Chapter 12, Problem 8E

(a)

To determine

The percentage by mass of each element in lime, ${\text{CaCO}}_3$.

Answer to Problem 8E

The percentage by mass of calcium $\left(\text{Ca}\right)$ is $40.0\text{ \%}$, the percentage by mass of carbon $\left(\text{C}\right)$ is $12.0\text{}\text{ \%}$ and percentage by mass of oxygen $\left(\text{O}\right)$ is $48.0\text{ \%}$ in lime, ${\text{CaCO}}_3$.

Explanation of Solution

The percentage by mass of an element in a compound is defined as the percentage of the 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 calcium $\left(\text{Ca}\right)$ is $40.\text{0 u}$.

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

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

The chemical formula of lime is ${\text{CaCO}}_3$, it consists of 1 atom of calcium, 1 atom of carbon and 3 atoms of oxygen.

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

$\text{total mass of O = 3}\left(\text{atomic mass of oxygen}\right)$

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

$\begin{array}{c}\text{total mass of O = 3}\left(16.0\text{ u}\right)\\ =48.0\text{ u}\end{array}$

The formula mass of lime, ${\text{CaCO}}_3$ is calculated as,

${\text{formula mass of CaCO}}_3=\left[\begin{array}{l}\left(\text{atomic mass}\text{}\text{of Ca}\right)+\left(\text{atomic mass}\text{}\text{of C}\right)\\ +3\left(\text{atomic mass of O}\right)\end{array}\right]$ (1)

Substitute $12.\text{0 u}$ for $\text{atomic mass of C}$, $40.\text{0 u}$ for $\text{atomic mass of Ca}$ and $16.0\text{ u}$ for $\text{atomic mass of O}$ in equation (1).

$\begin{array}{c}{\text{formula mass of CaCO}}_3=\left(40.\text{0 u}\right)+\left(12.\text{0 u}\right)+3\left(16.\text{0 u}\right)\\ =100.0\text{ u}\end{array}$

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

$\%\text{ Ca = }\left(\frac{\text{total mass of Ca}}{{\text{formula mass of CaCO}}_3}\right)100\%$ (2)

Substitute $40.\text{0 u}$ for $\text{total mass of Ca}$ and $100.0\text{ u}$ for ${\text{formula mass of CaCO}}_3$ in equation (2).

$\begin{array}{c}\%\text{ Ca = }\left(\frac{40.\text{0 u}}{100.0\text{ u}}\right)100\%\\ =40.0\text{\%}\end{array}$

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

$\%\text{ C = }\left(\frac{\text{total mass of C}}{{\text{formula mass of CaCO}}_3}\right)100\%$ (3)

Substitute $12.\text{0 u}$ for $\text{total mass of C}$ and $100.0\text{ u}$ for ${\text{total mass of CaCO}}_3$ in equation (4).

$\begin{array}{c}\%\text{ C = }\left(\frac{12.\text{0 u}}{100.0\text{ u}}\right)100\%\\ =12.0\text{\%}\end{array}$

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

$\%\text{ O = }\left(\frac{\text{total mass of O}}{{\text{ formula mass of CaCO}}_3}\right)100\%$ (4)

Substitute $48.0\text{ u}$ for $\text{total mass of O}$ and $100.0\text{ u}$ for ${\text{total mass of CaCO}}_3$ in equation (4).

$\begin{array}{c}\%\text{ O}=\left(\frac{48.0\text{ u}}{100.0\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= 48}\text{.0 \%}\end{array}$

Conclusion:

Therefore, the percentage by mass of calcium $\left(\text{Ca}\right)$ is $40.0\text{ \%}$, the percentage by mass of carbon $\left(\text{C}\right)$ is $12.0\text{}\text{ \%}$ and percentage by mass of oxygen $\left(\text{O}\right)$ is $48.0\text{ \%}$ in lime, ${\text{CaCO}}_3$.

(b)

To determine

The percentage by mass of each element in milk of magnesium, $\text{Mg}{\left(\text{OH}\right)}_2$.

Answer to Problem 8E

The percentage by mass of magnesium $\left(\text{Mg}\right)$ is $41.7\text{ \%}$, the percentage by mass of oxygen $\left(\text{O}\right)$ is $54.9\text{ \%}$ and percentage by mass of hydrogen $\left(\text{H}\right)$ is $3.4\text{ \%}$ in milk of magnesium, $\text{Mg}{\left(\text{OH}\right)}_2$.

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 magnesium $\left(\text{Mg}\right)$ is $24.3\text{ u}$.

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

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

The chemical formula of milk of magnesium is $\text{Mg}{\left(\text{OH}\right)}_2$, it consists of 1 atom of magnesium, 2 atoms of oxygen and 2 atoms of hydrogen.

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

$\text{total mass of O = 2}\left(\text{atomic mass of oxygen}\right)$

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

$\begin{array}{c}\text{total mass of O = 2}\left(16.0\text{ u}\right)\\ =32.0\text{ u}\end{array}$

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

$\text{total mass of H = 2}\left(\text{atomic mass of hydrogen}\right)$

Substitute $1.0\text{ u}$ for $\text{atomic mass of hydrogen}$.

$\begin{array}{c}\text{total mass of H = 2}\left(1.0\text{ u}\right)\\ =2.0\text{ u}\end{array}$

The formula mass of milk of magnesium, $\text{Mg}{\left(\text{OH}\right)}_2$ is calculated as,

$\text{formula mass of Mg}{\left(\text{OH}\right)}_2=\left[\begin{array}{l}\left(\text{atomic mass}\text{}\text{of Mg}\right)+2\left(\text{atomic mass}\text{}\text{of O}\right)\\ +2\left(\text{atomic mass of H}\right)\end{array}\right]$ (5)

Substitute $24.3\text{ u}$ for $\text{atomic mass of Mg}$, $16.0\text{ u}$ for $\text{atomic mass of O}$ and $1.\text{0 u}$ for $\text{atomic mass of H}$ in equation (5).

$\begin{array}{c}\text{formula mass of Mg}{\left(\text{OH}\right)}_2=\left(24.3\text{ u}\right)+2\left(16.\text{0 u}\right)+2\left(1.0\text{ u}\right)\\ =58.3\text{ u}\end{array}$

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

$\%\text{ Mg = }\left(\frac{\text{total mass of Mg}}{\text{formula mass of Mg}{\left(\text{OH}\right)}_2}\right)100\%$ (6)

Substitute $24.3\text{ u}$ for $\text{total mass of Mg}$ and $58.3\text{ u}$ for $\text{formula mass of Mg}{\left(\text{OH}\right)}_2$ in equation (6).

$\begin{array}{c}\%\text{ Mg = }\left(\frac{24.3\text{ u}}{58.3\text{ u}}\right)100\%\\ =41.7\text{ \%}\end{array}$

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

$\%\text{ O = }\left(\frac{\text{total mass of O}}{{\text{ formula mass of CaCO}}_3}\right)100\%$ (7)

Substitute $32.0\text{ u}$ for $\text{total mass of O}$ and $58.3\text{ u}$ for $\text{formula mass of Mg}{\left(\text{OH}\right)}_2$ in equation (7).

$\begin{array}{c}\%\text{ O}=\left(\frac{32.0\text{ u}}{58.3\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= }54.9\text{ \%}\end{array}$

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

$\%\text{ H = }\left(\frac{\text{total mass of H}}{\text{ formula mass of Mg}{\left(\text{OH}\right)}_2}\right)100\%$ (8)

Substitute $1.0\text{ u}$ for $\text{total mass of H}$ and $58.3\text{ u}$ for $\text{formula mass of Mg}{\left(\text{OH}\right)}_2$ in equation (8).

$\begin{array}{c}\%\text{ H}=\left(\frac{2.0\text{ u}}{58.3\text{ u}}\right)100\text{}\text{}\text{\%}\\ \text{= }3.4\text{ \%}\end{array}$

Conclusion:

Therefore, the percentage by mass of magnesium $\left(\text{Mg}\right)$ is $41.7\text{ \%}$, the percentage by mass of oxygen $\left(\text{O}\right)$ is $54.9\text{ \%}$ and percentage by mass of hydrogen $\left(\text{H}\right)$ is $3.4\text{ \%}$ in milk of magnesium, $\text{Mg}{\left(\text{OH}\right)}_2$.