Question

What is the empirical formula of a compound containing 24.56% potassium, 34.81% manganese, and 40.50% oxygen? 

Answer 1

Answer : The empirical formula of a compound is, $KMnO_4$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K = 24.56 g

Mass of Mn = 34.81 g

Mass of O = 40.50 g

Molar mass of K = 40 g/mole

Molar mass of Mn = 55 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of K = $\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{24.56g}{40g/mole}=0.614moles$

Moles of Mn = $\frac{\text{ given mass of Mn}}{\text{ molar mass of Mn}}= \frac{34.81g}{55g/mole}=0.633moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{40.50g}{16g/mole}=2.53moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = $\frac{0.614}{0.614}=1$

For Mn = $\frac{0.633}{0.614}=1.03\approx 1$

For O = $\frac{2.53}{0.614}=4.12\approx 4$

The ratio of K : Mn : O = 1 : 1 : 4

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $K_1Mn_1O_4=KMnO_4$

Therefore, the empirical formula of a compound is, $KMnO_4$

Answer 2

It would be the last anwer