Empirical and molecular formulas can seem daunting for chemistry students as it may feel like navigating through a maze of chemical symbols and equations. However, there are simple steps to follow that make this entire process much easier.

Understanding how empirical and molecular formulas work is the foundation to determining the composition of compounds, which are important in fields such as pharmaceuticals and forensic science.

In this article, we’ll break down the concepts of empirical and molecular formulas, and provide the steps to solving these questions. Chemistry teachers can use these steps to turn a daunting topic into a simple procedure.

Empirical Formulas:

Empirical formulas represent the simplest whole-number ratio of atoms in a compound. They provide essential information about the relative proportions of elements present, regardless of the actual number of atoms.

Let’s use this sample problem to begin: Consider a compound composed of 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find the empirical formula.

Here are the steps to solving empirical formula questions:

Step 1: Convert to grams

If the question provides percentages, convert them to grams by assuming a 100g sample. That means that in the sample problem, we have 40g of carbon, 6.7g of hydrogen and 53.3g of oxygen. If the question already gives you the information in grams, then you can skip this step.

Step 2: Convert to moles

To convert to moles, divide each mass by the element’s molar mass. The molar mass can be found on the periodic table.

• Carbon: 40g / 12.01g/mol = 3.33 moles
• Hydrogen: 6.7g / 1.01 g/mol = 6.65 moles
• Oxygen: 53.3g / 16 g/mol = 3.33 moles

Step 3: Find the simplest whole-number ratio

Divide each mole by the smallest mole value and if you get whole numbers, great. If you get a decimal, you’ll have to convert that to the closest whole number ratio by multiplying by a factor. For example, if one value ends in 0.5, multiply all values by 2. If one value ends in 0.33, multiply all values by 3. 0.95 can be rounded up and 0.05 can be rounded down.

In our sample problem, we already have whole number ratios from the initial division:

• Carbon: 3.33 / 3.33 = 1
• Hydrogen: 6.65 / 3.33 = 2
• Oxygen: 3.33 / 3.33 = 1

Step 4: Use these whole numbers as the formula subscripts

Therefore, the empirical formula is CH2O.

Analogy:

Analogous to baking a cake, where the recipe requires one cup of flour, two eggs, and one cup of sugar, regardless of whether you’re making a small or large cake. The ratio remains the same, just like the ratio of elements in an empirical formula remains constant.

Molecular Formulas:

Molecular formulas provide the actual number of atoms of each element in a compound. They are either equal to or multiplies of the empirical formula.

Sample problem continued: If the molar mass of the compound is found to be 180 g/mol, what is the molecular formula?

Here’s how to solve molecular formula questions:

Step 1: Find the molar mass of the empirical formula

In our sample problem, we figured out that the empirical formula is CH2O. To find its molar mass, look at the periodic table:

• 12.01 (C) + 2(1.01) (H) + 16 (O) = 30.02 g/mol

Step 2: Compare the molar mass of the empirical formula to the molar mass of the compound, given in the question.

The question tells us that the molar mass of the compound is 180g/mol, so we can calculate the ratio of the molar masses by dividing the empirical formula molar mass from the compound’s molar mass.

• 180 g/mol (molecular compound molar mass) / 30.02 g/mol (empirical formula molar mass) = approximately 6

Step 3: Multiply each of the subscripts in the empirical formula with this number.

The empirical formula is CH2O, therefore, the molecular formula is six times the empirical formula: C6H12O6.

Analogy:

Think of molecular formulas as different cake sizes. If the empirical formula is like a single-layer cake, the molecular formula could represent a double-layer cake or even a multi-tiered wedding cake. The ingredients (elements) remain the same, but the quantity (number of atoms) varies.

We hope you chemistry teachers and tutors found this article helpful! What other topics would you like to see covered in our future articles? Share your suggestions in the comments below!