Show the empirical formula manganese fluoride 591 Mn and 409

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.

Mn__________F__________

Solution

mass of Mn = 59.1 g

moles of Mn = 59.1 / 54.94 = 1.076

mass of F = 40.9 g

moles of F = 40.9 / 19 = 2.153

ratio :           Mn           F

             1.076           2.153

                     1                2

Emperical formula = MnF₂

coefficient of Mn = 1

coefficient of F = 2