Chapter 10, Problem 101A

Interpretation Introduction

Interpretation: The molecular formula of nitroglycerine needs to be determined.

Concept Introduction: For any compound, the empirical formula shows the simplest ratio of atoms of different elements present in the compound. On the other hand, the molecular formula shows the actual number of atoms of elements present in the compound.

Explanation of Solution

According to the question, in nitroglycerine, there are 60% as many carbon atoms as hydrogen atoms.

Let the number of hydrogen atoms be x thus, the number of carbon atoms will be:

  $\begin{array}{c}{n}_{\text{C}}=\frac{60}{100}x\\ =0.6x\end{array}$

Now, there are three times as many oxygen atoms as nitrogen atoms and there is the same number of carbon and nitrogen atoms.

The number of carbon atoms is 0. 6x; thus, the number of nitrogen atoms will also be the same. The number of oxygen atoms will be:

  $\begin{array}{c}{n}_{\text{O}}=3\times 0.6x\\ =1.8x\end{array}$

To determine the simplest ratio, consider the value of x equal to 1. Thus, the number of C, H, O, and N atoms will be:

  $\begin{array}{c}\text{C}:\text{H}:\text{O}:\text{N}=0.6x:x:1.8x:0.6x\\ =0.6:1:1.8:0.6\\ =1:1.67:3:1\\ =3:5:9:3\end{array}$

Thus, the empirical formula of nitroglycerine is ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{9}}$ . Calculate empirical mass as follows:

  $\begin{array}{c}M=3\left(12\text{ g/mol}\right)+5\left(1\text{ g/mol}\right)+3\left(14\text{ g/mol}\right)+9\left(16\text{ g/mol}\right)\\ =36+5+42+144\\ =227\text{ g/mol}\end{array}$

Now, calculate the molecular formula from the given mass and number of moles as follows:

  $M=\frac{m}{n}$

It is given that number of moles of nitroglycerine in 1.00 g is 0.00441 mol.

Thus,

  $\begin{array}{c}M=\frac{1.0\text{ g}}{0.00441\text{ mol}}\\ =226.75\text{ g/mol}\end{array}$

Since empirical mass is equal to the molecular mass; thus, the molecular formula will be equal to the empirical formula that is ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{9}}$ .