Determining Empirical Formulas — Calculation & Examples

How to Determine the Empirical Formula

What is the empirical formula?

• The empirical formula shows the whole number ratio of atoms in a molecule
• The empirical formula does NOT show the number of atoms in a molecule (in most cases)

Steps to determine the empirical formula:

1. Determine the number of moles of each atom in the molecule

2. Divide the each number of moles by the smallest number of moles calculated

3. Multiply by a factor to get a whole number for each element in the molecule

Practice Problem Walkthrough:

A scientist measures 3.7458g of a compound composed of hydrogen, carbon, and oxygen. There is 2.3450g of carbon in the compound, and 1.2440g of oxygen in the mixture. Determine the empirical formula of the compound.

1) Determine the grams of each element in the mixture

• We know the grams of carbon and oxygen, so we need to determine the number of grams of hydrogen

$3.7458-2.3450-1.2440=0.1568$ g Hydrogen

2) Use the molar mass of each element to determine the number of moles for each element in the compound

Carbon: $2.3450\cdot\frac{1}{12.011}=$ 0.195 mols carbon

Oxygen: $1.2440\cdot\frac{1}{15.999}=$ 0.0778 mols oxygen

Hydrogen: $0.1568\cdot\frac{1}{1.008}=$ 0.1556 mols hydrogen

3) Divide each calculated mole by the smallest mole

• In this case, oxygen has the smallest number of moles, 0.0778, compared to the other element, thus we must divide each number of moles by 0.0778

Carbon: $\frac{0.1952}{0.0778}=$ 2.5

Oxygen: $\frac{0.0778}{0.0778}=$ 1

Hydrogen: $\frac{0.1556}{0.0778}=$ 2

4) Multiply by a whole number factor to get each number to the nearest whole number. Sometimes you have to round up/down because it is impossible to get a perfect whole number!

• NOTE: you have to multiply every number by the same factor!

• We have 2.5, so we should multiply by a factor of 2 in order to get these to be whole numbers!

Carbon: $2.5\cdot2=$ 5 carbons

Oxygen: $1\cdot2=$ 2 oxygens

Hydrogen: $2\cdot2=$ 4 hydrogens

5) Write your new finished empirical formula!

$C_{5}H_{4}O_{2}$

You can also determine the empirical formula from mass percentages

• The steps are the same, you just need to first convert the percentages of each element, to moles of the element

• This is done by simply converting the percentage to grams by assuming the problem is out of 100 grams of the compound, thus the percentages are the same number as the grams of the element! (ex: 34.6% Oxygen = 34.6 g oxygen)

Now let's try a practice problem!

Solution to the problem above:

(Video) Empirical Formula & Molecular Formula Determination From Percent Composition

by The Organic Chemistry Tutor

In this video, the Organic Chemistry Tutor shows you how to find the empirical formula from the mass percent. He walks you through two example problems of organic molecules. You start by converting the mass percent to mass. Next, you can convert each element's mass to moles. The lowest whole number ratio tells you the subscripts for the empirical formula. The compound's molar mass allows you to convert the empirical formula to a molecular formula. If you're looking for help with empirical formula problems, this video is for you!

Empirical Formula Refresher

In a previous explanation, we examined empirical formulas. They're the simplest whole-number ratio of atoms within a molecule. The empirical formula is less informative than the molecular formula. But it's easy to obtain from experimental data. In fact, the word "empirical" means based on experiments.

Sometimes learning chemistry is so abstract. Generally, you're handed a piece of paper with test questions. Your teacher taught you the formulas and methods. So, you use them to answer the problems. But at its heart, chemistry is an experimental science. One common area of specialization is analytical chemistry. Analytical chemists use various techniques to determine a compound's composition. Many of these techniques give you a mass percentage composition. We can use it to determine the empirical formula!

Steps for Determining the Empirical Formula

• If given the mass percentage composition, assume you have a 100.00-gram sample. In this case, the percentages are equal to the number of grams.

• If you are told the mass of each component, move to the second step.

• Use the atomic masses to calculate the moles of each atom. You can find the atomic masses on the periodic table.

• Divide the moles of each atom by the smallest mole value from step two. By doing this, you calculate the relative mole ratio.

• Use the relative mole ratio to write the empirical formula.

• Sometimes, the relative mole ratio has a fractional component. Multiply each part of the ratio by the lowest common factor. The multiplication will give you whole numbers.

• If given the compound's molar mass or molecular mass, use the empirical formula to determine the molecular formula. Divide the compound's molar mass by the empirical formula's mass. This gives you the multiplication factor. Multiply the empirical mole ratio by the multiplication factor.

Common Test Question: Mass Percent Composition to Empirical Formula

An organic compound containing carbon, hydrogen, and oxygen has the following mass percentages: 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen. The compound's molar mass is 180.16 grams. Find the empirical and molecular formulas.

Step 1: Assume a 100.00-gram sample. This gives us the masses for each element present.

Carbon=40.00 g

Hydrogen = 6.67 g

Oxygen = 53.33 g

Step 2: Convert the mass of each element to moles.

Carbon: $40.00 g \times \frac{1mol}{12.01g}= 3.33 mols$

Hydrogen: $6.67g\times \frac{1mol}{1.01g}= 6.60 mols$

Oxygen: $53.33g\times \frac{1mol}{16.00g}=3.33 mols$

Step 3: Divide the number of moles by the smallest number to find the mole ratio.

Carbon: $\frac{3.33 mols}{3.33}=1$

Hydrogen: $\frac{6.60}{3.33}=1.98\rightarrow 2.00$

Oxygen: $\frac{3.33 mols}{3.33}=1$

We round hydrogen to the nearest whole number ratio. Any discrepancies are caused by experimental uncertainty.

Step 4: Write the empirical formula.

$CH_{2}O$

Step 5: Divide the mass of the empirical formula by the compound's molar mass.

$CH_{2}O: 12.01+2\times 1.01+ 16.00=30.03grams$

$\frac{180.16}{40.03}=5.99\rightarrow 6$

Step 6: Write the molecular formula.

$C_{6}H_{12}O_{6}$

Our compound is glucose!

Combustion Analysis

Earlier, I explained that chemistry is an experimental science. The empirical formula gets derived from experiments. But what experiments actually give us empirical formulas? One of the most common techniques is combustion analysis. Let's examine it in more detail!

As the name implies, combustion analysis takes advantage of combustion reactions. It's primarily used for hydrocarbons. Hydrocarbons are organic molecules that are predominately carbon and hydrogen. The hydrocarbon gets oxidized by oxygen gas. The products are carbon dioxide and water vapor.

So, how does the instrument work? In a combustion analyzer, a hydrocarbon gets placed in a compartment. The oxygen source is typically copper(II) oxide. A furnace begins to heat the container. At a very high temperature, the hydrocarbon combusts. Both products are gases. So, they travel into a set of secondary chambers. The first partition grabs the water vapor. We commonly use an ionic compound. Why? They absorb water to form hydrates. The standard material is calcium chloride. We can weigh the calcium chloride before and after the reaction. The difference tells us the water vapor's mass. The carbon dioxide passes through into another chamber. There, a strong base captures the carbon dioxide. Again, the mass differences tell us the amount of carbon absorbed.

Now, let's look at an example problem.

Let's say we have an organic compound composed of hydrogen, oxygen, and carbon. The mass of the combusted sample was 0.114 grams. The combustion analysis produced 0.22 grams of carbon dioxide and 0.12 grams of water. Determine the empirical formula.

Step 1: Calculate the grams of carbon and grams of hydrogen. Use dimensional analysis.

$g_{C}=.22g_{CO_{2}}\times\frac{1mol_{CO_{2}}}{44.0g_{CO_{2}}}\times\frac{1 mol_{C}}{1 mol_{CO_{2}}}\times \frac{12.0g_{C}}{1mol_{C}}=0.060g_{C}$

$g_{H}=.12g_{H_{2}O}\times\frac{1mol_{H_{2}O}}{18.0g_{H_{2}O}}\times\frac{2 mol_{H}}{1 mol_{H_{2}O}}\times \frac{1.01g_{H}}{1mol_{H}}=0.013g_{H}$

Step 2: Calculate the mass of oxygen. Subtract the sample's mass by that of carbon and hydrogen.

$g_{O_{2}}=0.114g-0.060g-0.015g=0.039g_{O_{2}}$

Step 3: Convert the mass of each element to the moles.

$mol_{C}=0.060g\times \frac{1mol}{12.0g}=.005mols$

$mol_{H}=0.015g\times \frac{1mol}{1.01g}=.0148mols$

$mol_{O}=0.039g\times \frac{1mol}{16.0g}=.00243$

Step 4: Divide the relative number of moles by the smallest value.

$C: \frac{.005mols}{.00243mols}=2.05 \rightarrow 2$

$H: \frac{.0148mols}{.00243mols}=6.09 \rightarrow 6$

$O: \frac{.00243mols}{.00243mols}=1$

Step 5: Write the empirical formula.

$C_{2}H_{6}O$

When I designed the problem, I was thinking of ethanol. So, it looks like we did it correctly. Note that there are slight differences in the numbers. We had to round to the nearest whole number. In actual practice, there is always experimental error and uncertainty. So, you seldom get perfect numbers.