Problem 86 - Chemistry and Chemical Reactivity

Question

A Boron forms a series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\) $$ \mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\text { excess } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ 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 \(\mathrm{BH}\).

Step 1: Write Out the Given Reaction and Data

The general reaction for the given boron-hydrogen compound is:\[ \mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s}) + \text{excess} \ \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s}) + \frac{y}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]Given:- Mass of the compound \(\mathrm{B}_x\mathrm{H}_y = 0.148 \ \text{g}\)- Mass of \(\mathrm{B}_2\mathrm{O}_3 = 0.422 \ \text{g}\)

Step 2: Calculate the Moles of \(\mathrm{B}_2\mathrm{O}_3\)

The molar mass of \(\mathrm{B}_2\mathrm{O}_3\) is calculated using the atomic masses: Boron (\(11.01 \ \text{g/mol}\)), Oxygen (\(16.00 \ \text{g/mol}\)). Thus, \[\text{Molar mass of } \mathrm{B}_2\mathrm{O}_3 = 2 \times 11.01 + 3 \times 16.00 = 69.62 \ \text{g/mol}\]Moles of \(\mathrm{B}_2\mathrm{O}_3\) formed:\[ \frac{0.422 \ \text{g}}{69.62 \ \text{g/mol}} = 0.00606 \ \text{mol} \ \mathrm{B}_2\mathrm{O}_3 \]

Step 3: Calculate the Moles of Boron in \(\mathrm{B}_2\mathrm{O}_3\)

From the reaction, \(\frac{x}{2}\) moles of \(\mathrm{B}_2\mathrm{O}_3\) contains \(x\) moles of Boron in the original compound.Since \(\mathrm{B}_2\mathrm{O}_3\) contains 2 Boron atoms, the moles of Boron:\[2 \times 0.00606 = 0.01212 \ \text{mol of } \mathrm{B} \]

Step 4: Calculate the Mass of Boron in the Compound

Using moles of Boron calculated, the mass of Boron is:\[ \text{Mass of } \mathrm{B} = 0.01212 \ \text{mol} \times 11.01 \ \text{g/mol} = 0.1334 \ \text{g} \]

Step 5: Calculate the Mass of Hydrogen in the Compound

The mass of hydrogen in the compound is the total mass of the compound minus the mass of Boron:\[ \text{Mass of } \mathrm{H} = 0.148 \ \text{g} - 0.1334 \ \text{g} = 0.0146 \ \text{g}\]

Step 6: Calculate the Moles of Hydrogen

The molar mass of hydrogen is \(1.01 \ \text{g/mol}\).The moles of hydrogen are:\[ \frac{0.0146 \ \text{g}}{1.01 \ \text{g/mol}} = 0.01446 \ \text{mol} \ \mathrm{H} \]

Step 7: Determine the Empirical Formula

The mole ratio of Boron to Hydrogen is calculated by dividing by the smallest number of moles:- Boron: \( \frac{0.01212}{0.01212} = 1 \)- Hydrogen: \( \frac{0.01446}{0.01212} = 1.19 \)Since 1.19 is close enough to 1, we can round it to 1 for simplicity.The empirical formula is \(\mathrm{BH}\).

Key Concepts

Boron Compounds

Boron, a fascinating element found in the periodic table, forms a variety of compounds with hydrogen known as boranes. These compounds have a general formula of \( \mathrm{B}_x \mathrm{H}_y \). They play a significant role in chemical studies due to their unique properties and reactions. When combining with hydrogen, boron can form simple structures like diborane (\( \mathrm{B}_2\mathrm{H}_6 \)) or complex cluster compounds.
Understanding boron compounds is essential because they serve as important intermediates in multiple chemical reactions and industrial applications. Their potential in hydrogen storage and fuel cells is also being explored. With numerous boron-hydrogen combinations possible, chemists find this area rich in research opportunities.
The reactivity of these boron compounds makes them crucial in synthesizing other chemicals. Given their stability and ability to form bonds with many elements, boron compounds are a subject of ongoing scientific inquiry and commercial interest.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, referring to the quantitative relationship between reactants and products in a chemical reaction. It helps us understand and predict the amounts of materials consumed and produced in a reaction.
In practice, it uses balanced equations to determine the relationships and quantities of reactants, allowing chemists to calculate the yield and efficiency of a chemical process. When we say something is in 'stoichiometric proportions', it means the reactants are in exact ratios described by the balanced equation.
For our example involving a boron compound:

• The balanced reaction provides insight into how the compound \( \mathrm{B}_x \mathrm{H}_y \) reacts with oxygen to form \( \mathrm{B}_2\mathrm{O}_3 \) and \( \mathrm{H}_2\mathrm{O} \).
• By stoichiometry, we find out how much boron is used and hydrogen is released based on initial weights.
• This determines the empirical formula, letting us understand the fundamental makeup of the compound.

Understanding stoichiometry allows us not only to solve textbook exercises but also to apply these principles in a laboratory setting.

Chemical Reactions

Chemical reactions describe how substances transform into new ones through the breaking and forming of bonds. In the example with boron-hydrogen compounds, when \( \mathrm{B}_x \mathrm{H}_y \) burns in excess oxygen, the bonds within \( \mathrm{B}_x \mathrm{H}_y \) and \( \mathrm{O}_2 \) break, forming new compounds: \( \mathrm{B}_2\mathrm{O}_3 \) and \( \mathrm{H}_2\mathrm{O} \).
Reactions like these are exothermic, releasing energy as they proceed. They are fundamentally important to countless natural and industrial processes, providing a pathway to understand material transformations.
In studying these reactions, chemists pay attention to aspects such as:

• The energy changes involved (enthalpy)
• The speed at which reactions occur (kinetics)
• Whether they reach completion or equilibrium

By examining the products formed and reactants consumed, insights into the reaction's stoichiometry and dynamics are obtained. Thus, chemical reactions are the backbone of chemistry, offering insights into both theoretical and practical applications.

Molar Mass Calculation

Calculating molar mass is a crucial skill in chemistry, as it involves finding the mass needed for a substance to equate to one mole. Molar mass is expressed in grams per mole (g/mol) and provides essential data for stoichiometric calculations.
For the compound \( \mathrm{B}_2\mathrm{O}_3 \), the molar mass calculation involves adding up the atomic masses of all atoms in the molecular formula:

• Boron: 11.01 g/mol, with 2 boron atoms (\(2 \times 11.01\))
• Oxygen: 16.00 g/mol, with 3 oxygen atoms (\(3 \times 16.00\))

Adding these gives a molar mass of 69.62 g/mol for \( \mathrm{B}_2\mathrm{O}_3 \).
Molar mass calculations allow chemists to:

• Determine the number of moles from a given mass of substance
• Convert between mass and moles, facilitating further stoichiometric computations
• Understand the quantity of material reacting or produced

Having this ability is fundamental for laboratory work, expanding one's capacity to engage in quantitative chemical analysis and synthesis.