Question

A piece of gold (Au) foil measuring \(0.25 \mathrm{mm} \times\) \(15 \mathrm{mm} \times 15 \mathrm{mm}\) is treated with fluorine gas. The treatment converts all the gold in the foil to \(1.400 \mathrm{g}\) of a gold fluoride. What is the formula and name of the fluoride? The density of gold is \(19.3 \mathrm{g} / \mathrm{cm}^{3}\)

Short answer

The formula of the gold fluoride is \(AuF3\) and its name is Gold(III) fluoride.

Step-by-step solution

Calculate the volume of gold

The volume can be calculated from the given dimensions of the gold foil. It is \(0.25 \mathrm{mm} \times15 \mathrm{mm} \times 15 \mathrm{mm}\) which translates to \(0.25 \mathrm{cm} \times1.5 \mathrm{cm} \times 1.5 \mathrm{cm}\) in cm since 1cm = 10mm. We proceed to multiply these measurements to get the volume of the bar.

Calculate the mass of the gold

This requires multiplying the volume of the gold bar calculated in the previous step by the given density of gold, \(19.3 \mathrm{g/cm}^{3}\) to obtain mass.

Find the mass of fluorine

The mass of fluorine can be calculated by subtracting the mass of gold derived in Step 2 from the total mass of the gold fluoride, 1.400g.

Convert the masses of gold and fluorine into moles

The molar mass of gold is approximately \(197.0 \mathrm{g/mol}\) and that of fluorine is approximately \(19.0 \mathrm{g/mol}\). Divide the mass of each by its molar mass.

Calculate the mole ratio

Still assuming that each gold atom reacts with just one fluorine atom, divide each number of moles by the smaller so that the empirical formula has the smallest whole numbers possible. Round off to the nearest whole numbers.

Deduce the formula and name of the fluoride

From the outcomes in step 5, construct the formula of the salt and give its name.

Key concepts

Chemical Formula Determination

Chemical formula determination is like revealing the recipe of a compound. For gold fluoride, our goal is to find out how many atoms of each element combine to create this compound. We start with known quantities and convert them into moles, which tells us how many atoms we have.
Once we know the moles of gold (Au) and fluorine (F), we can find the simplest ratio between them. This ratio helps to determine the empirical formula. For instance, if you have 1 mole of gold and 3 moles of fluorine, the formula could be AuF₃, indicating that one gold atom combines with three fluorine atoms.
So, in essence, determining a chemical formula involves these steps: calculating moles, finding ratios, and checking for the simplest whole number ratio. This process helps us reveal the true formula of a compound.

Stoichiometry

Stoichiometry is the study of the quantities of reactants and products in chemical reactions. It's like a balanced accounting system for chemistry where everything must add up correctly.
When dealing with gold fluoride, stoichiometry helps us understand how much fluorine reacts with a given amount of gold to create the compound. We use stoichiometric coefficients from balanced equations to calculate these amounts.
In our example, we can use stoichiometry to predict how much gold fluoride is formed from a known amount of gold and fluorine. By knowing the coefficient from the balanced chemical equation, we can accurately predict the mass of products formed from the reactants. This helps ensure that our calculated chemical formula is based on solid, empirical data.

Molar Mass Calculations

Molar mass calculations involve determining the mass of one mole of a given substance, which includes elementary atoms or molecules. This helps us translate between mass and moles—crucial for chemical calculations.
For gold, the molar mass is about 197.0 g/mol, and for fluorine, it's about 19.0 g/mol. To convert a mass into moles, you divide the substance's mass by its molar mass. For example, if you have a mass of gold from the foil, you use this formula: \(\text{Moles of Gold} = \frac{\text{Mass of Gold}}{197.0}\)
Knowing the molar mass lets us calculate how many moles of fluorine are needed to react with gold. It bridges the gap between the macroscopic world we measure and the microscopic world of atoms.

Chemical Reaction Analysis

Chemical reaction analysis is about studying how substances change during chemical reactions. This analysis gives insight into the reactants' and products' nature, properties, and composition.
In the case of gold fluoride formation, analyzing the chemical reaction can demonstrate how gold (Au) and fluorine (F) interact. One must consider conditions such as temperature and the physical state of reactants, ensuring the reaction is feasible.
A chemical reaction analysis involves:

• Identifying reactants and products
• Balancing chemical equations
• Understanding the reaction mechanism

By understanding these, we ensure the formation of the gold fluoride compound is thoroughly mapped out, confirming the practical aspects of chemical synthesis.