Question

If 1.25 g of aluminum metal is heated in an atmosphere of fluorine gas, 3.89 g of aluminum fluoride results. Determine the empirical formula of aluminum fluoride.

Short answer

The empirical formula of aluminum fluoride can be determined by converting the given masses to moles and finding the simplest ratio between them. Moles of aluminum (Al) can be calculated as \(\frac{1.25}{26.98}\), and the mass of fluorine (F) in aluminum fluoride is 3.89 g - 1.25 g = 2.64 g. Moles of fluorine can be calculated as \(\frac{2.64}{19}\). To determine the mole ratio between Al and F, divide each mole value by the smallest of the two mole values. Finally, round the ratios to the closest whole number and use them as subscripts in the empirical formula, Al_xF_y. If a mole ratio appears to be a simple fraction, it can be helpful to multiply both mole ratios by a small integer to convert them into whole numbers.

Step-by-step solution

Find the moles of aluminum (Al) and aluminum fluoride (AlFₓ)

Given the mass of aluminum is 1.25 g and the mass of aluminum fluoride is 3.89 g. We need to convert these masses to moles, using their respective molar masses. The molar mass of aluminum (Al) = 26.98 g/mol Moles of aluminum = \(\frac{1.25}{26.98}\) Now, let's find the mass of fluorine in the aluminum fluoride Mass of fluorine in aluminum fluoride = Mass of aluminum fluoride - Mass of aluminum Mass of fluorine = 3.89 g - 1.25 g = 2.64 g The molar mass of fluorine (F) = 19 g/mol Moles of fluorine = \(\frac{2.64}{19}\)

Calculate the mole ratio between Al and F

Now we'll determine the mole ratio between aluminum and fluorine, by dividing each mole value by the smallest of the two mole values. Mole ratio of Al: \(\frac{Moles\,of\,Aluminum}{Smallest\,mole\,value}\) Mole ratio of F: \(\frac{Moles\,of\,Fluorine}{Smallest\,mole\,value}\)

Determine the empirical formula of aluminum fluoride

Once the mole ratio between Al and F is calculated, we round the ratios to the closest whole number to find the empirical formula. If the mole ratio is already a whole number or very close to a whole number, then we don't need to round. The empirical formula of aluminum fluoride will be Al_xF_y, where x and y are the whole numbers representing the mole ratio between Al and F atoms. Note that sometimes, if a mole ratio appears to be a simple fraction (e.g., 0.5 or 1.5), it can be helpful to multiply both mole ratios by a small integer (e.g., 2) to convert them into whole numbers.

Key concepts

Molar mass

The concept of molar mass is an essential foundation in chemistry. It allows us to convert between mass and moles, facilitating various calculations in stoichiometry.
The molar mass is the weight of one mole of a given substance. It is expressed in grams per mole (g/mol) and is determined by the atomic weights of the constituent elements of a compound, as found on the periodic table. Here’s how it plays a role in determining empirical formulas:

• For individual elements, such as aluminum (Al), the molar mass can be directly read from the periodic table. Aluminum has a molar mass of 26.98 g/mol.
• For compounds, like aluminum fluoride (AlFₓ), you sum the molar masses of its elements. However, we initially use the individual molar masses before determining their proportion in the compound.

Calculating molar mass accurately is the first step in converting mass to moles, which is critical for mole ratio and empirical formula determination.

Mole ratio

Mole ratio is a core concept in determining the empirical formula as it shows the proportion of elements in a compound.
Once the moles of each element in a sample are calculated, the next step is to find out how these moles relate to each other. This is achieved by:

• Identifying the smallest number of moles found among the elements (this becomes the reference point).
• Dividing each element's mole count by the smallest mole count.

This helps in establishing a simple whole-number ratio of the elements involved. For example, if you have 1 mole of aluminum and 3 moles of fluorine, this indicates that the mole ratio is 1:3. Following these principles enables us to understand the basic structure of a compound through its simplest form, or empirical formula.

Chemical formula determination

Determining a chemical formula, especially an empirical formula, begins with experimental data regarding mass. With this data, we can calculate the mole ratio and derive the empirical formula.
The process includes several steps:

• Start by finding the mass of each element in the compound.
• Convert these masses into moles using their molar masses.
• Use the mole ratio to deduce the simplest whole-number ratio of the elements.
• Create the empirical formula based on these whole numbers (e.g., AlFₓ).

If the mole ratio is not a whole number immediately, it may be necessary to multiply through by a suitable factor to achieve whole numbers. This chemical formula gives insight into the elemental composition of the compound, making it a powerful tool in analytical chemistry.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate missing information and predict product mass, reactant quantities, and more.
Here’s how stoichiometry ties into empirical formula calculations:

• It fundamentally relies on the concept of moles as a bridge between particles and grams.
• For reactions, stoichiometry can predict how much product forms from given reactants. In empirical formula determination, it aids in back-calculating element ratios after a reaction.
• Understanding stoichiometry helps comprehend the mass conservation in reactions and the molecular make-up of products.

By mastering stoichiometry, a student gains the skills to navigate complex chemical equations and derive meaningful information from experimental data.