Problem 108 - General Chemistry: Principles and Modern Applications

Question

The atomic mass of \(\mathrm{Bi}\) is to be determined by converting the compound \(\mathrm{Bi}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3}\) to \(\mathrm{Bi}_{2} \mathrm{O}_{3} .\) If \(5.610 \mathrm{g}\) of \(\mathrm{Bi}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3}\) yields \(2.969 \mathrm{g} \mathrm{Bi}_{2} \mathrm{O}_{3},\) what is the atomic mass of Bi?

Step-by-Step Solution

Answer

The atomic mass of Bi is calculated to be approximately 208.98 g/mol after following above steps and performing the calculations.

Step 1: Calculate Molar Mass of Bismuth Trioxide (Bi2O3)

Firstly, determine the molar mass of \( \mathrm{Bi}_{2}\mathrm{O}_{3} \). Since it is composed of 2 atoms of Bi and 3 atoms of O, the formula would be: \(Molar mass = 2 × (atomic mass of Bi) + 3 × (atomic mass of O)\). Here, we know that the atomic mass of O is 16 amu from the periodic table, but we do not know the atomic mass of Bi yet.

Step 2: Establish Ratio of Bi in Bi(C6H5)3 to Bi in Bi2O3

The stoichiometry of the reaction indicates that each molecule of \( \mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3} \) generates half a molecule of \( \mathrm{Bi}_{2}\mathrm{O}_{3}\). Hence the ratio is 1:0.5.

Step 3: Calculate the Mass of Bi

We have 2.969g of \( \mathrm{Bi}_{2}\mathrm{O}_{3}\), and this contains 2 atoms of Bi. Therefore, the mass of Bi that came from the 5.61 g of \( \mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3} \) is 2/3 of 2.969g since Bi contributes 2 out of the 3 parts (2 from Bi, 1 from O3) to \( \mathrm{Bi}_{2}\mathrm{O}_{3}\).

Step 4: Calculate the Atomic Mass of Bi

From step 2, we know that the Bi in 5.61g of \( \mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3} \) is 2/3 of 2.969g. Hence, the atomic mass Bi can be obtained by diving this mass by the number of moles in 5.61g of \( \mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3} \). This gives the atomic mass in g/mol.

Key Concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the calculation of reactants and products in chemical reactions. It's like a recipe that tells you how much of each ingredient you need to make a certain amount of a chemical dish.
Stoichiometry is based on the conservation of mass, meaning the mass of the reactants must equal the mass of the products in a chemical reaction.
This balance allows us to predict how much product can be created from given reactants.
In the context of our exercise, stoichiometry helps us understand the relationship between the amounts of bismuth in the starting compound, \\(\mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3}\), and the resulting bismuth trioxide, \\(\mathrm{Bi}_{2}\mathrm{O}_{3}\).

• For every molecule of \\(\mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3}\), half a molecule of \\(\mathrm{Bi}_{2}\mathrm{O}_{3}\) is produced.
• This 1:0.5 ratio is derived from balancing the equation governing the conversion.

This information is crucial for calculating how much bismuth we get from the initial compound, an important step in determining the atomic mass of bismuth.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's essentially the mass of Avogadro's number (approximately \(6.022 \ imes 10^{23}\) entities) of the substance's molecules or atoms.
The molar mass of a compound is calculated by summing the atomic masses of all the atoms in its molecular formula. For our exercise, let's focus on calculating the molar mass of \\(\mathrm{Bi}_{2}\mathrm{O}_{3}\). This compound consists of:

• 2 atoms of Bismuth (\(\mathrm{Bi}\))
• 3 atoms of Oxygen (\(\mathrm{O}\))

To calculate its molar mass, we need to know the atomic mass of Bi and O. While the atomic mass of Oxygen is \(16\) amu, Bismuth's mass is what we are trying to determine. The formula would then be:\[\text{Molar mass of } \mathrm{Bi}_{2}\mathrm{O}_{3} = 2 \ imes (\text{atomic mass of } \mathrm{Bi}) + 3 \ imes 16\text{ amu}\] This molar mass calculation is pivotal for converting between mass and moles, especially when dealing with chemical reactions, as shown in our exercise.

Bismuth Trioxide (Bi2O3)

Bismuth trioxide (\(\mathrm{Bi}_{2}\mathrm{O}_{3}\)) is a chemical compound that plays an important role in various chemical processes. One primary application in our exercise is helping us determine the atomic mass of bismuth. Bismuth trioxide is composed of:

• Two atoms of Bismuth \((\mathrm{Bi})\)
• Three atoms of Oxygen \((\mathrm{O})\)

In the context of the reaction involved, transforming \\(\mathrm{Bi}(\mathrm{C}_{6}\mathrm{H}_{5})_{3}\) into \\(\mathrm{Bi}_{2}\mathrm{O}_{3}\) allows us to isolate and then quantify the amount of bismuth, thanks to stoichiometric conversion.By knowing the mass of the resulting \\(\mathrm{Bi}_{2}\mathrm{O}_{3}\) and understanding the mole ratio, we can work backwards to find the atomic mass of bismuth. This method of deduction highlights a critical process in chemistry where compounds help reveal the properties of elements contained within.