A 5.50 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.59 g fluorine, what is the mass of the KF in the mixture?

Question

Mass of KF in grams

Answer 1

The total mass of the mixture (LiF + KF) = 5.5 g

Let mass of KF (what we are solving for) = x

Then, LiF + x = 5.5 g

LiF = 5.5 - x

The moles of KF would be the mass divided by the molecular weight (58 g/mol)

moles KF = x /58

moles LiF = (5.5 - x) / 26

For every 1 mol of KF, we have 1 mol of K and 1 mol of F.

For every 1 mol of LiF, we have 1 mol of Li and 1 mol of F.

moles of F in KF = moles KF = x/58

moles of F in LiF = moles LiF = (5.5 - x) / 26

So, total moles of Fluorine would be adding those two:

total moles of F = (x/58) + ((5.5 - x) / 26)

We know the total mass of fluorine is 3.59 g, let's convert that to moles:

3.59 g / 19 g/mol = 0.189 moles

Set this equal to our equation for total moles of F

0.189 moles = (x/58) + ((5.5 - x) / 26)

(58)(26)(0.189) = 26x + (5.5)(58) - 58x

32x = 33.988

x = 1.06 grams