Question

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) A \(1.45-\mathrm{g}\) sample of the mixture is dissolved in water and an excess of \(\mathrm{NaOH}\) 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?

Short answer

The mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample is approximately \(16.15\%\).

Step-by-step solution

Calculate the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the precipitate.

Use the molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\) and the mass of the precipitate to find the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\). The molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(26.98+3\times(16.00+1.00)=78.00\) g/mol. Moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) = \(\frac{0.107\ \mathrm{g}}{78.00\ \mathrm{g/mol}}\) = \(1.371 \times 10^{-3}\) mol

Convert moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) to moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).

Since the molar ratio of \(\mathrm{Al}(\mathrm{OH})_{3}\) to \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is 2:1, you can find the moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample by dividing the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) by 2. Moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) = \(\frac{\mathrm{moles\ of\ Al(OH)_3}}{2}\) = \(\frac{1.371 \times 10^{-3}\ \mathrm{mol}}{2}\) = \(6.855\times 10^{-4}\) mol

Convert moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to mass.

Calculate the molar mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) = \(2\times26.98 + 3\times(4\times16.00 + 3\times1.00)\) = \(341.88\) g/mol, and then multiply it by the moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to find the mass in sample. Mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) = \(6.855\times 10^{-4}\ \mathrm{mol} \times 341.88\ \mathrm{g/mol}\) ≈ \(0.2342\) g

Calculate the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the mixture.

Divide the mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) by the total mass of the mixture and multiply by 100 to obtain the mass percent. Mass Percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) = \(\frac{0.2342\ \mathrm{g}}{1.45\ \mathrm{g}} \times 100\) ≈ \(16.15\%\) So, the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample is approximately \(16.15\%\).

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It's like a recipe for a chemical reaction that tells you how much of each substance is needed to react completely with other substances.

For instance, when working on a stoichiometry problem like our practice exercise, you determine the ratio of reactants to products. Often, this ratio comes from a balanced chemical equation. In the provided example, we had to understand the relationship between \(\mathrm{Al(OH)_3}\) and \(\mathrm{Al_2(SO_4)_3}\), which is vital in deducing the amount of \(\mathrm{Al_2(SO_4)_3}\) in the original mixture. Knowing how stoichiometry links the mass of reactants to the mass of products is crucial, as the concept was used to calculate the mass percent composition of the mixture.

One can also use stoichiometry to calculate the amounts of reactants required or products formed, which is often the cornerstone of chemical manufacturing and laboratory work.

Molar Mass

The molar mass is a property of a substance that tells us the mass in grams of one mole of that substance. It is the sum of the atomic masses of all the atoms in a molecule, measured in grams per mole (g/mol).

The calculation of molar mass is a fundamental step in stoichiometry because it allows the conversion between grams and moles, which is necessary for using the mole ratio from the balanced equation to calculate the amounts of reactants and products involved in a chemical reaction.

In our exercise, for example, calculating the molar mass of \(\mathrm{Al(OH)_3}\) correctly is essential as it leads to the determination of the moles of \(\mathrm{Al(OH)_3}\) from its given mass. Molar mass acts as a bridge between the mass of a substance and the number of moles, making it easier to compare the quantities of different substances involved in the chemical reaction.

Chemical Reaction

A chemical reaction is a process where reactants transform into products through breaking and forming of chemical bonds. This can be observed in various forms such as color change, temperature change, emission of gas, or formation of a solid (precipitate).

In our sample problem, the reaction involved the formation of a precipitate, \(\mathrm{Al(OH)_3}\), which can be filtered and weighed. This is a classic type of reaction called a 'precipitation reaction' used to isolate products in a solid form.

Understanding chemical reactions entails knowing how compounds interact with each other and what conditions are necessary for the reaction to occur – such as temperature, concentration, or the presence of catalysts. The knowledge of the specific sequence of steps in a reaction mechanism can also be useful for predicting the products and yields of reactions.