Question

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

Answer 1

It would be the last anwer

Answer 2

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}}= (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,
`KMnO_4`