Question

Question #fee41

Answer 1

It tells you the empirical formula of the substance — the relative numbers of each type of atom in a formula unit.

Example

A compound of nitrogen and oxygen contains 30.4 % nitrogen and 69.6 % oxygen by mass. What is its empirical formula?

Solution

Assume 100.0 g of the compound. Then we have 30.4 g nitrogen and 69.6 g oxygen.

Moles of N = 30.4 g N × `(1 mol N)/(14.01 g N)` = 2.17 mol N

Moles of O = 69.6 g O × `(1 mol O)/(16.00 g N)` = 4.35 mol O

Molar ratio N:O = 2.17 mol:4.35 mol = 1 mol:2.00 mol = 1:2

The ratio of the moles is the same as the ratio of the atoms. Therefore, there are two moles of O atoms for every one N atom.

The empirical formula is NO₂.

This is not necessarily the actual formula of the compound. The actual formula may be NO₂, N₂O₄, N₃O₆, etc. All of these have an N:O ratio of 1:2. We would have to do other experiments to determine the actual formula.

Answer 2

It tells you the empirical formula of the substance — the relative numbers of each type of atom in a formula unit.

Example

A compound of nitrogen and oxygen contains 30.4 % nitrogen and 69.6 % oxygen by mass. What is its empirical formula?

Solution

Assume 100.0 g of the compound. Then we have 30.4 g nitrogen and 69.6 g oxygen.

Moles of N = 30.4 g N × `(1 mol N)/(14.01 g N)` = 2.17 mol N

Moles of O = 69.6 g O × `(1 mol O)/(16.00 g O)` = 4.35 mol O

Molar ratio N:O = 2.17 mol:4.35 mol = 1 mol:2.00 mol = 1:2

The ratio of the moles is the same as the ratio of the atoms. Therefore, there are two moles of O atoms for every one N atom.

The empirical formula is NO₂.

This is not necessarily the actual formula of the compound. The actual formula may be NO₂, N₂O₄, N₃O₆, etc. All of these have an N:O ratio of 1:2. We would have to do other experiments to determine the actual formula.