Problem 88 - Chemistry and Chemical Reactivity

Question

Boron forms a series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\). $$\mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\text { excess } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{\gamma}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ If \(0.148 \mathrm{g}\) of one of these compounds gives \(0.422 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2},\) what is its empirical formula?

Step-by-Step Solution

Answer

The empirical formula is \(\mathrm{B}_2\mathrm{H}_3\).

Step 1: Determine moles of B2O3 produced

Given that the molar mass of \(\text{B}_2\text{O}_3\) is approximately \(69.62 \text{ g/mol}\), we can calculate the moles of \(\text{B}_2\text{O}_3\) by using the formula \( n = \frac{m}{M} \), where \(m\) is the mass and \(M\) is the molar mass. \[ n(\text{B}_2\text{O}_3) = \frac{0.422 \text{ g}}{69.62 \text{ g/mol}} \approx 0.00606 \text{ mol}\]

Step 2: Calculate moles of Boron used

Since \(\text{B}_2\text{O}_3\) contains two boron atoms, the moles of boron used in forming \(\text{B}_2\text{O}_3\) is twice the moles of \(\text{B}_2\text{O}_3\).\[ moles\,\text{of}\,\text{B} = 2 \times 0.00606 \text{ mol} = 0.01212 \text{ mol}\]

Step 3: Find mass of Boron used

The atomic mass of boron is approximately \(10.81 \text{ g/mol}\). To find the mass of boron used, use the formula \( m = n \times M \).\[ m(\text{B}) = 0.01212 \text{ mol} \times 10.81 \text{ g/mol} \approx 0.131 \text{ g}\]

Step 4: Determine mass of Hydrogen in compound

Given the total mass of the compound \(0.148\text{ g}\), calculate the mass of hydrogen by subtracting the mass of boron from the compound mass.\[ m(\text{H}) = 0.148 \text{ g} - 0.131 \text{ g} = 0.017 \text{ g}\]

Step 5: Calculate moles of Hydrogen

The atomic mass of hydrogen is approximately \(1.01 \text{ g/mol}\). Use \( n = \frac{m}{M} \) to find moles of hydrogen.\[ n(\text{H}) = \frac{0.017 \text{ g}}{1.01 \text{ g/mol}} \approx 0.01683 \text{ mol}\]

Step 6: Determine the mole ratio and empirical formula

Divide the moles of each element by the smallest number of moles to find the simplest whole-number ratio.\[ \text{Ratio of B} = \frac{0.01212}{0.01212} = 1\]\[ \text{Ratio of H} = \frac{0.01683}{0.01212} \approx 1.39 \approx 1.5\]Multiply the ratios by the smallest number to get whole numbers:\[ \text{B:H} = 1:1.5 \times 2 = 2:3\]Thus, the empirical formula is \(\text{B}_2\text{H}_3\).

Key Concepts

Mole Calculations

When we talk about mole calculations, we're referring to a method used by chemists to count particles, such as atoms or molecules, in a given sample. The mole is a fundamental unit in chemistry that makes these calculations manageable. Understanding the concept of the mole allows us to transition between the atomic scale and the macroscopic world. Here's a quick breakdown:

• One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles (atoms, molecules, etc.).
• The molar mass, which you can find on the periodic table, is the mass of one mole of any substance expressed in grams per mole.
• To calculate moles, use the formula \(n = \frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass in grams, and \(M\) is the molar mass.

In the exercise, we used these principles to calculate the number of moles of \(\text{B}_2\text{O}_3\) produced and subsequently determine the moles of boron and hydrogen in the compound.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. This transformation can be represented by a balanced chemical equation like the one provided in the exercise.
It's important to remember a few key aspects about chemical reactions:

• Reactions must obey the law of conservation of mass, meaning the mass of the products must equal the mass of the reactants.
• A balanced chemical equation gives the mole ratios of the reactants and products involved, which is crucial for stoichiometric calculations.
• In the provided reaction, \(\mathrm{B}_{x}\mathrm{H}_{y}\) reacts with oxygen to form \(\mathrm{B}_2\mathrm{O}_3\) and water. By studying this reaction, we derived the empirical formula of the compound based on the experimental data.

Understanding chemical reactions, including their representation and implications, is a cornerstone of chemistry, allowing us to predict and calculate the outcome of various processes.

Boron Hydride Compounds

Boron hydride compounds, known as boranes, are interesting due to their unique bonding and structure. These compounds consist of boron and hydrogen only, and they exhibit fascinating chemistry. Here are some insights about boron hydrides:

• They are classified by the number of boron and hydrogen atoms, represented as \(\text{B}_x \text{H}_y\).
• Boron acts as a bridge between hydrogen atoms, often forming unique cage-like or cluster structures.
• The empirical formula, such as \(\text{B}_2\text{H}_3\) in the exercise solution, gives the simplest whole-number ratio of the atoms in a compound.

Boron hydrides are still a topic of research due to their potential applications and chemical properties. Understanding these compounds helps in appreciating the complexity and diversity the elements of the periodic table can form.