Problem 36 - Chemistry and Chemical Reactivity

Question

A The aluminum in a 0.764 -g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH) \(_{3}\) which was then converted to \(\mathrm{Al}_{2} \mathrm{O}_{3}\) by heating strongly. If \(0.127 \mathrm{g}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is obtained from the \(0.764 \mathrm{-g}\) sample, what is the mass percent of aluminum in the sample?

Step-by-Step Solution

Answer

The mass percent of aluminum in the sample is approximately 8.79%.

Step 1: Write the Conversion Reaction

The conversion reaction of aluminum hydroxide to aluminum oxide is represented as follows:\[2 \text{Al(OH)}_3 \rightarrow \text{Al}_2\text{O}_3 + 3 \text{H}_2\text{O}\]This equation shows that two moles of aluminum hydroxide decompose to yield one mole of aluminum oxide.

Step 2: Calculate Molar Masses

Calculate the molar masses for \(\text{Al(OH)}_3\) and \(\text{Al}_2\text{O}_3\):- Molar mass of \(\text{Al(OH)}_3 = 78.00 \text{ g/mol}\).- Molar mass of \(\text{Al}_2\text{O}_3 = 101.96 \text{ g/mol}\).

Step 3: Calculate Moles of Al₂O₃

Using the mass of \(\text{Al}_2\text{O}_3\) obtained:\[\text{Moles of } \text{Al}_2\text{O}_3 = \frac{0.127 \text{ g}}{101.96 \text{ g/mol}} \approx 0.001245 \text{ mol}\]

Step 4: Convert to Moles of Al

Each mole of \(\text{Al}_2\text{O}_3\) contains 2 moles of Al. So,\[\text{Moles of Al} = 2 \times 0.001245 \text{ mol} \approx 0.00249 \text{ mol}\]

Step 5: Calculate Mass of Al

Calculate the mass of aluminum using its moles:\[\text{Molar mass of Al} = 26.98 \text{ g/mol}\]\[\text{Mass of Al} = 0.00249 \text{ mol} \times 26.98 \text{ g/mol} \approx 0.06714 \text{ g}\]

Step 6: Find Mass Percent of Al

The mass percent of aluminum in the sample is given by:\[\text{Mass percent of Al} = \left(\frac{0.06714 \text{ g}}{0.764 \text{ g}}\right) \times 100 \approx 8.79\%\]

Key Concepts

Molar Mass Calculations

Understanding molar mass is crucial in chemistry as it helps you convert between the mass of a substance and the number of moles, which are fundamental aspects of stoichiometry. The molar mass of a compound is the sum of the atomic masses of all atoms in its molecular formula.
For example, the molar mass of aluminum hydroxide, \(\text{Al(OH)}_3\), is calculated by adding the molar masses of aluminum (Al) and the hydroxide groups (OH):
- Aluminum (Al) has a molar mass of approximately \(26.98\, \text{g/mol}.\)
- Each hydroxide \((\text{OH})\) has a molar mass of \(17.01\, \text{g/mol}\) (Oxygen is \(16.00\, \text{g/mol}\) and Hydrogen is \(1.01\, \text{g/mol}\)).
The molar mass of \(\text{Al(OH)}_3\) is therefore \(26.98\, \text{g/mol} + (3 \times 17.01\, \text{g/mol}) = 78.00\, \text{g/mol}.\)

Similarly, the molar mass of \(\text{Al}_2\text{O}_3\) is calculated with:

• Two aluminum atoms: \(2 \times 26.98\, \text{g/mol} = 53.96\, \text{g/mol}.\)
• Three oxygen atoms: \(3 \times 16.00\, \text{g/mol} = 48.00\, \text{g/mol}.\)

Hence, the molar mass of \(\text{Al}_2\text{O}_3\) is \(53.96 + 48.00 = 101.96\, \text{g/mol}.\)
Calculating these values accurately is essential in determining the proportions of different reactants and products in a chemical reaction.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, known as precipitates. These reactions are fundamental in processes like water purification, chemical synthesis, and waste treatment.
In the given exercise, aluminum is precipitated from the solution as aluminum hydroxide, \(\text{Al(OH)}_3\).
This is a classic example of a precipitation reaction where a solid forms from the reaction of soluble components in solution.
Precipitates typically form when ionic compounds containing ions like Al\(^{3+}\) and OH\(^-\) meet and exceed their solubility limits in water.

• The balanced chemical equation given here shows the conversion: \(2\, \text{Al(OH)}_3 \rightarrow \text{Al}_2\text{O}_3 + 3\, \text{H}_2\text{O}.\)
• During this reaction, insoluble Al(OH)₃ is formed which can be isolated by filtration.

It's crucial to understand how precipitation reactions can be used to isolate specific compounds or elements from mixtures by forming a solid that is easily separated from the liquid phase.

Mass Percent Composition

Mass percent composition is a useful way to express the concentration of an element within a compound, or the concentration of a compound within a mixture. It tells you how much of the total mass of a mixture or compound is due to a specific component.
In the example, it's calculated to find out what proportion of the original sample is aluminum. The mass percent of aluminum in the original sample is found by:

• First, finding the mass of aluminum obtained: \(0.06714\, \text{g}.\)
• Then, dividing by the total mass of the sample: \(0.764\, \text{g}.\)
• This ratio is multiplied by 100 to convert it to a percentage.

This calculation gives us the mass percent of aluminum as approximately 8.79%.
Understanding mass percent composition is crucial for determining how much of a substance is in a mixture or a compound.
It allows chemists to identify the purity of samples and the effectiveness of their extraction or purification methods. Knowing this information helps in quality control and in verifying the accuracy of chemical processes.