Chapter 10, Problem 165A

Solution

Interpretation: The percent composition of phosphorus oxide and its empirical formula need to be determined.

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 the compound, the mass of the second element can be estimated.

Empirical formula is defined as a formula of a compound that shows the simplest ratio of atoms present in that compound.

43.63% of P, 56.37% of O and ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$

Given Information: The mass of phosphorus is 30.98 g and the mass of phosphorus oxide formed is 71.00 g.

According to the question, phosphorus reacts with oxygen to form phosphorus oxide. Mass of phosphorus is given 30.98 g and that of phosphorus oxide is given 71.00 g.

The mass percentage of phosphorus can be calculated as follows:

  $\%m_{\text{P}}=\frac{m_{\text{P}}}{m_{\text{Compound}}}\times 100\%$

Substituting values,

  $\begin{array}{c}\%m_{\text{P}}=\frac{30.98\text{ g}}{71.00\text{ g}}\times 100\%\\ =43.63\%\end{array}$

From the mass percentage of phosphorus, the mass percentage of oxygen can be calculated as follows:

  $\begin{array}{c}\%m_{\text{O}}=100\%-43.63\%\\ =56.37\%\end{array}$

Thus, the percentage composition of potassium oxide is 43.63% P and 56.36% O.

From the mass percentage of oxygen, the mass of oxygen can be calculated as follows:

  $\begin{array}{c}{m}_{\text{O}}=\frac{\%m_{\text{O}}\times m_{\text{compound}}}{100\%}\\ =\frac{56.37\%\times 71.00}{100\%}\\ =40.03\text{ g}\end{array}$

Now, the molar mass of oxygen is16 g/mol; thus, the number of moles of oxygen can be calculated as follows:

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{40.03\text{ g}}{16\text{ g/mol}}\\ =2.5\text{ mol}\end{array}$

Similarly, the number of moles of phosphorus can be calculated as follows:

  $\begin{array}{c}{n}_{\text{P}}=\frac{m_{\text{P}}}{M_{\text{P}}}\\ =\frac{30.98\text{ g}}{30.98\text{ g/mol}}\\ =1\text{ mol}\end{array}$

The empirical formula can be determined as follows:

  $\frac{n_{\text{O}}}{n_{\text{P}}}=\frac{2.5}{1}$

Now, the simplest ratio will be 5:2

Thus, the empirical formula is ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ .

Thus, the percentage composition of phosphorus oxide is 43.63% of P and 56.37% of O. Also, the empirical formula of the compound is ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ .