# Empirical Formula: Definition, Steps & Examples

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Lesson Transcript
Instructor: Nicola McDougal

Nicky has taught a variety of chemistry courses at college level. Nicky has a PhD in Physical Chemistry.

Ever wondered how we know the formula of a chemical compound? In this lesson, we will learn how to determine the empirical formula and the steps in calculating it.

## 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 the same as the compound's molecular formula, but not always. An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.

## Calculation of an Empirical Formula

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

M / least M value = Atomic Ratio (R)

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

R * whole number = Empirical Formula

## Examples of Empirical Formula Calculation

The absolute best way to learn how to figure out empirical formulas is to practice. Here are some examples to take you through, step-by-step.

#### Example 1

An oxide of aluminum is formed by the reaction of 4.151 g of aluminum and 3.692 g of oxygen. Calculate the empirical formula for this compound.

What do we know?

• The compound contains 4.151 g of aluminum and 3.692 g 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: 4.151 g of Al and 3.692 g of O

Step 2: Determine the number of moles by dividing the grams by the atomic mass

So let's do that now:

4.151 g Al x (1 mol Al / 26.98 g Al) = 0.1539 mol Al atoms

3.692 g O x (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

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 = Al2O3

#### Example 2

When a 0.3546 g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of 0.6330 g. What is the empirical formula of this vanadium oxide?

What do we know?

• The compound contains 0.3546 g of vanadium and a total mass of 0.6330 g
• We know the atomic masses of vanadium (50.94 g/mol) and oxygen (16.00 g/mol)

Let's go through the steps to solve:

Step 1: Determine the masses

We are given vanadium as 0.3546g and that must be present in the final mass.

mass of the oxygen = final mass - vanadium mass

0.6330 g - 0.3546 g = 0.2784 g

Step 2: Determine the number of moles by dividing the grams by the atomic mass

0.3456 g V x (1 mol V / 50.94 g V) = 0.006961 mol V atoms

0.2784 g O x (1 mol O /1 6.00 g O) = 0.01740 mol O atoms

Step 3: Divide the number of moles of each element by the smallest number of moles

0.006961 mol V / 0.006961 = 1.000 mol V atoms

0.01740 mol O / 0.006961 = 2.500 mol O atoms

Step 4: Convert numbers to whole numbers

1.000 V * 2 = 2.000 V atoms

2.500 O atoms * 2 = 5.000 O atoms

The compound contains 2 V atoms for every 5 O atoms

Empirical Formula = V2O5

#### Example 3

This example uses compounds containing more than three elements.

A chemist analyzes an unknown sample and finds it contains 0.8007 g of carbon, 0.9333 g of nitrogen, 0.2016 g of hydrogen, and 2.133 g of oxygen. What is the empirical formula of this compound?

What do we know?

• The compound contains 0.8007 g carbon, 0.9333 g nitrogen, 0.2016 g hydrogen, and 2.133 g oxygen
• We know the atomic masses of carbon (12.01 g/mol), nitrogen (14.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol)

Let's go through the steps to solve the problem:

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