Question

A Boron forms an extensive series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\) \(\mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\) excess \(\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{2}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If \(0.148 \mathrm{g}\) of \(\mathrm{B}_{x} \mathrm{H}_{y}\) gives \(0.422 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2},\) what is the empirical formula of \(\mathrm{B}_{x} \mathrm{H}_{y} ?\)

Short answer

The empirical formula is approximately \( \mathrm{B}_{5} \mathrm{H}_{7} \).

Step-by-step solution

Determine moles of B2O3 produced

Calculate the number of moles of \( \mathrm{B}_{2} \mathrm{O}_{3} \) produced by using its molar mass. The molar mass of \( \mathrm{B}_{2} \mathrm{O}_{3} \) is approximately 69.62 g/mol. \[\text{moles of } \mathrm{B}_{2} \mathrm{O}_{3} = \frac{0.422 \, \text{g}}{69.62 \, \text{g/mol}} \approx 0.00606 \, \text{mol}\]

Calculate moles of Boron (B) used

Since 1 mole of \( \mathrm{B}_{2} \mathrm{O}_{3} \) contains 2 moles of Boron, calculate the moles of Boron in the \( \mathrm{B}_{2} \mathrm{O}_{3} \):\[\text{moles of B} = 2 \times 0.00606 \, \text{mol} = 0.01212 \, \text{mol}\]

Determine mass of Boron in the sample

Use the molar mass of Boron (approximately 10.81 g/mol) to find the mass of Boron in the \( \mathrm{B}_{x} \mathrm{H}_{y} \):\[\text{mass of B} = 0.01212 \, \text{mol} \times 10.81 \, \text{g/mol} \approx 0.131 \, \text{g}\]

Determine mass of Hydrogen in the sample

Subtract the mass of Boron from the total mass of \( \mathrm{B}_{x} \mathrm{H}_{y} \) to find the mass of Hydrogen:\[\text{mass of H} = 0.148 \, \text{g} - 0.131 \, \text{g} = 0.017 \, \text{g}\]

Calculate moles of Hydrogen

The molar mass of Hydrogen (H) is approximately 1.01 g/mol. Use this to find the moles of Hydrogen:\[\text{moles of H} = \frac{0.017 \, \text{g}}{1.01 \, \text{g/mol}} \approx 0.0168 \, \text{mol}\]

Determine empirical formula

Divide the moles of Boron and Hydrogen by the smallest number of moles to find the simplest ratio:\[\text{Ratio of B: } \frac{0.01212}{0.01212} = 1\]\[\text{Ratio of H: } \frac{0.0168}{0.01212} \approx 1.39\]Since ratio sums to the nearest integer by roughly a factor of 2, the empirical formula is approximately \( \mathrm{B}_{5} \mathrm{H}_{7} \). This indicates some rounding due to empirical determination.

Key concepts

Boron Compounds

Boron is a fascinating element that forms a wide range of compounds, particularly with hydrogen, known as boranes. These compounds have general formulas like \( \text{B}_x\text{H}_y \), where \( x \) and \( y \) represent numbers of boron and hydrogen atoms, respectively. Boranes are unique because they can have different structures:

• They include simple ones like diborane, \( \text{B}_2\text{H}_6 \), where two boron atoms are bridged by hydrogen atoms.
• They can also form complex clusters that may include more than two or three borons.

These compounds are important in chemistry for their unusual bonding and as precursors in synthesizing other chemical compounds. Understanding boron compounds often involves exploring their structure, reactivity, and applications.
Boron forms strong bonds with hydrogen due to its high affinity for forming stable structures. The empirical formula of a boron hydride helps in identifying the simplest ratio of boron to hydrogen within the compound. This is essential for predicting reactivity and stability during chemical reactions. Thus, boron compounds are central in advancing knowledge in inorganic chemistry and material sciences.

Stoichiometry

Stoichiometry is a key concept in chemistry that involves calculations based on the reactants and products in chemical reactions. It allows chemists to predict the quantities of substances consumed and produced. The empirical formula calculation is a process rooted in stoichiometry, providing the simplest whole number ratio of the elements in a compound. In exercises like determining the empirical formula of boron hydrides:

• We begin by finding the mole ratio of each element in the compound using the given mass and molar masses.
• This requires converting the mass of boron and hydrogen into moles.
• Finally, simplify these ratios to the lowest whole numbers to deduce the empirical formula.

Stoichiometry also involves using balanced chemical equations. These help us relate the amounts of reactants used and the products formed, as highlighted by our chemical reaction involving \( \text{B}_x\text{H}_y \) and \( \text{O}_2 \) to form \( \text{B}_2\text{O}_3 \). By understanding stoichiometry, students gain insights into how elements interact and transform during chemical reactions, laying the groundwork for more complex chemical analyses.

Chemical Reactions

Chemical reactions describe the transformation of substances through the breaking and forming of bonds to create new products. In our example, the burning of boranes in the presence of oxygen is a classic oxidation reaction. Here, \( \text{B}_x\text{H}_y \) reacts with \( \text{O}_2 \) to form \( \text{B}_2\text{O}_3 \) and \( \text{H}_2\text{O} \). This reaction demonstrates key chemical principles:

• Combustion reactions require oxygen, leading to oxidation of the initial compound.
• The reaction produces thermal energy and often light, characteristic of combustion.

Chemical reactions are described by balanced chemical equations, which ensure mass and charge are conserved. In educational exercises, focus is often on understanding these conservation principles through calculations determining empirical formulas, predicting the products of reactions, and assessing reactivity.
Understanding chemical reactions is crucial, not just for predicting products, but also for designing processes in industry, such as in the production of chemicals or material development. Therefore, the study of chemical reactions is essential in both academia and practical applications in chemistry.