Lesson Video: Empirical and Molecular Formulas | Nagwa Lesson Video: Empirical and Molecular Formulas | Nagwa

Lesson Video: Empirical and Molecular Formulas Chemistry • First Year of Secondary School

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In this video, we will learn about two ways to express the chemical formula of a compound: the empirical formula and the molecular formula. And we will learn how to define, determine, and convert between a compound’s empirical and molecular formulas.

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Video Transcript

In this video, we will learn about two ways to express the chemical formula of a compound, the empirical formula and the molecular formula. And we’ll learn how to define, determine, and convert between a compound’s empirical and molecular formulas.

Have you ever wondered how scientists know the chemical formulas of all the substances we encounter in chemistry? How did we ever figure out the difference between carbon monoxide gas and carbon dioxide gas? Both colorless, odorless gases containing carbon and oxygen, but one contains one carbon atom and one oxygen atom, and the other contains one carbon atom and two oxygen atoms. To determine the composition of a substance, scientists use a series of techniques that are collectively known as elemental analysis.

Elemental analysis was originally invented in the late 1700s by the French scientist Antoine Lavoisier to assess the chemical composition of a compound. Though we now know the chemical formula for hundreds of different substances, elemental analysis is still used all the time to determine the chemical formula of newly discovered compounds in chemistry research, such as determining the composition of newly discovered proteins in the human body. There are many different ways to perform elemental analysis. In some techniques, the sample is burned; in other techniques, the sample is exposed to X-ray radiation. And there are many other techniques that are commonly used in chemistry.

But no matter the technique, we can use elemental analysis to determine the masses or the mass percentages of each element that’s in a compound. For example, let’s take a look at this molecule. It’s composed of one carbon atom and four hydrogen atoms, and it’s called methane. We can use elemental analysis on methane by burning a sample of it in the presence of oxygen, which would produce carbon dioxide and water vapor. We could then use special chambers to trap the carbon dioxide and water vapor that was produced as a result of this reaction. We could also have other chambers to trap other elements if our sample contained more than just carbon and hydrogen.

Since all of the carbon in the carbon dioxide came from methane, as well as all the hydrogen in the water, we can determine the amount of carbon and hydrogen that was originally in our methane sample. If we actually did elemental analysis on methane, we determine that it was 74.9 percent carbon and 25.1 percent hydrogen by mass. So, how can we use this experimental results to come up with the chemical formula of methane? We’re essentially aiming to come up with a ratio of the amount of carbon to the amount of hydrogen that’s in the methane sample.

Let’s say we had 100 grams of methane to start off with. 74.9 percent of 100 grams is 74.9 grams. So, we’d have that much carbon and we’d have 25.1 grams of hydrogen in our methane sample. We can’t use these masses to create a ratio, since there are a different amount of hydrogen and carbon atoms in one gram of each element. So, we’re going to convert these masses into moles. We can do this by dividing each mass by the molar mass of the element. The molar mass of carbon is 12.01 grams per mole, and the molar mass of hydrogen is 1.008 grams per mole. If we perform that calculation, we’ll find that we have 6.2365 moles of carbon and 24.9001 moles of hydrogen in our methane sample.

Now that we have our amounts, we now know the ratio of carbon to hydrogen in methane, which we can use to create a formula. But this formula is kind of gross to work with. So, what if we reduce it to the simplest whole number ratios? We can do this by dividing the smallest number of moles into each value that we obtained. 6.2365 divided by itself is one and 24.9001 divided by 6.2365 gives us 3.99, which we can round up to four. This gives us much nicer looking numbers.

And now, we can express the ratio of carbon to hydrogen and methane using simple whole numbers. And we can use that to create a formula for methane, CH4, which conveniently matches the formula for methane that we started off with. This chemical formula that we’ve ended up with that expresses the simplest whole number ratio of atoms in a compound is called the empirical formula because it was empirically determined from experimental data.

With this example, our simple ratio of atoms in the molecule conveniently also expressed the actual number of atoms of each element in the compound. We have one carbon atom and four hydrogen atoms, and that’s what our formula reflects. But the results of elemental analysis aren’t always so convenient. For example, there are several compounds made of sulfur and oxygen. Here are two of them. One has one sulfur atom and one oxygen atom, and the other has two sulfur atoms and two oxygen atoms.

Since, again, the empirical formula expresses the simplest whole number ratio of atoms, the top compound would have one sulfur atom for every one oxygen atom, meaning that its empirical formula is SO. The molecule in the bottom has two sulfur atoms for every two oxygen atoms. But since we’re aiming for the simplest whole number ratio of atoms, we can reduce this ratio further by dividing it by two, which would give us one to one, which would again give us the empirical formula SO. So both of them would have the same empirical formula SO, even though they’re different molecules with different amounts of sulfur and oxygen atoms in each molecule.

So it seems that the empirical formula, though useful, is not always going to give us enough information to tell us the precise atomic makeup of a substance. To address this, we have a second kind of chemical formula, the molecular formula. While the empirical formula just tells us the ratio of elements in a compound, the molecular formula tells us the amount of each type of atom in a molecule. So while both of these compounds have the same empirical formula, SO, they will have different molecular formulas that reflect the actual amount of atoms in each molecule.

This molecule on the top has one sulfur atom and one oxygen atom. So its molecular formula is SO. And this one on the bottom has two of each atom in a molecule, so its molecular formula is S2O2. So now, there’s one last thing to address here: how to convert between an empirical formula and a molecular formula. That is, how do we obtain the molecular formula once we’ve determined the empirical formula of a compound through elemental analysis? To address this, let’s take a look at the molecule hydrogen peroxide.

As we can see, hydrogen peroxide has two hydrogen atoms and two oxygen atoms in a molecule. So, the ratio of hydrogen to oxygen here is two to two, which we can reduce further by dividing by two to give us the ratio of one to one. Which means that the empirical formula of hydrogen peroxide is HO. As we can see from looking at the molecule, it has two hydrogen atoms and two oxygen atoms. So, its molecular formula must be H2O2. But what if we didn’t know what a molecule of hydrogen peroxide look like and we had determine the empirical formula of hydrogen peroxide from the results of elemental analysis like we did with methane at the start of this video? Well, if we know the molar mass of the compound, which we could determine through other experimental techniques like using a mass spectrometer, we’ll be able to work out the molecular formula from the empirical formula.

The molar mass of hydrogen peroxide is 34.0147 grams per mole. The next thing we’ll need is the mass of the empirical formula. Hydrogen has a mass of 1.008 grams per mole and oxygen has a mass of 15.99 grams per mole. Adding those together gives us the mass of 16.998 grams per mole for the mass of the empirical formula. Now, we just need to divide the molar mass of hydrogen peroxide by the mass of its empirical formula. 34.0147 grams per mole divided by 16.998 grams per mole is equal to two, which means that the mass of the molecular formula is twice the mass of the empirical formula.

So, we should multiply the empirical formula by two to give us the molecular formula, giving us H2O2 which is the same as the molecular formula that we got from looking at the cartoon of the molecule. One thing that we’ll notice from this example is that the molecular formula is always a whole number multiple of the empirical formula.

So now that we know what empirical formulas and molecular formulas are and how to obtain both of them from experimental evidence, let’s take a look at some example problems.

In a chemical compound containing carbon, oxygen, and hydrogen, there are equal numbers of carbon and hydrogen atoms. However, the number of oxygen atoms is twice the number of hydrogen atoms. What is the empirical formula of the compound?

The empirical formula of a compound tells us the simplest whole number ratio of atoms that are in the compound. Our compound contains carbon, oxygen, and hydrogen. So, we need to figure out the ratio of carbon to oxygen to hydrogen so we can determine the empirical formula of this compound. We’re told that there are an equal number of carbon and hydrogen atoms, which means that the ratio of carbon to hydrogen in our compound is one to one. Then the problem says that the number of oxygen atoms is twice the number of hydrogen atoms. In other words, the ratio of oxygen to hydrogen in the compound is two to one.

If we put both of these pieces of information together, that means that the ratio of carbon to oxygen to hydrogen that’s in this compound is one to two to one. We can’t reduce this ratio any further, which means that it’s already expressed in the simplest whole numbers. So, we can use this ratio to create the empirical formula for the compound, CO2H. Because this formula just tells us the ratio of carbon to oxygen to hydrogen in the compound, that doesn’t necessarily mean that there’s one carbon atom, two oxygen atoms, and one hydrogen atom in the compound.

The type of chemical formula that does tell us that information is the molecular formula, which tells us the amount of each type of atom in the compound, not just the simplest whole number ratio of atoms. But we were just tasked with finding the empirical formula of this compound, which is CO2H.

Now, let’s look at a problem that’s a little bit harder.

Oxalic acid is an organic compound found in many plants, including rhubarb. It is found to contain 26.7 percent carbon and 71.1 percent oxygen, with the rest being hydrogen. If the molar mass of oxalic acid is 90 grams per mole, what is its molecular formula?

The molecular formula tells us the amount of each type of atom that’s in a molecule. This problem gives us the percentage of carbon, oxygen, and hydrogen in the compound, which we’ll be able to use to create a ratio of carbon to oxygen to hydrogen. If we express this in the simplest whole number ratio of atoms, we would have the compound’s empirical formula. And then, we could use the molar mass of the compound to determine the molecular formula from the empirical formula.

So to solve this problem, we’ll first find the ratio of carbon to hydrogen to oxygen that’s in the compound, which we’ll use to create the empirical formula. And then, finally, we’ll find the molecular formula. The problem tells us that our sample is 26.7 percent carbon and 71.1 percent oxygen. Since the rest of the compound is hydrogen, we can find the percentage of hydrogen in our compound by subtracting the percentage of carbon and oxygen from 100 percent, which gives us 2.2. So, our sample is 2.2 percent hydrogen.

If we assume that our sample is 100 grams, we can easily go from these mass percentages into the masses of each element. 26.7 percent carbon corresponds to 26.7 grams of carbon, since 26.7 percent of 100 is 26.7. Now, we want the ratio of carbon to hydrogen to oxygen, which we can’t find using these masses, since there’s a different amount of carbon atoms and hydrogen atoms in one gram of each element. So, we’re going to need to convert these masses into moles, which we can do by dividing each path by the molar mass of the element.

The molar mass of carbon is 12.01 grams per mole. 26.7 divided by 12.01 is 2.223. The molar mass of hydrogen is 1.008 and 2.2 divided by 1.008 gives us 2.183 moles of hydrogen. Similarly, the molar mass of oxygen is 15.99. So, we have 4.447 moles of oxygen in our sample. Now that we have the amount of carbon, oxygen, and hydrogen in our sample in moles, we need to figure out how to express this in the simplest whole number ratio.

The easiest way to do this is by dividing by the number that’s the smallest amount of moles, which is the moles of hydrogen, 2.183. Dividing this number into the moles of carbon gives us 1.02 which is very close to one, but not quite likely due to some experimental error associated with finding these percentages. 2.183 divided by 2.183 gives us one. And finally, 4.447 divided by 2.183 gives us 2.04 which is slightly above two, again likely due to some experimental error. But we can still safely round it down to two.

So, the ratio of carbon to hydrogen to oxygen in our sample is one to one to two. Now that we have this ratio, we can find the empirical formula. Since the empirical formula expresses the simplest whole number ratio of atoms in a compound, the empirical formula is going to be CHO2.

Now, we can move on to finding the molecular formula. We can do this by taking the molar mass and dividing it by the empirical formula mass. The molar mass of the compound is given in the problem. It’s 90 grams per mole, so we just need to calculate the mass of this empirical formula. So, we can calculate the mass of the empirical formula by summing the molar masses of each atom in the compound, which gives us a total of 44.998 grams per mole. So now dividing 90 by 44.998 gives us two, which means that the molecular formula has twice the mass of the empirical formula. So, we can find the molecular formula by multiplying the empirical formula by two, giving us C2H2O4.

So now that we’ve learned about empirical and molecular formulas and worked some problems, let’s take a look at the key points for this lesson. The empirical formula expresses the simplest whole number ratio of atoms in a compound, while the molecular formula expresses the amount of each type of atom that’s in a molecule. If we know the molar mass of a compound, we can convert from the empirical formula to the molecular formula.

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