Question

CHEGG A 7.43 g sample of a compound contains 2.37 g of potassium, K , 2.15 g of chlorine, Cl , and oxygen, O . Calculate the empirical formula.

Answer 1

Answer: The empirical formula of the given compound is $KClO_3$

Explanation:

We are given:

Mass of K = 2.37 g

Mass of Cl = 2.15 g

Mass of O = [7.43 - (2.37 + 2.15)] = 2.91 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Potassium =$\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{2.37g}{39g/mole}=0.0608moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{2.15g}{35.5g/mole}=0.0606moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{2.91g}{16g/mole}=0.182moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0606 moles.

For Potassium = $\frac{0.0608}{0.0606}=1$

For Chlorine = $\frac{0.0606}{0.0606}=1$

For Oxygen  = $\frac{0.182}{0.0606}=3.003\approx 3$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of K : Cl : O = 1 : 1 : 3

Hence, the empirical formula of the given compound is $KClO_3$