What is the empirical formula of a compound containing 24.56% potassium, 34.81% manganese, and 40.50% oxygen? KMnO KMnO2 K0.63Mn0.63O2.53 KMnO4 K3MnO2

Question

asked Jul 8, 2019 117k views

2 Answers

Konsolebox · Answer 1 · 2019-07-14T15:43:19+0000

Answer : The empirical formula of a compound is,

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

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

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

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

For K =
(0.614)/(0.614)=1

For Mn =
$(0.633)/(0.614)=1.03\approx 1$

For O =
$(2.53)/(0.614)=4.12\approx 4$

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,