Question

Boron forms an extensive series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\). To analyze one of these compounds, you burn it in air and isolate the boron in the form of \(\mathrm{B}_{2} \mathrm{O}_{3}\) and the hydrogen in the form of water. You find that \(0.1482 \mathrm{~g} \mathrm{~B}_{x} \mathrm{H}_{y}\) gives \(0.4221 \mathrm{~g}\) \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2} .\) Determine the empirical formula of \(\mathrm{B}_{x} \mathrm{H}_{y}\)

Short answer

The empirical formula is B$_2$H$_3$.

Step-by-step solution

Determine Moles of B$_2$O$_3$

First, calculate the moles of \( \mathrm{B}_2\mathrm{O}_3 \) formed. The molar mass of \( \mathrm{B}_2\mathrm{O}_3 \) is approximately 69.62 g/mol.\[\text{Moles of } \mathrm{B}_2\mathrm{O}_3 = \frac{0.4221 \text{ g}}{69.62 \text{ g/mol}} \approx 0.00606 \text{ mol}\]

Calculate Moles of Boron in B$_2$O$_3$

In \( \mathrm{B}_2\mathrm{O}_3 \), there are 2 moles of boron for every mole of \( \mathrm{B}_2\mathrm{O}_3 \).\[\text{Moles of Boron} = 2 \times 0.00606 \text{ mol} = 0.01212 \text{ mol}\]

Calculate Mass of Boron

Determine the mass of boron in the compound using its atomic mass.Atomic mass of boron is approximately 10.81 g/mol.\[\text{Mass of Boron} = 0.01212 \text{ mol} \times 10.81 \text{ g/mol} = 0.1310 \text{ g}\]

Determine Mass of Hydrogen in B$_x$H$_y$

Subtract the mass of boron from the original compound mass to find the mass of hydrogen in \( \mathrm{B}_x\mathrm{H}_y \).\[\text{Mass of Hydrogen} = 0.1482 \text{ g} - 0.1310 \text{ g} = 0.0172 \text{ g}\]

Calculate Moles of Hydrogen

Convert the mass of hydrogen to moles using the atomic mass of hydrogen (1.008 g/mol).\[\text{Moles of Hydrogen} = \frac{0.0172 \text{ g}}{1.008 \text{ g/mol}} \approx 0.01706 \text{ mol}\]

Determine Empirical Formula Ratios

Calculate the mole ratio of boron to hydrogen to determine the empirical formula. Divide by the smallest number of moles calculated.\[\text{Ratio of Boron:} \frac{0.01212}{0.01212} = 1\]\[\text{Ratio of Hydrogen:} \frac{0.01706}{0.01212} \approx 1.41\]

Finalize Empirical Formula

Since the calculated ratio of hydrogen is closest to 1.5, we multiply all subscripts by 2 to clear the fraction.\[\text{Empirical Formula: } \mathrm{B}_2\mathrm{H}_3\]

Key concepts

Boron Hydrides

Boron hydrides are fascinating compounds formed by boron and hydrogen with the general formula \( \mathrm{B}_x \mathrm{H}_y \). These compounds are known for their intricate structures and interesting properties. They vary in the number of boron and hydrogen atoms, resulting in various hydride forms such as diborane \( \mathrm{B}_2\mathrm{H}_6 \) and more complex structures. These compounds are significant in chemistry due to their application in materials science and potential use in hydrogen storage. Understanding and analyzing boron hydrides often involves determining their chemical formulas through several methods, including combustion analysis. In this process, the compound is burned to ascertain its empirical formula, which gives insight into the elemental composition and ratios of boron to hydrogen in the compound.

Combustion Analysis

Combustion analysis is a widely used technique in chemistry to determine the elemental composition of a compound. It involves burning the compound in the presence of oxygen and measuring the products formed. For boron hydrides, the compound is burned, and the boron and hydrogen are isolated as \( \mathrm{B}_2 \mathrm{O}_3 \) and \( \mathrm{H}_2 \mathrm{O} \) respectively. By measuring the mass of these products, one can deduce the amount of each element in the original compound. This data is crucial for calculating the empirical formula.

To carry out the analysis:

• Burn the compound in excess \( \mathrm{O}_2 \) to ensure complete combustion.
• Isolate and measure the mass of \( \mathrm{B}_2 \mathrm{O}_3 \) and \( \mathrm{H}_2 \mathrm{O} \).
• Use the mass and the known molar masses to calculate moles of boron and hydrogen.

This process, although detailed, provides an accurate representation of the elemental makeup of the compound, essential for empirical formula calculations.

Stoichiometry

Stoichiometry is the mathematical foundation of chemical reactions and formulas, focusing on the quantitative relationships between reactants and products. In the context of deriving an empirical formula from combustion analysis, stoichiometry is invaluable. It allows chemists to convert measured masses to moles using molar masses, revealing the proportions of elements in a compound.

In the analysis of boron hydrides, stoichiometry is applied to:

• Convert the measured mass of \( \mathrm{B}_2 \mathrm{O}_3 \) to moles of boron.
• Determine the mass of hydrogen directly from the difference between the original sample mass and the boron mass.
• Convert this hydrogen mass to moles.
• Calculate the boron to hydrogen mole ratio to deduce the empirical formula.

Understanding stoichiometry is crucial for balancing chemical equations and deriving meaningful chemical formulas from empirical data. It provides the necessary calculations that transform experimental observations into accurate chemical representations.