Chapter 6, Problem 37A

Interpretation Introduction

Interpretation: The empirical formula of calcium oxide 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 37A

CaO.

Explanation of Solution

The mass of sample of calcium is 2.514 g. It is heated with pure oxygen and mass increased by 1.004 g.

Since, mass is conserved in a chemical reaction thus, the increased mass will be equal to the mass of oxygen.

Therefore, mass of oxygen is 1.004 g.

The number of moles of Ca and O can be calculated from their masses and molar masses as follows:

  $n=\frac{m}{M}$

Molar mass of Ca and O is 40.078 g/mol and 15.99 g/mol respectively.

Thus,

  $\begin{array}{l}{n}_{Ca}=\frac{2.514\text{ g}}{40.078\text{ g/mol}}=0.0627\text{ mol}\\ n_O=\frac{1.004\text{ g/mol}}{15.99\text{ g/mol}}=0.0627\text{ mol}\end{array}$

Thus, the ratio of number of moles of Ca and O will be 0.0627:0.0627 or 1:1.

The empirical formula of calcium oxide will be CaO.

Conclusion

Therefore, the empirical formula of calcium oxide will be CaO.