Problem 51 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

Step-by-Step Solution

Answer

The percentage of free \(\mathrm{SiO}_{2}\) in the rock is (c) \(0.50\%\).

Step 1: Analyze the composition of Kaolin

Understand that kaolin has a chemical formula of the form \(2\mathrm{SiO}_{2} \cdot \mathrm{Al}_{2}\mathrm{O}_{3}\). This means for each mole of \(\mathrm{Al}_{2}\mathrm{O}_{3}\), there are two moles of \(\mathrm{SiO}_{2}\). We will use this ratio to find out how much \(\mathrm{SiO}_{2}\) comes from kaolin in the extract.

Step 2: Calculate the moles of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) in the extract

Using the percentage found and the molar mass, calculate the moles of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) in the extract. The molar mass of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) is \(2(27) + 3(16) = 102 \, \text{g/mol}\). Therefore, \(1.02\%\) of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) corresponds to \(\frac{1.02}{102} \, \text{mol}\) per \(100 \, \text{g}\) of extract.

Step 3: Calculate the equivalent moles of \(\mathrm{SiO}_{2}\) from kaolin

For every mole of \(\mathrm{Al}_{2}\mathrm{O}_{3}\), there are two moles of \(\mathrm{SiO}_{2}\) in kaolin. This means \(2 \times \frac{1.02}{102} = \frac{2.04}{102} \, \text{mol}\) of \(\mathrm{SiO}_{2}\) per \(100 \, \text{g}\) of extract is from kaolin.

Step 4: Calculate the percentage of \(\mathrm{SiO}_{2}\) from kaolin

We need to convert moles of \(\mathrm{SiO}_{2}\) to mass and then to percentage. The molar mass of \(\mathrm{SiO}_{2}\) is \(28 + 2(16) = 60 \, \text{g/mol}\). Therefore, the weight of \(\mathrm{SiO}_{2}\) from kaolin is \(\frac{2.04}{102} \times 60 \, \text{g}\) and the corresponding percentage is \(\frac{\frac{2.04}{102} \times 60}{100} \%\).

Step 5: Calculate the percentage of free \(\mathrm{SiO}_{2}\) in the extract

Subtract the percentage of \(\mathrm{SiO}_{2}\) from kaolin found in Step 4 from the total percentage of \(\mathrm{SiO}_{2}\) in the extract to get the percentage of free \(\mathrm{SiO}_{2}\). So, free \(\mathrm{SiO}_{2} \% = 1.52\% - (\text{percentage of }\mathrm{SiO}_{2}\text{ from kaolin})\).

Step 6: Solve for the percentage of free \(\mathrm{SiO}_{2}\)

Apply the results from Step 2-5 to find the percentage of free \(\mathrm{SiO}_{2}\). This will help us determine the correct option among the ones given.

Key Concepts

Chemical Composition

In chemistry, understanding the chemical composition of substances is fundamental. It tells us what elements are present in a compound and in what ratios. Taking the substance kaolin, with the formula \(2\mathrm{SiO}_{2} \cdot \mathrm{Al}_{2}\mathrm{O}_{3}\), as an example, we can see that for every one part of aluminum oxide (\(\mathrm{Al}_{2}\mathrm{O}_{3}\)), there are two parts of silicon dioxide (\(\mathrm{SiO}_{2}\)).

In practical terms, when analyzing solubility or determining the free amount of a compound in a mixture, like free \(\mathrm{SiO}_{2}\) in a rock, chemical composition guides us to not only the presence of specific components but also to the contribution of each substance within the mixture. When these components bind in specific ratios inherent to their molecular structures, gleaming insights from their compositions impact how we analyze and quantify them in mixtures or solutions.

Molar Mass Calculation

Molar mass calculation is the process of determining the mass of one mole of a substance. The molar mass of an element is determined by its atomic weight, while for a compound, it is the sum total of the atomic weights of the atoms that make up the molecule. For instance, the molar mass of \(\mathrm{Al}_{2}\mathrm{O}_{3}\) is calculated by adding the atomic masses of two aluminum atoms (each with an approximate atomic mass of 27 g/mol) and three oxygen atoms (each with an approximate atomic mass of 16 g/mol), yielding a molar mass of 102 g/mol.

To complete accurate solubility analysis, it's essential to work out the molar mass as it allows you to convert from mass to moles or vice versa. This conversion is a cornerstone of quantitative chemistry, enabling comparisons and calculations that help deduce various attributes of the substances involved, such as finding out how much \(\mathrm{SiO}_{2}\) originated from kaolin in a given sample.

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is based on the law of conservation of mass and the concept of the mole. Stoichiometry requires a balanced chemical equation and using ratios derived from that equation to compute desired qualities.

In the context of our exercise, stoichiometry illuminates the quantitative relationship between \(\mathrm{SiO}_{2}\) and \(\mathrm{Al}_{2}\mathrm{O}_{3}\) in kaolin. With the stoichiometric ratio, chemists can reveal not just the 'what' but the 'how much,' which is crucial for separating and identifying the amounts of free \(\mathrm{SiO}_{2}\) versus that bound in kaolin within the rock sample. Ultimately, stoichiometry allows calculating unknowns in the context of the known, guiding students in uncovering precise percentages as seen in the step-by-step solution provided for our exercise.