Chapter 18.1, Problem 7R

To determine

To calculate: the percentage oxygen in the compound $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ .

Answer to Problem 7R

The percentage of oxygen in $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is 54.87 %.

Explanation of Solution

Given:

The given compound is $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ .

Molecular mass of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is 58.32 amu.

Atomic mass of an atom of oxygen is 15.999 amu.

Formula used:

The percentage of oxygen in $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is given by

  $P_O=\frac{{\left(m_{mol}\right)}_{\text{O}}}{{\left(m_{mol}\right)}_{\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}}\times 100$.... (1)

Here, ${\left(m_{mol}\right)}_{\text{O}}$ is the molar mass of oxygen and ${\left(m_{mol}\right)}_{\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}$ is molar mass of magnesium hydroxide.

Calculation:

The molecular mass of 2 unit of oxygen is

  $\begin{array}{c}{\left(m_{mol}\right)}_O=2\times 15.999\\ =31.998\text{amu}\end{array}$

Thus, the molar mass of oxygen is 31.998 amu.

Percentage of oxygen in magnesium hydroxide $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ can be calculated by substitute 31.998 amu for ${\left(m_{mol}\right)}_{\text{O}}$ and 58.32 amu for ${\left(m_{mol}\right)}_{\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}$ in equation (1)

  $\begin{array}{l}{P}_O=\frac{31.998\text{amu}}{58.32\text{amu}}\times 100\\ P_O=\frac{32\text{amu}}{58.32\text{amu}}\times 100\\ \boxed{P_O=54.87\%}\end{array}$

Conclusion:

Hence, the percentage of oxygen in $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is 54.87 %.