Question

What is the empirical formula of a compound composed of o and mn in equal weight ratio

Answer 1

Answer:

Step-by-step explanation:

Let us assume the mass of O and Mn in a compound is x each. So the number of mols of O will be x/16 (Molar mass if O is 16) and the number of mols of Mn will be x/55 (Molar mass of Mn is approximately 55).

The molar ratio of Mn and O in the compound will be (x/55)/(x/16) which comes out to be 16:55. So the empirical formula will be Mn16O55 or Mn0.29 so...appx

MnO3

also

% of both is 50 and 50

for o= 50/16=3.125

for Mn=50/55=0.909

since 0.909<3.125

so...for O=3.125/0.909=3.43 appx=3

for. Mn=0.909/0.909=1

empirical formula=MnO3

Answer 2

Answer:

Step-by-step explanation:

Step 1 : let us assume the mass of O and Mn in a compound is X each.

So the number of moles of O will be X/16 (Molar mass if O is 16 )

The number of moles of Mn will be X/55(Molar mass of Mn is approximately 55)

Step 2 :The molar ratio of Mn and O in the compound will be (X/55)/(X/16) Which gives the result is to be 16 : 55.

Hence , the empirical formula will be Mn 16 and O 55.

Thank you