Empirical and molecular formula

An empirical formula is the simplest ratio of whole-numbered atoms in a compound, which represents the relative proportions of the different elements in a molecule. It is determined based on the percentages or mass of each element present in a substance, and it does not provide information about the actual number of atoms or the molecular structure of the compound.

Example

A compound has the following percentage composition:

carbon: 15.87%, hydrogen: 2.22%, nitrogen: 18.50%, oxygen: 63.41%

Determine the empirical formula of the compound

Assume 100% = 100g

Use this gram equivalence to calculate the moles

C: 15.87 ÷ 12.01 = 1.321

H: 2.22 ÷ 1.01 = 2.198

N: 18.50 ÷ 14.01 = 1.320

O: 63.41 ÷ 16.0 = 3.963

Divide through by the smallest number

C: 1.321 ÷ 1.32 = 1

H: 2.198 ÷ 1.32 = 1.66

N: 1.32 ÷ 1.32 = 1

O: 3.963 ÷ 1.32 = 3

Many students make the mistake of rounding the 1.66 to 2. The best way is to turn it into an equivalent fraction and then multiply through by it’s denominator. 1.66 can be thought of as 5/3.

C: 1 x 3 = 3

H: 5/3 x 3 = 5

N: 1 x 3 = 3

O: 3 x 3 = 9

The empirical formula is...

Example

A compound contains 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. What is the empirical formula?

The compound's molecular weight is 102.2 g/mol. What is its molecular formula?

By assuming that there is 100g of the compound, we can use grams instead of percentage. Work out the number of moles and then find the simplest ratio.

S: 31.42 ÷ 32 = 0.982

O: 31.35 ÷ 16 = 1.96

F: 37.23 ÷ 19 = 1.96

Ratio:

S: 0.982 ÷ 0.982 = 1

O: 1.96 ÷ 0.982 = 2

F: 1.96 ÷ 0.982 = 2

empirical formula…

The empirical formula mass works out as: 32 + 32 + 38 = 102g which is the same as the molecular mass

The molecular formula is also…