An empirical formula of a substance is found using the masses and relative atomic masses of the elements it contains. The law of conservation of mass applies to closed and non-enclosed systems.

The empirical formula of a substance is the simplest whole number ratio of the atoms of each element present.

The molecular formula of ethane is C_{2}H_{6}. It shows the actual number of atoms of each element in a molecule of ethane. This formula does not show the simplest whole number ratio because each number can be divided by two. This gives the empirical formula of ethane: CH_{3}.

The molecular formula and empirical formula of some substances are the same. For example, both types of formula for carbon dioxide are CO_{2}.

The formulae given for ionic compounds, giant molecules and metals are all empirical formulae. This is because the actual numbers of ions and atoms they contain is so huge.

The molecular formula for a substance can be worked out using:

- the empirical formula.
- the relative formula mass,
*M*_{r}

The empirical formula for a compound is CH_{2} and its relative formula mass is 42. Deduce its molecular formula. (*A*_{r} of C = 12, *A*_{r} of H = 1)

*M*_{r} of CH_{2} = 12 + (2 × 1) = 14

Factor to apply = 42 ÷ 14 = 3

Multiply the numbers in the empirical formula by the factor 3:

Molecular formula = C_{3}H_{6}

- Question
The empirical formula for a compound is C

_{2}H_{5}and its relative formula mass is 58. Deduce its molecular formula. (*A*_{r}of C = 12,*A*_{r}of H = 1)*M*_{r}of C_{2}H_{5}= (2 × 12) + (5 × 1) = 29Factor to apply = 58 ÷ 29 = 2

Multiply the numbers in the empirical formula by the factor 2:

Molecular formula = C

_{4}H_{10}

Information about reacting masses is used to calculate empirical formulae. This is obtained from experiments.

A hydrocarbon is found to contain 4.8 g of carbon and 1.0 g of hydrogen. Calculate its empirical formula. (*A*_{r} of C = 12, *A*_{r} of H = 1)

Step | Action | Result | Result |
---|---|---|---|

1 | Write the element symbols | C | H |

2 | Write the masses | 4.8 g | 1.0 g |

3 | Write the A_{r} values | 12 | 1 |

4 | Divide masses by A_{r} | 4.8 ÷ 12 = 0.4 | 1.0 ÷ 1 = 1 |

5 | Divide by the smallest number | 0.4 ÷ 0.4 = 1 | 1.0 ÷ 0.4 = 2.5 |

6 | Multiply (if needed) to get whole numbers | 1 × 2 = 2 | 2.5 × 2 = 5 |

7 | Write the formula | C_{2}H_{5} | C_{2}H_{5} |

The action at step 5 usually gives you the simplest whole number ratio straightaway. Sometimes it does not, so you may need to multiply both numbers to get a whole number (step 6). For example, by 2 if you have .5, by 3 if you have .33, or by 4 if you have .25 in a number.

- Question
3.2 g of sulfur reacts completely with oxygen to produce 6.4 g of an oxide of sulfur. Calculate the empirical formula of the oxide of sulfur. (

*A*_{r}of S = 32,*A*_{r}of O = 16)Step Action Result Result 1 Write the element symbols **S****O**2 Write the masses **3.2 g**6.4 g - 3.2 g = **3.2 g**3 Write the *A*_{r}values**32****16**4 Divide masses by *A*_{r}3.2 ÷ 32 = **0.1**3.2 ÷ 16 = **0.2**5 Divide by the smallest number 0.1 ÷ 0.1 = **1**0.2 ÷ 0.1 = **2**6 Multiply (if needed) to get whole numbers (not needed: already have whole numbers) (not needed: already have whole numbers) 7 Write the formula **SO**_{2}**SO**_{2}