Show the empirical formula manganese fluoride; 59.1% mn and 40.9% f. fill in the subscripts on the formula below. make sure to have a whole number of each subscript, even if it is a 1.

Answer :  The empirical formula of manganese fluoride is .

Solution :  Given,

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 Mn = 59.1 g

Mass of F = 40.9 g

Molar mass of Mn = 54.94 g/mole

Molar mass of F = 18.99 g/mole

Step 1 : convert given masses into moles.

Moles of Mn = = = 1.076 moles

Moles of F = = = 2.154 moles

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

For Mn = 1.076/1.076 = 1

For F = 2.154/1.076 ≈ 2

Mn : F = 1 : 2 

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

Therefore, the Empirical formula =

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