5.12: Moles, Molecular Formulas, and Calculating Empirical Formulas for Compounds

Learning Objectives

• Define molecular formula.
• Define empirical formula.
• Identify the difference between molecular and empirical formulas.
• Determine empirical formula from percent composition of a compound.

When studying chemical formulas, they may be written in several different ways. The most common way is via a molecular formula. Molecular formulas tell us the number of atoms of each type of element in a molecule or compound. This formula does not necessarily tell us the order in which the atoms are connected, but will provide the total number of each type of atom. Take the molecular formula for sugar C₆H₁₂O₆. This is the formula for both glucose and galactose. Both sugars have the same molecular formula, but different structures. In section 5.11 we covered calculating formula mass values. The mass of any compound represents the mass of all atoms of the substance in 1 mole of a substance. We will cover more about moles later, but for now it is important to understand that moles are the way we quantify a grouping of atoms for a specific substance. A great example of this is taking a carton of eggs. We measure eggs by the dozen. It is commonly accepted that one dozen eggs is the same as 12 eggs. When calculating the mass of a substance, it is accepted that the mass of each element shown on the periodic table is the mass for one mole (or one grouping of atoms) for that substance. The units used here for mass will be grams. Hence, if you look at any element on the periodic table the mass shown is the number of grams of atoms in one mole of that substance.

Another way in which a formula may be written is called an empirical formula. Empirical formulas do not tell you the number of teach type of atom that is in a substance, but rather only the types of atoms involved in a substance. Often times empirical formulas are found, and then through taking measurements the molecular formula is determined. We will cover the concept of empirical formulas in this unit.

Determining Empirical Formulas

An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H₂O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H₂O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.

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.

Note

Steps to determine empirical formula.

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 \(\PageIndex{1}\)

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

Steps for Problem Solving	Find the empirical formula of a compound of \(69.94\%\) iron and \(30.06\%\) oxygen.
Identify the "given"information and what the problem is asking you to "find."	Given: \(\%\) of \(\ce{Fe} = 69.94\%\) \(\%\) of \(\ce{O} = 30.06\%\) Find: Empirical formula \(= \ce{Fe}_?\ce{O}_?\)
Calculate
a. Assume a \(100 \: \text{g}\) sample, convert the same % values to grams.	\[69.94 \: \text{g} \: \ce{Fe} \nonumber\] \[30.06 \: \text{g} \: \ce{O} \nonumber\]
b. Convert to moles.	\[69.94 \: \text{g} \: \ce{Fe} \times \frac{1 \: \text{mol} \: \ce{Fe}}{55.85 \: \text{g} \: \ce{Fe}} = 1.252 \: \text{mol} \: \ce{Fe} \nonumber\] \[30.06 \: \text{g} \: \ce{O} \times \frac{1 \: \text{mol} \: \ce{O}}{16.00 \: \text{g} \: \ce{O}} = 1.879 \: \text{mol} \: \ce{O} \nonumber\]
c. Divide both moles by the smallest of the results.	\(\mathrm{Fe:\:\dfrac{1.252\:mol}{1.252}}\) \(\mathrm{O:\:\dfrac{1.879\:mol}{1.252}}\) The "non- whole number" empirical formula of the compound is \(\ce{Fe_1O}_{1.5}\)
Multiply each of the moles by the smallest whole number that will convert each into a whole number.	Fe:O = 2 (1:1.5) = 2:3 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.
Write the empirical formula.	The empirical formula of the compound is \(\ce{Fe_2O_3}\).
Think about your result.	The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The compound is the ionic compound iron (III) oxide.

 

Exercise \(\PageIndex{1}\)

Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?

Answer

HgCl₂



Summary

• A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.

Contributions & Attributions

This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:

• Marisa Alviar-Agnew (Sacramento City College)

• Henry Agnew (UC Davis)