Question

A compound is found to contain 63.65 % nitrogen and 36.35 % oxygen by mass. what is the empirical formula for this compound?

Answer 1

Let the total mass of the compound is 100 g.

Mass of N = 63.65 % = 63.65 g

Mass of O = 36.35 % = 36.35 g

Molar mass of N = 14.01 g/mol

Molar mass of O = 16.00 g/mol

Number of moles (mol) = mass (g) / molar mass(g/mol)

Number of Moles of N = 63.65 g / 14.01 g/mol = 4.54 mol

Number of Moles of O = 36.35 g / 16.00 g/mol = 2.27 mol

Divide the Number of moles of each element with the lowest number of moles, here lowest number of moles is 2.27 .

Dividing number of the moles of each element by lowest number of moles to get the mole ratio between elements.

N = 4.54 mol / 2.27 mol = 2

O = 2.27 mol / 2.27 mol = 1

So, Empirical formula = N₂O₁ = N₂O

Answer 2

Let's find the empirical formula stepwise.

1st step - Get the mass of each element of the compound.

Here, the masses of N and O are given as percentage.

Hence, let's assume that total mass of the compound is 100 g. Then we can get the masses in grams.

Mass of N = 63.65 % = 63.65 g

Mass of O = 36.35 % = 36.35 g

2nd step - Get the molar mass of each element .

Molar mass of N = 14.01 g/mol

Molar mass of O = 16.00 g/mol

3rd step - find the moles of each element.

We know that,

moles (mol) = mass (g) / molar mass(g/mol)

Let's apply that formula to find the moles.

Moles of N = 63.65 g / 14.01 g/mol = 4.54 mol

Moles of O = 36.35 g / 16.00 g/mol = 2.27 mol

4th step - find the lowest number of moles

According to the above calculations, O has the lowest number of moles as 2.27 mol.

5th step - Divide the moles of each element by lowest number of moles to get the mole ratio between elements.

N = 4.54 mol / 2.27 mol = 2

O = 2.27 mol / 2.27 mol = 1

6th step - write the formula by using mole ratios.

Empirical formula = N₂O₁ = N₂O