Question

A 9.75 g sample of a compound contains 3.11of potassium, K, 2.82g of chlorine, Cl, and oxygen, O. Calculate the empirical formula.

Insert subscripts as needed.

Answer 1

The mass of oxygen remaining is 9.75 - 3.11 - 2.82 = 3.82 g

Amounts in moles are:

$n(K)=3.11\ g\times\frac{1\ mol}{39.10\ g}=0.0795\ mol$

$n(Cl)=2.82\ g\times\frac{1\ mol}{35.45\ g}=0.0795\ mol$

$n(O)=3.82\ g\times\frac{1\ mol}{16.00\ g}=0.239\ mol$

The lowest whole number ratio of K : Cl : O is 1 : 1: 3. Therefore, the empirical formula is KClO₃.

Answer: KClO₃