Chapter 6, Problem 29A

(a)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of methane, CH₄ needs to be determined.

Concept Introduction:

The percentage by mass of any component in a compound is calculated as follows:

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

Answer to Problem 29A

74.83 %

Explanation of Solution

Mass of C = 12.011 g Molar mass of CH₄ =16.05g/mol

Thus,

  $\begin{array}{l}\%bymass=\left[6\times \frac{C}{(C{H}_4)}\right]\times 100\\ =\left[6\times \frac{12.011}{16.05}\right]\times 100\\ =74.83\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound NaNO₃ needs to be determined.

Concept Introduction:

The percentage by mass of any component in a compound is calculated as follows:

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

Answer to Problem 29A

27.06 %

Explanation of Solution

Mass of Na= 22.99 g

The molar mass of the given compound = 84.97 g/mol

  $\begin{array}{l}\%bymass=\left[\frac{Na}{(NaN{O}_3)}\right]\times 100\\ =\left[\frac{22.99}{84.97}\right]\times 100\\ =27.06\%\end{array}$

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

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

Answer to Problem 29A

42.89 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 28.0 g/mol

Thus,

  $\begin{array}{l}\%bymass=\left[\frac{C}{(CO)}\right]\times 100\\ =\left[\frac{12.01}{28}\right]\times 100\\ =42.89\%\end{array}$

(d)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound NO₂ needs to be determined.

Concept Introduction:

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

Answer to Problem 29A

30.46 %

Explanation of Solution

Mass of N = 14.01 g

The molar mass of given compound = 45.99 g/mol

  $\begin{array}{l}\%bymass=\left[\frac{N}{(N{O}_2)}\right]\times 100\\ =\left[\frac{14.01}{45.99}\right]\times 100\\ =30.46\%\end{array}$

(e)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound1-octanol, C₈H₁₈O needs to be determined.

Concept Introduction:

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

Answer to Problem 29A

73.77 %

Explanation of Solution

Mass of C = 12.011 g

Molar mass of given compound = 130.25 g/mol

  $\begin{array}{l}\%bymass=\left[8\times \frac{C}{C_8{H}_{18}O}\right]\times 100\\ =\left[8\times \frac{12.011}{130.25}\right]\times 100\\ =73.77\%\end{array}$

(f)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound calcium phosphate, Ca₃(PO₄)₂ needs to be determined.

ConceptIntroduction:

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

Answer to Problem 29A

38.77 %

Explanation of Solution

Mass of Ca = 40.08 g

Molar mass of given compound = 310.1g/mol

  $\begin{array}{l} \hfill \\ C{a}_3{(P{O}_4)}_2\hfill \\ \hfill \\ \begin{array}{l}\%bymass=\left[3\times \frac{Ca}{C{a}_3{(P{O}_4)}_2}\right]\times 100\\ =\left[3\times \frac{40.08}{310.1}\right]\times 100\\ =38.77\%\end{array}\hfill \end{array}$

(g)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound 3-phenylphenol, C₁₂H₁₀O needs to be determined.

Concept Introduction:

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

Answer to Problem 29A

84.67 %

Explanation of Solution

Mass of C = 12.011 g

The molar mass of given compound = 170.21 g/mol

  $\begin{array}{l}{C}_{12}{H}_{10}O\hfill \\ \hfill \\ \begin{array}{l}\%bymass=\left[12\times \frac{C}{C_{12}{H}_{10}O}\right]\times 100\\ =\left[12\times \frac{12.011}{170.21}\right]\times 100\\ =84.67\%\end{array}\hfill \end{array}$

(h)

Interpretation Introduction

Interpretation:

The percentage of the mass of the element listed first in the formula of the given compound aluminum acetate, Al(C₂H₃O₂)₃ needs to be determined.

Concept Introduction:

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

Answer to Problem 29A

13.22 %

Explanation of Solution

Mass of Al = 26.98 g

The molar mass of given compound = 204.07 g/mol

  $\begin{array}{l}Al{\left(C_2{H}_3{O}_2\right)}_3\hfill \\ \hfill \\ \begin{array}{l}\%bymass=\left[\frac{Al}{Al{\left(C_2{H}_3{O}_2\right)}_3}\right]\times 100\\ =\left[2\times \frac{26.98}{204.07}\right]\times 100\\ =13.22\%\end{array}\hfill \end{array}$