Problem 98 - Chemistry An Atoms First Approach

Question

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). A 1.45 -g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\). The precipitate is filtered, dried, and weighed. The mass of the precipitate is \(0.107 \mathrm{g} .\) What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

Step-by-Step Solution

Answer

The mass percent of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ in the 1.45-g sample is approximately 16.14%.

Step 1: Calculate the moles of Al(OH)3 formed

First, we need to find the moles of Al(OH)3 formed in the reaction. We can do this by using the molar mass of Al(OH)3 and the given mass of the precipitate. The molar mass of Al(OH)3 is: Al: 26.98 g/mol O: 16.00 g/mol H: 1.01 g/mol Molar mass of Al(OH)3 = Al + 3(O + 3H) = 26.98 + 3(16.00 + 3(1.01)) ≈ 78.00 g/mol Now we can calculate the moles of Al(OH)3 formed: Moles of Al(OH)3 = mass of precipitate / molar mass of Al(OH)3 Moles of Al(OH)3 = 0.107 g / 78.00 g/mol ≈ 0.00137 mol

Step 2: Calculate the moles of Al2(SO4)3 in the sample

Since there is a 1:1 ratio between moles of Al(OH)3 and moles of Al2(SO4)3, we can simply use the moles of Al(OH)3 to find the moles of Al2(SO4)3 in the sample. Moles of Al2(SO4)3 = Moles of Al(OH)3 / 2 = 0.00137 mol / 2 ≈ 0.000685 mol

Step 3: Calculate the mass of Al2(SO4)3 in the sample

Now, we can find the mass of Al2(SO4)3 in the sample by multiplying the moles of Al2(SO4)3 by the molar mass of Al2(SO4)3. The molar mass of Al2(SO4)3 is: Al: 26.98 g/mol S: 32.07 g/mol O: 16.00 g/mol Molar mass of Al2(SO4)3 = 2(Al) + 3(2S + 4O) = 2(26.98) + 3(2(32.07) + 4(16.00)) ≈ 342.15 g/mol Mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3 Mass of Al2(SO4)3 = 0.000685 mol * 342.15 g/mol ≈ 0.234 g

Step 4: Calculate the mass percent of Al2(SO4)3 in the sample

Finally, we can find the mass percent of Al2(SO4)3 in the sample by dividing the mass of Al2(SO4)3 by the total mass of the sample and multiplying by 100. Mass percent of Al2(SO4)3 = (Mass of Al2(SO4)3 / Mass of the sample) * 100 Mass percent of Al2(SO4)3 = (0.234 g / 1.45 g) * 100 ≈ 16.14 % Thus, the mass percent of Al2(SO4)3 in the sample is approximately 16.14%.

Key Concepts

Molar Mass

The concept of molar mass is fundamental in chemistry as it allows us to calculate the amount of substance present in a given mass. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). To determine the molar mass of a compound, you simply add up the atomic masses of all the elements present in the compound, according to their proportion. For example, for \(\mathrm{Al(OH)_3}\), you calculate the molar mass by summing up:

• One aluminum atom: 26.98 g/mol
• Three oxygen atoms: 3 \(\times\) 16.00 g/mol = 48.00 g/mol
• Three hydrogen atoms: 3 \(\times\) 1.01 g/mol = 3.03 g/mol

This results in a total molar mass of approximately 78.00 g/mol for \(\mathrm{Al(OH)_3}\). Similarly, for \(\mathrm{Al_2(SO_4)_3}\), the molar mass is calculated by adding:

• Two aluminum atoms
• Three sulfate ions (each consisting of one sulfur and four oxygen atoms)

This calculation gives a molar mass of approximately 342.15 g/mol.

Precipitation Reaction

A precipitation reaction occurs when two soluble substances in a solution react to form an insoluble solid product, known as the precipitate. In our exercise, when an excess of \(\mathrm{NaOH}\) is added to a solution containing \(\mathrm{Al_2(SO_4)_3}\), \(\mathrm{Al(OH)_3}\) is formed as a precipitate. Precipitation reactions are frequently utilized to isolate certain elements or compounds from a mixture.
The key to understanding these reactions is recognizing the solubility rules of the compounds. \(\mathrm{Al(OH)_3}\) is insoluble in water, causing it to separate out from the solution upon formation. The equation for the precipitation reaction is a clean representation of this process:

• \(\text{\(\mathrm{Al_2(SO_4)_3} + 6\mathrm{NaOH} \rightarrow 2\mathrm{Al(OH)_3} + 3\mathrm{Na_2SO_4}\)}\)

Here, the solid \(\mathrm{Al(OH)_3}\) can be collected by filtration.

Mass Percent

Mass percent is a way to express the concentration of an element in a mixture or compound. It is calculated by dividing the mass of the component of interest by the total mass of the mixture, then multiplying the result by 100 to get a percentage. In the given exercise, to find the mass percent of \(\mathrm{Al_2(SO_4)_3}\) in the mixture, you use the formula:

• \((\text{mass of component}/\text{total mass of mixture}) \times 100\)

Substituting the respective values:

• Mass of \(\mathrm{Al_2(SO_4)_3}\) is approximately 0.234 g
• Total mass of the sample is 1.45 g
• Mass percent = \((0.234 g / 1.45 g) \times 100 \approx 16.14\%)\)

This means that about 16.14% of the sample is made up of \(\mathrm{Al_2(SO_4)_3}\). Calculating mass percent is essential in chemical analysis and ensures accurate formulation of solutions.

Chemical Reaction

A chemical reaction involves the rearrangement of atoms to transform reactants into products. In our exercise, the chemical reaction can be expressed as:

• \(\text{\(\mathrm{Al_2(SO_4)_3} + 6\mathrm{NaOH} \rightarrow 2\mathrm{Al(OH)_3} + 3\mathrm{Na_2SO_4}\)}\)

In this reaction:

• Reactants: \(\mathrm{Al_2(SO_4)_3}\) and \(\mathrm{NaOH}\)
• Products: \(\mathrm{Al(OH)_3}\) and \(\mathrm{Na_2SO_4}\)

The precipitate \(\mathrm{Al(OH)_3}\) confirms that a chemical reaction has occurred. During a reaction, bonds between atoms in the reactants are broken and new bonds form in the products. Balancing chemical equations ensures the same number of each type of atom on both sides of the equation, obeying the law of conservation of mass. Understanding chemical reactions is vital for predicting the outcome of a chemical process and designing experiments accordingly.