times 2 = 3 \: \text{mol} \: \ce{O}$

The empirical formula of the compound is $$\ce{Fe_2O_3}$$.