Question

What is the empirical formula of a compound composed of 32.3 g potassium (K) and 6.61 g oxygen (O)?

Answer 1

To calculate the empirical formula of a compound, we have to follow some steps:

Step 1: Calculate the number of moles of K and O.

For this we need to know the molar mass of K and O.

Molar mass of K=39.0983 g/mol

Molar mass of O=16 g/mol

Given mass of K= 32.3 g

Given mass of O= 6.61 g

Number of moles=Mass /molar mass

So, Number of moles of O=Mass /molar mass

=6.61/16

=0.41

Number of moles of K =Mass /molar mass

=32.3/39

=0.82

Step 2: Find the lowest number of moles and divide each element's moles by it:

Here the lowest number of moles are of oxygen(O) so dividing each element's moles by its mole number will give:

K =0.82/0.41

= 2

O = 0.41/0.41

= 1

So the empirical formula of the compound should be:

K₂O.

Answer 2

Let's find the empirical formula step wise.

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

Here, the masses of K and O are given in grams.

Mass of K = 32.3 g

Mass of O = 6.61 g

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

Molar mass of K = 39.10 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 K = 32.3 g / 39.10 g/mol = 0.83 mol

Moles of O = 6.61 g / 16.00 g/mol = 0.41 mol

4th step - find the lowest number of moles

According to the above calculations, O has the lowest number of moles as 0.41 mol.

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

K = 0.83 mol / 0.41 mol = 2.02 = 2

O = 0.41 mol / 0.41 mol = 1

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

Empirical formula = K₂O₁ = K₂O