Hey there!:
Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.
Molar mass of Mn= 54.938 g/mol
Moles of Mn = mass / molar mass
59.1 /54.938 => 1.07 ≈ 1 mol.
and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.
Molar mass of F=18.998 g/mol
Moles of F :
40.9 / 18.999 => 2.15 mol ≈ 2 mol.
The mole ratio between Mn:F= 1 : 2
Therefore the empirical formula of manganese fluoride:
=> MnF2=Mn1F2
Hope that helps!
Answer : The empirical formula of manganese fluoride is [tex]Mn_1F_2[/tex].
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 = [tex]\frac{\text{ given mass of Mn}}{\text{ molar mass of Mn}}\times 1\text{ mole of Mn}[/tex] = [tex]\frac{59.1g}{54.94g/mole}\times 1\text{ mole of Mn}[/tex] = 1.076 moles
Moles of F = [tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}\times 1\text{ mole of F}[/tex] = [tex]\frac{40.9g}{18.99g/mole}\times 1\text{ mole of F}[/tex] = 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 = [tex]Mn_1F_2[/tex]
