A 5.93 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.94 g fluorine, what is the mass of the KF in the mixture?

The mass of KF in the mixture is 1.067 grams

Explanation:

Step 1: Data given

Total weight of mixture (LiF+KF) = 5.93 grams

Mixture contains 3.94 grams of fluorine

Molar mass K = 39.10 g/mol

Molar mass F = 19.00 g/mol

Molar mass KF = 58.10 g/mol

Molar mass of lithium = 6.94 g/mol

Step 2: mass of KF and LiF

Mass KF = X grams

Mass LiF = 5.97 - X grams

Step 3: Calculate moles

Moles KF = x grams / 58.10 g/mol

Moles LiF = (5.97-X grams) / 25.94 g/mol

1 mol of KF contains 1 mol of F atoms.

1 mol of LiF contains 1 mol of F atoms.

moles of F in KF = moles of KF = x/58.10 g/mol

moles of F in LiF = moles of LiF = (5.97-x) / 25.94g/mol

Total moles of Fluorine = (x/58.10) + ((5.97-x) / 25.94)

Step 4: Calculate mass

The total weight of Fluorine in sample

3.94 grams = ((x/58.10) + ((5.97-x) / 25.94)) * 19g/mol

x = 1.067 grams

The mass of KF in the mixture is 1.067 grams