Chapter 3, Problem 80A

Solution

Interpretation: The mass percentage of oxygen and the mass of the second element in the compound in the given table needs to be calculated to complete the table.

Concept Introduction: For any given compound, if the mass of the compound is known and the mass of an element (A) is known, its percentage can be calculated as follows:

  $\%\text{A}=\frac{m_{\text{A}}}{m_{\text{compound}}}\times 100\%$

Here, $m_{\text{A}}$ is the mass of element A.

To calculate the mass of the second element in the compound, the percentage of the second element can be calculated first, and then using the mass of a compound, the mass of the second element can be estimated.

The following formula can be used to calculate the mass percentage of oxygen from the mass of the compound and the mass of oxygen given:

  $\text{Mass \% of oxygen}=\frac{m_{{\text{O}}_2}}{m_{\text{compound}}}\times 100\%$

To calculate the mass of the second element, the mass percentage of the second element needs to be calculated first.

In case 1, the second element is Cu. Its mass percentage will be:

  $\text{Mass \% of Cu}=100-\text{Mass \% of O}$

The mass of Cu can be calculated as follows:

  $m_{\text{Cu}}=\frac{\text{Mass \% of Cu}\times m_{\text{compound}}}{100\%}$

Compound	Mass of compound (g)	Mass of oxygen (g)	Mass % of oxygen	Mass % of second element	Mass of second element in the compound (g)
$\text{CuO}$	80.0	16	$\begin{array}{c}\text{\%}{m}_{\text{O}}=\frac{m_{\text{O}}}{m_{\text{compound}}}\times 100\%\\ =\frac{16}{80.0}\\ =20\%\end{array}$	$\begin{array}{c}\text{\%}{m}_{\text{Cu}}=100-\text{\%}{m}_{\text{O}}\\ \text{=100}-20\\ =80\%\end{array}$	$\begin{array}{c}{m}_{\text{Cu}}=\frac{\%m_{\text{Cu}}\times m_{\text{compound}}}{100\%}\\ =\frac{80\%\times 18.0}{100\%}\\ =14.4\text{ g}\end{array}$
${\text{H}}_{\text{2}}\text{O}$	18.0	16	$\begin{array}{c}\text{\%}{m}_{\text{O}}=\frac{m_{\text{O}}}{m_{\text{compound}}}\times 100\%\\ =\frac{16}{18.0}\\ =88.89\%\end{array}$	$\begin{array}{c}\text{\%}{m}_{\text{H}}=100-\text{\%}{m}_{\text{O}}\\ =100-88.89\\ =11.11\%\end{array}$	$\begin{array}{c}{m}_{\text{H}}=\frac{\%m_{\text{H}}\times m_{\text{compound}}}{100\%}\\ =\frac{11.11\%\times 16.0}{100\%}\\ =1.78\text{ g}\end{array}$
${\text{H}}_{\text{2}}{\text{O}}_2$	34.0	32	$\begin{array}{c}\text{\%}{m}_{\text{O}}=\frac{m_{\text{O}}}{m_{\text{compound}}}\times 100\%\\ =\frac{32}{34.0}\\ =94.12\%\end{array}$	$\begin{array}{c}\text{\%}{m}_{\text{H}}=100-\text{\%}{m}_{\text{O}}\\ =100-94.12\\ =5.88\%\end{array}$	$\begin{array}{c}{m}_{\text{H}}=\frac{\%m_{\text{H}}\times m_{\text{compound}}}{100\%}\\ =\frac{5.88\%\times 34.0}{100\%}\\ =2.0\text{ g}\end{array}$
$\mathrm{CO}$	28.0	16	$\begin{array}{c}\text{\%}{m}_{\text{O}}=\frac{m_{\text{O}}}{m_{\text{compound}}}\times 100\%\\ =\frac{16}{28.0}\\ =57.14\%\end{array}$	$\begin{array}{c}\text{\%}{m}_{\text{C}}=100-\text{\%}{m}_{\text{O}}\\ =100-57.14\\ =42.86\%\end{array}$	$\begin{array}{c}{m}_{\text{C}}=\frac{\%m_{\text{C}}\times m_{\text{compound}}}{100\%}\\ =\frac{57.14\%\times 28.0}{100\%}\\ =16.0\text{ g}\end{array}$
${\mathrm{CO}}_2$	44.0	32	$\begin{array}{c}\text{\%}{m}_{\text{O}}=\frac{m_{\text{O}}}{m_{\text{compound}}}\times 100\%\\ =\frac{32}{44.0}\\ =72.73\%\end{array}$	$\begin{array}{c}\text{\%}{m}_{\text{C}}=100-\text{\%}{m}_{\text{O}}\\ =100-72.73\\ =27.27\%\end{array}$	$\begin{array}{c}{m}_{\text{C}}=\frac{\%m_{\text{C}}\times m_{\text{compound}}}{100\%}\\ =\frac{72.73\%\times 44.0}{100\%}\\ =32.0\text{ g}\end{array}$