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 formed in the reaction is LiF.

Step-by-step solution

Convert masses to moles

Given, the mass of lithium metal is 1.00 mg and the mass of fluoride salt is 3.73 mg. Since the compound formed is lithium fluoride, the mass difference is due to the reaction with fluorine gas. Thus, the mass of fluorine reacted is 3.73 mg - 1.00 mg = 2.73 mg. Now, we will convert these masses into moles: Lithium: Mass = 1.00 mg Molar mass of Li = 6.94 g/mol (from the periodic table) Convert mass to grams: 1.00 mg × (1 g/1000 mg) = 0.00100 g Now, moles of Li = mass / molar mass Moles of Li = 0.00100 g / 6.94 g/mol = 1.44×10^(-4) mol Fluorine: Mass = 2.73 mg Molar mass of F = 19.00 g/mol (from the periodic table) Convert mass to grams: 2.73 mg × (1 g/1000 mg) = 0.00273 g Now, moles of F = mass / molar mass Moles of F = 0.00273 g / 19.00 g/mol = 1.44×10^(-4) mol

Calculate the mole ratio

Now that we have the moles of lithium and fluorine, we can find the ratio between them: Mole ratio (Li:F) = Moles of Li / Moles of F Mole ratio (Li:F) = (1.44×10^(-4) mol) / (1.44×10^(-4) mol) = 1:1

Determine the empirical formula

Since the mole ratio of lithium and fluorine is 1:1, the empirical formula of lithium fluoride is LiF (one Li atom combined with one F atom). So, the empirical formula of lithium fluoride formed in the reaction is LiF.

Key concepts

Chemical Reactions

Understanding chemical reactions is vital for solving problems like finding the empirical formula. A chemical reaction occurs when substances interact to form new materials. In our example, lithium (\(Li\)) and fluorine gas (\(F_2\)) combine to form lithium fluoride (\(LiF\)). A few key points about chemical reactions:

• Reactions involve the breaking and forming of chemical bonds.
• They include reactants (starting materials) and products (newly formed substances).
• Conservation of mass means the mass of the reactants equals the mass of the products.

Here, lithium reacts with fluorine to produce lithium fluoride. The empirical formula we calculate reflects the simplest ratio of the atoms in the final product, not their molecular complexity.

Stoichiometry

Stoichiometry is the calculation based on balanced chemical equations. It helps determine the relationships between the quantities of reactants and products. This is crucial when calculating the empirical formula, as it involves mass-to-mole conversions based on atomic weights. In this exercise:

• Determine the mass of each element involved.
• Convert these masses into moles using their atomic masses. Lithium has a molar mass of 6.94 g/mol and fluorine 19.00 g/mol.
• Establish the simplest mole ratio of the elements to find the empirical formula.

By understanding stoichiometry, we apply a systematic approach to solve this problem, ensuring each elemental atom's balance based on the chemical equation.

Mole Concept

The mole concept is all about quantifying atoms and molecules. It represents a number - specifically, Avogadro's number, which is approximately \(6.022 \times 10^{23}\) units of any substance. This concept is used to relate masses of substances in chemical reactions to the number of molecules involved.In our example, we:

• Convert mass (mg) into grams for easier calculations.
• Use the molar mass to convert grams into moles, allowing us to relate this to the number of atoms.
• Determine the mole ratio between lithium and fluorine based on measured values to find the empirical formula.

Using the mole concept helps us compute accurately and find the \(1:1\) mole ratio, confirming that \(LiF\) is indeed the empirical formula of lithium fluoride, reflecting the simplest ratio of elements involved.