# Empirical Formula: Definition and Examples

How to read the element ratio in an empirical formula

The empirical formula of a compound is defined as the formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols.

Also Known As: The empirical formula is also known as the simplest formula because the subscripts are the smallest whole numbers that indicate the ratio of elements.

### Key Takeaways: Empirical Formula

• The empirical formula is also known as the simplest formula in chemistry.
• It gives the smallest whole number ratio of elements in a compound using subscripts following element symbols.
• In some cases, the empirical formula is the same as the molecular formula, which gives the actual number of atoms in a compound (e.g., H2O).
• Otherwise, the molecular formula is a multiple of the empirical formula (e.g., CH2O is the empirical formula for glucose, C6H12O6).

## Empirical Formula Examples

Glucose has a molecular formula of C6H12O6. It contains 2 moles of hydrogen for every mole of carbon and oxygen. The empirical formula for glucose is CH2O.

The molecular formula of ribose is C5H10O5, which can be reduced to the empirical formula CH2O.

## How to Determine Empirical Formula

1. Begin with the number of grams of each element, which you usually find in an experiment or have given in a problem.
2. To make the calculation easier, assume the total mass of a sample is 100 grams, so you can work with simple percentages. In other words, set the mass of each element equal to the percent. The total should be 100 percent.
3. 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.
4. Divide each mole value by the small number of moles you obtained from your calculation.
5. 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 formula to find the molecular formula if you know the molar mass of the compound. To do this, calculate the empirical formula mass and then divide the compound molar mass by the empirical formula mass. 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.

## Empirical Formula Example Calculation

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

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.

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

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

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

Next, divide each mole amount by the smallest number or 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:

CaO2H2

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

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