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# Empirical and Molecular Formula

We have talked a lot about molecules. You may have seen drawings of the structural formula of a molecule, like the one for benzene below. Fig. 1 - There are a few ways to draw the structural formula of benzeneThere are two more ways we can represent molecules: the empirical formula and the molecular formula.We will discuss what we mean by…

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# Empirical and Molecular Formula

Empirical and Molecular Formula
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Nie wieder prokastinieren mit unseren Lernerinnerungen. We have talked a lot about molecules. You may have seen drawings of the structural formula of a molecule, like the one for benzene below. Fig. 1 - There are a few ways to draw the structural formula of benzene

There are two more ways we can represent molecules: the empirical formula and the molecular formula.

• We will discuss what we mean by empirical and molecular formulae.
• You will learn two ways to find the empirical formula: by using relative atomic mass and by using the percent composition.
• You will also learn how to find the molecular formula by using the relative formula mass.

## What are the empirical and molecular formulae?

The molecular formula shows the actual number of atoms of each element in a molecule.

The empirical formula shows the simplest whole-number molar ratio of each element in a compound.

## How to write the empirical and molecular formula

Have a look at the table below.

 Molecular Empirical Benzene $$C_6H_6$$ $$CH$$ Water $$H_2O$$ \begin {align} H_2O \end {align} Sulfur $$S_8$$ $$S$$ Glucose $$C_6H_{12}O_6$$ $$CH_2O$$

Did you notice that the empirical formula simplifies the molecular formula? The molecular formula represents how many of each atom is in a molecule. The empirical formula shows the ratio or proportion of each atom in a molecule.

For example, we can see from the table that benzene has the molecular formula $$C_6H_6$$. That means that for every one carbon atom in benzene, there is one hydrogen atom. So we write the empirical formula of benzene as $$CH$$

As another example, let's look at phosphorus oxide $$P_4O_{10}$$

Find the empirical formula of phosphorus oxide.

Empirical formula of phosphorus oxide = $$P_2O_5$$

For every two phosphorus atoms, there are five oxygen atoms.

Here's a Tip:

You can discover the empirical formula by counting the number of each atom in a compound and dividing it by the lowest number.

In the phosphorus oxide example ( $$P_4O_{10}$$ ) the lowest number is 4.

4 ÷ 4 = 1

10 ÷ 4 = 2.5

Since the empirical formula must be a whole number, you must choose a factor to multiply them by that will give a whole number.

1 x 2 = 2

2.5 x 2 = 5

$$P_4O_{10}$$ → $$P_2O_5$$

Sometimes the molecular and empirical formulas are identical, like in the case of water ( $$H_2O$$ ). You can also get the same empirical formula from different molecular formulas.

## How to find the empirical formula

When scientists discover new materials, they want to know their molecular and empirical formulae too! You can find the empirical formula by using the relative mass and the percent composition of each element in the compound.

### Empirical formula from relative mass

Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of oxygen.

Find the atomic mass of oxygen and hydrogen

O = 16

H = 1

Divide the masses of each element by their atomic mass to find the number of moles.

80g ÷ 16g = 5 mol. of oxygen

10g ÷ 1g = 10 mol. of hydrogen

Divide the number of moles by the lowest figure to get the ratio.

5 ÷ 5 = 1

10 ÷ 5 = 2

Empirical formula = $$H_2O$$

0.273g of Mg is heated in a Nitrogen ($$N_2$$) environment. The product of the reaction has a mass of 0.378g . Calculate the empirical formula.

Find the mass percentage of the elements in the compound.

N = 0.3789 - 0.273g = 0.105g

N = (0.105 ÷ 0.378) x 100 = 27.77%

Mg = (0.273 ÷ 0.378) x 100 = 77.23%

Change the percent composition to grams.

27.77% → 27.77g

77.23% → 77.23g

Divide the percent compositions by their atomic mass.

N = 14g

27.77g ÷ 14g = 1.98 mol

Mg = 24.31g

77.23g ÷ 24.31g = 2.97 mol

Divide the number of moles by the smallest number.

1.98 ÷ 1.98 = 1

2.97 ÷ 1.98 = 1.5

Remember we need whole number ratios, choose a factor to multiply that will give a whole number.

1 x 2 = 2

1.5 x 2 = 3

Empirical formula = $$Mg_3N_2$$ [Magnesium Nitride]

### Empirical formula from percent composition

Determine the empirical formula of a compound that contains 85.7% carbon and 14.3% hydrogen.

% mass C = 85.7

% mass H = 14.3

Divide the percentages by the atomic mass.

C = 12

H = 1

85.7 ÷ 12 = 7.142 mol

14.3 ÷ 1 = 14.3 mol

Divide by the lowest number.

7.142 ÷ 7.142 = 1

14.3 ÷ 7.142 = 2

Empirical formula = $$CH_2$$

## How to find the molecular formula

You can convert the empirical formula to the molecular formula if you know the relative formula mass or the molar mass.

### Molecular formula from relative formula mass

A substance has the empirical formula $$C_4H_{10}S$$ and relative formula mass (Mr) of 180. What is its molecular formula?

Find the relative formula mass (Mr) of $$C_4H_{10}S$$ (the empirical formula).

Ar of C = 12

Ar of H = 1

Ar of S = 32

Mr = (12 x 4) + (10 x 1) + 32 = 90

Divide the Mr of the molecular formula by the Mr of the empirical formula.

180 ÷ 90 = 2

The ratio between the Mr of the substance and the empirical formula is 2.

Multiply each number of elements by two.

(C4 x 2 H10 x 2 S1 x2)

Molecular formula = $$C_8H_{10}S_2$$

A substance has the empirical formula $$C_2H_6O$$ and a molar mass of 46g.

Find the mass of one mole of the empirical formula.

(Carbon 12 x 2) + (Hydrogen 1 x 2) + (Oxygen 16) = 46g

The molar mass of the empirical formula and the molecular formula are the same. The molecular formula must be the same as the empirical formula.

Molecular formula = $$C_2H_6O$$

## Empirical and Molecular Formula - Key takeaways

• The molecular formula shows the actual number of atoms of each element in a molecule.
• The empirical formula shows the simplest whole number molar ratio of each element in a compound.
• You can find the empirical formula by using the relative atomic mass and the mass percentage of each element.
• You can find the molecular formula by using relative formula mass.

The empirical formula shows the simplest whole-number molar ratio of each element in a compound.

An example of an empirical formula would be benzene (C6H6). A benzene molecule has six carbon atoms and six hydrogen atoms. This means the ratio of the atoms in a benzene molecule is one carbon to one hydrogen. So the empirical formula of benzene is simply CH.

The empirical formula shows the ratio of the atoms in a molecule. The molecular formula shows the actual number of atoms of each element in a molecule. Sometimes the empirical and molecular formulas are identical because the ratio of atoms cannot be simplified further.

Take a look at water as an example. Water has the molecular formula . This means in every molecule of water there are two hydrogen atoms for every one oxygen atom. This ratio cannot be made any simpler so the empirical formula for water is also . You can also get the same empirical formula from different molecular formulas.

## Empirical and Molecular Formula Quiz - Teste dein Wissen

Question

What is the empirical formula?

The empirical formula shows the simplest whole-number molar ratio of each element in a compound.

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Question

What is molecular formula?

The molecular formula shows the actual number of atoms of each element in a molecule.

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Question

Why are the molecular and empirical formulas sometimes identical?

Sometimes the empirical and molecular formulas are identical because the ratio of atoms in a species cannot be simplified further.

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Question

Which of these do you need to find the empirical formula of a substance?

Relative atomic mass

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Question

Which of these do you need to find the molecular formula of a substance?

Relative formula mass

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Question

A hydrocarbon has a mass that is 83.7 percent carbon and a relative atomic mass of 86.

Find the empirical formula and then the molecular formula.

100 - 83.7 = 16.3

83.7 ÷ 12 = 6.975

16.3 ÷ 1 = 16.3

6.975 ÷ 6.975 = 1

16.3 ÷ 6.975 = 2.337

*find ratio to the nearest whole number

2.337 x 3 = 7.011

Empirical formula = C3H7

Arof empirical formula

(C 12 x 3) + (H 1 x 7) = 43

86 ÷ 43 = 2

Molecular formula:

C(3x2)H(7x2)

C6H14

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