Chapter 6, Problem 35A

Interpretation Introduction

Interpretation: The empirical formula of 0.4791 g of compound with given mass of elements needs to be determined.

• C = 0.1929 g
• H = 0.01079 g
• O = 0.08566 g
• Cl = 0.1898 g

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Here the molecular formula represents the exact number of all bonded atoms in the compound.

Whereas empirical formula can be defined as the simplest positive integer ratio of all bonded atoms in the compound.

Answer to Problem 35A

The empirical formula must be ${\text{C}}_{\text{3}}{\text{H}}_{\text{2}}\text{OCl}$ .

Explanation of Solution

Given information:

• C = 0.1929 g
• H = 0.01079 g
• O = 0.08566 g
• Cl = 0.1898 g
• Mass of compound = 0.4791 g

Given:

Calculate mass percentage of each element:

  $\begin{array}{l}\text{C = }\frac{0.1929}{0.4791}\times 100=40.26\%\\ \text{H = }\frac{0.01079}{0.4791}\times 100=2.252\%\\ \text{O = }\frac{0.08566}{0.4791}\times 100=17.88\%\\ \text{Cl = }\frac{0.1898}{0.4791}\times 100=39.62\%\end{array}$

Calculate moles of each with the help of atomic mass:

  $\begin{array}{l}\text{O = }\frac{\text{17}\text{.88 g}}{\text{16}\text{.00 g/mol }}\text{=1}\text{.1175 moles}\\ \text{C = }\frac{\text{40}\text{.26 g}}{\text{12}\text{.0 g/mol }}\text{= 3}\text{.355 moles}\\ \text{H =}\frac{\text{2}\text{.252 g}}{1.001}\text{= 2}\text{.250 moles}\\ \text{Cl = }\frac{\text{39}\text{.62 g}}{\text{35}\text{.5 g/mol }}\text{=1}\text{.116 moles}\end{array}$

Divide by least moles to get the whole number ratio:

  $\begin{array}{l}\text{O = }\frac{\text{1}\text{.1175}}{\text{1}\text{.116}}\text{= }1\\ \text{C = }\frac{\text{3}\text{.355}}{\text{1}\text{.116}}\text{= 3}\\ \text{H =}\frac{\text{2}\text{.225}}{\text{1}\text{.116}}\text{= 2}\\ \text{Cl =}\frac{1.116}{\text{1}\text{.116}}\text{= 1}\end{array}$

Thus the empirical formula must be ${\text{C}}_{\text{3}}{\text{H}}_{\text{2}}\text{OCl}$ .

Conclusion

The empirical formula must be ${\text{C}}_{\text{3}}{\text{H}}_{\text{2}}\text{OCl}$ .