Empirical Formula: Study with Video Lessons, Practice Problems & Examples

The empirical formula represents the simplest whole number ratio of atoms in a compound, derived from the mass percentage of its elements using the mole concept. In contrast, the molecular formula indicates the actual number of each type of atom present. For example, C₃H₆O₃ simplifies to CH₂O, while C₁₂H₂₂O₁₁ remains unchanged. Understanding these distinctions is crucial for analyzing chemical compounds and their properties.

Empirical Formula

In this video, we're going to take a look at the difference between our empirical formula and molecular formula. First of all, when it comes to the empirical formula of a compound, we're going to say it's related to the mass percentage of its constituent elements using the mole concept. Remember, the mole concept has to do with us going between grams, moles, and units such as molecules, ions, or atoms. We're going to say here that when it comes to our compound, the empirical formula gives us the relative number of atoms and represents the most simplified form. So by convention, we're going to say any formula must contain whole numbers of each atom in what's called the whole number ratio. Now let's put this to practice and really understand what's the difference between molecular formula and empirical formula. Here, we have the molecular formulas of different compounds. This is how many actual elements of each type that are present within that compound. The empirical formula can be thought of as its reduced or simplified form. So if we take a look here, we have C₃H₆O₃. Realize that here 3, 6, and 3 are all divisible by 3. So if I divide this by 3, what I'll have left is CH₂O. That CH₂O represents the reduced simplified form or the empirical formula. If we look at the next one, we have C₁₀H₁₄N₂. All these numbers here are divisible by 2. So when we divide everything by 2, we're going to get C₅H₇N. That represents the empirical formula. And then finally, we have C₁₂H₂₂O₁₁. They don't share any number in common where we can reduce it to a simplified form, so we'd actually have the same empirical formula as molecular formula, and we'll see from time to time that this is true. So just remember that your molecular formula is the actual number of each of the elements within a compound, and the empirical formula is the reduced form of that compound. Keep this in mind when comparing molecular formula versus empirical formula.

Empirical Formula Example 1

We're now going to calculate the empirical formula for a given compound. The empirical formula can be calculated from the masses or percentages of elements within a compound. In this example question, it says, "Determine the empirical formula of a compound that is 68.40 percent chromium and 31.60% oxygen."

Step 1, you're going to write down the symbols for each element in the question. We're dealing with chromium and its symbol as Cr and oxygen with its symbol as just O.

Step 2, we're going to write down the masses in grams of each element given. Notice here in this question that they're not giving us masses. What they're giving us are percentages. We're going to convert all the percentages into grams by assuming there are 100 grams of the compound. If you add the percentages up, 68.40 percent plus 31.60 percent, you'll see that you get 100% of your total compound. This makes sense because if our compound is made up of chromium and only oxygen, together they should add up to the complete compound. So we're going to convert 68.40% to 68.40 grams of Cr, and 31.60 grams of just O.

Step 3, convert all the masses into moles. This requires some mole concept theory on your part. For chromium, 68.40 grams of Cr requires conversion, using that 1 mole of Cr weighs 51.996 grams, giving us 1.3155 moles of Cr. For oxygen, 31.60 grams of oxygen requires conversion using that 1 mole of oxygen weighs 16.00 grams, resulting in 1.9750 moles of oxygen. To avoid rounding errors, ensure the values have at least 4 decimal places.

Step 4, you're going to divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. Here we have 1.3155 moles of Cr and 1.9750 moles of O. Dividing both by 1.3155, we get 1 chromium and 1.5 oxygens.

Step 5, after dividing by the smallest number, if you encounter a fraction such as 0.5, which can't simply be rounded, we multiply by a factor to create whole numbers. In this case, multiplying by 2, we get 2 chromiums and 3 oxygens, rendering the empirical formula as Cr₂O₃.

These steps must be employed to determine the empirical formula when given information on percentages or mass of different elements within a compound.

A compound that contains only carbon, hydrogen, and oxygen is composed of 48.64% C and 43.2% O by mass. What is the empirical formula of this compound?

• A. C₃H₆O₂
• B. CH₂O₂
• C. CH₄O
• D. C₂H₆O₂

Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula of the compound?

• A. NaClO
• B. NaClO₂
• C. NaClO₃
• D. NaClO₄

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 10²³ hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?

Here the practice question states: A compound composed of carbon, hydrogen, and chlorine contains 4.19 \times 10^{23} hydrogen atoms. If 9 grams of the compound also contains 55% chlorine by mass, what is the empirical formula? Now, sometimes you'll run into a question and based on its wording it may seem very confusing. But remember, just first look to see what they are asking you to find. They want you to find the empirical formula. Then rely on the steps that you've learned to help calculate the empirical formula. Remember, Step 2, we need to find the masses in grams of each of the elements. So let's do that first. Here we have hydrogen atoms. So what we need to do here is we need to convert those hydrogen atoms into grams. So we have 4.19 \times 10^{23} hydrogen atoms. Atoms mean we're talking about a single atomic element so it's just H by itself. What we're going to do next is we're going to go from atoms to moles. Remember that means that we have to use Avogadro's number. So 1 mole of hydrogen is 6.022 \times 10^{23} hydrogen atoms. And remember, I've said this in earlier videos. When it comes to working things out like this, if you have something written in scientific notation, it's best to put it in parentheses in your calculator because based on the model that you're using, you may get an entirely wrong answer. So now, atoms cancel out. We have moles of hydrogen and we're going to say that for every 1 mole of hydrogen, it's 1.008 \, \text{grams hydrogen}. So moles of H cancel out and we have our grams as 0.7013 \, \text{grams hydrogen}. Alright. So we have hydrogen figured out. Remember this compound is composed of carbon and chlorine as well. They're telling us that our sample weighs, or compound weighs 9 grams and 55% of that total mass is chlorine. So we'd say that we have 9 grams total. We cannot use the percentage form when it comes to math, we have to use the decimal form. So you divide this by 100 and that gives me 0.55. So it's 0.55 \times 9, so that gives me 4.95 \, \text{grams of chlorine}. So now we've found the grams of hydrogen, the grams of chlorine. How do we figure out the grams of carbon? Well, remember we said that our compound weighs 9 grams. So those 9 grams are of the carbon, hydrogen, and chlorine altogether. We just found out the masses of hydrogen and chlorine, so subtract those from 9 and the difference which you have left will have to represent the grams So we So we have left at the end is 3.3487 \, \text{grams of carbon}. So this originally difficult-looking question is pretty approachable now because we have the grams of each of the elements. So all we're going to do is follow the steps that we know. We're going to change those grams into moles, divide them all by the smallest moles and hopefully be able to get our empirical formula at the end. Alright. So, we're gonna start out with carbon. So we have those grams of carbon. So we have 3.3487 \, \text{grams carbon}. We have 0.7013 \, \text{grams hydrogen}. And we have 4.95 \, \text{grams chlorine}. I'm writing in this order because that's the order, they're given to us within the question. Carbon, hydrogen then chlorine. Alright. So, 1 mole of carbon weighs 12.01 \, \text{grams carbon}. One mole of H is 1.008 \, \text{grams H}. And then, 1 mole of chlorine is 35.45 \, \text{grams chlorine}. All the grams will cancel out, and we'll have moles at the end. Remember, at this point we need to make sure that our moles had at least 4 decimal places carbon. Hydrogen comes out to be 0.6958 \, \text{moles H}. And then chlorine comes out to be 0.1396 \, \text{moles of chlorine}. Next we divide all the moles that we've gotten by the smallest mole answer. Which would be the 0.1396. So that would give us 1 chlorine. And then when we divide hydrogen by that number, we get 4.984. And because it's 0.9, 9, we can round up to 5, so that's 5 \, \text{H}. And then carbon, when we divide it by that, we get 1.997 \, \text{carbon}. Again, it's 0.9 so I can round up to 2. So we have 2 carbons, 5 hydrogens, and one chlorine. That means that our empirical formula here would be \text{C}_{2}\text{H}_{5}\text{Cl}. So again, there will be times when you're gonna run into a question that seems very confusing by the way it's written. First, find out what exactly are they asking you to find and then rely on the steps that you've learned to help you get to the answer. Here, all we have to do is convert all the information we have to the grams of each of the elements. They only gave us information on carbon, chlorine, and hydrogen. We found their grams. We know that the complete mass of the compound was 9 grams so we knew that, the grams of chlorine and hydrogen if we subtract it from the 9 grams, that would give us the mass of the missing carbon. Once you have the grams of everything, we follow the basic steps that we've learned to find our empirical formula. So keep this in mind and remember the best way to approach any question that seems confusing is to first find out what they're asking you to do. And then if you know any steps, follow those steps as closely as you can to get to your final answer.

Here’s what students ask on this topic:

What is the difference between empirical and molecular formulas?

The empirical formula represents the simplest whole number ratio of atoms in a compound, derived from the mass percentage of its elements using the mole concept. For example, the empirical formula for glucose (C₆H₁₂O₆) is CH₂O. In contrast, the molecular formula indicates the actual number of each type of atom present in a molecule. For glucose, the molecular formula is C₆H₁₂O₆. While the empirical formula is a simplified version, the molecular formula provides the exact composition of the molecule.

How do you determine the empirical formula from mass percentages?

To determine the empirical formula from mass percentages, follow these steps: 1) Convert the mass percentages to grams. 2) Convert grams to moles using the molar mass of each element. 3) Divide the moles of each element by the smallest number of moles calculated. 4) Round to the nearest whole number to get the simplest ratio. For example, if you have 40% carbon, 6.7% hydrogen, and 53.3% oxygen, you would convert these to moles and find the simplest ratio to get the empirical formula CH₂O.

Can the empirical formula and molecular formula be the same?

Yes, the empirical formula and molecular formula can be the same if the compound's simplest ratio of atoms is also its actual number of atoms. For example, in the case of formaldehyde, both the empirical and molecular formulas are CH₂O. This occurs when the molecular formula cannot be simplified further into a smaller whole number ratio.

Why is the empirical formula important in chemistry?

The empirical formula is important in chemistry because it provides the simplest representation of a compound's composition, which is essential for understanding its basic properties and reactions. It helps chemists determine the relative proportions of elements in a compound, which is crucial for stoichiometric calculations, predicting reaction outcomes, and analyzing chemical behavior. Additionally, it serves as a foundational step in determining the molecular formula of a compound.

How do you convert an empirical formula to a molecular formula?

To convert an empirical formula to a molecular formula, follow these steps: 1) Determine the empirical formula mass by summing the atomic masses of all atoms in the empirical formula. 2) Divide the compound's molar mass (given or determined experimentally) by the empirical formula mass to find the multiplication factor (n). 3) Multiply each subscript in the empirical formula by this factor (n) to get the molecular formula. For example, if the empirical formula is CH₂O and the molar mass is 180 g/mol, the factor n is 6, resulting in the molecular formula C₆H₁₂O₆.