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

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

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