Chapter 5, Problem 5.93P

(a)

Interpretation Introduction

Interpretation:

The molar mass of each liquid in all the samples at $70.00°\text{C}$ and $750.0\text{ mL}$ is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows:

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the moles of gas is as follows:

  $\text{Moles of gas}=\frac{\text{Mass of gas}}{\text{Molar mass of gas}}$

Answer to Problem 5.93P

The molar mass of liquid in sample I is $\text{63}\text{.09 g/mol}$, the molar mass of liquid in sample II is $\text{53}\text{.29 g/mol}$ and the molar mass of liquid in sample III is $\text{65}\text{.10 g/mol}$.

Explanation of Solution

The formula to convert $°\text{C}$ to Kelvin is:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (1)

Substitute $\text{70 }°\text{C}$ for $\text{T }°\text{C}$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=70°\text{C}+273.15\\ =343.15\text{ K}\end{array}$

The expression to calculate the molar mass of sample I is as follows:

  $PV=\frac{m}{M}RT$        (2)

Here, $m$ is the mass of the sample and $M$ is the molar mass.

Rearrange equation (2) for $M$ as follows:

  $M=\frac{mRT}{PV}$        (3)

Substitute the value $\text{0}\text{.05951 atm}$ for $P$, $0.1000\text{ g}$ for the mass of sample $\text{343}\text{.15 K}$ for $T$, $\text{750 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (3).

  $\begin{array}{c}M=\frac{\left(0.1000\text{ g}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{343}\text{.15 K}\right)}{\left(\text{0}\text{.05951 atm}\right)\left(\text{0}\text{.7500 L}\right)}\\ =63.0905\text{ g/mol}\end{array}$

Substitute the value $\text{0}\text{.07045 atm}$ for $P$, $0.1000\text{ g}$ for the mass of sample $\text{343}\text{.15 K}$ for $T$, $\text{750 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (3) to calculate molar mass of sample II.

  $\begin{array}{c}M=\frac{\left(0.1000\text{ g}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{343}\text{.15 K}\right)}{\left(\text{0}\text{.07045 atm}\right)\left(\text{0}\text{.7500 L}\right)}\\ =53.293\text{ g/mol}\\ \approx 53.29\text{ g/mol}\end{array}$

Substitute the value $\text{0}\text{.05767 atm}$ for $P$, $0.1000\text{ g}$ for the mass of sample $\text{343}\text{.15 K}$ for $T$, $\text{750 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (3) to calculate molar mass of sample III.

  $\begin{array}{c}M=\frac{\left(0.1000\text{ g}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{343}\text{.15 K}\right)}{\left(\text{0}\text{.05767 atm}\right)\left(\text{0}\text{.7500 L}\right)}\\ =65.10349\text{ g/mol}\\ \approx 65.10\text{ g/mol}\text{.}\end{array}$

Conclusion

The molar mass of liquid in sample III is greater than the molar mass of liquid in sample I and II.

(b)

Interpretation Introduction

Interpretation:

From the mass percent of boron in each sample, the molecular formula for each sample is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (4)

An empirical formula gives the simplest whole number ratio of atoms of each element present in a compound. The molecular formula tells the exact number of atoms of each element present in a compound.

Following are the steps to determine the molecular formula of a compound.

Step 1: Add the molar mass of each element multiplied by its number of atoms present in the empirical formula to obtain the empirical formula mass for the compound.

Step 2: Divide the molar mass of the compound by its empirical formula mass to obtain the whole number. The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$        (5)

Step 3: Multiply the whole number with the subscript of each element present in the empirical formula. This gives the molecular formula of the compound.

Answer to Problem 5.93P

The molecular formula for the liquid in sample I is ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}$, in sample II is ${\text{B}}_4{\text{H}}_{10}$ and in sample III is ${\text{B}}_{\text{5}}{\text{H}}_{11}$.

Explanation of Solution

The expression to calculate the percentage of hydrogen in the sample I is as follows:

  $\%\text{ H}=\left(100\text{ \%}\right)-\left(\text{Percentage of boron}\right)$        (6)

Substitute $85.63\text{ \%}$ for the percentage of boron in the equation (6).

  $\begin{array}{c}\%\text{ H}=\left(100\text{ \%}\right)-\left(85.63\text{ \%}\right)\\ =14.37\%\end{array}$

Substitute $81.10\text{ \%}$ for the percentage of boron in equation (6) to calculate the percentage of hydrogen in sample II.

  $\begin{array}{c}\%\text{ H}=\left(100\text{ \%}\right)-\left(81.10\text{ \%}\right)\\ =18.90\%\end{array}$

Substitute $82.98\text{ \%}$ for the percentage of boron in equation (6) to calculate the percentage of hydrogen in sample III.

  $\begin{array}{c}\%\text{ H}=\left(100\text{ \%}\right)-\left(82.98\text{ \%}\right)\\ =17.02\%\end{array}$

Consider $100\text{ g}$ of each sample.

For sample I:

Calculate the mass of $\text{B}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of B}=\left(100\text{ g of sample}\right)\left(\frac{85.63\%\text{ B}}{100\text{ \% sample}}\right)\\ =85.63\text{ g B}\end{array}$

Substitute $85.63\text{ g}$ for mass and $10.81\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of $\text{B}$.

  $\begin{array}{c}\text{Moles of B}=\frac{85.63\text{ g}}{10.81\text{ g/mol}}\\ =7.921369\text{ mol}\end{array}$

Calculate the mass of hydrogen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of H}=\left(100\text{ g of sample}\right)\left(\frac{14.37\%\text{ H}}{100\text{ \% sample}}\right)\\ =14.37\text{ g H}\end{array}$

Substitute $14.37\text{ g}$ for mass and $1.008\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of hydrogen.

  $\begin{array}{c}\text{Moles of H}=\frac{\text{ 14}\text{.37 g}}{1.008\text{ g/mol}}\\ =14.25595\text{ mol}\end{array}$

Construct the preliminary formula and use the values $7.921369\text{ mol}$ and $14.25595\text{ mol}$ directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  $\text{Preliminary formula}={\text{B}}_{7.921369\text{ mol}}{\text{H}}_{14.25595\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply with 5 for B and H to make the whole number. Now, construct the empirical formula.

  $\begin{array}{c}\text{Empirical formula}={\text{B}}_{\frac{7.921369\text{ mol}}{7.921369\text{ mol}}}{\text{H}}_{\frac{14.25595\text{ mol}}{7.921369\text{ mol}}}\\ ={\text{B}}_{\left(1\right)\left(5\right)}{\text{H}}_{\left(1.7997\right)\left(2\right)}\\ ={\text{B}}_{\text{5}}{\text{H}}_{\text{9}}\end{array}$

The expression to calculate the empirical formula mass of ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}$ is as follows:

  ${\text{empirical formula mass of B}}_{\text{5}}{\text{H}}_{\text{9}}=\left[\left(5\right)\left(\text{M of B}\right)+\left(9\right)\left(\text{M of H}\right)\right]$        (7)

Substitute $10.01\text{ g/mol}$ for $\text{M}$ of $\text{B}$ and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ in equation (7).

  $\begin{array}{c}{\text{Empirical formula mass of B}}_{\text{5}}{\text{H}}_{\text{9}}=\left[\left(5\right)\left(10.81\text{ g/mol}\right)+\left(9\right)\left(1.008\text{ g/mol}\right)\right]\\ =63.12\text{ g/mol}\end{array}$

Substitute $63.09\text{ g/mol}$ for molar mass of the compound and $63.12\text{ g/mol}$ for empirical formula mass in equation (5) to find molecular formula.

  $\begin{array}{c}\text{Whole number multiple}=\left(\frac{63.12\text{ g/mol}}{63.09\text{ g/mol}}\right)\\ \approx 1\end{array}$

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 1. The molecular formula is ${\text{B}}_{\text{5}}{\text{H}}_{\text{9}}$.

For sample II:

Calculate the mass of $\text{B}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of B}=\left(100\text{ g of sample}\right)\left(\frac{81.10\%\text{ B}}{100\text{ \% sample}}\right)\\ =81.10\text{ g B}\end{array}$

Substitute $81.10\text{ g}$ for mass and $10.81\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of $\text{B}$.

  $\begin{array}{c}\text{Moles of B}=\frac{81.10\text{ g}}{10.81\text{ g/mol}}\\ =7.50321\text{ mol}\end{array}$

Calculate the mass of hydrogen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of H}=\left(100\text{ g of sample}\right)\left(\frac{18.90\%\text{ H}}{100\text{ \% sample}}\right)\\ =18.90\text{ g H}\end{array}$

Substitute $18.90\text{ g}$ for mass and $1.008\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of hydrogen.

  $\begin{array}{c}\text{Moles of H}=\frac{18.90\text{ g}}{1.008\text{ g/mol}}\\ =18.750\text{ mol}\end{array}$

Construct the preliminary formula and use the values $7.50231\text{ mol}$ and $18.750\text{ mol}$ directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  $\text{Preliminary formula}={\text{B}}_{7.50231\text{ mol}}{\text{H}}_{18.750\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply with 2 for B and H to make the whole number. Now, construct the empirical formula.

  $\begin{array}{c}\text{Empirical formula}={\text{B}}_{\frac{7.50231\text{ mol}}{7.50231\text{ mol}}}{\text{H}}_{\frac{18.750\text{ mol}}{7.50231\text{ mol}}}\\ ={\text{B}}_{\left(1\right)\left(2\right)}{\text{H}}_{\left(2.4992\right)\left(2\right)}\\ ={\text{B}}_2{\text{H}}_5\end{array}$

The expression to calculate the empirical formula mass of ${\text{B}}_2{\text{H}}_5$ is as follows:

  ${\text{empirical formula mass of B}}_2{\text{H}}_5=\left[\left(2\right)\left(\text{M of B}\right)+\left(5\right)\left(\text{M of H}\right)\right]$        (8)

Substitute $10.01\text{ g/mol}$ for $\text{M}$ of $\text{B}$ and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ in equation (8).

  $\begin{array}{c}{\text{Empirical formula mass of B}}_2{\text{H}}_5=\left[\left(2\right)\left(10.81\text{ g/mol}\right)+\left(5\right)\left(1.008\text{ g/mol}\right)\right]\\ =26.66\text{ g/mol}\end{array}$

Substitute $53.29\text{ g/mol}$ for molar mass of the compound and $26.66\text{ g/mol}$ for empirical formula mass in equation (5) to find molecular formula.

  $\begin{array}{c}\text{Whole number multiple}=\left(\frac{53.29\text{ g/mol}}{26.66\text{ g/mol}}\right)\\ =2\end{array}$

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 2. The molecular formula is ${\text{B}}_4{\text{H}}_{\text{10}}$.

For sample III,

Calculate the mass of $\text{B}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of B}=\left(100\text{ g of sample}\right)\left(\frac{82.98\%\text{ B}}{100\text{ \% sample}}\right)\\ =82.98\text{ g B}\end{array}$

Substitute $82.98\text{ g}$ for mass and $10.81\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of $\text{B}$.

  $\begin{array}{c}\text{Moles of B}=\frac{82.98\text{ g}}{10.81\text{ g/mol}}\\ =7.6762\text{ mol}\end{array}$

Calculate the mass of hydrogen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of H}=\left(100\text{ g of sample}\right)\left(\frac{17.02\%\text{ H}}{100\text{ \% sample}}\right)\\ =17.02\text{ g H}\end{array}$

Substitute $17.02\text{ g}$ for mass and $1.008\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of hydrogen.

  $\begin{array}{c}\text{Moles of H}=\frac{17.02\text{ g}}{1.008\text{ g/mol}}\\ =16.8849\text{ mol}\end{array}$

Construct the preliminary formula and use the values $7.50231\text{ mol}$ and $16.8849\text{ mol}$ directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  $\text{Preliminary formula}={\text{B}}_{7.50231\text{ mol}}{\text{H}}_{16.8849\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply with 5 for $\text{B}$ and $\text{H}$ to make the whole number. Now, construct the empirical formula.

  $\begin{array}{c}\text{Empirical formula}={\text{B}}_{\frac{7.50231\text{ mol}}{7.50231\text{ mol}}}{\text{H}}_{\frac{16.8849\text{ mol}}{7.50231\text{ mol}}}\\ ={\text{B}}_{\left(1\right)\left(5\right)}{\text{H}}_{\left(2.2\right)\left(5\right)}\\ ={\text{B}}_{\text{5}}{\text{H}}_{11}\end{array}$

The expression to calculate the empirical formula mass of ${\text{B}}_{\text{5}}{\text{H}}_{11}$ is as follows:

  ${\text{empirical formula mass of B}}_{\text{5}}{\text{H}}_{11}=\left[\left(5\right)\left(\text{M of B}\right)+\left(11\right)\left(\text{M of H}\right)\right]$        (9)

Substitute $10.01\text{ g/mol}$ for $\text{M}$ of $\text{B}$ and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ in equation (9).

  $\begin{array}{c}{\text{Empirical formula mass of B}}_{\text{5}}{\text{H}}_{11}=\left[\left(5\right)\left(10.81\text{ g/mol}\right)+\left(11\right)\left(1.008\text{ g/mol}\right)\right]\\ =65.14\text{ g/mol}\end{array}$

Substitute $65.14\text{ g/mol}$ for molar mass of the compound and $65.14\text{ g/mol}$ for empirical formula mass in equation (9) to find molecular formula.

  $\begin{array}{c}\text{Whole number multiple}=\left(\frac{65.14\text{ g/mol}}{65.14\text{ g/mol}}\right)\\ =1.\end{array}$

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 1. The molecular formula is ${\text{B}}_{\text{5}}{\text{H}}_{11}$.

Conclusion

The molecular formula of the liquid in three samples indicates that the number of boron atoms in sample I and III is 5 and in sample II is 4. The number of hydrogen also varies.

(c)

Interpretation Introduction

Interpretation:

The molecular formula of sample IV that contains $78.14\text{ \%}$ boron is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  $\frac{\text{Rate of A}}{\text{Rate of B}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}$

Here,

$M_{\text{B}}$ is the molar mass of gas $\text{B}$

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.

Answer to Problem 5.93P

The molecular formula of sample IV is ${\text{B}}_2{\text{H}}_6$.

Explanation of Solution

The expression to calculate the molar mass of the unknown gas is as follows:

  $\frac{{\text{Rate of SO}}_2}{\text{Rate of sample IV}}=\frac{\sqrt{\text{Molar mass of sample IV}}}{\sqrt{\text{Molar mass}{\text{of SO}}_2}}$        (10)

Rearrange equation (10) for the molar mass of sample IV as follows:

  $\text{Molar mass of sample IV}=\left[\begin{array}{l}{\left(\frac{{\text{Rate of SO}}_2}{\text{Rate of sample IV}}\right)}^2\\ \left(\text{Molar mass}{\text{of SO}}_2\right)\end{array}\right]$        (11)

Rearrange equation (11) for the rate of ${\text{SO}}_{\text{2}}$ and rate of sample IV in the term of volume per unit time as follows:

  $\left(\begin{array}{l}\text{Molar mass }\\ \text{of sample IV}\end{array}\right)=\left[{\left(\frac{\frac{{\text{ Volume of SO}}_2}{{\text{time of SO}}_2}}{\frac{\text{Volume of sample IV}}{\text{time of sample IV}}}\right)}^2\left(\text{Molar mass}{\text{of SO}}_2\right)\right]$        (12)

Substitute $250\text{ mL}$ for the volume of ${\text{SO}}_{\text{2}}$, $13.04\text{ min}$ for the time of ${\text{SO}}_{\text{2}}$, $350\text{ mL}$ for the volume of sample IV, $12.0\text{ min}$ for the time of sample IV and $64.06\text{ g/mol}$ for molar mass of ${\text{SO}}_{\text{2}}$ in the equation (12).

  $\begin{array}{c}\left(\begin{array}{l}\text{Molar mass }\\ \text{of sample IV}\end{array}\right)=\left[{\left(\frac{\frac{\text{ 250 mL}}{\text{13}\text{.04 min }}}{\frac{350\text{ mL}}{12\text{ min}}}\right)}^2\left(64.06\text{ g/mol}\right)\right]\\ ={\left(0.657318\right)}^2\left(64.06\text{ g/mol}\right)\\ =27.6782\text{ g/mol}\\ \approx 27.68\text{ g/mol}\end{array}$

Substitute $78.14\text{ \%}$ for the percentage of boron in equation (4) to calculate the percentage of hydrogen in sample IV.

  $\begin{array}{c}\%\text{ H}=\left(100\text{ \%}\right)-\left(78.14\text{ \%}\right)\\ =27.68\%\end{array}$

For sample IV:

Calculate the mass of $\text{B}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of B}=\left(100\text{ g of sample}\right)\left(\frac{\text{78}\text{.14 }\%\text{ B}}{100\text{ \% sample}}\right)\\ =\text{78}\text{.14 g B}\end{array}$

Substitute $78.14\text{ g}$ for mass and $10.81\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of $\text{B}$.

  $\begin{array}{c}\text{Moles of B}=\frac{78.14\text{ g}}{10.81\text{ g/mol}}\\ =7.22849\text{ mol}\end{array}$

Calculate the mass of hydrogen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of H}=\left(100\text{ g of sample}\right)\left(\frac{\text{21}\text{.86 }\%\text{ H}}{100\text{ \% sample}}\right)\\ =\text{21}\text{.86 g H}\end{array}$

Substitute $\text{21}\text{.86 g}$ for mass and $1.008\text{ g/mol}$ for molecular mass in equation (4) to calculate moles of hydrogen.

  $\begin{array}{c}\text{Moles of H}=\frac{\text{ 21}\text{.86 g}}{1.008\text{ g/mol}}\\ =21.6865\text{ mol}\end{array}$

Construct the preliminary formula and use the values $7.22849\text{ mol}$ and $21.6865\text{ mol}$ directly in the preliminary formula as subscripts of boron and hydrogen element symbols respectively.

  $\text{Preliminary formula}={\text{B}}_{7.22849\text{ mol}}{\text{H}}_{21.6865\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply with 5 to make the whole number. Now, construct the empirical formula.

  $\begin{array}{c}\text{Empirical formula}={\text{B}}_{\frac{7.22849\text{ mol}}{7.22849\text{ mol}}}{\text{H}}_{\frac{21.6865\text{ mol}}{7.22849\text{ mol}}}\\ ={\text{BH}}_3\end{array}$

The expression to calculate the empirical formula mass of ${\text{BH}}_3$ is as follows:

  ${\text{empirical formula mass of BH}}_3=\left[\left(1\right)\left(\text{M of B}\right)+\left(3\right)\left(\text{M of H}\right)\right]$        (13)

Substitute $10.01\text{ g/mol}$ for $\text{M}$ of $\text{B}$ and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ in equation (13).

  $\begin{array}{c}{\text{Empirical formula mass of BH}}_3=\left[\left(10.81\text{ g/mol}\right)+\left(3\right)\left(1.008\text{ g/mol}\right)\right]\\ =13.83\text{ g/mol}\end{array}$

Substitute $27.68\text{ g/mol}$ for molar mass of the compound and $13.83\text{ g/mol}$ for empirical formula mass in equation (2) to find molecular formula.

  $\begin{array}{c}\text{Whole number multiple}=\left(\frac{\text{27}\text{.68 g/mol}}{\text{13}\text{.83 g/mol}}\right)\\ =2.\end{array}$

The empirical formula and the molecular formulas are the same as the value of the whole number multiple is 2. The molecular formula is ${\text{B}}_2{\text{H}}_6$.

Conclusion

The molecular formula of sample IV is ${\text{B}}_2{\text{H}}_6$. One formula unit contains two boron atoms and six hydrogen atoms.