Chapter 6, Problem 41A

Interpretation Introduction

Interpretation: The empirical formula of calcium 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 41A

CaCl₂

Explanation of Solution

The mass of calcium is 2.461 g and that of chlorine gas is 4.353 g. From the given masses, number of moles of calcium and chlorine gas can be calculated as follows:

  $n=\frac{m}{M}$

Molar mass of Ca and Cl₂ gas is 40.078 g/mol and 70.906 g/mol respectively.

The number of moles will be:

  $\begin{array}{l}{n}_{Ca}=\frac{m}{M}=\frac{2.461\text{ g}}{40.078\text{ g/mol}}=0.0614\text{ mol}\\ n_{C{l}_2}=\frac{4.353\text{ g}}{70.906\text{ g/mol}}=0.0614\text{ mol}\end{array}$

Thus, the ratio of Ca to Cl₂ will be 1:1 and empirical formula will be CaCl₂.

Conclusion

Therefore, the empirical formula will be CaCl₂.