Chapter 12, Problem 12.77P

Interpretation Introduction

Interpretation:

The mass percent of sulfur in $\text{5}\text{.63 g}$ sample has to be calculated.

Concept Introduction:

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Mass percent is calculated by the formula as follows:

  $\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Total Mass }}\left(100\text{ \%}\right)$

Answer to Problem 12.77P

The mass percent of sulfur is $1.22\text{ \%}$.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{21}\text{.35 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{21}\text{.35 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.02135\text{ L}\end{array}$

The formula to evaluate moles from molarity is given as follows:

  $n=MV$        (1)

Substitute $0.1006\text{ mol/L}$ for $M$  and $0.02135\text{ L}$ for $V$ in equation (1).

  $\begin{array}{c}n=\left(0.1006\text{ mol/L}\right)\left(0.02135\text{ L}\right)\\ =0.0021478\text{ mol}\end{array}$

The balanced chemical reaction when ${\text{SO}}_{\text{2}}$ combines with ${\text{H}}_2{\text{O}}_2$ is given as follows:

  ${\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)\to {\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{H}}_2\left(g\right)$        (2)

A similar reaction occurs when ${\text{SO}}_3$ combines with ${\text{H}}_2{\text{O}}_2$ given as follows:

  ${\text{2SO}}_3\left(g\right)+2{\text{H}}_2{\text{O}}_2\left(aq\right)\to 2{\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{O}}_2\left(g\right)$        (3)

The reaction when ${\text{H}}_2{\text{SO}}_{\text{4}}$ is neutralized by $\text{NaOH}$ is given as follows:

  ${\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)+2\text{NaOH}\left(aq\right)\to {\text{Na}}_2{\text{SO}}_{\text{4}}\left(aq\right)+2{\text{H}}_2\text{O}\left(g\right)$        (4)

According to the stoichiometry of the balanced equation (4), $2\text{ mol NaOH}$ neutralizes ${\text{1 mol H}}_2{\text{SO}}_{\text{4}}$, therefore moles of ${\text{H}}_2{\text{SO}}_{\text{4}}$ that is neutralized by $0.0021478\text{ mol NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_{\text{4}}=\left(0.0021478\text{ mol NaOH}\right)\left(\frac{1{\text{ mol H}}_2{\text{SO}}_{\text{4}}}{\text{2 mol NaOH}}\right)\\ =0.0010739{\text{ mol H}}_2{\text{SO}}_{\text{4}}\end{array}$

As per reaction (3), ${\text{2 mol SO}}_3$ are required to form ${\text{2 mol H}}_2{\text{SO}}_{\text{4}}$, thus moles of ${\text{SO}}_3$ that would form $0.0010739{\text{ mol H}}_2{\text{SO}}_{\text{4}}$ is $0.0010739\text{ mol}$.

Similarly in accordance with reaction (2), moles of ${\text{SO}}_2$ in exhaust fuel gases is $0.0010739\text{ mol}$.

Thus total moles of gases are calculated as follows:

  $\begin{array}{c}\text{Moles of exhaust gases}=0.0010739\text{ mol}+0.0010739\text{ mol}\\ =\text{0}\text{.0021478 mol}\end{array}$

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (5)

Substitute $\text{0}\text{.0021478 mol}$ for moles and $\text{32}\text{.064 g/mol}$ for molar mass of $\text{S}$ in equation (5).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.0021478 mol}\right)\left(\text{32}\text{.064 g/mol}\right)\\ =0.068867\text{ g}\end{array}$

Mass percent is calculated by the formula as follows:

  $\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Total Mass }}\left(100\text{ \%}\right)$        (6)

Substitute $0.068867\text{ g}$ for mass of solute, $\text{5}\text{.63 g}$ for total mass in equation (6) to calculate mass percent of $\text{S}$.

  $\begin{array}{c}\text{Mass percent}=\frac{0.068867\text{ g}}{\text{5}\text{.63 g}}\left(100\text{ \%}\right)\\ =1.22\text{ \%}\end{array}$