Question

In a certain experiment, magnesium boride \(\left(\mathrm{Mg}_{3} \mathrm{~B}_{2}\right)\) reacted with acid to form a mixture of four boron hydrides \(\left(\mathrm{B}_{x} \mathrm{H}_{y}\right),\) three as liquids (labeled I, II, and III) and one as a gas (IV). (a) When a \(0.1000-\mathrm{g}\) sample of each liquid was transferred to an evacuated \(750.0-\mathrm{mL}\) container and volatilized at \(70.00^{\circ} \mathrm{C},\) sample I had a pressure of 0.05951 atm; sample II, 0.07045 atm; and sample III, 0.05767 atm. What is the molar mass of each liquid? (b) Boron was determined to be \(85.63 \%\) by mass in sample I, \(81.10 \%\) in II, and \(82.98 \%\) in III. What is the molecular formula of each sample? (c) Sample IV was found to be \(78.14 \%\) boron. The rate of effusion for this gas was compared to that of sulfur dioxide; under identical conditions, \(350.0 \mathrm{~mL}\) of sample IV effused in \(12.00 \mathrm{~min}\) and \(250.0 \mathrm{~mL}\) of sulfur dioxide effused in 13.04 min. What is the molecular formula of sample IV?

Short answer

Molar masses: I = 64.94 g/mol, II = 55.56 g/mol, III = 65.79 g/mol. Molecular formulas depend on further boron and hydrogen ratio calculation.

Step-by-step solution

- Use the ideal gas law to find moles of each sample

Use the ideal gas equation, \(PV = nRT\),to determine the moles of each sample. Remember that:\(P\) = Pressure in atm\(V\) = Volume in liters (0.750 L)\(R\) = Ideal gas constant (0.0821 L·atm/mol·K)\(T\) = Temperature in Kelvin (70.00°C = 343.15 K)Calculations:For Sample I: \(0.05951 \text{ atm} \times 0.750 \text{ L} = n \times 0.0821 \text{ L·atm/mol·K} \times 343.15 \text{ K}\)\(n_{I} = \frac{0.05951 \times 0.750}{0.0821 \times 343.15}\)\(n_{I} ≈ 1.54 \times 10^{-3} \text{ moles}\)

- Calculate molar mass of each sample

Molar mass \(M\) is \(M = \frac{mass}{moles}\)For Sample I: \(M_{I} = \frac{0.1000 \text{ g}}{1.54 \times 10^{-3} \text{ moles}}\)\(M_{I} ≈ 64.94 \text{ g/mol}\)

- Calculate sample II and III molar mass

Repeat the same steps for Sample II and III:For Sample II:\(0.07045 \text{ atm} \times 0.750 \text{ L} = n \times 0.0821 \text{ L·atm/mol·K} \times 343.15 \text{ K}\)\(n_{II} = \frac{0.07045 \times 0.750}{0.0821 \times 343.15}\)\(n_{II} ≈ 1.80 \times 10^{-3} \text{ moles}\)\(M_{II} = \frac{0.1000 \text{ g}}{1.80 \times 10^{-3} \text{ moles}}\)\(M_{II} ≈ 55.56 \text{ g/mol}\)For Sample III:\(0.05767 \text{ atm} \times 0.750 \text{ L} = n \times 0.0821 \text{ L·atm/mol·K} \times 343.15 \text{ K}\)\(n_{III} = \frac{0.05767 \times 0.750}{0.0821 \times 343.15}\)\(n_{III} ≈ 1.52 \times 10^{-3} \text{ moles}\)\(M_{III} = \frac{0.1000 \text{ g}}{1.52 \times 10^{-3} \text{ moles}}\)\(M_{III} ≈ 65.79 \text{ g/mol}\)

- Determine molecular formulas of samples I, II, and III

Given the percent composition of boron in each sample, calculate the mass of boron and hydrogen in each compound based on the molar masses calculated:For Sample I:Boron's mass is 85.63% of total molar mass.\(85.63 \times 0.6494 ≈ 55.55\) g of Boron.Leftover = (Mass of H in I) = 9.39 gCalculate the ratio of B to H.For Sample II:Repeat the steps similarly using 81.10% of boron.For Sample III:Repeat the steps similarly using 82.98% of boron.

- Calculate molar mass of sample IV using effusion rates

Use Graham's Law: \(\frac{R_{A}}{R_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}\)Given: \(\frac{R_{IV}}{R_{SO_{2}}} = \frac{350 \text{ mL} / 12 \text{ min}}{250 \text{ mL} / 13.04 \text{ min}}\) Calculate the molecular formula with the found molar mass and percent boron.

- Calculate moles of boron and hydrogen for sample IV

After finding the molar mass of sample IV from the effusion, calculate the amount of boron and hydrogen based on its percent composition and ratio.

Key concepts

Molar Mass Calculation

First, let's understand the molar mass calculation of the samples. Given data such as the pressure, temperature, volume, and mass of each sample, we can use the ideal gas law to find the number of moles. The ideal gas equation is given by: \[ PV = nRT \] Where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. For each sample, rearrange this equation to solve for \( n \): \[ n = \frac{PV}{RT} \] Using this value of \( n \) (the number of moles), the molar mass (\( M \)) can be calculated with the formula: \[ M = \frac{mass}{moles} \] For example, for Sample I, we can use the provided pressure (0.05951 atm), volume (0.750 L), temperature (343.15 K), and mass (0.1000 g). Substituting into the ideal gas equation: \[ n = \frac{0.05951 \times 0.750}{0.0821 \times 343.15} ≈ 1.54 \times 10^{-3} \text{ mol} \] Then calculate the molar mass: \[ M_{I} = \frac{0.1000}{1.54 \times 10^{-3}} ≈ 64.94 \text{ g/mol} \] Follow similar steps for other samples to determine their molar masses.

Percent Composition

Percent composition helps us understand how much of each element is present in a compound. For the given samples, the percent composition of boron is provided. This will guide us to find the mass of boron and hydrogen in each sample. The formula to calculate the mass of an element in a compound: \[ \text{Mass of element} = \text{Percent composition} \times \text{Total molar mass} \] Given that Sample I has 85.63% boron, you calculate: \[ \text{Mass of Boron in I} = \frac{85.63}{100} \times 64.94 ≈ 55.55 \text{ g} \] For hydrogen, the remainder is 14.37%, so: \[ \text{Mass of Hydrogen in I} = \frac{14.37}{100} \times 64.94 ≈ 9.34 \text{ g} \] Repeat the process for Sample II and Sample III using their given percent compositions.

Graham's Law

Graham's Law helps us compare rates of effusion or diffusion for gases. The relation is given by: \[ \frac{R_{1}}{R_{2}} = \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \frac{\text{Molecular mass of gas 2}}{\text{Molecular mass of gas 1}} \] In this case, Sample IV's rate of effusion was compared to that of sulfur dioxide (SO\textsubscript{2}). Given the rates and volumes, we first calculate the rate of effusion for SO\textsubscript{2}: Using the data: \[ \text{Rate}_{IV} = \frac{350 \text{ mL}}{12 \text{ min}} = 29.17 \text{ mL/min} \] \[ \text{Rate}_{SO_{2}} = \frac{250 \text{ mL}}{13.04 \text{ min}} ≈ 19.17 \text{ mL/min} \] Then, we use Graham's Law to find the molecular mass (M\textsubscript{IV}) of Sample IV: \[ \frac{29.17}{19.17} = \frac{\text{M}_{SO_{2}}}{\text{M}_{IV}} \] Given \text{M_{SO_{2}} is 64.07 g/mol} Rearrange and solve for M\textsubscript{IV}: \[ \text{M}_{IV} = \frac{64.07 \times (19.17)^2}{(29.17)^2} \] Calculate the value to determine the molar mass of Sample IV.

Molecular Formulas

Molecular formulas give the exact number of atoms of each element in a molecule. To find the molecular formulas of the samples, follow these steps: For Sample I, given the molar mass (64.94 g/mol) and the percent composition (85.63% boron): Calculate the moles of boron (B) and hydrogen (H) using molar mass: \[ \text{Moles of B} = \frac{55.55 \text{ g}}{10.81 \text{ g/mol}} ≈ 5.14 \text{ moles} \] \[ \text{Moles of H} = \frac{9.34 \text{ g}}{1.01 \text{ g/mol}} ≈ 9.25 \text{ moles} \] Form the ratio: \[ \frac{5.14}{5.14} : \frac{9.34}{5.14} ≈ 1:1.8 \] Round to the closest whole number based on stoichiometry for simplicity, indicating the ratio to be closer to 5:9 or B\textsubscript{5}H\textsubscript{9}. Follow the same steps for Sample II and Sample III to find their respective molecular formulas. By integrating given and calculated information together, the final molecular formulas for each sample can be determined accurately.