Question

When \(1.00 \mathrm{mg}\) of lithium metal is reacted with fluorine gas \(\left(\mathrm{F}_{2}\right),\) the resulting fluoride salt has a mass of \(3.73 \mathrm{mg}\). Calculate the empirical formula of lithium fluoride.

Short answer

The empirical formula of lithium fluoride is LiF since the ratio of Li and F in the compound is approximately 1:1.

Step-by-step solution

Recall the molar masses of lithium and fluorine

To calculate the moles of each element, we need the molar masses of lithium (Li) and fluorine (F). - Molar mass of Li = 6.94 g/mol - Molar mass of F = 19.00 g/mol

Calculate the mass of fluorine in the lithium fluoride

Since the mass of the lithium fluoride is 3.73 mg and the mass of lithium is 1.00 mg, we can determine the mass of fluorine in the compound: Mass of F = Total mass of lithium fluoride - Mass of Li Mass of F = 3.73 mg - 1.00 mg = 2.73 mg

Convert masses to moles

Now, we need to convert the masses of lithium and fluorine to moles using their respective molar masses: Moles of Li = (1.00 mg * (1 g/1000 mg)) / 6.94 g/mol = 1.44 x 10^(-4) mol Moles of F = (2.73 mg * (1 g/1000 mg)) / 19.00 g/mol = 1.43 x 10^(-4) mol

Find ratio between moles of Li and F

Divide the moles of Li and F by the smallest value to find the simplest whole number ratio: Li ratio = Moles of Li / smallest value = 1.44 x 10^(-4) mol / 1.43 x 10^(-4) mol ≈ 1.01 F ratio = Moles of F / smallest value = 1.43 x 10^(-4) mol / 1.43 x 10^(-4) mol = 1

Determine the empirical formula

Since the ratio of Li and F in the compound is approximately 1:1, the empirical formula for lithium fluoride is: LiF