## Finding Empirical Formula And Molecular Formula Of Compounds

**Empirical formula** is a formula that shows the simplest ratio in which elements are combined in a compound. For example, the empirical formula of a compound that has a chemical formula of C_{2}H_{6} is CH_{3} because carbon and hydrogen are combined in the ratio 1:3.

Empirical formula can be calculated using either of the following:

- the actual masses of the combined elements.
- the percentage composition by mass of the combined elements.
- experimental data from a chemical reaction.

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### Calculating empirical formula of a compound given the masses of combined elements

Given the masses of the combined elements in a compound, we can calculate the empirical formulae of the compound as follows:

- Change the given masses to moles of their atoms by dividing the masses with molar mass of the atoms.
- Find the simplest molar ratio of the combined atoms.
- Use the molar ratio to write down the empirical formula.

#### Example: 56 g of iron combine with 32 g of sulphur to form iron sulphide. Find the empirical formula for iron sulphide.

To make the calculation easier, you tabulate the data as follows:

Iron | Sulphur | |
---|---|---|

mass | 56 g | 32 g |

find the number of moles | 56/56 = 1 | 32/32 = 1 |

express the moles as a simplest ratio | 1 | 1 |

Since the ratio of Fe:S is 1:1, the empirical formula is FeS.

### Empirical formula of a compound given the percentage composition by mass of the combined elements.

Given the percentages by mass of the combined elements in a compound, we can calculate the empirical formulae of the compound as follows:

- Treat the percentages as masses
- Change the given masses to moles of their atoms by dividing the masses with molar mass of the atoms.
- Find the simplest molar ratio of the combined atoms.
- Use the molar ratio to write down the empirical formula.

#### Example: An oxide of sulphur contains 40% sulphur and 60% oxygen by mass. What is its empirical formula?

When given percentages we treat 100% as 100 g, which means the mass of sulphur becomes 40 g and the mass of oxygen 60 g.

Sulphur | Oxygen | |
---|---|---|

mass | 40 g | 60 g |

find the number of moles | 40/32 = 1.25 | 60/16 = 3.75 |

express the moles as a simplest ratio by diving with the smallest of the two number | 1.25/1.25 = 1 | 3.75/1.25 = 3 |

The simples ratio of sulphur : oxygen = 1:3

Therefore the empirical formula = SO_{3}.

### Calculating molecular formula, given empirical formula and M_{r}

When can easily convert empirical formula to molecular using the following formula:

n(empirical mass) = molecular mass, whereby n is the formula multiplier.

#### Determine the empirical formula of an organic hydrocarbon compound which contains 80% by mass of carbon and 20% by mass of hydrogen. If the M_{r} of the compound is 30, what is its molecular formula?

First we find the empirical formula using the given percentages:

carbon | hydrogen | |
---|---|---|

mass | 80 g | 20 g |

find the number of moles | 80/12 = 6.67 | 20/1 = 20 |

express the moles as a simplest ratio | 6.67/6.67 = 1 | 20/6.67 = 3 |

The simples ratio of carbon : hydrogen = 1:3

Therefore the empirical formula = CH_{3}.

To find the molecular formula we first calculate the empirical mass the ratios given in the empirical formula

Empirical mass = (12× 1) + (1 × 3)

= 12 + 3

= 15

Then we find the formula multiplier

$latex n = \frac{Molecular\ mass}{Empirical\ mass}$

$latex n = \frac{30}{15}$

n = 2

Then multiply all the elements in the empirical formula by 2.

2(CH_{3}) = C_{2}H_{6}

Therefore the molecular mass is C_{2}H_{6}

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