Chapter 6, Problem 38A

Interpretation Introduction

Interpretation: The empirical formula of aluminium chloride needs to be determined.

Concept Introduction: The empirical formula is the simplest ratio of atoms of elements present in the compound. It does not tell about the actual number of atoms present in it. Some molecules have empirical formula equal to their molecular formula.

Answer to Problem 38A

AlCl_3.

Explanation of Solution

The mass of aluminium metal is 1.271 g. when it is heated in the chlorine gas atmosphere; aluminium chloride is produced of mass 6.280 g.

For any reaction, mass is conserved thus, the difference in the mass of aluminium chloride and aluminium metal will be the mass of chlorine gas used.

The mass of chlorine gas will be:

  $m_{C{l}_2}=6.280-1.271=5.009\text{ g}$

The number of moles of Al and Cl₂ can be calculated as follows:

  $n=\frac{m}{M}$

The molar mass of Al and Cl₂ is 26.98 g/mol and 70.906 g/mol respectively.

The number of moles of Al and Cl₂ will be:

  $\begin{array}{l}{n}_{\text{Al}}=\frac{m}{M}=\frac{1.271\text{ g}}{26.98\text{ g/mol}}=0.047\\ n_{{\text{Cl}}_{\text{2}}}=\frac{m}{M}=\frac{5.009\text{ g}}{70.906\text{ g/mol}}=0.0706\end{array}$

Thus, the number of moles of Cl will be 0.1412.

The ratio of number of moles of Al and Cl will be 0.047:0.1412 or 1:3.

The empirical formula of aluminium chloride will be AlCl₃.

Conclusion

Therefore, the empirical formula of aluminium chloride will be AlCl₃.