Question 8 if the mass percentage composition of a compound is 72.1% mn and 27.9% o, its empirical formula is selected correct answer. d. mn304 d. mn304

The compound is

Explanation:

The mass percentage of Mn is 72.1% and the mass percentage of O is 27.9%.

The mass percentage of a compound is given by:

where:

n is its coefitient in the compund formula

MMa=Molar mass of the element A

MMc=Molar mass of the compound

So, we can figure out which compound is by dividing the percentage by its molar mass

Mn=72.1÷54.938045=1.31239

O=27.9÷15.9994=1.74382

Then, we divide each result by the smaller one (Mn)

Mn=1.31239÷1.31239=1

O=1.74382÷1.31239=1.3287

Each the realation of Mn:O is 1:1.3287

Then we multiply each result by 3:

Mn=1×3=3

O=1.3287×3=3.986≈4

Finally we figure out that the compound has 3 atoms of Mn and 4 atoms O. Result=

Mn3O4 is sometimes used as a starting material in the production of soft ferrites e.g. manganese zinc ferrite, and lithium manganese oxide, used in lithium batteries.

the answers are

c and c