Chapter 19, Problem 18STP

Interpretation Introduction

Interpretation: The flask containing 18.6% oxygen out of the given five chemicals is to be determined.

Concept introduction:

% composition can be calculated by the formula:

$\text{\%}\text{c}\text{o}\text{m}\text{p}\text{o}\text{s}\text{i}\text{t}\text{i}\text{o}\text{n}=\frac{\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{a}\text{t}\text{o}\text{m}}{\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{c}\text{o}\text{m}\text{p}\text{o}\text{u}\text{n}\text{d}}\times 100$

Answer to Problem 18STP

$\text{B}\text{a}{\left(\text{O}\text{H}\right)}_2$ contains 18.6% oxygen.

Explanation of Solution

$\begin{array}{l}{\text{M}}_{\text{K}\text{C}\text{l}}=19+35.5\end{array}$

$\begin{array}{c}=54.5\text{g}/\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}{\text{M}}_{\text{C}{\text{H}}_3\text{O}\text{H}}=12+3\times 1+16+1\end{array}$

  $\begin{array}{c}=32\text{g}/\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}{\text{M}}_{\text{B}\text{a}{\left(\text{O}\text{H}\right)}_2}=137+2\times \left(16+1\right)\end{array}$

  $\begin{array}{c}=171\text{g}/\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}{\text{M}}_{\text{C}{\text{H}}_3\text{C}\text{O}\text{O}\text{H}}=12\times 2+16\times 2+1\times 4\end{array}$

  $\begin{array}{c}=60\text{g}/\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{c}{\text{M}}_{\text{N}\text{a}\text{O}\text{H}}=23+16+1\end{array}$

 $\begin{array}{c}=40\text{g}/\text{m}\text{o}\text{l}\end{array}$

$\begin{array}{l}\text{\%}\text{o}\text{f}\text{O}\text{i}\text{n}\text{C}{\text{H}}_3\text{O}\text{H}=\frac{16}{32}\times 100\end{array}$

 $\begin{array}{c}=50\text{\%}\end{array}$

 $\begin{array}{c}\text{\%}\text{o}\text{f}\text{O}\text{i}\text{n}\text{B}\text{a}{\left(\text{O}\text{H}\right)}_2=\frac{32}{137}\times 100\end{array}$

   $\begin{array}{c}=18.6\text{\%}\end{array}$

$\begin{array}{c}\text{\%}\text{o}\text{f}\text{O}\text{i}\text{n}\text{C}{\text{H}}_3\text{C}\text{O}\text{O}\text{H}=\frac{32}{60}\times 100\end{array}$

   $\begin{array}{c}=53.33\text{\%}\end{array}$

  $\begin{array}{c}\text{\%}\text{o}\text{f}\text{O}\text{i}\text{n}\text{N}\text{a}\text{O}\text{H}=\frac{16}{40}\times 100\end{array}$

   $\begin{array}{c}=40\text{\%}\end{array}$

Conclusion

Thus the flask containing 18.6 % oxygen is $\text{B}\text{a}{\left(\text{O}\text{H}\right)}_2$ .