Lesson Explainer: Empirical and Molecular Formulas Chemistry

In this explainer, we will learn how to define, determine, and convert between a compound’s empirical and molecular formulas.

The composition of a molecule can be expressed as a molecular formula. The molecular formula tells us the exact number of atoms of each element in a molecule. For example, the molecular formula of water, HO2, indicates that every water molecule will contain two hydrogen atoms and one oxygen atom. The molecular formula of glucose, CHO6126, indicates that every glucose molecule will contain six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.

Definition: Molecular Formula

A molecular formula is a chemical formula expressing the exact number of atoms of each element in a molecule.

The ratio of carbon atoms to hydrogen atoms to oxygen atoms in a molecule of glucose is 6126. This ratio can be simplified to 121. We can write the simplified ratio as a chemical formula, CHO2. This is the empirical formula of glucose. The empirical formula tells us the simplest whole number ratio of the elements in a molecule.

Definition: Empirical Formula

An empirical formula is a chemical formula that indicates the elements in a molecule or ionic compound and the simplest whole number ratio of atoms of each element.

Empirical formulas are also used to represent the chemical formulas of ionic compounds because ionic compounds exist as a three-dimensional network of ions rather than a discrete molecular unit. The sodium chloride crystal lattice, shown below, must have equal numbers of Na+ and Cl in order for the compound to be electrically neutral. Therefore, the simplified ratio of sodium to chlorine in sodium chloride is 11 and the empirical formula is NaCl.

We can easily determine the empirical formula from the molecular formula by dividing the subscripts in the molecular formula by their greatest common factor.

Example 1: Determining an Empirical Formula from a Molecular Formula

Caffeine has the chemical formula CHNO81042. What is its empirical formula?

Answer

The ratio of carbon atoms to hydrogen atoms to nitrogen atoms to oxygen atoms in the molecular formula is 81042. The greatest common factor of these values is 2. We can divide each value by the greatest common factor to simplify the ratio to 4521.

We can replace the subscripts in the molecular formula with the simplified ratio to give us the empirical formula of CHNO452.

It is possible for the empirical formula and molecular formula to be the same. For example, formaldehyde, a biological specimen preservative, has the molecular formula CHO2 and the empirical formula CHO2. It is also possible for multiple molecules to have the same empirical formula. Glucose and formaldehyde both have an empirical formula of CHO2.

Empirical formulas are often derived from experimental data. In fact, the word empirical means based on observation or experience. In order to determine the empirical formula, we will need to experimentally determine the percent composition or measure the mass of each element in a pure sample. In either case, we will need to calculate the number of moles of each element in order to determine the ratio.

Definition: Percent Composition

Percent composition is the percent by mass of each element in a compound.

Let us examine how to determine the empirical formula from the mass of each element in a sample. A sample is composed of 42.2 grams of carbon and 112.0 grams of oxygen. Dividing the mass of each element by its molar mass will give us the number of moles of each element in the sample: CggmolmolOggmolmol42.212/=3.517,112.016/=7.

This gives us a ratio of carbon to oxygen of 3.5177. However, an empirical formula represents the simplest whole number ratio. We can divide the number of moles of each element by the smallest number of moles to try to convert the ratio to whole numbers: CmolmolOmolmol3.5173.517=1,73.517=1.9902.

Decimal values that are close to whole numbers can be rounded to the nearest whole number. The sign means “approximately equal to.” Once the simplest whole number ratio has been determined, it can be written as a chemical formula. Therefore, the empirical formula of this sample is CO2.

In this example, we knew the mass of each element in the sample. If we knew the percent composition of each element, for example, 27.3% carbon and 72.7% oxygen, we would begin by assuming that the sample size is 100 grams. This would allow us to represent the amount of each element in grams27.3 grams of carbon and 72.7 grams of oxygen—so that we could convert it into moles and find the molar ratio.

The complete process for determining the empirical formula is shown below.

Example 2: Determining an Empirical Formula from Experimental Data

In an experiment, mercury oxide is heated in a container, producing oxygen gas and mercury in its pure elemental form. The results of the experiment are shown below. [Hg = 201 g/mol, O = 16 g/mol]

Mass of Empty ContainerMass of Empty Container + Mercury OxideMass of Container + Mercury
30.0 g73.4 g70.2 g

  1. What is the mass of oxygen gas lost?
  2. What is the mass of mercury produced?
  3. What is the empirical formula for the compound of mercury oxide?

Answer

Part 1

As the mercury oxide is heated, the oxygen is driven off leaving behind the pure mercury metal. The difference between the mass of the container with the mercury oxide and the mass of the container with the mercury is the mass of oxygen gas lost. Therefore, 73.470.2=3.2.gggofoxygengas

Part 2

We know the mass of the container with the mercury. To find the mass of the mercury alone, we simply subtract the mass of the empty container. Therefore, 70.230.0=40.2.gggofmercury

Part 3

To determine the empirical formula, we begin by dividing the total mass of each element by its molar mass: HgggmolmolOggmolmol40.2201/=0.2,3.216/=0.2.

This tells us that the ratio of mercury to oxygen is 0.20.2. Now, we can divide the number of moles of each element by the smallest number of moles: HgmolmolOmolmol0.20.2=1,0.20.2=1.

This tells us that the simplest whole number ratio of mercury to oxygen is 11. We can write this as the empirical formula HgO.

There will be occasions when solving for the empirical formula where a whole number ratio is not produced when dividing by the smallest number of moles. Let us consider the following mathematical steps: StepCggmolmolHggmolmolOggmolmolStepCmolmolHmolmolOmolmol140.9212/=3.414.581/=4.5854.5016/=3.4123.413.41=14.583.41=1.343.413.41=1

In step 1, the mass of carbon, hydrogen, and oxygen in a sample was divided by the molar mass of the element. In step 2, the number of moles of each element was divided by the smallest number of moles. This gave us a ratio of carbon to hydrogen to oxygen of 11.341. In cases like the above, we will need to multiply the ratio by the smallest integer that will produce a ratio with whole numbers.

We can multiply the entire ratio in the example by 3 to give a ratio of carbon to hydrogen to oxygen of 343. This gives us an empirical formula of CHO343.

Example 3: Determining an Empirical Formula from Percent Composition

A vanadium oxide compound is found to be 68% vanadium and 32% oxygen by mass. What is the empirical formula of this compound? Take the atomic masses of vanadium and oxygen to be 51 and 16, respectively.

Answer

First, we need to assume that the sample size is 100 grams. By making this assumption, we know that 68 grams of the sample is vanadium and 32 grams is oxygen.

Next, we can convert the mass of each element into moles by dividing by the molar mass: VggmolmolOggmolmol6851/=1.333,3216/=2.

Now, we can determine the ratio by dividing the number of moles of each element by the smallest number of moles: VmolmolOmolmol1.3331.333=1,21.333=1.5.

This step did not produce a ratio with whole numbers. We will need to multiply the ratio by 2, the smallest integer that will produce a whole number ratio: VO1×2=2,1.5×2=3.

We can write the whole number ratio as a chemical formula, giving us the empirical formula VO23.

The chemical formula determined from the percent composition will always be the empirical formula. In order to determine the molecular formula, the molar mass of the molecule must be known. As the molecular formula is a multiple of the empirical formula, the molar mass of the molecule will be a multiple of the molar mass of the empirical formula. Let us examine how to determine the molecular formula from the empirical formula:

Naphthalene is a compound containing carbon and hydrogen that is often used as a moth repellent. The molar mass of naphthalene is 128 g/mol and its empirical formula is CH54. What is the molecular formula of naphthalene?

We can follow these steps to determine the molecular formula from the molar mass and empirical formula. First, we will calculate the molar mass of the empirical formula: molarmassofempiricalformulamolarmassofcarbonmolarmassofhydrogenmolarmassofempiricalformulagmolgmolgmol=(5×)+(4×)=(5×12/)+(4×1/)=64/.

Then, we will divide the molar mass of the molecule by the molar mass of the empirical formula. The answer should be a whole number or very close to a whole number: molarmassofnaphthalenemolarmassofnaphthalenesempiricalformulagmolgmol=128/64/=2.

We can multiply the subscripts in the empirical formula by this integer to determine the molecular formula: CHCH54108×2=.

The molecular formula of naphthalene is CH108. The process for determining the molecular formula is shown below.

Example 4: Determining a Molecular Formula from an Empirical Formula

A compound with the empirical formula CHNO572 has a molar mass of 339 g/mol. What is the molecular formula of this compound? [H = 1 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol]

Answer

In order to determine the molecular formula, we will begin by calculating the molar mass of the empirical formula: empiricalformulamolarmassmolarmassofCmolarmassofHmolarmassofNmolarmassofOempiricalformulamolarmassgmolgmolgmolgmolgmol=(5×)+(7×)+()+(2×)=(5×12/)+(7×1/)+14/+(2×16/)=113/.

Next, we will divide the molar mass of the compound by the molar mass of the empirical formula to produce an integer: molarmassofcompoundmolarmassofcompoundsempiricalformulagmolgmol=339/113/=3.

This value indicates that the molecular formula is three times larger than the empirical formula. We can multiply the subscripts in the empirical formula by the integer to determine the molecular formula: CHNOCHNO572152136×3=.

The molecular formula of the compound is CHNO152136.

Example 5: Determining the Molecular Formula from the Percent Composition

Oxalic acid is an organic compound found in many plants, including rhubarb. It is found to contain 26.7% carbon and 71.1% oxygen, with the rest being hydrogen. If the molar mass of oxalic acid is 90 g/mol, what is its molecular formula? [H = 1 g/mol, C = 12 g/mol, O = 16 g/mol]

Answer

In order to determine the molecular formula, we need to know the molar mass of oxalic acid (90 g/mol) and the empirical formula. The empirical formula was not provided, but we can determine it from the percent composition data.

First, we will assume that the sample size is 100 grams. This means that the sample is composed of 26.7 grams of carbon, 71.1 grams of oxygen, and 2.2 grams of hydrogen. The hydrogen mass was calculated by subtracting the mass of carbon and oxygen from 100 grams.

Next, we will convert the mass of each element into moles by dividing by the molar mass: CggmolmolOggmolmolHggmolmol26.712/=2.225,71.116/=4.444,2.21/=2.2.

This tells us that the molar ratio of carbon to oxygen to hydrogen is 2.2254.4442.2. We can divide the number of moles of each element by the smallest number of moles to try to convert the ratio to whole numbers: CmolmolOmolmolHmolmol2.2252.2=1.0111,4.4442.2=2.022,2.22.2=1.

We substitute the ratio as subscripts in a chemical formula, giving us an empirical formula of CHO2.

Now that we know the molar mass of oxalic acid and its empirical formula, we can determine the molecular formula. We begin by calculating the molar mass of the empirical formula: empiricalformulamolarmassmolarmassofCmolarmassofHmolarmassofOgmolgmolgmolgmol=()+(2×)+()=12/+(2×1/)+16/=30/.

Then, we divide the molar mass of oxalic acid by the molar mass of the empirical formula: molarmassofoxalicacidmolarmassofoxalicacidsempiricalformulagmolgmol=90/30/=3.

This value indicates that the molecular formula of oxalic acid is three times larger than the empirical formula. We can multiply the subscripts in the empirical formula by three to determine the molecular formula: CHOCHO2363×3=.

The molecular formula of oxalic acid is CHO363.

Key Points

  • A molecular formula indicates the exact number of atoms of each element in a molecule.
  • An empirical formula indicates the simplest whole number ratio of atoms of each element in a molecule or ionic compound.
  • The empirical formula can be determined from the experimental data.
  • The percent composition by mass can be used to calculate the empirical formula.
  • The molecular formula can be determined from the empirical formula as long as the molar mass of the molecule is known.

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