An empirical formula is the lowest ratio of the atoms within a molecule. The empirical formula accurately describes ionic compounds, which cannot be broken into a single molecule unit.

But when describing covalent compounds, we use a molecular formula that describes the atoms within a single molecule. And the ratio of atoms within a molecular formula is the same as that in the empirical formula, but it is not reduced. So molecules with the same empirical formulas have the same percent composition.

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## Definition of Empirical Formula

The empirical formula of a chemical compound represents the simplest whole-number ratio between the elements comprising the compound.

In chemistry, the empirical formulas of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

A simple example of this concept is that the empirical formula of sulfur monoxide, or SO, would be SO, as is the empirical formula of disulfur dioxide, S_{2}O_{2}.

Thus, sulfur monoxide and disulfur dioxide, both Sulfur and oxygen compounds, and have the same empirical formula.

However, their molecular formulas, which express the number of atoms in each molecule of a chemical compound, are not the same.

## Other Definitions of Empirical Formula

Here are some more definitions of empirical formula:

- An empirical formula makes no mention of the arrangement or number of atoms. It is standard for many ionic compounds, like calcium chloride (CaCl2), and for macromolecules, such as silicon dioxide (SiO
_{2}). - On the other hand the molecular formula, shows the number of each type of atom in a molecule. Because the structural formula shows the arrangement of the molecule. It is also possible for different types of compounds to have equal empirical formulas. Samples are analyzed in specific elemental analysis tests to determine what percent of a particular element the sample is composed of.
- The empirical formula is the simplest formula for a compound, defined as the ratio of subscripts of the smallest possible whole number of the elements present in the formula. And it is also known as the simplest formula.
- An empirical formula for a compound is the formula of a substance written with the smallest integer subscript.
- The empirical formula gives information about the ratio of numbers of atoms in the compound. The percent composition of a compound directly leads to its empirical formula.

## How to Determine Empirical Formula ?

To determine the formula we have to follow some steps :

- Begin with the number of grams of each element, which you usually find in an experiment or have given in a problem.
- To make the calculation easier, assume the total mass of a sample is 100 grams, so you can work with simple percentages. And in other words, set the mass of each element equal to the percent. The total should be 100 percent.
- Use the molar mass you get by adding up the atomic weight of the elements from the periodic table to convert the mass of each element into moles.
- Divide each mole value by the small number of moles you obtained from your calculation.
- Round each number, you get to the nearest whole number. The whole numbers are the mole ratio of elements in the compound. Which are the subscript numbers that follow the element symbol in the chemical formula.

Sometimes determining the whole number ratio is tricky. And you’ll need to use trial and error to get the correct value. For values close to x.5, you’ll multiply each value by the same factor to obtain the smallest whole number multiple.

**For example**, if you get 1.5 for a solution, multiply each number in the problem by 2 to make the 1.5 into 3. If you get a value of 1.25, multiply each value by 4 to turn the 1.25 into 5.

## Using Empirical Formula to Find Molecular Formula

You can use the empirical formulas to find the molecular formula if you know the molar mass of the compound.

To do this, calculate the empirical formulas mass and then divide the compound molar mass by the empirical formula mass. And this gives you the ratio between the molecular and empirical formulas. Multiply all of the subscripts in the empirical formula by this ratio to get the subscripts for the molecular formula.

## Calculating Empirical Formula with Examples

Here are some solved examples to know how to calculate the empirical formula:

### Example-1

A compound is analyzed and calculated to consist of 13.5g Ca, 10.8g O, and 0.675g H. Find the compound’s empirical formula.

Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H.

13.5g Ca x (1 mol Ca / 40.1g Ca) = 0.337 mol Ca

10.8g O x (1 mol O / 16.0g O) = 0.675 mol O

0.675g H x (1 mol H / 1.01g H) = 0.668 mol H

Next, divide each mole amount by the smallest number of moles (which is 0.337 for calcium) and round to the nearest whole number:

0.337 mol Ca / 0.337 = 1.00 mol Ca

0.675 mol O / 0.337 = 2.00 mol O

0.668 mol H / 0.337 = 1.98 mol H which rounds up to 2.00

Now you have the subscripts for the atoms in the empirical formula:

CaO_{2} H_{2}

Finally, apply the rules of writing formulas to present the formula correctly. And the cation of the compound is written first, followed by the anion. The empirical formula is properly written as Ca(OH)^{2}.

#### Summary:

Empirical formulas represent the simplest notational form. They provide the lowest ratio of whole numbers between the compound elements. They will not have information on the total number of atoms in a single compound molecule instead of molecular formulae.

Keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance.

Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles.

These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance.

Consider a sample of a compound determined to contain 1.71g C and 0.287g H.

The corresponding numbers of atoms (in moles) are:

1.71g C×1mol C/12.01g C=0.142mol C

0.287g H×1mol H/1.008g H=0.284mol H

Thus, this compound may be represented by the formula C_{0.142} H_{0.284}. Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

C _{0.142/0.142} | H _{0.284/0.142} or CH_{2} |

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

So the empirical formula for this compound is CH_{2}. This may or not be the compound’s molecular formula as well. However, additional information is needed to make that determination (as discussed later in this section).

### Example-2

Consider as another example a sample of compound determined to contain 5.31g Cl and 8.40g O. Following the same approach yields a tentative empirical formula of:

Cl_{o.150} O_{0.525}

= Cl_{0.150/0.150} O_{0.525/0.150}

= Cl_{O3.5}

and in this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula.

To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl_{2}O_{7} as the final empirical formula.

#### In summary, empirical formulas are derived from experimentally measured element masses by:

- Deriving the number of moles of each element from its mass.
- Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formulas.
- Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained.

### Example-3

A compound contains 88.79% oxygen (O) and 11.19% hydrogen (H). Compute the empirical formula of the compound

**Solution :**

- Assume 100.0g of substance. We see that the percentage of each element matches the grams of each element

11.19g H

88.79g O

2. Convert grams of each element to moles

H: (11.19/1.008) = 11.10 mol H atoms [molar mass of H=1.008g/mol]

O: (88.79/16.00) = 5.549 mol O atoms [molar mass of O= 16.00g/mol]

The formula could be articulated as H11.10O5.549. However, it’s usual to use the smallest whole number ratio of atoms.

3. By dividing the lowest number alter the numbers to whole numbers

H =11.10/ 5.549 = 2.000

O = 5.549/ 5.549= 1.000

The simplest ratio of H to O is 2:1

Empirical formula = H_{2}O

### Example-4

A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?

**Solution:**

- As the mass of each element is known, we use them directly

2. Convert grams of each element to moles

Fe: (2.233g /55.85g) = 0.03998 mol Fe atoms [molar mass of Fe =1.008g/mol]

S: (1.926 /32.07) = 0.06006 mol S atoms [molar mass of S =32.07g/mol]

3. By dividing by the smallest number, change the numbers to whole numbers.

Fe =0.03998/0.03998 = 1.000

S = 0.06006/0.03998mol = 1.502

4. As we still have not reached a ratio that gives whole numbers in the formula so we multiply by a number that will give us whole numbers.

Fe: (1.000)2 = 2.000

S: (1.502)2 = 3.004

Empirical formula = Fe_{2}S_{3}.

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