Empirical Formula - Online Tutor, Practice Problems & Exam Prep

The empirical formula represents the simplest whole number ratio of atoms in a compound, derived from the mass percentages of its elements. 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 formulas is crucial for grasping concepts like molecular weight and stoichiometry in chemical reactions.

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 C1H2O. That C1H2O 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 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 doing a comparison of 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 itself can be calculated from the masses or percentages of elements within a compound. We're going to say in this example question, it says, determine the empirical formula of a compound that is 68.40 percent chromium and 31.60% oxygen. So 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 O. Now 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. So, we're going to convert all the percentages into grams by assuming there are 100 grams of the compound. And also, if you look at the percentages and you add them up, so 68.40 percent plus 31.60 percent, you'll see that you get a 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 O.

Step 3, convert all the masses into moles. Now, this requires some mole concept theory on your part. So remember, this is just a simple conversion. So we have 68.40 grams of Cr, we want to get rid of grams of Cr, so we put grams of Cr on the bottom, We put moles on top. We're gonna say 1 mole of Cr. If you look at your periodic table, you'll see that 1 mole of Cr weighs 51.996 grams. So grams here will cancel out, And here's the thing, we're gonna say, to avoid rounding errors, make sure the values have at least 4 decimal places. So plugging this into our calculator, we get 1.3155 moles of Cr. So again, to avoid rounding errors, it's best to have at least 4 decimal places. We converted grams of chromium into moles. Now let's do the same with grams of oxygen. 31.60 grams of oxygen. We want to get rid of grams of oxygen, so place it on the bottom. We want to convert it into moles of oxygen, so put 1 mole of oxygen on top. Look at the periodic table. The mass of oxygen on the periodic table is 16. So here grams of oxygen cancel out, and when we do that we're going to get 1.9750 moles of oxygen.

Next, this is important, step 4, you're going to divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. So here we have 1.3155 moles of Cr and 1.9750 moles of O. The smaller number is 1.3155, so both will get divided by that number. When we do that, we're gonna get 1 chromium, and when we do that here, we're gonna get 1.5 oxygens. Okay. So we've run into an issue here. Step 5, after dividing by the smallest number, if you get a value that is 0.1, some number 0.1, and some number 0.9, then you can round to the nearest whole number. Unfortunately, we got 0.5 here. We can't just simply round that. So what do we do instead? Well, in these situations, if you can't round, we multiply by a factor to create whole numbers. So think to yourself, 1.5, what number can I multiply that by to get a whole number? If we start out by multiplying by 2, that's gonna give me 3. 3 is a whole number, so we'd have 3 oxygens. But here's the thing, if I multiply the number of oxygens by 2, then I have to multiply the number of chromiums also by 2. So the empirical formula contains 2 chromium atoms and 3 oxygen atoms. So that would mean that the empirical formula here is Cr2O3. So this would be my empirical formula for this question. These are the steps you have to employ to get the empirical formula when given information on the percentages or mass of different elements within.

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 to the 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 may seem very confusing. But remember, just first look to see what they are asking me to find. They want me 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 to the 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 to the 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, again, if numbers are written in scientific notation, it's scientific notation, so a number times 10 to any power, put it in parentheses. 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 grams hydrogen. So moles of h cancel out and we have our grams as 0.7013 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 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 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. Oops. So we have 3.3487 grams of carbon. We have 0.7013 grams hydrogen. And we have 4.95 grams chlorine. I'm writing in this order because that's the order they're given to us within the question. Carbon, hydrogen and then chlorine. Alright. So, 1 mole of carbon weighs 12.01 grams of carbon. One mole of H is 1.008 grams H. And then, 1 mole of chlorine is 35.45 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 have at least 4 decimal places carbon. Hydrogen comes out to be 0.6958 moles H. And then chlorine comes out to be 0.1396 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 H. And then carbon, when we divide it by that, we get 1.97 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 C2H5Cl. 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 percentages of its elements. 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. Understanding these formulas is crucial for grasping concepts like molecular weight and stoichiometry in chemical reactions.

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, the molecular formula for benzene is C₆H₆, which simplifies to the same empirical formula CH. This occurs when the molecular formula cannot be reduced further into a simpler ratio.

Why is the empirical formula important in chemistry?

The empirical formula is important in chemistry because it provides the simplest ratio of elements in a compound, which is essential for understanding the composition and properties of the substance. It is used in stoichiometry to calculate reactants and products in chemical reactions and helps in determining molecular formulas when combined with molecular weight data. This foundational knowledge is crucial for various applications, including material science, pharmaceuticals, and chemical engineering.

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

To convert an empirical formula to a molecular formula, you need the compound's molar mass. Follow these steps: 1) Calculate the molar mass of the empirical formula. 2) Divide the compound's molar mass by the empirical formula's molar mass to find a multiplication factor. 3) Multiply the subscripts in the empirical formula by this factor to get the molecular formula. For example, if the empirical formula is CH₂O and the molar mass is 180 g/mol, the factor is 6 (180/30), giving the molecular formula C₆H₁₂O₆.