Question

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____.

Answer 1

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 2

Answer : The empirical formula of manganese fluoride is
`Mn_1F_2`.

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 =
`\frac{\text{ given mass of Mn}}{\text{ molar mass of Mn}}* 1\text{ mole of Mn}` =
`(59.1g)/(54.94g/mole)* 1\text{ mole of Mn}` = 1.076 moles

Moles of F =
`\frac{\text{ given mass of F}}{\text{ molar mass of F}}* 1\text{ mole of F}` =
`(40.9g)/(18.99g/mole)* 1\text{ mole 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 =
`Mn_1F_2`