Chapter 13, Problem 44E

Interpretation Introduction

(a)

Interpretation:

The mass percentage of $\text{20}\text{.0}\text{g}\text{KI}$ in $\text{100}\text{.0}\text{g}$ of water is to be stated.

Concept introduction:

The percentage composition of a compound is calculated by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 44E

The mass percentage of $\text{20}\text{.0}\text{g}\text{KI}$ in $\text{100}\text{.0}\text{g}$ of water is $\text{16}\text{.67}\%$.

Explanation of Solution

The value of mass of $\text{KI}$ is $20.0\text{g}$. The value of mass of water is $\text{100}\text{g}$.

The mass percentage is calculated by the formula as shown below.

$\text{Mass}\%=\frac{{\text{Mass}}_{\text{KI}}}{{\text{Mass}}_{\text{KI}}+{\text{Mass}}_{\text{water}}}\times 100\%$ …(1)

Substitute the value of mass of $\text{KI}$ and mass of water in the equation (1).

$\begin{array}{rll}\text{Mass}\%& =& \frac{{\text{Mass}}_{\text{KI}}}{{\text{Mass}}_{\text{KI}}+{\text{Mass}}_{\text{water}}}\times 100\%\\ & =& \frac{\text{20}\text{g}}{\text{20}\text{g}+\text{100}\text{g}}\times 100\%\\ & =& \text{16}\text{.67}\%\end{array}$

Therefore, the mass percentage of $\text{20}\text{.0}\text{g}\text{KI}$ in $\text{100}\text{.0}\text{g}$ of water is $\text{16}\text{.67}\%$.

Conclusion

The mass percentage of $\text{20}\text{.0}\text{g}\text{KI}$ in $\text{100}\text{.0}\text{g}$ of water is $\text{16}\text{.67}\%$.

Interpretation Introduction

(b)

Interpretation:

The mass percentage of $\text{2}\text{.50}\text{g}{\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{95}\text{.0}\text{g}$ of water is to be stated.

Concept introduction:

The percentage composition of a compound is found by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 44E

The mass percentage of $\text{2}\text{.50}\text{g}{\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{95}\text{.0}\text{g}$ of water is $\text{2}\text{.56}\%$.

Explanation of Solution

The mass percentage is calculated by the formula as shown below.

$\text{Mass}\%=\frac{{\text{Mass}}_{{\text{AgC}}_2{\text{H}}_3{\text{O}}_2}}{{\text{Mass}}_{{\text{AgC}}_2{\text{H}}_3{\text{O}}_2}+{\text{Mass}}_{\text{water}}}\times 100\%$ …(1)

The value of mass of ${\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ is $2.50\text{g}$. The value of mass of water is $\text{95}\text{.0}\text{g}$.

Substitute the value of mass of ${\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ and mass of water in equation (1).

$\begin{array}{rll}\text{Mass}\%& =& \frac{{\text{Mass}}_{{\text{AgC}}_2{\text{H}}_3{\text{O}}_2}}{{\text{Mass}}_{{\text{AgC}}_2{\text{H}}_3{\text{O}}_2}+{\text{Mass}}_{\text{water}}}\times 100\%\\ & =& \frac{2.50\text{g}}{2.50\text{g}+\text{95}\text{.0}\text{g}}\times 100\%\\ & =& \text{2}\text{.56}\%\end{array}$

Therefore, the mass percentage of $\text{2}\text{.50}\text{g}{\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{95}\text{.0}\text{g}$ of water is $\text{2}\text{.56}\%$.

Conclusion

The mass percentage of $\text{2}\text{.50}\text{g}{\text{AgC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{95}\text{.0}\text{g}$ of water is $\text{2}\text{.56}\%$.

Interpretation Introduction

(c)

Interpretation:

The mass percentage of $\text{5}\text{.57}\text{g}{\text{SrCl}}_2$ in $\text{225}\text{.0}\text{g}$ of water is to be stated.

Concept introduction:

The percentage composition of a compound is calculated by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 44E

The mass percentage of $\text{5}\text{.57}\text{g}{\text{SrCl}}_2$ in $\text{225}\text{.0}\text{g}$ of water is $\text{2}\text{.42}\%$.

Explanation of Solution

The mass percentage is calculated by the formula as shown below.

$\text{Mass}\%=\frac{{\text{Mass}}_{{\text{SrCl}}_2}}{{\text{Mass}}_{{\text{SrCl}}_2}+{\text{Mass}}_{\text{water}}}\times 100\%$ …(1)

The value of mass of ${\text{SrCl}}_2$ is $\text{5}\text{.57}\text{g}$. The value of mass of water is $\text{225}\text{g}$.

Substitute the value of mass of ${\text{SrCl}}_2$ and mass of water in equation (1).

$\begin{array}{rll}\text{Mass}\%& =& \frac{{\text{Mass}}_{{\text{SrCl}}_2}}{{\text{Mass}}_{{\text{SrCl}}_2}+{\text{Mass}}_{\text{water}}}\times 100\%\\ & =& \frac{\text{5}\text{.57}\text{g}}{\text{5}\text{.57}\text{g}+225\text{g}}\times 100\%\\ & =& \text{2}\text{.42}\%\end{array}$

Therefore, the mass percentage of $\text{5}\text{.57}\text{g}{\text{SrCl}}_2$ in $\text{225}\text{.0}\text{g}$ of water is $\text{2}\text{.42}\%$.

Conclusion

The mass percentage of $\text{5}\text{.57}\text{g}{\text{SrCl}}_2$ in $\text{225}\text{.0}\text{g}$ of water is $\text{2}\text{.42}\%$.

Interpretation Introduction

(d)

Interpretation:

The mass percentage of $\text{50}\text{.0}\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in $\text{200}\text{.0}\text{g}$ of water is to be calculated.

Concept introduction:

The percentage composition of a compound is calculated by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 44E

The mass percentage of $\text{50}\text{.0}\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in $\text{200}\text{.0}\text{g}$ of water is $\text{20}\%$.

Explanation of Solution

The mass percentage is calculated by the formula as shown below.

$\text{Mass}\%=\frac{{\text{Mass}}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}{{\text{Mass}}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}+{\text{Mass}}_{\text{water}}}\times 100\%$ …(1)

The value of mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is $50.0\text{g}$. The value of mass of water is $\text{200}\text{g}$.

Substitute the value of mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and mass of water in equation (1).

$\begin{array}{rll}\text{Mass}\%& =& \frac{{\text{Mass}}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}{{\text{Mass}}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}+{\text{Mass}}_{\text{water}}}\times 100\%\\ & =& \frac{\text{50}\text{g}}{\text{50}\text{g}+\text{200}\text{g}}\times 100\%\\ & =& \text{20}\%\end{array}$

Therefore, the mass percentage of $\text{50}\text{.0}\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in $\text{200}\text{.0}\text{g}$ of water is $\text{20}\%$.

Conclusion

The mass percentage of $\text{50}\text{.0}\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in $\text{200}\text{.0}\text{g}$ of water is $\text{20}\%$.