Login

Empirical Formula: Definition, Steps & Examples

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Excited State in Chemistry: Definition & Overview

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:05 Empirical Formula Definition
  • 0:19 Calculation of an…
  • 1:34 Examples of Empirical…
  • 11:45 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Login or Sign up

Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!
Create An Account
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:

Step 1: Determine the masses

We are given these as 0.8007 g C, 0.9333 g N, 0.2016 g H and 2.133 g O.

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

0.8007 g C x (1 mol C / 12.01 g C) = 0.06667 mol C atoms

0.9333 g N x (1 mol N / 14.01 g N) = 0.06662 mol N atoms

0.2016 g H x (1 mol H / 1.008 g H) = 0.2000 mol H atoms

2.133 g O x (1 mol O / 16.00 g O) = 0.1333 mol O atoms

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

0.06667 mol C / 0.06662 = 1.001 mol C atoms

0.06662 mol N / 0.06662 = 1.000 mol N atoms

0.2000 mol H / 0.06662 = 3.002 mol H atoms

0.1333 mol O / 0.06662 = 2.001 mol O atoms

Step 4: Convert numbers to whole numbers

To unlock this lesson you must be a Study.com Member.
Create your account

Register for a free trial

Are you a student or a teacher?
I am a teacher

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back

Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it free for 5 days!
Create An Account
Support