# Empirical Formula: Definition, Steps & Examples Chapter In this lesson, discover what the empirical formula for a compound is and how it differs from the molecular formula. We will then go on and learn how to calculate the empirical formula for a given compound as well.
Empirical Formula Definition
The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. It can be calculated from information about the mass of each element in a compound or from the percentage composition.
Visually, the empirical formula looks similar to the molecular formula, which gives the number of atoms in a single molecule of a compound. In fact, a compound’s empirical formula can end up being the same as its molecular formula, but this doesn’t always happen.
In this image we can see how empirical and molecular formulas can differ by comparing both formulas for dinitrogen tetroxide.
empirical formula versus molecular formula
How to Calculate
To calculate the empirical formula, you must first determine the relative masses of the various elements present. You can either use mass data in grams or percent composition.
For percent composition, we assume the total percent of a compound is equal to 100% and the percent composition is the same in grams. For example, the total mass of the compound is 100 grams. If a compound contained 68% carbon, 9% hydrogen, and 23% oxygen, we would assume 68 grams of carbon, 9 grams of hydrogen, and 23 grams of oxygen.
The steps for determining the empirical formula of a compound are as follows:
Step 1: Obtain the mass of each element present in grams.
Element % = mass in g = m
Step 2: Determine the number of moles of each type of atom present.
m / atomic mass = Molar amount (M)
Step 3: Divide the number of moles of each element by the smallest number of moles that you found in the previous step.
M / least M value = Atomic Ratio (R)
Step 4: Convert numbers to whole numbers by multiplying each of them by the smallest number that will end up with only whole numbers as a result. These whole numbers are the subscripts used in the empirical formula.
R * whole number = Empirical Formula
Example One
The absolute best way to learn how to figure out empirical formulas is to practice. Here are a couple of examples to work through, step by step.
An oxide of aluminum is formed by the reaction of 4.151g of aluminum and 3.692g of oxygen. Calculate the empirical formula for this compound.
What do we know?
The compound contains 4.151g of aluminum and 3.692g of oxygen.
We also know the atomic masses by looking them up on the periodic table: aluminum (26.98 g/mol) and oxygen (16.00 g/mol).
Let’s go through the steps to solve this:
Step 1: Determine the masses.
We have these: 4.151g of Al and 3.692g of O
Step 2: Determine the number of moles by dividing the grams by the atomic mass.
4.151 g Al * (1 mol Al / 26.98 g Al) = 0.1539 mol Al atoms
3.692 g O * (1 mol O / 16.00 g O) = 0.2398 mol O atoms
Step 3: Divide the number of moles of each element by the smallest number of moles.
0.1539 mol Al / 0.1539 = 1.000 mol Al atoms
0.2398 mol O / 0.1539 = 1.500 mol O atoms
Step 4: Convert numbers to whole numbers.
The smallest number we can multiply both 1.5 and 1 by to get whole numbers as a result is 2.
1.000 Al * 2 = 2.000 Al atoms and 1.500 O atoms * 2 = 3.000 O atoms
The compound contains 2 Al atoms for every 3 O atoms.
Empirical formula = Al2 O3

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