Chapter 6, Problem 29A

(a)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundC₆H₁₀O₄ needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

49.31 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 146.14 g/mol

C₆H₁₀O₄

Molar mass of C₆H₁₀O₄ =146.14

  $\begin{array}{c}\text{\% by mass }=\left[\text{6 }\times \frac{\text{C}}{{\text{C}}_{\text{6}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}}\right]\times 100\text{ \%}\end{array}$

        $\begin{array}{c}=\left[6\times \frac{\text{ 12}\text{.011}}{146.14}\right]\times 100\end{array}$

        $\begin{array}{c}=\text{ 49}\text{.31 }\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundNH₄NO₃ needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

35.00 %

Explanation of Solution

Mass of N= 14.067 g

The molar mass of given compound = 80.043 g/mol

  $\begin{array}{c}\text{\% by mass }=\left[\frac{\text{N}}{{\text{(NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{)}}\right]\times 100\%\end{array}$

        $\begin{array}{c}=\left[2\times \frac{\text{ 14}\text{.0067}}{80.043}\right]\times 100\%\end{array}$

        $\begin{array}{c}=\text{ 35}\text{.00 }\%\end{array}$

(c)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundC₈H₁₀N₄O₂ needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

49.48 %

Explanation of Solution

Mass of C = 12.011 g

The molar mass of given compound = 194.19 g/mol

  $\begin{array}{c}\text{\% by mass = }\left[\text{8 × }\frac{\text{C}}{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= }\left[\text{8×}\frac{\text{ 12}\text{.011}}{\text{194}\text{.19}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= 49}\text{.48 \%}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundClO₂ needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

52.56 %

Explanation of Solution

Mass of Cl = 35.453 g

The molar mass of given compound = 67.45 g/mol

  $\begin{array}{c}\text{\% by mass = }\left[\frac{\text{Cl}}{{\text{(ClO}}_{\text{2}}\text{)}}\right]\text{ × 100\%}\end{array}$

        $\begin{array}{c}\text{= }\left[\frac{\text{ 35}\text{.453}}{\text{67}\text{.45}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= 52}\text{.56 \%}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundC₆H₁₁OH needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

71.95 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 100.16 g/mol

  $\begin{array}{c}\text{\% by mass = }\left[\text{6× }\frac{\text{C}}{{\text{C}}_{\text{6}}{\text{H}}_{\text{11}}\text{OH}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= }\left[\text{8×}\frac{\text{ 12}\text{.011}}{\text{100}\text{.16}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= 71}\text{.95 \%}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundC₆H₁₂O₆ needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

40.00 %

Explanation of Solution

Mass of C = 12.011 g

The molar mass of given compound = 180.156 g/mol

  $\begin{array}{c}\text{\% by mass = }\left[\text{6 × }\frac{\text{C}}{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= }\left[\text{6 ×}\frac{\text{ 12}\text{.011}}{\text{180}\text{.156}}\right]\text{ × 100\% }\end{array}$

        $\begin{array}{c}\text{= 40}\text{.00 \%}\end{array}$

(g)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element C₂₀H₄₂ listed first in the formula of the given compound needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

85.02 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 282.5457 g/mol

  $\begin{array}{l}{\text{C}}_{\text{20}}{\text{H}}_{\text{42}}\hfill \\ \hfill \\ \begin{array}{c}\text{\% by mass = }\left[\text{20 × }\frac{\text{C}}{{\text{C}}_{\text{20}}{\text{H}}_{\text{42}}}\right]\text{ × 100 }\end{array}\hfill \end{array}$

        $\begin{array}{c}\text{= }\left[\text{20 × }\frac{\text{12}\text{.011}}{\text{282}\text{.5475}}\right]\text{ × 100 }\end{array}$

        $\begin{array}{c}\text{= 85}\text{.02 \%}\end{array}$

(h)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compoundC₂H₅OH needs to be determined.

Concept Introduction :

The mass percentage of any component can be calculated as follows:

  $\text{\% by mass}=\frac{\text{mass of component}}{\text{Total mass}}\times \text{100\%}$

Answer to Problem 29A

52.14 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 46.07 g/mol

  $\begin{array}{c}\text{\% by mass = }\left[\text{2 × }\frac{\text{C}}{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\right]\text{ ×100\%}\end{array}$

        $\begin{array}{c}\text{ = }\left[\text{2 × }\frac{\text{12}\text{.011}}{\text{46}\text{.07}}\right]\text{× 100\% }\end{array}$

        $\begin{array}{c}\text{= 52}\text{.14 \%}\end{array}$