Chapter 6, Problem 43A

Interpretation Introduction

Interpretation:The empirical formula of two binary compounds of nitrogen and sodium with given percentage of the elements needs to be determined.

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 43A

Compound 1 = ${\text{Na}}_{\text{3}}\text{N}$ (sodium nitride)

Compound 2 = ${\text{NaN}}_{\text{3}}$ (sodium azide)

Explanation of Solution

Compound 1:

Calculate moles of each with the help of atomic mass:

  $\begin{array}{l}\text{Na = }\frac{\text{83}\text{.12g}}{\text{23}\text{.0g/mol}}\text{=3}\text{.614moles}\\ \text{N = }\frac{\text{16}\text{.88 g}}{\text{14}\text{.00 g/mol }}\text{= 1}\text{.206 moles }\end{array}$

Divide by least moles to get the whole number ratio:

  $\begin{array}{l}\text{Na = }\frac{\text{3}\text{.614}}{\text{1}\text{.206}}\text{=}2.99\approx 3\\ \text{N = }\frac{\text{1}\text{.206}}{\text{1}\text{.206}}\text{= 1}\end{array}$

Thus, the empirical formula must be ${\text{Na}}_{\text{3}}\text{N}$ .

Compound 2:

Calculate moles of each with the help of atomic mass:

  $\begin{array}{l}\text{Na = }\frac{\text{35}\text{.36 g}}{\text{23}\text{.0g/mol}}\text{= 1}\text{.537 moles}\\ \text{N = }\frac{\text{64}\text{.64 g}}{\text{14}\text{.00 g/mol }}\text{= 4}\text{.617 moles }\end{array}$

Divide by least moles to get the whole number ratio:

  $\begin{array}{l}\text{Na = }\frac{\text{1}\text{.537}}{\text{1}\text{.537}}\text{=}1\\ \text{N = }\frac{\text{4}\text{.617}}{\text{1}\text{.537}}\text{= 3}\end{array}$

Thus, the empirical formula must be ${\text{NaN}}_{\text{3}}$ .

Conclusion

Sodium can form two types of nitrogen containing binary compounds.