Chapter 12, Problem 7E

(a)

To determine

The percentage by mass of each element in salt, $\text{NaCl}$.

Answer to Problem 7E

The percentage by mass of sodium $\left(\text{Na}\right)$ is $39.3\text{}\text{ \%}$ and the percentage by mass of chlorine $\left(\text{Cl}\right)$ is $60.7\text{}\text{ \%}$ in salt, $\text{NaCl}$.

Explanation of Solution

The percentage by mass of an element in a compound is defined as the percentage of total mass of that element present in the total mass or formula mass of the compound.  Mass percent of any element B in compound AB is given as,

$\%\text{ B component = }\left(\frac{\text{total mass of component B}}{\text{formula mass of compound AB}}\right)100\%$

Atomic mass of sodium $\left(\text{Na}\right)$ is $23.\text{0 u}$ and atomic mass of chlorine $\left(\text{Cl}\right)$ is $35.5\text{ u}$.

The formula mass of $\text{NaCl}$ is calculated as,

$\text{Formula mass of NaCl}=\text{atomic mass of Na}+\text{atomic mass of Cl}$

Substitute $23.\text{0 u}$ for $\text{atomic mass of Na}$ and $35.5\text{ u}$ for $\text{atomic mass of Cl}$ in the above equation.

$\begin{array}{c}\text{Formula mass of NaCl}=23.0\text{ u}+35.5\text{ u}\\ =58.5\text{}\text{ u}\end{array}$

The percentage by mass of sodium $\left(\text{Na}\right)$ is given as,

$\%\text{ Na = }\left(\frac{\text{total mass of Na}}{\text{formula mass of NaCl}}\right)100\%$ (1)

Substitute $23.\text{0 u}$ for $\text{total mass of Na}$ and $58.5\text{}\text{ u}$ for $\text{formula mass of NaCl}$ in equation (1).

$\begin{array}{c}\%\text{ Na}=\left(\frac{23.0\text{}\text{ u}}{58.5\text{}\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= 39}\text{.3 \%}\end{array}$

The percentage by mass of chlorine $\left(\text{Cl}\right)$ is given as,

$\%\text{ Cl = }\left(\frac{\text{total mass of Cl}}{\text{formula mass of NaCl}}\right)100\%$ (2)

Substitute $35.5\text{ u}$ for $\text{total mass of Cl}$ and $58.5\text{}\text{ u}$ for $\text{formula mass of NaCl}$ in equation (2).

$\begin{array}{c}\%\text{ Cl}=\left(\frac{35.5\text{ u}}{58.5\text{}\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= }60.7\text{}\text{ \% }\end{array}$

Conclusion:

Therefore, the percentage by mass of sodium $\left(\text{Na}\right)$ is $39.3\text{}\text{ \%}$ and percentage by mass of chlorine $\left(\text{Cl}\right)$ is $60.7\text{}\text{ \%}$ in salt, $\text{NaCl}$.

(b)

To determine

The percentage by mass of each element in sucrose, ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.

Answer to Problem 7E

The percentage by mass of carbon $\left(\text{C}\right)$ is $42.1\text{}\text{ \%}$, the percentage by mass of hydrogen $\left(\text{H}\right)$ is $6.4\text{\%}$ and percentage by mass of oxygen $\left(\text{O}\right)$ is $51.5\text{ \%}$ in sucrose, ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.

Explanation of Solution

The percentage by mass of an element in a compound is defined as the percentage of total mass of that element present in the total mass or formula mass of the compound.  Mass percent of any element B in compound AB is given as,

$\%\text{ B component = }\left(\frac{\text{total mass of component B}}{\text{formula mass of compound AB}}\right)100\%$

Atomic mass of carbon $\left(\text{C}\right)$ is $12.\text{0 u}$.

Atomic mass of hydrogen $\left(\text{H}\right)$ is $1.\text{0 u}$.

Atomic mass of oxygen $\left(\text{O}\right)$ is $16.0\text{ u}$.

The chemical formula of sucrose is ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$, it consists of 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen.

The formula mass of sucrose, ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is calculated as,

${\text{Formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\left[\begin{array}{l}12\left(\text{atomic mass}\text{}\text{of C}\right)+22\text{}\left(\text{atomic mass of H}\right)\\ +11\left(\text{atomic mass of O}\right)\end{array}\right]$ (3)

Substitute $12.\text{0 u}$ for $\text{atomic mass of C}$, $1.\text{0 u}$ for $\text{atomic mass of H}$ and $16.0\text{ u}$ for $\text{atomic mass of O}$ in equation (3).

$\begin{array}{c}{\text{Formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=12\left(12.\text{0 u}\right)+22\text{}\left(1.\text{0 u}\right)+11\left(16.\text{0 u}\right)\\ =342.0\text{ u}\end{array}$.

The total mass of carbon $\left(\text{C}\right)$ atoms is,

$\text{total mass of C = 12}\left(\text{atomic mass of carbon}\right)$

Substitute $12.\text{0 u}$ for $\text{atomic mass of carbon}$.

$\begin{array}{c}\text{total mass of C = 12}\left(12.\text{0 u}\right)\\ =144.0\text{ u}\end{array}$

The total mass of hydrogen $\left(\text{H}\right)$ atoms is,

$\text{total mass of H = 22}\left(\text{atomic mass of hydrogen}\right)$

Substitute $1.\text{0 u}$ for $\text{atomic mass of hydrogen}$.

$\begin{array}{c}\text{total mass of C = 22}\left(1.\text{0 u}\right)\\ =22.0\text{ u}\end{array}$

The total mass of oxygen $\left(\text{O}\right)$ atoms is,

$\text{total mass of O = 12}\left(\text{atomic mass of oxygen}\right)$

Substitute $16.0\text{ u}$ for $\text{atomic mass of oxygen}$.

$\begin{array}{c}\text{total mass of O = 11}\left(16.0\text{ u}\right)\\ =176.0\text{ u}\end{array}$

The percentage by mass of carbon $\left(\text{C}\right)$ is given as,

$\%\text{ C = }\left(\frac{\text{total mass of C}}{{\text{formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\right)100\%$ (4)

Substitute $144.\text{0 u}$ for $\text{total mass of C}$ and $342.0\text{ u}$ for ${\text{formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in equation (4).

$\begin{array}{c}\%\text{ C = }\left(\frac{144.\text{0 u}}{342.0\text{ u}}\right)100\%\\ =42.1\text{\%}\end{array}$

The percentage by mass of hydrogen $\left(\text{H}\right)$ is given as,

$\%\text{ H = }\left(\frac{\text{total mass of H}}{{\text{ formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\right)100\%$ (5)

Substitute $22.0\text{ u}$ for $\text{total mass of H}$ and $342.0\text{ u}$ for ${\text{formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in equation (5).

$\begin{array}{c}\%\text{ Cl}=\left(\frac{22.0\text{ u}}{342.0\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= 6}\text{.4 \%}\end{array}$

The percentage by mass of oxygen $\left(\text{O}\right)$ is given as,

$\%\text{ O = }\left(\frac{\text{total mass of O}}{{\text{ formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\right)100\%$ (6)

Substitute $176.0\text{ u}$ for $\text{total mass of O}$ and $342.0\text{ u}$ for ${\text{formula mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in equation (6).

$\begin{array}{c}\%\text{ O}=\left(\frac{176.0\text{ u}}{342.0\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= 51}\text{.5 \%}\end{array}$

Conclusion:

Therefore, the percentage by mass of carbon $\left(\text{C}\right)$ is $42.1\text{}\text{ \%}$, the percentage by mass of hydrogen $\left(\text{H}\right)$ is $6.4\text{\%}$ and percentage by mass of oxygen $\left(\text{O}\right)$ is $51.5\text{ \%}$ in sucrose, ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.