Bauxite, the principal ore used in the production of aluminum, has a molecular formula of \(\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). The \(\cdot \mathrm{H}_{2} \mathrm{O}\) in the formula are called waters of hydration. Each formula unit of the compound contains two water molecules. a. What is the molar mass of bauxite? b. What is the mass of aluminum in \(0.58\) mole of bauxite? c. How many atoms of aluminum are in \(0.58\) mole of bauxite? d. What is the mass of \(2.1 \times 10^{24}\) formula units of bauxite?

Short Answer

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a. The molar mass of bauxite is 138.00 g/mol. b. The mass of aluminum in 0.58 mole of bauxite is 31.28 g. c. There are \(6.98 \times 10^{23}\) atoms of aluminum in 0.58 mole of bauxite. d. The mass of \(2.1 \times 10^{24}\) formula units of bauxite is 481.62 g.

Step by step solution

01

a. Finding the molar mass of bauxite

In order to find the molar mass of bauxite, we first need to determine the molar mass of each element in the compound and then find the sum of the individual molar masses. The molecular formula of bauxite is \(\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Molar mass of \(\mathrm{Al}\) (aluminum) = 26.98 g/mol Molar mass of \(\mathrm{O}\) (oxygen) = 16.00 g/mol Molar mass of \(\mathrm{H}\) (hydrogen) = 1.01 g/mol Now, we can find the molar mass of bauxite: \( Molar \ mass \ of \ bauxite = (2 \cdot 26.98) + (5 \cdot 16.00) + (4 \cdot 1.01) \) Calculate the sum: \( Molar \ mass \ of \ bauxite = 53.96 + 80.00 + 4.04 \) Finally: \( Molar \ mass \ of \ bauxite = 138.00 \ g/mol \)
02

b. Finding the mass of aluminum in 0.58 mole of bauxite

We are given that there are 0.58 moles of bauxite. Since bauxite contains two aluminum atoms per molecule, the number of moles of aluminum within 0.58 moles of bauxite will also be twice. Moles of aluminum in bauxite = 2 x 0.58 moles = 1.16 moles Now, we can find the mass of aluminum by multiplying the moles of aluminum by its molar mass: \( Mass \ of \ aluminum = 1.16 \ moles \times 26.98 \ g/mol \) Calculate the product: \( Mass \ of \ aluminum = 31.28 \ g \)
03

c. Finding the number of aluminum atoms in 0.58 mole of bauxite

We know the moles of aluminum in 0.58 mole of bauxite is 1.16 moles (from the previous part). To find the number of aluminum atoms in these moles, we will simply multiply the moles of aluminum by Avogadro's number, which is approximately \( 6.022 \times 10^{23} \ atoms/mol \). Number of aluminum atoms = 1.16 moles x \( 6.022 \times 10^{23} \ atoms/mol \) Calculate the product: Number of aluminum atoms = \( 6.98 \times 10^{23} \ atoms \)
04

d. Finding the mass of 2.1 x 10^24 formula units of bauxite

We are given that there are \(2.1 \times 10^{24} \) formula units of bauxite. To find the mass of these formula units, we first need to convert the number of formula units to moles. We do this by dividing the number of formula units by Avogadro's number: Moles of bauxite = \( \frac{2.1 \times 10^{24} \ formula \ units}{6.022 \times 10^{23} \ formula \ units/mol} \) Calculate the quotient: Moles of bauxite = 3.49 moles Now, we can find the mass of bauxite by multiplying the moles of bauxite by its molar mass (138.00 g/mol): \( Mass \ of \ bauxite = 3.49 \ moles \times 138.00 \ g/mol \) Calculate the product: \( Mass \ of \ bauxite = 481.62 \ g \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Waters of Hydration
Understanding waters of hydration is essential when studying compounds like bauxite, the primary ore of aluminum.
Most students are taught that salts can contain water molecules that are an integral part of their crystal structure. These water molecules are known as waters of hydration or water of crystallization.

In the case of bauxite, with the molecular formula \(\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{H}_{2} \mathrm{O}\), the dot (\cdot) indicates that two molecules of water are bound to the compound. Therefore, for each unit of bauxite, there are two water molecules incorporated into its structure. This knowledge is crucial when calculating the molar mass of a hydrated compound, as it involves counting the mass of the waters of hydration along with the anhydrous salt.
Molecular Formula Calculation
The process of molecular formula calculation involves determining the composition of a chemical compound in terms of the elements present and their respective numbers.
To calculate the molar mass of a compound like bauxite, one must consider the atomic masses of each element and their quantity within the formula. For bauxite, the molar mass accounts for aluminum (Al), oxygen (O), and the waters of hydration, each involving their own molar mass multiplied by the number of atoms present in the formula. Adding these values gives the total molar mass of the compound.

This is crucial for subsequent calculations, such as finding the mass of a certain element within a given number of moles of the compound or converting between moles and mass for the entire compound. Problem-solving in chemistry frequently entails such computations, reinforcing the importance of accurately determining the molecular formula of substances.
Avogadro's Number
A key concept in chemistry is Avogadro's number, which helps students transition between the microscopic world of atoms and the macroscopic world we can measure.
Avogadro's number, approximately \(6.022 \times 10^{23}\), represents the number of units (be they atoms, molecules, or formula units) in one mole of a substance.

In the example of bauxite, understanding Avogadro's number allows us to calculate the actual number of aluminum atoms in a given sample by multiplying the number of moles by Avogadro's number. Similarly, the opposite is true when we have a specific number of formula units and wish to calculate the mass of the sample; we simply convert the number of units to moles using Avogadro's number, then use the molar mass to transition to grams. Being able to manipulate these quantities is fundamental in chemistry, as it bridges the gap between theoretical concepts and practical measurements.

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Most popular questions from this chapter

In the production of printed circuit boards for the electronics industry, a \(0.60\) -mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{~cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

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A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

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